Step 1: Understanding the Concept:

The question asks for the total charge in Faradays (F) required to reduce one mole of permanganate ion (MnO\(_4^-\)) to manganese dioxide (MnO\(_2\)). This involves determining the number of moles of electrons transferred in the redox reaction. One Faraday is the charge of one mole of electrons, approximately 96500 Coulombs.


Step 2: Key Formula or Approach:

1. Determine the oxidation state of Manganese (Mn) in the reactant (MnO\(_4^-\)) and the product (MnO\(_2\)).

2. Calculate the change in oxidation state, which gives the number of electrons (n) gained per mole of MnO\(_4^-\).

3. The total charge required is given by Q = nF, where 'n' is the number of moles of electrons and 'F' is the Faraday constant.


Step 3: Detailed Explanation:

First, let's find the oxidation state of Mn in MnO\(_4^-\).

Let the oxidation state of Mn be \(x\). The oxidation state of oxygen is -2.

The overall charge on the ion is -1.
\[ x + 4(-2) = -1 \] \[ x - 8 = -1 \] \[ x = +7 \]
So, the oxidation state of Mn in MnO\(_4^-\) is +7.


Next, let's find the oxidation state of Mn in MnO\(_2\).

Let the oxidation state of Mn be \(y\). The oxidation state of oxygen is -2.

The overall charge on the molecule is 0.
\[ y + 2(-2) = 0 \] \[ y - 4 = 0 \] \[ y = +4 \]
So, the oxidation state of Mn in MnO\(_2\) is +4.


Now, we calculate the change in the oxidation state of Mn during the reduction.

Change in oxidation state = Initial state - Final state = (+7) - (+4) = +3.

This means that 3 electrons are gained for each MnO\(_4^-\) ion reduced.

MnO\(_4^-\) + 3e\(^-\) \(\rightarrow\) MnO\(_2\) (in a balanced half-reaction).

For the reduction of 1 mole of MnO\(_4^-\), 3 moles of electrons are required.


The charge of 1 mole of electrons is 1 Faraday (1 F).

Therefore, the charge required for 3 moles of electrons is 3 F.


Step 4: Final Answer:

The charge required for the reduction of 1 mol of MnO\(_4^-\) to MnO\(_2\) is 3 F.
Quick Tip: To solve such electrochemistry problems quickly, first balance the oxidation numbers. The change in the oxidation number of the central atom directly gives you the number of moles of electrons transferred per mole of the substance. This number is equal to the required charge in Faradays.

Step 1: Understanding the Concept:

This question tests the fundamental concepts of chemical kinetics, including reaction order, half-life, and molecularity. We need to evaluate each statement to identify the one that is incorrect.


Step 2: Detailed Explanation:

Let's analyze each statement:


(A) Rate of zero order reaction is independent of initial concentration of reactant.

For a zero-order reaction, the rate law is given by: Rate = k[A]\(^0\).

Since any quantity raised to the power of zero is 1, the rate law simplifies to Rate = k.

This shows that the rate of the reaction is constant and does not depend on the concentration of the reactant A. This statement is true.


(B) Half-life of a zero order reaction is inversely proportional to the rate constant.

The formula for the half-life (t\(_{1/2}\)) of a zero-order reaction is:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
Here, [A]\(_0\) is the initial concentration and k is the rate constant.

From the formula, it is clear that t\(_{1/2}\) is inversely proportional to the rate constant k (t\(_{1/2} \propto 1/k\)). This statement is true.


(C) Molecularity of a reaction may be zero.

Molecularity is defined as the number of reacting species (atoms, ions, or molecules) that must collide simultaneously in an elementary reaction to bring about a chemical change.

For a reaction to occur, at least one molecule must be present. Therefore, molecularity must be a positive integer (typically 1, 2, or rarely 3). It cannot be zero, fractional, or negative. This statement is false.


(D) For a first order reaction, t\(_{1/2}\) = 0.693/k.

The formula for the half-life (t\(_{1/2}\)) of a first-order reaction is:
\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \]
This is a standard and correct formula in chemical kinetics. This statement is true.


Step 3: Final Answer:

The false statement is (C) because molecularity represents the number of colliding molecules and cannot be zero.
Quick Tip: Remember the key differences: 'Order of reaction' is an experimental quantity that can be zero, fractional, or an integer. 'Molecularity' is a theoretical concept for elementary reactions and must be a positive integer. This distinction is a common point of confusion and a frequent topic in exams.

Step 1: Understanding the Concept:

This question asks for the term that defines the number of reacting species in a single-step (elementary) reaction. This is a direct definition-based question from chemical kinetics.


Step 2: Detailed Explanation:

Let's analyze the given options:

(A) Activation energy of the reaction: This is the minimum amount of energy required for reactants to transform into products. It relates to the reaction rate but not the number of reacting molecules.


(B) Stoichiometry of the reaction: This refers to the quantitative relationship between reactants and products in a balanced chemical equation for an overall reaction. While it can sometimes be the same as molecularity for an elementary reaction, it is not the definition itself. Stoichiometry applies to complex reactions as well.


(C) Molecularity of the reaction: This is the precise definition. Molecularity is the number of molecules, atoms, or ions that come together to react in an elementary (single-step) reaction. For example, in the elementary reaction A + B \(\rightarrow\) P, the molecularity is 2.


(D) Order of the reaction: This is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. It describes how the rate is affected by the concentration of reactants and is not necessarily equal to the number of reacting molecules, especially in complex reactions.


Step 3: Final Answer:

The definition provided in the question perfectly matches the definition of "molecularity of the reaction".
Quick Tip: For an elementary (single-step) reaction, the order of the reaction with respect to a reactant is equal to its stoichiometric coefficient, and the overall order is equal to the molecularity. However, for complex (multi-step) reactions, this is not true. The order is determined experimentally, while molecularity is a theoretical concept applied only to elementary steps.

Step 1: Understanding the Concept:

The question asks to identify an element based on its electronic configuration. We need to find the atomic number (Z) corresponding to the given configuration and then identify the element.


Step 2: Key Formula or Approach:

The total number of electrons in an atom is equal to its atomic number. We can calculate the total number of electrons from the given configuration.

The notation [Ar] represents the electronic configuration of Argon, which has 18 electrons.

Total electrons = (electrons in [Ar]) + (electrons in 3d orbital) + (electrons in 4s orbital).


Step 3: Detailed Explanation:

The given electronic configuration is [Ar]3d\(^{10}\)4s\(^1\).

Number of electrons in Argon (Ar) core = 18.

Number of electrons in the 3d subshell = 10.

Number of electrons in the 4s subshell = 1.


Total number of electrons = 18 + 10 + 1 = 29.


The atomic number (Z) of the element is 29.

Now we identify the element with Z = 29 from the options:

(A) Cu (Copper) has Z = 29.

(B) Zn (Zinc) has Z = 30.

(C) Cr (Chromium) has Z = 24.

(D) Mn (Manganese) has Z = 25.


The element with atomic number 29 is Copper (Cu). This configuration is an exception to the Aufbau principle. The expected configuration for Cu would be [Ar]3d\(^9\)4s\(^2\), but a completely filled d-subshell (3d\(^{10}\)) is more stable, so an electron from the 4s orbital moves to the 3d orbital.


Step 4: Final Answer:

The element with the electronic configuration [Ar]3d\(^{10}\)4s\(^1\) is Copper (Cu).
Quick Tip: Memorize the two major exceptions to the Aufbau principle in the first transition series: Chromium (Cr, Z=24) is [Ar]3d\(^5\)4s\(^1\) (half-filled d-subshell stability) and Copper (Cu, Z=29) is [Ar]3d\(^{10}\)4s\(^1\) (fully-filled d-subshell stability). These are very frequently asked in exams.

Step 1: Understanding the Concept:

This question deals with structural isomerism in coordination compounds. We need to identify the specific type of isomerism exhibited by the two given complex ions.


Step 2: Detailed Explanation:

Let's analyze the structure of the two complex ions:

1. [Co(NH\(_3\))\(_5\)(NO\(_2\))]\(^{2+}\): In this complex, the ligand NO\(_2^-\) is bonded to the central cobalt (Co) atom through the nitrogen atom. This is called the nitro complex.

2. [Co(NH\(_3\))\(_5\)(ONO)]\(^{2+}\): In this complex, the ligand ONO\(^-\) is bonded to the central cobalt (Co) atom through an oxygen atom. This is called the nitrito complex.


The molecular formula for both complexes is the same. The central metal ion and the other ligands (NH\(_3\)) are also the same. The only difference is the point of attachment of the NO\(_2^-\) ligand.


The ligand NO\(_2^-\) is an ambidentate ligand, meaning it has two different donor atoms (N and O) through which it can coordinate to the central metal ion.


Now let's review the types of isomers given in the options:

(A) Ionization isomers: These isomers have the same composition but yield different ions in solution. This occurs when a counter ion and a ligand exchange places. This is not the case here.

(B) Linkage isomers: This type of isomerism arises in a coordination compound containing an ambidentate ligand. The isomers differ in which atom of the ambidentate ligand is bonded to the metal ion. This perfectly describes the given pair of complexes.

(C) Co-ordination isomers: This occurs in compounds containing complex cations and complex anions, where there is an interchange of ligands between the cationic and anionic coordination spheres. This is not applicable here.

(D) Geometrical isomers: These are stereoisomers that have the same chemical formula and bonds but different spatial arrangements of atoms (e.g., cis-trans isomers). This is not the difference between the two given ions.


Step 3: Final Answer:

The two complex ions are examples of linkage isomers because they differ in the donor atom of the ambidentate ligand NO\(_2^-\).
Quick Tip: Whenever you see a coordination compound with an ambidentate ligand, immediately think of linkage isomerism. Common ambidentate ligands include NO\(_2^-\) (nitro/nitrito), SCN\(^-\) (thiocyanato/isothiocyanato), and CN\(^-\) (cyano/isocyano).

Step 1: Understanding the Concept:

A diamagnetic species is one that has no unpaired electrons. All its electrons are paired. A paramagnetic species has one or more unpaired electrons. To determine if a complex is diamagnetic, we need to find the electronic configuration of the central metal ion and consider the effect of the ligands (strong-field or weak-field) on the d-electron arrangement.


Step 2: Detailed Explanation:

We will analyze each complex:


(A) [Ni(CN)\(_4\)]\(^{2-}\):

- Atomic number of Ni is 28. Electronic configuration: [Ar]3d\(^8\)4s\(^2\).

- In this complex, Ni is in the +2 oxidation state (since CN has a -1 charge, x + 4(-1) = -2 \(\Rightarrow\) x = +2).

- Ni\(^{2+}\) configuration: [Ar]3d\(^8\).

- CN\(^-\) is a strong-field ligand. It will cause the pairing of electrons in the 3d orbitals.

- The 8 electrons in the 3d orbitals will pair up to occupy 4 orbitals, leaving one 3d orbital empty.

- The hybridization will be dsp\(^2\) (using one 3d, one 4s, and two 4p orbitals), resulting in a square planar geometry.

- Since all 8 electrons are paired, there are no unpaired electrons. Thus, [Ni(CN)\(_4\)]\(^{2-}\) is diamagnetic.


(B) [NiCl\(_4\)]\(^{2-}\):

- Ni is in the +2 oxidation state. Ni\(^{2+}\) configuration: [Ar]3d\(^8\).

- Cl\(^-\) is a weak-field ligand. It does not cause pairing of electrons.

- The 8 electrons in the 3d orbitals will be arranged according to Hund's rule: three orbitals will be fully filled and two will have one electron each.

- There are 2 unpaired electrons.

- The hybridization is sp\(^3\) (using one 4s and three 4p orbitals), resulting in a tetrahedral geometry.

- The complex is paramagnetic.


(C) [Fe(CN)\(_6\)]\(^{3-}\):

- Atomic number of Fe is 26. Electronic configuration: [Ar]3d\(^6\)4s\(^2\).

- In this complex, Fe is in the +3 oxidation state (x + 6(-1) = -3 \(\Rightarrow\) x = +3).

- Fe\(^{3+}\) configuration: [Ar]3d\(^5\).

- CN\(^-\) is a strong-field ligand, causing pairing.

- The 5 electrons will be arranged in the lower energy t\(_{2g}\) orbitals. Two orbitals will be paired, and one will be unpaired (t\(_{2g}^5\)).

- There is 1 unpaired electron. The complex is paramagnetic.


(D) [CoF\(_6\)]\(^{3-}\):

- Atomic number of Co is 27. Electronic configuration: [Ar]3d\(^7\)4s\(^2\).

- In this complex, Co is in the +3 oxidation state (x + 6(-1) = -3 \(\Rightarrow\) x = +3).

- Co\(^{3+}\) configuration: [Ar]3d\(^6\).

- F\(^-\) is a weak-field ligand, so no pairing occurs.

- The 6 electrons will be arranged in the d orbitals according to Hund's rule (t\(_{2g}^4\) e\(_g^2\)).

- There will be 4 unpaired electrons. The complex is paramagnetic.


Step 3: Final Answer:

Based on the analysis, only [Ni(CN)\(_4\)]\(^{2-}\) has no unpaired electrons and is therefore diamagnetic.
Quick Tip: To quickly determine magnetic properties, remember the spectrochemical series. Ligands like CN\(^-\) and CO are strong-field and cause pairing. Halides (F\(^-\), Cl\(^-\), etc.) and H\(_2\)O are generally weak-field and don't cause pairing for first-row transition metals. Checking the ligand type is the first crucial step.

Step 1: Understanding the Concept:

This question requires applying the IUPAC nomenclature rules for substituted benzene rings. The rules involve identifying the parent compound and numbering the substituents to give them the lowest possible locants, often following alphabetical order.


Step 2: Detailed Explanation:

The given structure is a benzene ring with two substituents: a methyl group (-CH\(_3\)) and a chlorine atom (-Cl). They are at para positions (positions 1 and 4) to each other.


There are two ways to name this compound according to IUPAC rules:

Method 1: Using a common name as the parent.

Methylbenzene is commonly known as Toluene. If we consider Toluene as the parent compound, the methyl group is at position 1. The chlorine atom is then at position 4. The name would be 4-Chlorotoluene. This is a correct IUPAC-accepted name. However, it's not listed as an option in this exact form.


Method 2: Treating benzene as the parent and listing substituents.

When substituents are different, they are listed in alphabetical order. The substituents are 'Chloro' and 'Methyl'.

Alphabetically, 'Chloro' comes before 'Methyl'.

According to the rule of first point of difference (lowest locant set), we can number in two ways:

- Start numbering from the carbon with the Chloro group: 1-Chloro, 4-Methyl. The locant set is (1, 4).

- Start numbering from the carbon with the Methyl group: 1-Methyl, 4-Chloro. The locant set is (1, 4).

Since both numbering schemes give the same locant set (1, 4), we use the alphabetical order to decide which substituent gets the lower number. Since 'Chloro' comes first alphabetically, it gets the number 1.

Therefore, the correct IUPAC name is 1-Chloro-4-Methylbenzene.


Let's evaluate the options:

(A) Methylchlorobenzene: This name is ambiguous as it doesn't specify the positions of the substituents.

(B) Toluene: This is the name for methylbenzene, not the given compound.

(C) 1-Chloro-4-Methylbenzene: This name correctly follows the IUPAC rule of assigning the lower number to the substituent that comes first alphabetically when locant sets are otherwise tied.

(D) 1-Methyl-4-Chlorobenzene: This name is incorrect because it violates the alphabetical priority rule for numbering when locants are tied. 'Chloro' should get the lower number than 'Methyl'.


Step 3: Final Answer:

The most systematic and correct IUPAC name among the choices is 1-Chloro-4-Methylbenzene.
Quick Tip: For polysubstituted benzenes, remember the priority order for numbering: 1. Give the principal functional group the lowest number. 2. Number to give the lowest possible locants to all substituents. 3. If there's still a tie, give the lowest number to the substituent that comes first alphabetically. In this case, Chloro vs Methyl, the tie is broken by giving 'C' the priority over 'M'.

Step 1: Understanding the Concept:

This question tests the knowledge of oxidation reactions of alcohols. The product of oxidation depends on the type of alcohol (primary, secondary, or tertiary) and the strength of the oxidizing agent.


Step 2: Key Formula or Approach:

The general reactions for alcohol oxidation are:

- Primary alcohol (\(RCH_2OH\)) \(\xrightarrow{Mild oxidant}\) Aldehyde (\(RCHO\)) \(\xrightarrow{Strong oxidant}\) Carboxylic acid (\(RCOOH\)).

- Secondary alcohol (\(R_2CHOH\)) \(\xrightarrow{Oxidant}\) Ketone (\(R_2CO\)). Further oxidation is difficult under normal conditions.

- Tertiary alcohol (\(R_3COH\)) does not undergo oxidation easily without breaking C-C bonds.

Chromic anhydride (CrO\(_3\)) is a strong oxidizing agent.


Step 3: Detailed Explanation:

The question specifies a secondary alcohol. A secondary alcohol has the general formula R-CH(OH)-R', where the carbon atom bonded to the -OH group is also bonded to two other carbon atoms.

When a secondary alcohol is oxidized, the hydrogen atom from the hydroxyl group and the hydrogen atom from the carbon atom bearing the hydroxyl group are removed. This results in the formation of a carbon-oxygen double bond (a carbonyl group).

The reaction is:
\[ R-CH(OH)-R' + [O] \xrightarrow{CrO_3} R-C(=O)-R' + H_2O \]
The product, R-C(=O)-R', which has a carbonyl group bonded to two alkyl groups, is a ketone.

Even though CrO\(_3\) is a strong oxidizing agent, ketones are resistant to further oxidation under normal conditions. Therefore, the reaction stops at the ketone stage.


Step 4: Final Answer:

The oxidation of a secondary alcohol with chromic anhydride yields a ketone.
Quick Tip: A simple way to remember the oxidation products of alcohols: - **1° alcohol** → Aldehyde → Carboxylic acid - **2° alcohol** → Ketone (and stops there) - **3° alcohol** → No reaction (under mild conditions) This pattern is fundamental to organic chemistry and is very useful in synthesis and identification problems.

Step 1: Understanding the Concept:

This question asks to identify the specific named reaction used to synthesize salicylic acid (2-hydroxybenzoic acid) from phenol. This requires knowledge of common named reactions in organic chemistry involving phenols.


Step 2: Detailed Explanation:

Let's review the reactions listed:

(A) Reimer-Tiemann reaction: In this reaction, phenol is treated with chloroform (CHCl\(_3\)) in the presence of a strong base like sodium hydroxide (NaOH). The product is primarily salicylaldehyde (2-hydroxybenzaldehyde). This is not salicylic acid.


(B) Friedel-Crafts reaction: This reaction involves the alkylation or acylation of an aromatic ring using an alkyl halide or acyl halide in the presence of a Lewis acid catalyst (like AlCl\(_3\)). Phenols are not ideal substrates as the -OH group coordinates with the Lewis acid, deactivating the ring. It is not used to make salicylic acid.


(C) Kolbe's reaction (or Kolbe-Schmitt reaction): This is the correct reaction. First, phenol is converted to its more reactive phenoxide ion by treating it with NaOH. The sodium phenoxide is then heated with carbon dioxide (CO\(_2\)) under pressure (around 125°C and 4-7 atm). This results in electrophilic substitution of the phenoxide ring by CO\(_2\), primarily at the ortho position, a process called carboxylation. Subsequent acidification of the product yields salicylic acid.

Reaction Scheme:
1. Phenol + NaOH \(\rightarrow\) Sodium phenoxide
2. Sodium phenoxide + CO\(_2\) (heat, pressure) \(\rightarrow\) Sodium salicylate
3. Sodium salicylate + H\(^+\) \(\rightarrow\) Salicylic acid


(D) Coupling reaction: This typically refers to the reaction of a diazonium salt with an activated aromatic compound (like phenol or aniline) to form an azo compound (a dye). This does not produce salicylic acid.


Step 3: Final Answer:

The conversion of phenol to salicylic acid is achieved through the Kolbe's reaction.
Quick Tip: Associate key reagents and products with named reactions for phenols: - **Phenol + CHCl\(_3\)/NaOH \(\rightarrow\) Reimer-Tiemann \(\rightarrow\) Salicylaldehyde.** - **Phenol + NaOH then CO\(_2\)/H\(^+\) \(\rightarrow\) Kolbe's \(\rightarrow\) Salicylic acid.** This direct association helps in quickly answering multiple-choice questions.

Step 1: Understanding the Concept:

Denaturation of protein is a process in which the protein loses its native three-dimensional structure (quaternary, tertiary, and secondary structures) due to the application of some external stress or compound, such as a strong acid or base, a concentrated inorganic salt, an organic solvent, or heat. The primary structure (sequence of amino acids) remains intact.


Step 2: Detailed Explanation:

Let's analyze the given options:

(A) Coagulation of egg white: Egg white is primarily a protein called albumin. When an egg is heated (e.g., boiled or fried), the heat disrupts the hydrogen bonds and other non-covalent interactions that maintain the specific folded structure of albumin. The unfolded protein chains then aggregate and precipitate, causing the egg white to turn from a clear liquid to an opaque solid. This is a classic example of denaturation by heat.


(B) Curdling of milk: Milk contains a protein called casein. When the pH of milk is lowered (by adding an acid like lemon juice or by bacterial action that produces lactic acid), the negative charges on the surface of casein micelles are neutralized. This disrupts the electrostatic repulsions that keep the micelles suspended, causing them to clump together and precipitate, forming curd. This is an example of denaturation by change in pH.


(C) Clotting of blood: Blood clotting is a complex enzymatic cascade involving multiple proteins (clotting factors). The final step involves the conversion of the soluble protein fibrinogen into insoluble fibrin strands, which form a mesh to trap blood cells. While it involves a change in protein structure, it is a highly specific, regulated physiological process, not a general disruption of structure like denaturation. It is generally not considered a standard example of denaturation in the same context as (A) and (B).


Since both coagulation of egg white and curdling of milk are clear examples of protein denaturation, the most appropriate answer is (D).


Step 3: Final Answer:

Both the coagulation of egg white and the curdling of milk are prime examples of the denaturation of proteins.
Quick Tip: Denaturation usually results in the loss of the protein's biological activity. Think of it as "unfolding" the protein. Common causes are heat, extreme pH, heavy metal ions, and organic solvents. The two most common textbook examples are boiling an egg and curdling milk.

Step 1: Understanding the Concept:

This question asks about the type of covalent bond that connects individual nucleotide units to form a polynucleotide chain, such as in DNA or RNA.


Step 2: Detailed Explanation:

Let's break down the structure of a polynucleotide and the function of each type of bond mentioned:

- A nucleotide consists of three components: a pentose sugar (deoxyribose in DNA, ribose in RNA), a nitrogenous base, and a phosphate group.


(A) Glycosidic linkage: This is the covalent bond that connects the nitrogenous base to the 1' carbon of the pentose sugar within a single nucleotide. It does not join two different nucleotides together.


(B) Peptide linkage: This is an amide bond (-CO-NH-) that joins amino acids together to form polypeptide chains (proteins). It is not found in nucleic acids.


(C) Hydrogen bonding: These are non-covalent bonds. In DNA, hydrogen bonds form between the complementary nitrogenous bases (A with T, and G with C) to hold the two strands of the double helix together. They do not form the backbone of the chain.


(D) Phosphodiester linkage: This is the correct answer. A phosphodiester bond is a strong covalent bond that forms the backbone of DNA and RNA strands. It connects the 3' carbon of one sugar molecule to the 5' carbon of another through a phosphate group. Specifically, the phosphate group forms an ester bond with the 5'-OH group of one nucleotide and another ester bond with the 3'-OH group of the adjacent nucleotide.


Step 3: Final Answer:

The linkage that joins nucleotides together to form a polymer is the phosphodiester linkage.
Quick Tip: Visualize the "ladder" structure of DNA: - The "rungs" of the ladder are the base pairs connected by **hydrogen bonds**. - The "side rails" or the backbone of the ladder are made of sugar and phosphate units connected by **phosphodiester bonds**. - Within each nucleotide, the base is attached to the sugar by a **glycosidic bond**.

Step 1: Understanding the Concept:

This is a knowledge-based question from the topic of biomolecules, specifically vitamins and their deficiency diseases. Scurvy is a well-known deficiency disease.


Step 2: Detailed Explanation:

Let's identify the vitamin associated with each option and its deficiency disease:

(A) Vitamin B1 (Thiamine): Its deficiency causes the disease Beriberi, which affects the nervous and cardiovascular systems.


(B) Vitamin B2 (Riboflavin): Its deficiency can lead to ariboflavinosis, with symptoms like sore throat, inflammation of the tongue (glossitis), and cracks at the corners of the mouth (cheilosis).


(C) Ascorbic acid: This is the chemical name for Vitamin C. A deficiency of Vitamin C causes Scurvy. The symptoms of scurvy include fatigue, bleeding gums, joint pain, and poor wound healing, which are related to the impaired synthesis of collagen, a protein for which Vitamin C is an essential cofactor.


(D) Glutamic acid: This is a non-essential amino acid, a building block of proteins. It is not a vitamin, and its deficiency is not associated with scurvy.


Step 3: Final Answer:

Scurvy is caused by the deficiency of Ascorbic acid (Vitamin C).
Quick Tip: Creating a simple table of essential vitamins, their chemical names, and their deficiency diseases is an effective way to study for exams. For example: - Vit A (Retinol) → Night Blindness - Vit B1 (Thiamine) → Beriberi - Vit C (Ascorbic Acid) → Scurvy - Vit D (Calciferol) → Rickets This makes recall much faster.

Step 1: Understanding the Concept:

This question tests the relationship between half-life and initial concentration for a first-order reaction. It is an Assertion-Reason type question, where we must evaluate the truthfulness of both statements and the causal link between them.


Step 2: Key Formula or Approach:

The half-life (t\(_{1/2}\)) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \]
where k is the rate constant.


Step 3: Detailed Explanation:

Evaluate Assertion (A):

The assertion states that for a first-order reaction, doubling the initial concentration of the reactant doubles its half-life.

Looking at the formula \( t_{1/2} = 0.693/k \), we can see that the initial concentration term, [A]\(_0\), is not present. This means that the half-life of a first-order reaction is independent of the initial concentration. Therefore, doubling the concentration will have no effect on the half-life.

Thus, Assertion (A) is false.


Evaluate Reason (R):

The reason states that the half-life of a first-order reaction does not depend upon the initial concentration of the reactant.

As established from the formula \( t_{1/2} = 0.693/k \), this statement is correct. The half-life only depends on the rate constant, k.

Thus, Reason (R) is true.


Step 4: Final Answer:

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: Remember the dependence of half-life on initial concentration for different orders: - **Zero-order:** \(t_{1/2} \propto [A]_0\) (directly proportional) - **First-order:** \(t_{1/2}\) is independent of \([A]_0\) - **Second-order:** \(t_{1/2} \propto 1/[A]_0\) (inversely proportional) This summary helps to quickly solve such Assertion-Reason questions.

Step 1: Understanding the Concept:

This question relates the reactivity of a metal with dilute acids to its position in the electrochemical series, which is quantified by its standard electrode potential (E°).


Step 2: Detailed Explanation:

Evaluate Assertion (A):

The reaction for a metal (M) liberating hydrogen from a dilute acid (like HCl) is:

M(s) + 2H\(^+\)(aq) \(\rightarrow\) M\(^{2+}\)(aq) + H\(_2\)(g)

For this reaction to be spontaneous, the metal M must be more reactive than hydrogen. This means the metal must be able to reduce H\(^+\) ions to H\(_2\) gas. In the electrochemical series, metals that are placed above hydrogen have negative standard reduction potentials and can displace hydrogen from dilute acids. Metals below hydrogen have positive standard reduction potentials and cannot.

Copper (Cu) is placed below hydrogen in the electrochemical series. Therefore, it is less reactive than hydrogen and cannot liberate H\(_2\) from dilute non-oxidizing mineral acids like HCl or H\(_2\)SO\(_4\).

Thus, Assertion (A) is true.


Evaluate Reason (R):

The reason states that Cu has a positive electrode potential. The standard reduction potential for the half-reaction Cu\(^{2+}\)(aq) + 2e\(^-\) \(\rightarrow\) Cu(s) is E° = +0.34 V.

The standard reduction potential for the hydrogen electrode (2H\(^+\) + 2e\(^-\) \(\rightarrow\) H\(_2\)) is defined as E° = 0.00 V.

Since the E° value for copper is positive, it means that Cu\(^{2+}\) has a greater tendency to be reduced than H\(^+\). Conversely, Cu metal has a lesser tendency to be oxidized than H\(_2\) gas. Therefore, Cu cannot oxidize H\(^+\) to H\(_2\).
Thus, Reason (R) is true.


Connecting Assertion and Reason:

The reason (positive electrode potential) is the fundamental electrochemical principle that explains the assertion (inability to liberate H\(_2\)). Because copper's reduction potential is positive (greater than hydrogen's), it cannot displace hydrogen. The reason correctly explains the assertion.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: Remember the rule for the electrochemical series: Any metal higher in the series (more negative E°) can displace any metal lower in the series (more positive E°) from its salt solution. Since hydrogen (E°=0) is above copper (E°=+0.34V), copper cannot displace hydrogen from acid.

Step 1: Understanding the Concept:

This question probes the scope and limitations of the Gabriel Phthalimide synthesis, a method for preparing primary amines. It specifically asks why this method fails for aromatic primary amines.


Step 2: Detailed Explanation:

Evaluate Assertion (A):

The Gabriel Phthalimide synthesis is a two-step method to prepare primary amines.

Step 1: Phthalimide is treated with a base (like KOH) to form the phthalimide anion, which is a good nucleophile.

Step 2: This anion then attacks an alkyl halide in a nucleophilic substitution (S\(_N\)2) reaction to form an N-alkylphthalimide.

Step 3: Hydrolysis or hydrazinolysis of the N-alkylphthalimide yields a pure primary alkyl amine.

To prepare an aromatic primary amine (like aniline), the required substrate in Step 2 would be an aryl halide (like chlorobenzene). However, the synthesis does not work with aryl halides. Therefore, aromatic primary amines cannot be prepared using this method.

Thus, Assertion (A) is true.


Evaluate Reason (R):

The reason states that aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

This is correct. Aryl halides are very unreactive towards nucleophilic substitution reactions for two main reasons:

1. Resonance: The lone pair of electrons on the halogen atom participates in resonance with the benzene ring, giving the Carbon-Halogen (C-X) bond a partial double-bond character. This makes the bond stronger and harder to break.

2. Hybridization: The carbon atom of the C-X bond in an aryl halide is sp\(^2\) hybridized, which is more electronegative than the sp\(^3\) carbon in an alkyl halide. This holds the electrons of the C-X bond more tightly, making it shorter and stronger.

Because of this low reactivity, the phthalimide anion (a nucleophile) cannot displace the halide from the benzene ring under the normal conditions of the Gabriel synthesis.

Thus, Reason (R) is true.


Connecting Assertion and Reason:

The reason (unreactivity of aryl halides in nucleophilic substitution) is the precise explanation for why the Gabriel synthesis (which relies on this type of reaction) fails to produce aromatic primary amines.


Step 3: Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
Quick Tip: Remember that Gabriel Phthalimide synthesis is exclusively for the preparation of primary **aliphatic** amines. It fails for aromatic amines due to the low reactivity of aryl halides, and it fails for tertiary amines due to steric hindrance.

Step 1: Understanding the Concept:

This question deals with the classification and storage of vitamins in the human body. Vitamins are classified as either fat-soluble or water-soluble, which determines how they are absorbed, stored, and excreted.


Step 2: Detailed Explanation:

Evaluate Assertion (A):

The assertion states that Vitamin D cannot be stored in our body.

Vitamins are categorized into two groups:

- Fat-soluble vitamins: A, D, E, and K. These vitamins dissolve in fat and can be stored in the body's fatty tissues (adipose tissue) and the liver. Because they can be stored, they do not need to be consumed every day.

- Water-soluble vitamins: Vitamin C and the B-complex vitamins. These dissolve in water and are not stored in the body in significant amounts. Any excess is usually excreted in the urine.

Since Vitamin D is a fat-soluble vitamin, it can be stored in the body.

Thus, Assertion (A) is false.


Evaluate Reason (R):

The reason states that Vitamin D is a fat-soluble vitamin and is not excreted from the body in urine.

This statement is correct. As mentioned above, Vitamin D belongs to the group of fat-soluble vitamins. Because it is not soluble in water, it is not readily excreted through the kidneys in urine. Instead, it is stored in fat depots.

Thus, Reason (R) is true.


Step 3: Final Answer:

Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: Use the mnemonic "ADEK" to remember the fat-soluble vitamins (A, D, E, K). All other essential vitamins (B-complex and C) are water-soluble. Knowing this simple classification helps answer many questions about vitamin storage and toxicity.

Step 1: Understanding the Concept:

The question asks for the time required for a concentration change in a zero-order reaction. We need to use the integrated rate law for a zero-order reaction.


Step 2: Key Formula or Approach:

The integrated rate law for a zero-order reaction is given by:
\[ [A]_t = -kt + [A]_0 \]
where:
\( [A]_t \) = concentration of reactant A at time t
\( [A]_0 \) = initial concentration of reactant A

k = rate constant

t = time

Rearranging the formula to solve for time (t):
\[ t = \frac{[A]_0 - [A]_t}{k} \]

Step 3: Detailed Explanation:

Given values are:

Rate constant, \( k = 0.0030 \) mol L\(^{-1}\)s\(^{-1}\)

Initial concentration, \( [A]_0 = 0.10 \) M

Final concentration, \( [A]_t = 0.075 \) M


Substitute these values into the rearranged formula:
\[ t = \frac{0.10 - 0.075}{0.0030} \] \[ t = \frac{0.025}{0.0030} \] \[ t = \frac{25}{3} \] \[ t \approx 8.33 s \]

Step 4: Final Answer:

It will take approximately 8.33 seconds for the initial concentration of A to fall from 0.10 M to 0.075 M.
Quick Tip: For a zero-order reaction, the rate is constant (Rate = k). This means the concentration decreases linearly with time. Remember the units of the rate constant for different orders: Zero order: mol L\(^{-1}\)s\(^{-1}\) First order: s\(^{-1}\) Second order: L mol\(^{-1}\)s\(^{-1}\) The units given in the question (mol L\(^{-1}\)s\(^{-1}\)) confirm it is a zero-order reaction.

Step 1: Understanding the Concept:

The question relates the rate of a zero-order reaction to the rates of formation of its products. For a zero-order reaction, the rate of reaction is equal to the rate constant. The rates of appearance of products are related to the overall reaction rate by their stoichiometric coefficients.


Step 2: Key Formula or Approach:

First, write the balanced chemical equation for the decomposition of ammonia (NH\(_3\)).
\[ 2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g) \]
The rate of the reaction can be expressed in terms of the change in concentration of reactants and products:
\[ Rate = -\frac{1}{2}\frac{d[NH_3]}{dt} = +\frac{d[N_2]}{dt} = +\frac{1}{3}\frac{d[H_2]}{dt} \]
For a zero-order reaction, Rate = k.

The rate of production of N\(_2\) is \( \frac{d[N_2]}{dt} \).

The rate of production of H\(_2\) is \( \frac{d[H_2]}{dt} \).


Step 3: Detailed Explanation:

Given that the reaction is zero order and the rate constant \( k = 2.5 \times 10^{-4} \) mol L\(^{-1}\) s\(^{-1}\).

So, the rate of the reaction is:
\[ Rate = k = 2.5 \times 10^{-4} mol L^{-1} s^{-1} \]
Now, we can find the rates of production of N\(_2\) and H\(_2\).


Rate of production of N\(_2\):

From the rate expression, \( \frac{d[N_2]}{dt} = Rate \).
\[ \frac{d[N_2]}{dt} = k = 2.5 \times 10^{-4} mol L^{-1} s^{-1} \]

Rate of production of H\(_2\):

From the rate expression, \( \frac{1}{3}\frac{d[H_2]}{dt} = Rate \).
\[ \frac{d[H_2]}{dt} = 3 \times Rate \] \[ \frac{d[H_2]}{dt} = 3 \times k = 3 \times (2.5 \times 10^{-4}) \] \[ \frac{d[H_2]}{dt} = 7.5 \times 10^{-4} mol L^{-1} s^{-1} \]

Step 4: Final Answer:

The rate of production of N\(_2\) is 2.5 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\).

The rate of production of H\(_2\) is 7.5 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\).
Quick Tip: Always start by writing the balanced chemical equation. The stoichiometric coefficients are crucial for relating the overall reaction rate to the rate of consumption of reactants or formation of products. Remember, for products, the rate is positive (appearance), and for reactants, it's negative (disappearance).

Step 1: Understanding the Concept:

The term "pseudo" means false or appearing to be something it is not. A pseudo first-order reaction is a chemical reaction that appears to be first order, but its true mechanism involves more than one species in the rate-determining step.


Step 2: Detailed Explanation:

Consider a bimolecular reaction:
\[ A + B \rightarrow P \]
The rate law for this reaction would typically be:
\[ Rate = k[A][B] \]
This is a second-order reaction.

However, if one of the reactants, say B, is taken in very large excess compared to A (e.g., B is the solvent), its concentration [B] will not change significantly as the reaction proceeds. We can consider [B] to be a constant.

So, the rate law can be rewritten as:
\[ Rate = (k[B])[A] \]
Since k and [B] are both constants, we can combine them into a new constant, k'.
\[ k' = k[B] \]
The rate law simplifies to:
\[ Rate = k'[A] \]
This rate law is in the form of a first-order reaction. Such a reaction is called a pseudo first-order reaction.


A classic example is the acid-catalyzed hydrolysis of an ester, like ethyl acetate:
\[ CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH \]
The reaction takes place in an aqueous solution where water (H\(_2\)O) is the solvent and is present in a very large excess. The rate depends only on the concentration of the ester.
\[ Rate = k'[CH_3COOC_2H_5] \] Quick Tip: Look for reactions where one of the reactants is a solvent (like water in hydrolysis reactions). These are the most common examples of pseudo first-order reactions asked in exams. The concentration of the solvent is so large that it is considered constant.

Step 1: Understanding the Concept:

The half-life is a characteristic time scale for a chemical reaction. It provides a measure of how fast a reaction occurs. A shorter half-life indicates a faster reaction.


Step 2: Detailed Explanation:

The definition can be expressed mathematically. If \( [A]_0 \) is the initial concentration of a reactant at time t=0, then the half-life, \( t_{1/2} \), is the time at which the concentration becomes \( [A]_t = \frac{[A]_0}{2} \).

The formula for half-life depends on the order of the reaction:


For a zero-order reaction: The half-life is directly proportional to the initial concentration.

\[ t_{1/2} = \frac{[A]_0}{2k} \]
For a first-order reaction: The half-life is independent of the initial concentration. It is a constant for a given reaction at a specific temperature.

\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \]
For a second-order reaction: The half-life is inversely proportional to the initial concentration.

\[ t_{1/2} = \frac{1}{k[A]_0} \]

The concept of half-life is particularly important for first-order processes, such as radioactive decay.
Quick Tip: A key distinction to remember for exams is how half-life depends on initial concentration. For first-order reactions, \(t_{1/2}\) is constant. This is a unique and frequently tested property. For zero-order, it decreases as the reaction proceeds (since [A] decreases), and for second-order, it increases.

Step 1: Understanding the Concept:

The colour of transition metal compounds (ions in solution or solid state) is a characteristic property that arises from their electronic structure, specifically the electrons in their d-orbitals.


Step 2: Detailed Explanation:

The phenomenon of colour in transition metal compounds can be explained by the following points:


Presence of Partially Filled d-Orbitals: Most transition metal ions have incompletely filled d-orbitals.

Splitting of d-Orbitals: In the presence of ligands (in a complex) or counter-ions, the five degenerate (same energy) d-orbitals of the metal ion split into two or more sets of orbitals with different energy levels. For an octahedral complex, they split into a lower energy t\(_{2g}\) set and a higher energy e\(_g\) set.

d-d Transition: When visible light falls on the compound, an electron from a lower energy d-orbital can absorb energy corresponding to a specific wavelength of light and get promoted to a higher energy d-orbital. This process is called a d-d transition.

Complementary Colour: The amount of energy required for this transition falls in the visible region of the electromagnetic spectrum. The compound absorbs a particular colour (wavelength) from the white light to cause this transition. The remaining light is transmitted or reflected, and this transmitted light is the colour we perceive, which is the complementary colour of the light absorbed. For example, if a compound absorbs orange light, it appears blue.

Ions with empty d-orbitals (like Sc\(^{3+}\), Ti\(^{4+}\)) or completely filled d-orbitals (like Zn\(^{2+}\), Cu\(^+\)) cannot undergo d-d transitions and are therefore generally colourless.
Quick Tip: To predict if a transition metal ion will be coloured, check its d-electron count. If it's d\(^1\) through d\(^9\), it's likely to be coloured. If it's d\(^0\) or d\(^{10}\), it will be colourless or white. This is a quick check for many exam questions.

Step 1: Understanding the Concept:

The definition of a transition element is specific and based on the electronic configuration of the d-subshell. We need to check if zinc (Zn) fits this definition.


Step 2: Detailed Explanation:


Definition of a Transition Element: A transition element is defined as an element that has an incompletely filled d subshell (i.e., d\(^1\) to d\(^9\) configuration) in its ground state or in any of its stable oxidation states.

Electronic Configuration of Zinc (Zn):

The atomic number of zinc is 30.
Its ground state electronic configuration is [Ar] 3d\(^{10}\) 4s\(^2\).
In this state, the 3d subshell is completely filled (d\(^{10}\)).

Electronic Configuration of Zinc Ion (Zn\(^{2+}\)):

The only common and stable oxidation state for zinc is +2.
To form the Zn\(^{2+}\) ion, two electrons are lost from the outermost 4s orbital.
The electronic configuration of Zn\(^{2+}\) is [Ar] 3d\(^{10}\).
In its common oxidation state, the 3d subshell is also completely filled.

Conclusion: Since zinc has a completely filled d-orbital in both its elemental ground state and its only common oxidation state, it does not meet the definition of a transition element. The same logic applies to the other elements in Group 12, Cadmium (Cd) and Mercury (Hg). Quick Tip: Remember that the elements of Group 12 (Zn, Cd, Hg) are considered d-block elements because their last electron enters the d-orbital, but they are not considered transition elements due to their completely filled d-orbitals. This is a very common point of confusion tested in exams.

Step 1: Understanding the Concept:

This question requires the application of IUPAC rules for naming coordination compounds. This involves identifying the cation and anion, naming the ligands, naming the central metal, and determining its oxidation state.


Step 2: Detailed Explanation:

The formula is [Co(NH\(_3\))\(_4\)(H\(_2\)O)Cl]Cl\(_2\).


Identify Cation and Anion: The part in the square brackets is the complex cation, [Co(NH\(_3\))\(_4\)(H\(_2\)O)Cl]\(^{2+}\), and the two Cl ions outside are the counter-anions. The cation is named first.

Name the Ligands: Identify all ligands inside the coordination sphere and name them in alphabetical order.

NH\(_3\): ammine
H\(_2\)O: aqua
Cl: chlorido

Alphabetical order: ammine, aqua, chlorido.

Use Prefixes for Ligand Numbers: Use prefixes like di-, tri-, tetra- to indicate the number of each simple ligand.

Four NH\(_3\) groups: tetraammine
One H\(_2\)O group: aqua (mono- is usually omitted)
One Cl group: chlorido

The ligand part of the name is: tetraammineaquachlorido.

Determine the Oxidation State of the Central Metal (Co):
Let the oxidation state of Cobalt be 'x'. The charge of NH\(_3\) and H\(_2\)O is 0, and the charge of Cl is -1. The overall charge of the complex ion must balance the two Cl\(^-\) counter-ions, so the complex has a +2 charge.

\[ x + 4(0) + 1(0) + 1(-1) = +2 \]
\[ x - 1 = +2 \]
\[ x = +3 \]
The oxidation state is +3, which is written as (III) in Roman numerals.

Name the Central Metal: Since the complex is a cation, the metal name is used as is: cobalt. So, the full name of the cation is Tetraammineaquachloridocobalt(III).

Name the Anion: The counter-ion is Cl\(^-\), which is named chloride. The prefix 'di-' is not used for counter-ions.

Combine the Names: Combine the cation and anion names (with a space in between).

Tetraammineaquachloridocobalt(III) chloride Quick Tip: When alphabetizing ligands, ignore the numerical prefixes (di-, tri-, tetra-). For example, in dichlorido and ammine, 'a' of ammine comes before 'c' of chlorido.

Step 1: Understanding the Concept:

This question requires naming a coordination compound containing a bidentate ligand (en). Special rules for naming complex ligands apply.


Step 2: Detailed Explanation:

The formula is [CrCl\(_2\)(en)\(_2\)]Cl.


Identify Cation and Anion: The complex cation is [CrCl\(_2\)(en)\(_2\)]\(^+\) and the counter-anion is Cl\(^-\).

Name the Ligands:

Cl: chlorido
en: ethylenediamine (This is a neutral, bidentate ligand)

Alphabetical order: chlorido, ethylenediamine.

Use Prefixes for Ligand Numbers:

Two Cl groups: dichlorido
Two 'en' groups: Since the ligand name 'ethylenediamine' already contains a numerical term ('di'), we use the special prefixes bis for two, tris for three, etc., and enclose the ligand name in parentheses. So, it is bis(ethylenediamine).

The ligand part of the name is: Dichloridobis(ethylenediamine).

Determine the Oxidation State of the Central Metal (Cr):
Let the oxidation state of Chromium be 'x'. The charge of Cl is -1, and the charge of ethylenediamine (en) is 0. The complex has a +1 charge to balance the single Cl\(^-\) counter-ion.

\[ x + 2(-1) + 2(0) = +1 \]
\[ x - 2 = +1 \]
\[ x = +3 \]
The oxidation state is +3, written as (III).

Name the Central Metal: The complex is a cation, so the metal is named chromium. The full cation name is Dichloridobis(ethylenediamine)chromium(III).

Name the Anion: The counter-ion is Cl\(^-\), named chloride.

Combine the Names:

Dichloridobis(ethylenediamine)chromium(III) chloride Quick Tip: Always use prefixes like 'bis', 'tris', and 'tetrakis' for ligands whose names already contain numerical prefixes (e.g., ethylenediamine, diethylamine) or are otherwise complex. Enclose the ligand name in parentheses when using these prefixes.

Step 1: Understanding the Concept:

The rate of an S\(_N\)1 (unimolecular nucleophilic substitution) reaction is determined by the stability of the carbocation intermediate formed in the rate-determining step. The more stable the carbocation, the faster the reaction.


Step 2: Key Formula or Approach:

The stability order of carbocations is:

Tertiary (3°) \textgreater Secondary (2°) \textgreater Primary (1°) \textgreater Methyl

This stability is due to two main effects:


Inductive Effect (+I): Alkyl groups are electron-donating and help disperse the positive charge on the carbocation, stabilizing it.
Hyperconjugation: The overlap of C-H \(\sigma\)-bonds with the empty p-orbital of the carbocationic carbon delocalizes the positive charge. More alkyl groups provide more C-H bonds for hyperconjugation.


Step 3: Detailed Explanation:

Let's analyze the two compounds:


2-chloro-2-methylpropane (tert-butyl chloride):

Structure: (CH\(_3\))\(_3\)C-Cl. This is a tertiary (3°) alkyl halide.

In the first step of an S\(_N\)1 reaction, the C-Cl bond breaks to form a carbocation:

\[ (CH_3)_3C-Cl \rightarrow (CH_3)_3C^+ + Cl^- \]
The intermediate is a tertiary carbocation (tert-butyl carbocation). This carbocation is highly stabilized by the +I effect and hyperconjugation from the nine \(\alpha\)-hydrogens on the three methyl groups.


2-chlorobutane:

Structure: CH\(_3\)-CHCl-CH\(_2\)-CH\(_3\). This is a secondary (2°) alkyl halide.

It forms a secondary carbocation upon ionization:

\[ CH_3-CHCl-CH_2-CH_3 \rightarrow CH_3-CH^+-CH_2-CH_3 + Cl^- \]
The intermediate is a secondary carbocation. It is stabilized by the +I effect and hyperconjugation from the five \(\alpha\)-hydrogens (three on the left methyl, two on the right methylene).


Conclusion:

The tertiary carbocation formed from 2-chloro-2-methylpropane is significantly more stable than the secondary carbocation formed from 2-chlorobutane. Therefore, 2-chloro-2-methylpropane will have a lower activation energy for the rate-determining step and will undergo the S\(_N\)1 reaction much faster.
Quick Tip: For S\(_N\)1 reactions, think "carbocation stability". The order of reactivity is 3° \textgreater 2° \textgreater 1°. For S\(_N\)2 reactions, think "steric hindrance". The order of reactivity is 1° \textgreater 2° \textgreater 3°. This simple mnemonic helps solve most comparison problems.

Step 1: Understanding the Concept:

The S\(_N\)2 (bimolecular nucleophilic substitution) reaction involves a single, concerted step where a nucleophile attacks the carbon atom bearing the leaving group from the backside. The rate of an S\(_N\)2 reaction is highly dependent on steric hindrance. Less sterically hindered alkyl halides react faster.


Step 2: Key Formula or Approach:

The general order of reactivity for alkyl halides in S\(_N\)2 reactions is:

Methyl \textgreater Primary (1°) \textgreater Secondary (2°) \textgreater Tertiary (3°) (due to increasing steric hindrance)

We need to classify each given compound as primary, secondary, or tertiary.


Step 3: Detailed Explanation:

Let's analyze the structure and classify each compound:


1-Bromopentane:

Structure: CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH\(_2\)Br. The bromine atom is attached to a carbon that is bonded to only one other carbon atom. This is a primary (1°) alkyl halide. It has the least steric hindrance around the reaction center.


2-Bromopentane:

Structure: CH\(_3\)CH(Br)CH\(_2\)CH\(_2\)CH\(_3\). The bromine atom is attached to a carbon that is bonded to two other carbon atoms. This is a secondary (2°) alkyl halide. It has more steric hindrance than the primary halide.


2-Bromo-2-methylbutane:

Structure: (CH\(_3\))\(_2\)C(Br)CH\(_2\)CH\(_3\). The bromine atom is attached to a carbon that is bonded to three other carbon atoms. This is a tertiary (3°) alkyl halide. It is the most sterically hindered of the three, making backside attack by a nucleophile very difficult.


Arranging in Increasing Order of Reactivity:

Based on the principle that reactivity in S\(_N\)2 reactions decreases with increasing steric hindrance (3° \textless 2° \textless 1°), we can arrange the compounds.

- Least Reactive: 2-Bromo-2-methylbutane (tertiary, most hindered)

- Moderately Reactive: 2-Bromopentane (secondary, intermediate hindrance)

- Most Reactive: 1-Bromopentane (primary, least hindered)


The increasing order of reactivity towards S\(_N\)2 displacement is:

2-Bromo-2-methylbutane \(\textless\) 2-Bromopentane \(\textless\) 1-Bromopentane
Quick Tip: To quickly assess steric hindrance for S\(_N\)2 reactions, simply classify the alkyl halide as primary (1°), secondary (2°), or tertiary (3°). A primary halide will almost always be more reactive than a secondary, which is far more reactive than a tertiary (which usually doesn't react via S\(_N\)2 at all).

Step 1: Understanding the Concept:

This problem involves Raoult's law for solutions containing a non-volatile solute (urea). Raoult's law states that the relative lowering of vapour pressure of a dilute solution is equal to the mole fraction of the solute.


Step 2: Key Formula or Approach:

According to Raoult's law:
\[ \frac{P^o - P_s}{P^o} = X_{solute} \]
where:
\( P^o \) = Vapour pressure of the pure solvent (water)
\( P_s \) = Vapour pressure of the solution
\( X_{solute} \) = Mole fraction of the solute (urea)

The mole fraction of the solute is calculated as:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]

Step 3: Detailed Explanation:

Given:
\( P^o = 24 \) mm Hg

Molar mass of urea (solute), \( M_{urea} = 60 \) g mol\(^{-1}\)

Molar mass of water (solvent), \( M_{water} = 18 \) g mol\(^{-1}\)


A 5% aqueous solution of urea means 5 g of urea is present in 100 g of the solution.

Mass of urea (\(w_{urea}\)) = 5 g

Mass of water (\(w_{water}\)) = Mass of solution - Mass of urea = 100 g - 5 g = 95 g


Now, calculate the number of moles of urea and water:

Moles of urea, \( n_{urea} = \frac{w_{urea}}{M_{urea}} = \frac{5 g}{60 g mol^{-1}} \approx 0.0833 mol \)

Moles of water, \( n_{water} = \frac{w_{water}}{M_{water}} = \frac{95 g}{18 g mol^{-1}} \approx 5.2778 mol \)


Next, calculate the mole fraction of urea:
\[ X_{urea} = \frac{n_{urea}}{n_{urea} + n_{water}} = \frac{0.0833}{0.0833 + 5.2778} = \frac{0.0833}{5.3611} \approx 0.01554 \]

Now, apply Raoult's law to find the vapour pressure of the solution, \( P_s \):
\[ \frac{24 - P_s}{24} = 0.01554 \] \[ 24 - P_s = 24 \times 0.01554 \] \[ 24 - P_s \approx 0.373 \] \[ P_s = 24 - 0.373 \approx 23.627 mm Hg \]

Step 4: Final Answer:

The saturated vapour pressure of the 5% aqueous solution of urea is approximately 23.63 mm Hg.
Quick Tip: When dealing with percentage concentrations, assume a 100 g sample of the solution. This makes it easy to determine the mass of the solute and the solvent. Also, for very dilute solutions, you can approximate \(n_{solute} + n_{solvent} \approx n_{solvent}\), but it's always safer to use the full formula unless specified.

Step 1: Understanding the Concept:

This problem requires the calculation of three key electrochemical properties: resistivity (\(\rho\)), conductivity (\(\kappa\)), and molar conductivity (\(\Lambda_m\)) from the given resistance and cell dimensions.


Step 2: Key Formula or Approach:

The formulas required are:

1. Resistivity (\(\rho\)): \( \rho = R \frac{A}{l} \)

2. Conductivity (\(\kappa\)): \( \kappa = \frac{1}{\rho} \)

3. Molar Conductivity (\(\Lambda_m\)): \( \Lambda_m = \frac{\kappa \times 1000}{M} \)

where:

R = Resistance (\(\Omega\))

A = Area of cross-section (cm\(^2\))

l = Length of the column (cm)

M = Molarity (mol L\(^{-1}\))


Step 3: Detailed Explanation:

Given values:

Molarity, M = 0.05 M

Resistance, R = 5 \(\times\) 10\(^3\) \(\Omega\)

Area, A = 0.8 cm\(^2\)

Length, l = 40 cm


Calculation of Resistivity (\(\rho\)):
\[ \rho = R \frac{A}{l} = (5 \times 10^3 \, \Omega) \times \frac{0.8 \, cm^2}{40 \, cm} \] \[ \rho = (5 \times 10^3) \times 0.02 \, \Omega \cdot cm \] \[ \rho = 100 \, \Omega \cdot cm \]

Calculation of Conductivity (\(\kappa\)):

Conductivity is the reciprocal of resistivity.
\[ \kappa = \frac{1}{\rho} = \frac{1}{100 \, \Omega \cdot cm} \] \[ \kappa = 0.01 \, \Omega^{-1} cm^{-1} \quad or \quad 0.01 \, S cm^{-1} \]
(Note: Siemens, S = \(\Omega^{-1}\))


Calculation of Molar Conductivity (\(\Lambda_m\)):
\[ \Lambda_m = \frac{\kappa \times 1000}{M} \]
Here, the factor of 1000 is used to convert the volume from L to cm\(^3\) (since 1 L = 1000 cm\(^3\)).
\[ \Lambda_m = \frac{(0.01 \, S cm^{-1}) \times 1000 \, (cm^3 L^{-1})}{0.05 \, (mol L^{-1})} \] \[ \Lambda_m = \frac{10}{0.05} \, S cm^2 mol^{-1} \] \[ \Lambda_m = 200 \, S cm^2 mol^{-1} \]

Step 4: Final Answer:

The resistivity is 100 \(\Omega\) cm.

The conductivity is 0.01 S cm\(^{-1}\).

The molar conductivity is 200 S cm\(^2\) mol\(^{-1}\).
Quick Tip: Pay close attention to units. The most common mistake in these calculations is mixing up units (e.g., using meters instead of centimeters). The formula \( \Lambda_m = \frac{\kappa \times 1000}{M} \) is specifically for when \(\kappa\) is in S cm\(^{-1}\) and M is in mol L\(^{-1}\).

Step 1: Understanding the Concept:

This is a redox reaction in an acidic medium. We need to balance it using the ion-electron method (half-reaction method).


Step 2: Detailed Explanation:

1. Identify Half-Reactions:
- Reduction: The oxidation state of Mn in MnO\(_4^-\) is +7, and it gets reduced to Mn\(^{2+}\) (oxidation state +2) in acidic medium.
\[ MnO_4^- \rightarrow Mn^{2+} \]
- Oxidation: The oxidation state of C in C\(_2\)O\(_4^{2-}\) (oxalate) is +3, and it gets oxidized to CO\(_2\) (oxidation state +4).
\[ C_2O_4^{2-} \rightarrow CO_2 \]

2. Balance Atoms other than O and H:
- Reduction: Mn is already balanced.
- Oxidation: Balance Carbon atoms.
\[ C_2O_4^{2-} \rightarrow 2CO_2 \]

3. Balance Oxygen Atoms by adding H\(_2\)O:
- Reduction: Add 4 H\(_2\)O to the right side.
\[ MnO_4^- \rightarrow Mn^{2+} + 4H_2O \]
- Oxidation: Oxygen is already balanced.

4. Balance Hydrogen Atoms by adding H\(^+\):
- Reduction: Add 8 H\(^+\) to the left side.
\[ MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O \]
- Oxidation: No H atoms to balance.

5. Balance Charge by adding electrons (e\(^-\)):
- Reduction: Left side charge = (-1) + 8(+1) = +7. Right side charge = +2. Add 5e\(^-\) to the left side.
\[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \]
- Oxidation: Left side charge = -2. Right side charge = 0. Add 2e\(^-\) to the right side.
\[ C_2O_4^{2-} \rightarrow 2CO_2 + 2e^- \]

6. Equalize Electrons:
- Multiply the reduction half-reaction by 2.
- Multiply the oxidation half-reaction by 5.
\[ 2(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O) \]
\[ 5(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-) \]
This gives:
\[ 2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O \]
\[ 5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^- \]

7. Add the Half-Reactions and cancel the electrons:
\[ 2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O \]

Step 3: Final Answer:

The complete balanced equation is:

2MnO\(_4^-\) + 5C\(_2\)O\(_4^{2-}\) + 16H\(^+\) \(\rightarrow\) 2Mn\(^{2+}\) + 10CO\(_2\) + 8H\(_2\)O
Quick Tip: For permanganate (MnO\(_4^-\)) in acidic medium, a common reaction is its reduction to Mn\(^{2+}\). This involves a transfer of 5 electrons. Memorizing this can speed up the balancing process.

Step 1: Understanding the Concept:

This equation represents the thermal decomposition of potassium permanganate (KMnO\(_4\)). When heated, it decomposes into potassium manganate (K\(_2\)MnO\(_4\)), manganese dioxide (MnO\(_2\)), and oxygen gas (O\(_2\)).


Step 2: Detailed Explanation:

The unbalanced reaction is:
\[ KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \]
To balance this equation, we can use the inspection method or track oxidation states.

- On the left, we have 1 K, 1 Mn, 4 O.

- On the right, we have 2 K, 2 Mn, 4+2+2 = 8 O.

Let's start by balancing K. Place a '2' in front of KMnO\(_4\):
\[ 2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \]
Now let's check the atoms again:

- Left side: 2 K, 2 Mn, 8 O.

- Right side: 2 K, (1+1) = 2 Mn, (4+2+2) = 8 O.

The equation is now balanced.


Step 3: Final Answer:

The complete balanced equation for the thermal decomposition of KMnO\(_4\) is:

2KMnO\(_4\)(s) \(\xrightarrow{\Delta}\) K\(_2\)MnO\(_4\)(s) + MnO\(_2\)(s) + O\(_2\)(g)
Quick Tip: The thermal decomposition of potassium permanganate is a standard laboratory method for preparing small amounts of pure oxygen gas. It's a key reaction to memorize for exams related to p-block and d-block elements.

Step 1: Understanding the Concept:

This is a redox reaction in an acidic medium where dichromate ion (Cr\(_2\)O\(_7^{2-}\)) acts as an oxidizing agent and hydrogen sulfide (H\(_2\)S) acts as a reducing agent.


Step 2: Detailed Explanation:

1. Identify Half-Reactions:
- Reduction: The oxidation state of Cr in Cr\(_2\)O\(_7^{2-}\) is +6, and it gets reduced to Cr\(^{3+}\).
\[ Cr_2O_7^{2-} \rightarrow Cr^{3+} \]
- Oxidation: The oxidation state of S in H\(_2\)S is -2, and it gets oxidized to elemental sulfur (S) with an oxidation state of 0.
\[ H_2S \rightarrow S \]

2. Balance Atoms other than O and H:
- Reduction: Balance Cr atoms.
\[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \]
- Oxidation: S is already balanced.

3. Balance Oxygen Atoms by adding H\(_2\)O:
- Reduction: Add 7 H\(_2\)O to the right side.
\[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \]
- Oxidation: No O atoms.

4. Balance Hydrogen Atoms by adding H\(^+\):
- Reduction: Add 14 H\(^+\) to the left side.
\[ Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O \]
- Oxidation: Add 2 H\(^+\) to the right side.
\[ H_2S \rightarrow S + 2H^+ \]

5. Balance Charge by adding electrons (e\(^-\)):
- Reduction: Left side charge = (-2) + 14(+1) = +12. Right side charge = 2(+3) = +6. Add 6e\(^-\) to the left side.
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
- Oxidation: Left side charge = 0. Right side charge = +2. Add 2e\(^-\) to the right side.
\[ H_2S \rightarrow S + 2H^+ + 2e^- \]

6. Equalize Electrons:
- Multiply the oxidation half-reaction by 3.
\[ 3(H_2S \rightarrow S + 2H^+ + 2e^-) \]
This gives:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
\[ 3H_2S \rightarrow 3S + 6H^+ + 6e^- \]

7. Add the Half-Reactions and cancel common species (electrons and H\(^+\)):
\[ Cr_2O_7^{2-} + 3H_2S + 14H^+ \rightarrow 2Cr^{3+} + 3S + 7H_2O + 6H^+ \]
Cancel 6H\(^+\) from both sides:
\[ Cr_2O_7^{2-} + 3H_2S + 8H^+ \rightarrow 2Cr^{3+} + 3S + 7H_2O \]

Step 3: Final Answer:

The complete balanced equation is:

Cr\(_2\)O\(_7^{2-}\) + 3H\(_2\)S + 8H\(^+\) \(\rightarrow\) 2Cr\(^{3+}\) + 3S + 7H\(_2\)O
Quick Tip: The change in color from orange (Cr\(_2\)O\(_7^{2-}\)) to green (Cr\(^{3+}\)) is a characteristic test for reducing agents. This is a very common redox couple in inorganic chemistry.

Step 1: Understanding the Concept:

Valence Bond Theory (VBT) explains the formation of coordinate bonds in terms of orbital hybridization and overlap. We need to determine the hybridization of the central metal ion and its magnetic properties based on the arrangement of electrons.


Step 2: Detailed Explanation:

1. Determine the oxidation state of Cobalt (Co):

Let the oxidation state of Co be 'x'. Ammonia (NH\(_3\)) is a neutral ligand (charge = 0). The overall charge of the complex is +3.

\[ x + 6(0) = +3 \implies x = +3 \]
So, we have the Co\(^{3+}\) ion.


2. Write the electronic configuration of Co and Co\(^{3+}\):

- Atomic number of Co is 27.
- Ground state configuration of Co: [Ar] 3d\(^7\) 4s\(^2\).
- Configuration of Co\(^{3+}\) (loss of 2 electrons from 4s and 1 from 3d): [Ar] 3d\(^6\).


3. Consider the ligand and its effect:

- The ligand is ammonia (NH\(_3\)), which is a strong-field ligand.
- A strong-field ligand forces the d-electrons of the central metal ion to pair up against Hund's rule.


4. Arrange electrons and determine hybridization:

- For Co\(^{3+}\) (3d\(^6\)), the electrons are arranged in the 3d orbitals.
- Due to the strong-field nature of NH\(_3\), the six 3d electrons will pair up in the first three 3d orbitals.
- This leaves two 3d orbitals, one 4s orbital, and three 4p orbitals empty.
- To accommodate the six lone pairs from the six NH\(_3\) ligands, the central ion uses these six empty orbitals for hybridization.
- The hybridization is therefore d\(^2\)sp\(^3\). This corresponds to an octahedral geometry.
- Since the inner d-orbitals (3d) are used, it is an inner orbital or low spin complex.


5. Determine the magnetic character:

- After the pairing of electrons, there are no unpaired electrons in the 3d orbitals.
- A substance with no unpaired electrons is repelled by a magnetic field and is called diamagnetic.


Step 3: Final Answer:

The complex [Co(NH\(_3\))\(_6\)]\(^{3+}\) has d\(^2\)sp\(^3\) hybridization and is diamagnetic.
Quick Tip: Remember the spectrochemical series to identify strong and weak field ligands. For first-row transition metals, ligands like NH\(_3\), en, CN\(^-\), and CO are generally strong-field, causing electron pairing. Halides and H\(_2\)O are generally weak-field.

Step 1: Understanding the Concept:

We will apply Valence Bond Theory to the metal carbonyl complex, tetracarbonylnickel(0), to determine its geometry, hybridization, and magnetic properties.


Step 2: Detailed Explanation:

1. Determine the oxidation state of Nickel (Ni):

- Carbon monoxide (CO) is a neutral ligand (charge = 0). The complex is also neutral.
- Let the oxidation state of Ni be 'x'.
\[ x + 4(0) = 0 \implies x = 0 \]
So, Nickel is in its elemental state.


2. Write the electronic configuration of Ni(0):

- Atomic number of Ni is 28.
- Ground state configuration of Ni: [Ar] 3d\(^8\) 4s\(^2\).


3. Consider the ligand and its effect:

- The ligand is carbon monoxide (CO), which is a very strong-field ligand.
- Strong-field ligands can cause the pairing of electrons, and in the case of metal carbonyls, they force the electrons from the outer s-orbital to shift into the d-orbitals.


4. Arrange electrons and determine hybridization:

- In the presence of the strong CO ligands, the two 4s electrons are pushed into the 3d orbitals to pair up with the existing 3d electrons.
- The configuration of Ni becomes [Ar] 3d\(^{10}\) 4s\(^0\).
- Now, the 3d subshell is completely filled.
- To accommodate the four lone pairs from the four CO ligands, the central Ni atom uses its empty outermost orbitals, which are the one 4s orbital and the three 4p orbitals.
- The hybridization is therefore sp\(^3\). This corresponds to a tetrahedral geometry.


5. Determine the magnetic character:

- In the 3d\(^{10}\) configuration, all electrons are paired.
- Since there are no unpaired electrons, the complex is diamagnetic.


Step 3: Final Answer:

The complex [Ni(CO)\(_4\)] has sp\(^3\) hybridization and is diamagnetic.
Quick Tip: Metal carbonyls with zero oxidation state, like [Ni(CO)\(_4\)] and [Fe(CO)\(_5\)], are an important class of compounds. Remember that the strong CO ligand typically forces all valence electrons of the metal into the d-orbitals, leading to diamagnetic complexes.

Step 1: Understanding the Concept:

Enantiomers are a specific type of stereoisomer related by a mirror image relationship. The key properties are chirality and optical activity.


Step 2: Detailed Explanation:


Stereoisomers: They have the same molecular formula and the same connectivity of atoms, but differ in the spatial arrangement of atoms.
Chirality: The most common cause of enantiomerism is the presence of a chiral center (an atom, usually carbon, bonded to four different groups). A molecule that is chiral is not superimposable on its mirror image. Our hands are a classic example of chiral objects.
Non-Superimposable Mirror Images: If you place one enantiomer in front of a mirror, the reflection you see is the other enantiomer. No matter how you rotate or move them, you can never make them coincide perfectly.
Optical Activity: Enantiomers have identical physical properties (e.g., melting point, boiling point, solubility) except for their interaction with plane-polarized light.

One enantiomer will rotate the plane of polarized light in a clockwise direction; it is called dextrorotatory (+).
The other enantiomer will rotate the plane of polarized light by the exact same angle but in a counter-clockwise direction; it is called levorotatory (-).


For example, the two enantiomers of 2-butanol are non-superimposable mirror images.
Quick Tip: To quickly check for enantiomers, find a chiral center and draw its mirror image. Then, mentally try to rotate the mirror image to see if it can be perfectly aligned with the original molecule. If it cannot, they are enantiomers.

Step 1: Understanding the Concept:

A racemic mixture (or racemate) is a specific mixture of enantiomers that results in a loss of optical activity for the bulk sample.


Step 2: Detailed Explanation:


Equimolar Mixture: A racemic mixture contains exactly equal amounts (50:50) of the dextrorotatory (+) and levorotatory (-) enantiomers of a chiral compound.
Optical Inactivity: The defining characteristic of a racemic mixture is that it does not rotate plane-polarized light, i.e., it is optically inactive.
External Compensation: This optical inactivity is not because the molecules themselves are achiral (like a meso compound), but because the effect of one enantiomer is cancelled out by the effect of the other. For every molecule that rotates light by +x degrees, there is another molecule that rotates it by -x degrees. The net rotation is zero. This phenomenon is called external compensation.
Notation: A racemic mixture is often denoted by the prefix (\(\pm\))- or (dl)- before the name of the compound, for example, (\(\pm\))-2-butanol.
Physical Properties: The physical properties (like melting point and solubility) of a racemic mixture can be different from those of the pure enantiomers. Quick Tip: Distinguish between external compensation (racemic mixture) and internal compensation (meso compound). A meso compound is a single achiral molecule with chiral centers, which is optically inactive because of an internal plane of symmetry. A racemic mixture is a mix of two different chiral molecules that is optically inactive.

Step 1: Understanding the Concept:

Aryl halides, such as chlorobenzene, are significantly less reactive than alkyl halides towards nucleophilic substitution. This low reactivity can be explained by several electronic and structural factors.


Step 2: Detailed Explanation:

There are four main reasons for the low reactivity of chlorobenzene:


Resonance Effect: The lone pair of electrons on the chlorine atom can participate in resonance with the \(\pi\)-electrons of the benzene ring. This delocalization of electrons creates a partial double-bond character between the carbon and chlorine atoms.

A double bond is stronger and shorter than a single bond, making the C-Cl bond in chlorobenzene more difficult to break compared to the C-Cl bond in an alkyl chloride.


Hybridization of Carbon Atom: The carbon atom bonded to the chlorine in chlorobenzene is sp\(^2\) hybridized. In contrast, the carbon in an alkyl halide is sp\(^3\) hybridized. An sp\(^2\) orbital has more s-character (33.3%) than an sp\(^3\) orbital (25%). This makes the sp\(^2\) carbon more electronegative, allowing it to hold the electrons of the C-Cl bond more tightly. This results in a shorter, stronger bond that is harder to break.


Instability of Phenyl Cation: If the reaction were to proceed via an S\(_N\)1 mechanism, it would involve the formation of a phenyl cation after the C-Cl bond breaks. The phenyl cation is highly unstable because the positive charge is on an sp\(^2\) carbon, and the vacant p-orbital is perpendicular to the \(\pi\)-system of the ring, preventing stabilization by resonance.


Nucleophilic Repulsion: The benzene ring is an electron-rich system due to its delocalized \(\pi\)-electron cloud. An incoming nucleophile, which is also electron-rich, will experience electrostatic repulsion from the ring, making the attack difficult. Quick Tip: The most important reason to cite in an exam is the resonance effect leading to partial double-bond character. This is the primary explanation for the enhanced strength of the C-X bond in aryl halides.

Step 1: Understanding the Concept:

This reaction is a specific method for the deoxygenation of aldehydes and ketones to form corresponding alkanes. It is particularly useful for compounds that are sensitive to acid, as the reaction is carried out under basic conditions.


Step 2: Reaction Mechanism and Equation:

The reaction proceeds in two main steps:

1. Formation of a hydrazone: The aldehyde or ketone reacts with hydrazine to form a hydrazone intermediate.

2. Elimination of N\(_2\): In the presence of a strong base and heat, the hydrazone loses a molecule of nitrogen gas to form the alkane.


General Chemical Equation:
\[ \chemfig{R-C(=[2]O)-R'} \xrightarrow[ethylene glycol, heat]{1. H_2N-NH_2 \quad 2. KOH/\Delta} \chemfig{R-CH_2-R'} + N_2 \]
where R and R' can be alkyl groups or hydrogen atoms.


Example (Reduction of Acetone):

Acetone (a ketone) is reduced to propane (an alkane).
\[ \chemfig{CH_3-C(=[2]O)-CH_3} \xrightarrow[ethylene glycol, heat]{H_2N-NH_2, KOH} \chemfig{CH_3-CH_2-CH_3} + N_2 + H_2O \] Quick Tip: Remember the two main reactions for reducing carbonyls to alkanes: \textbf{Wolff-Kishner Reduction:} Uses hydrazine (H\(_2\)N-NH\(_2\)) and a strong base (KOH). Best for molecules that are stable in basic conditions. \textbf{Clemmensen Reduction:} Uses zinc amalgam (Zn-Hg) and concentrated HCl. Best for molecules that are stable in acidic conditions.

Step 1: Understanding the Concept:

This is a named reaction for the synthesis of aromatic aldehydes from alkylbenzenes. The key feature is the use of chromyl chloride, a mild oxidizing agent that stops the oxidation at the aldehyde stage, preventing further oxidation to a carboxylic acid.


Step 2: Reaction Mechanism and Equation:

1. The reaction begins with the addition of chromyl chloride to toluene, which forms a brown, solid chromium complex intermediate.
2. This intermediate is then hydrolyzed (treated with water) to yield the final product, benzaldehyde.


General Chemical Equation:
\[ \chemfig{Ar-CH_3} \xrightarrow[2. H_3O^+]{1. CrO_2Cl_2, CS_2} \chemfig{Ar-CHO} \]

Example (Oxidation of Toluene):

Toluene is oxidized to benzaldehyde.
\[ \chemfig{C_6H_5-CH_3} \xrightarrow[2. H_3O^+]{1. CrO_2Cl_2, CS_2} \chemfig{C_6H_5-CHO} \] Quick Tip: The Etard reaction is a classic method for preparing benzaldehyde from toluene. Another common method is the Gattermann-Koch reaction, which synthesizes benzaldehyde from benzene using CO and HCl. Knowing multiple synthesis routes is key for organic chemistry questions.

Step 1: Understanding the Concept:

This reaction is characteristic of aldehydes that do not have any hydrogen atoms on the carbon adjacent to the carbonyl group (the \(\alpha\)-carbon). In this reaction, one molecule of the aldehyde is oxidized to a carboxylic acid, and another molecule is reduced to an alcohol.


Step 2: Reaction Mechanism and Equation:

1. The hydroxide ion (OH\(^-\)) from the strong base attacks the carbonyl carbon of one aldehyde molecule.
2. A hydride ion (H\(^-\)) is then transferred from this intermediate to a second molecule of the aldehyde.
3. This results in one molecule being oxidized (to the carboxylate anion) and the other being reduced (to the alkoxide ion).
4. Acid-base exchange then yields the final products.


General Chemical Equation:

(For an aldehyde RCHO with no \(\alpha\)-H) \[ 2 RCHO + conc. NaOH \rightarrow RCH_2OH + RCOONa \]

Example (Reaction of Formaldehyde):

Formaldehyde (HCHO) has no \(\alpha\)-carbon and thus no \(\alpha\)-hydrogens. \[ 2 HCHO + conc. NaOH \rightarrow CH_3OH + HCOONa \]
(Methanol and Sodium formate are produced)
Quick Tip: Remember the key condition for the Cannizzaro reaction: the aldehyde must **lack** \(\alpha\)-hydrogens (e.g., formaldehyde, benzaldehyde). If the aldehyde **has** \(\alpha\)-hydrogens, it will undergo the Aldol condensation reaction in the presence of a base instead.

Step 1: Identify Reaction 1 (Formation of A)
\[ CH_3COOH \xrightarrow{SOCl_2} A \]
This reaction involves a carboxylic acid (acetic acid) and thionyl chloride (SOCl\(_2\)). This is a standard method for converting a carboxylic acid into an acid chloride. The -OH group is replaced by -Cl.

A is Acetyl chloride, CH\(_3\)COCl.


Step 2: Identify Reaction 2 (Formation of B)
\[ CH_3COCl \xrightarrow{H_2, Pd-BaSO_4} B \]
This is the Rosenmund reduction. It is a catalytic hydrogenation process that reduces an acid chloride to an aldehyde. The catalyst is palladium on barium sulfate, which is partially poisoned (e.g., with sulfur or quinoline) to prevent over-reduction of the aldehyde to a primary alcohol.

B is Acetaldehyde, CH\(_3\)CHO.


Step 3: Identify Reaction 3 (Formation of C)
\[ CH_3CHO \xrightarrow{H_2N-NH_2} C \]
This is the reaction of an aldehyde (acetaldehyde) with hydrazine (H\(_2\)N-NH\(_2\)). This is a condensation reaction that forms a hydrazone. The oxygen atom of the carbonyl group and two hydrogen atoms from the -NH\(_2\) group are eliminated as a molecule of water.

C is Acetaldehyde hydrazone, CH\(_3\)CH=NNH\(_2\).
Quick Tip: The Rosenmund reduction (acid chloride \(\rightarrow\) aldehyde) is a very important named reaction. The key is the poisoned catalyst (Pd/BaSO\(_4\)). Without the poison, the acid chloride would be reduced all the way to a primary alcohol.

Step 1: Identify Reaction 1 (Formation of A)
\[ CH_3CN \xrightarrow{1.(DIBAL-H) 2. H_2O} A \]
This is the reduction of a nitrile (acetonitrile) using Diisobutylaluminium hydride (DIBAL-H). DIBAL-H is a reducing agent that can selectively reduce nitriles to imines. The imine intermediate is then hydrolyzed upon workup with water (H\(_2\)O) to yield an aldehyde.

A is Acetaldehyde, CH\(_3\)CHO.

This reaction is also known as the Stephen reduction when SnCl\(_2\)/HCl is used.


Step 2: Identify Reaction 2 (Formation of B)
\[ CH_3CHO \xrightarrow{Dil. NaOH} B \]
This is the Aldol Addition reaction. Acetaldehyde has \(\alpha\)-hydrogens. In the presence of a dilute base (like NaOH), one molecule of acetaldehyde acts as a nucleophile (forming an enolate) and attacks the carbonyl carbon of a second molecule. The product is a \(\beta\)-hydroxy aldehyde, also known as an aldol.

B is 3-Hydroxybutanal, CH\(_3\)CH(OH)CH\(_2\)CHO.


Step 3: Identify Reaction 3 (Formation of C)
\[ CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} C \]
This is the dehydration step of the Aldol Condensation. When the aldol addition product (B) is heated, it loses a molecule of water (dehydration) to form an \(\alpha\), \(\beta\)-unsaturated aldehyde.

C is But-2-enal, CH\(_3\)CH=CHCHO.
Quick Tip: Recognize the two stages of the Aldol reaction. "Aldol Addition" (dilute base, often cold) gives the \(\beta\)-hydroxy aldehyde/ketone. "Aldol Condensation" (addition followed by heat) gives the \(\alpha\), \(\beta\)-unsaturated product. The question clearly separates these two steps.

Step 1: Understanding the Concept:

Glycosidic linkages are the fundamental bonds that connect monosaccharide units to form larger carbohydrate structures like disaccharides, oligosaccharides, and polysaccharides.


Step 2: Detailed Explanation:

- When two monosaccharides join, the bond is formed between the anomeric carbon (the carbon that was part of the carbonyl group in the open-chain form, usually C-1 for aldoses) of one sugar and a hydroxyl (-OH) group of the other sugar.

- For example, in the formation of sucrose, a glycosidic bond is formed between the C-1 of \(\alpha\)-glucose and the C-2 of \(\beta\)-fructose.

- This bond is essentially an ether linkage (-O-).

- The orientation of the bond at the anomeric carbon can be either alpha (\(\alpha\)) or beta (\(\beta\)), which has significant structural and biological consequences (e.g., starch vs. cellulose).
Quick Tip: Think of glycosidic linkages as the carbohydrate equivalent of peptide bonds in proteins or phosphodiester bonds in nucleic acids. They are the essential covalent bonds that create the polymer chains.

Step 1: Understanding the Concept:

Invert sugar is a specific product derived from sucrose (common table sugar). Its name comes from a change in an optical property.


Step 2: Detailed Explanation:

- Source: Sucrose, a disaccharide, is composed of one glucose unit and one fructose unit.
- Reaction (Hydrolysis): \[ Sucrose + H_2O \xrightarrow{Acid or Invertase} D-Glucose + D-Fructose \]
- Optical Rotation:
- Sucrose is dextrorotatory, with a specific rotation of +66.5°.
- D-Glucose is also dextrorotatory, with a specific rotation of +52.5°.
- D-Fructose is strongly levorotatory, with a specific rotation of -92.4°.
- Inversion: When sucrose is hydrolyzed, the resulting mixture contains equal moles of glucose and fructose. The strong levorotation of fructose outweighs the dextrorotation of glucose. The net specific rotation of the mixture is approximately -19.9°.
- The direction of rotation "inverts" from positive to negative during the hydrolysis. This is the origin of the name "invert sugar".
- Invert sugar is sweeter than sucrose and is commonly found in honey and is used in the food industry to prevent crystallization.
Quick Tip: The key to "invert sugar" is the inversion of optical rotation. Remember that fructose is highly levorotatory, and this property is strong enough to make the final mixture levorotatory, even though glucose is dextrorotatory.

Step 1: Understanding the Concept:

Oligosaccharides are polymers of sugars that are intermediate in size between the simple monosaccharides and the very large polysaccharides.


Step 2: Detailed Explanation:

- The prefix "oligo-" means "a few".
- Based on the number of monosaccharide units they produce upon hydrolysis, they are further classified:
- Disaccharides: Yield two monosaccharide units. Examples include sucrose (glucose + fructose), lactose (glucose + galactose), and maltose (glucose + glucose).
- Trisaccharides: Yield three monosaccharide units. An example is raffinose (glucose + fructose + galactose).
- Tetrasaccharides: Yield four monosaccharide units. An example is stachyose.
- Oligosaccharides are crystalline solids, soluble in water, and are often sweet in taste.
- They play important roles in biology, particularly in cell recognition, where they are often attached to proteins (glycoproteins) or lipids (glycolipids) on the cell surface.
Quick Tip: Remember the classification of carbohydrates based on hydrolysis products: \textbf{Monosaccharides:} Cannot be hydrolyzed further (e.g., glucose, fructose). \textbf{Oligosaccharides:} Yield 2-10 monosaccharide units (e.g., sucrose, lactose). \textbf{Polysaccharides:} Yield a large number of monosaccharide units (e.g., starch, cellulose).

Step 1: Understanding the Concept:

The question asks for the definition of reverse osmosis (RO) and an example of a semipermeable membrane (SPM) used in this process. The definition is provided in the given passage.


Step 2: Detailed Explanation:

Definition of Reverse Osmosis:

Based on the passage, normal osmosis is the flow of solvent from a dilute solution to a concentrated solution. Osmotic pressure is the pressure required to stop this flow. Reverse osmosis occurs when the applied external pressure on the solution side is greater than the osmotic pressure. This high pressure overcomes the natural osmotic flow and forces the solvent molecules to move in the opposite direction - from the concentrated solution to the dilute solution (or pure solvent) through the semipermeable membrane. This process is widely used for the desalination of seawater to produce fresh drinking water.


Example of an SPM:

A semipermeable membrane for RO needs to be strong enough to withstand high pressures while being permeable to the solvent (water) but impermeable to the solute (salts). A commonly used material for this purpose is a thin film of cellulose acetate placed over a suitable support. Other modern membranes are often made of polyamides.
Quick Tip: Remember the key condition for reverse osmosis: Applied Pressure (\(P_{ext\)) \textgreater Osmotic Pressure (\(\Pi\)). This is the fundamental principle behind water purification systems that use RO technology.

Step 1: Understanding the Concept:

This question deals with the effect of osmosis on biological cells, specifically red blood cells (RBCs). The outcome depends on the tonicity (relative solute concentration) of the external solution compared to the fluid inside the RBCs.


Step 2: Detailed Explanation:

1. Isotonic Solution: The fluid inside red blood cells has a solute concentration that is equivalent to a 0.9% (mass/volume) NaCl solution. This is called an isotonic solution. When RBCs are placed in an isotonic solution, there is no net flow of water, and the cells maintain their shape.

2. Hypotonic Solution: The given solution is 0.5% NaCl. Since its concentration (0.5%) is lower than the concentration inside the RBCs (0.9%), it is a hypotonic solution relative to the cells.

3. Osmosis: According to the principle of osmosis, water will move from a region of lower solute concentration (the 0.5% NaCl solution) to a region of higher solute concentration (the inside of the RBCs) through the semipermeable cell membrane.

4. Result: This net influx of water into the RBCs will cause them to swell up. Since RBCs do not have a rigid cell wall, they can swell to a certain point and then burst, a process known as hemolysis.
Quick Tip: Remember the three scenarios for cells in solutions: \textbf{Isotonic (same concentration):} No change. For RBCs, this is 0.9% NaCl. \textbf{Hypotonic (lower concentration):} Cell swells and may burst. \textbf{Hypertonic (higher concentration):} Cell shrinks (crenation).

Step 1: Understanding the Concept:

Osmotic pressure is a colligative property, meaning it depends on the number of solute particles in the solution. For electrolytes, we must consider their dissociation into ions, which is quantified by the van't Hoff factor (i).


Step 2: Key Formula or Approach:

The formula for osmotic pressure (\(\Pi\)) is: \[ \Pi = i \times C \times R \times T \]
where:
\(i\) = van't Hoff factor (number of particles formed per formula unit)
\(C\) = Molar concentration
\(R\) = Gas constant
\(T\) = Temperature


Step 3: Detailed Explanation:

1. For 1 M urea solution: Urea (NH\(_2\)CONH\(_2\)) is a non-electrolyte. It does not dissociate in solution. Therefore, its van't Hoff factor is \(i = 1\).

\[ \Pi_{urea} = 1 \times C \times R \times T \]

2. For 1 M KCl solution: Potassium chloride (KCl) is a strong electrolyte. It dissociates completely in solution into two ions: one K\(^+\) ion and one Cl\(^-\) ion.

\[ KCl(aq) \rightarrow K^+(aq) + Cl^-(aq) \]
Therefore, its van't Hoff factor is \(i = 2\).

\[ \Pi_{KCl} = 2 \times C \times R \times T \]

3. Comparison: Since the molar concentration (C), gas constant (R), and temperature (T) are the same for both solutions, the osmotic pressure is directly proportional to the van't Hoff factor (i). As \(i_{KCl} (2) \textgreater i_{urea} (1)\), the osmotic pressure of the 1 M KCl solution will be approximately double that of the 1 M urea solution.
Quick Tip: When comparing colligative properties of solutions with the same molarity, always check if the solutes are electrolytes or non-electrolytes. Electrolytes will always have a greater effect on colligative properties due to dissociation (i \textgreater 1).

Step 1: Understanding the Concept:

The question asks for the definition of a colligative property and how osmotic pressure fits this definition.


Step 2: Detailed Explanation:

1. Definition of Colligative Property: A colligative property is a physical property of a solution that depends on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. The four main colligative properties are:

Relative lowering of vapour pressure
Elevation in boiling point
Depression in freezing point
Osmotic pressure

2. Osmotic Pressure's Dependence: The formula for osmotic pressure, \(\Pi = iCRT\), explicitly shows this dependence. The term 'C' represents the molar concentration of the solute, and 'i' (the van't Hoff factor) accounts for the total number of particles after any dissociation. The product 'iC' gives the total molar concentration of all solute particles. The formula does not include any term that relates to the size, charge, or chemical identity of the solute, only the number of particles.
3. Conclusion: Because osmotic pressure is directly proportional to the total concentration of solute particles, it perfectly fits the definition of a colligative property.
Quick Tip: The word "colligative" comes from the Latin 'colligatus', which means 'bound together'. It refers to properties that are bound together by the common feature of depending on the number of particles.

Step 1: Understanding the Concept:

The basicity of aromatic amines (anilines) depends on the availability of the lone pair of electrons on the nitrogen atom for donation to a proton. This availability is influenced by substituent groups on the benzene ring. Electron-donating groups (EDGs) increase basicity, while electron-withdrawing groups (EWGs) decrease it.


Step 2: Detailed Explanation:

Let's analyze the effect of the substituent in each compound:

1. Aniline (\chemfig{C_6H_5-NH_2}): This is our reference compound. The lone pair on the nitrogen is delocalized into the benzene ring through resonance, which decreases its availability and makes aniline less basic than aliphatic amines.


2. p-toluidine (p-CH\(_3\)-C\(_6\)H\(_4\)-NH\(_2\)): The methyl group (-CH\(_3\)) is an electron-donating group. It exerts a positive inductive effect (+I) and a hyperconjugation effect. Both effects push electron density into the benzene ring, which in turn increases the electron density on the nitrogen atom. This makes the lone pair more available for protonation, so p-toluidine is more basic than aniline.


3. p-nitroaniline (p-NO\(_2\)-C\(_6\)H\(_4\)-NH\(_2\)): The nitro group (-NO\(_2\)) is a strong electron-withdrawing group. It exerts a strong negative inductive effect (-I) and a strong negative resonance effect (-R). Both effects pull electron density away from the benzene ring and, consequently, from the nitrogen atom. This makes the lone pair significantly less available for protonation, so p-nitroaniline is much less basic than aniline.


Conclusion:

Comparing the three, the EWG (-NO\(_2\)) decreases basicity the most, while the EDG (-CH\(_3\)) increases it. Therefore, the increasing order of basic character is:

p-nitroaniline \textless Aniline \textless p-toluidine
Quick Tip: To compare the basicity of substituted anilines, first identify the substituent as either an electron-donating group (EDG) or an electron-withdrawing group (EWG). The order of basicity will generally be: EDG-substituted \textgreater unsubstituted \textgreater EWG-substituted.

Step 1: Understanding the Concept:

The question asks us to compare the basic strength of an aromatic amine (aniline) and an aliphatic amine (methylamine). We must remember the relationship between basic strength (K\(_b\)) and pK\(_b\): \( pK_b = -\log(K_b) \). A stronger base has a higher K\(_b\) and a lower pK\(_b\). Therefore, the question is effectively asking why aniline is a weaker base than methylamine.


Step 2: Detailed Explanation:

1. Aniline (\chemfig{C_6H_5-NH_2}):

Aniline is an aromatic amine. The lone pair of electrons on the nitrogen atom is not localized on the nitrogen. Instead, it is delocalized into the \(\pi\)-electron system of the benzene ring through resonance. This can be shown by its resonance structures.


This delocalization makes the lone pair less available to accept a proton (H\(^+\)). Consequently, aniline is a relatively weak base.


2. Methylamine (\chemfig{CH_3-NH_2}):

Methylamine is a primary aliphatic amine. The methyl group (-CH\(_3\)) is an electron-donating group. It exerts a positive inductive effect (+I), which pushes electron density towards the nitrogen atom. This increases the electron density on the nitrogen, making the lone pair more readily available for donation to a proton. Consequently, methylamine is a relatively strong base.


Conclusion:

Due to the electron-withdrawing resonance effect in aniline and the electron-donating inductive effect in methylamine, methylamine is a significantly stronger base than aniline. A stronger base has a lower pK\(_b\) value. Therefore, the pK\(_b\) of aniline (a weaker base) is more than that of methylamine (a stronger base). (pK\(_b\) of aniline \(\approx\) 9.38; pK\(_b\) of methylamine \(\approx\) 3.38).
Quick Tip: As a general rule, aliphatic amines are stronger bases than ammonia, and aromatic amines are weaker bases than ammonia. This provides a quick way to rank the basicity of different classes of amines.

Step 1: Understanding the Concept:

The basicity of aliphatic amines in an aqueous solution is a complex phenomenon determined by a combination of three factors:


Inductive Effect (+I): Alkyl groups are electron-donating, which increases the electron density on the nitrogen atom, making the lone pair more available for protonation.
Solvation Effect (Hydration): The protonated amine (conjugate acid, R-NH\(_{3}^{+}\)) is stabilized by hydrogen bonding with water molecules. More hydrogen atoms on the nitrogen lead to stronger solvation and greater stability of the conjugate acid, which increases the basicity of the amine.
Steric Hindrance: Bulky alkyl groups around the nitrogen atom can hinder the approach of a proton and also impede the solvation of the conjugate acid.


Step 2: Detailed Explanation:

Let's analyze the effects for each amine:

- Ammonia (NH\(_3\)): Has no alkyl groups, so no +I effect. Its conjugate acid (NH\(_4^+\)) has four H-atoms and is well-solvated. It serves as our baseline.

- Methylamine (CH\(_3\)NH\(_2\), Primary 1\(^\circ\)): One methyl group provides a +I effect. Its conjugate acid (CH\(_3\)NH\(_3^+\)) has three H-atoms for strong solvation.

- Dimethylamine ((CH\(_3\))\(_2\)NH, Secondary 2\(^\circ\)): Two methyl groups provide a stronger +I effect than in methylamine. Its conjugate acid ((CH\(_3\))\(_2\)NH\(_2^+\)) has two H-atoms for good solvation.

- Trimethylamine ((CH\(_3\))\(_3\)N, Tertiary 3\(^\circ\)): Three methyl groups provide the strongest +I effect. However, its conjugate acid ((CH\(_3\))\(_3\)NH\(^+\)) has only one H-atom for solvation, leading to poor stabilization by water. Also, the three methyl groups cause significant steric hindrance.


Conclusion:

In an aqueous solution, the combined influence of these factors leads to a non-linear trend.

- Dimethylamine (2\(^\circ\)) is the strongest base because it has the best balance between a strong +I effect and good solvation.

- Methylamine (1\(^\circ\)) is the next strongest, with a moderate +I effect and excellent solvation.

- Trimethylamine (3\(^\circ\)) is weaker than both 1\(^\circ\) and 2\(^\circ\) amines because the poor solvation and steric hindrance of its conjugate acid outweigh its strong +I effect. It is, however, slightly more basic than ammonia.

- Ammonia (NH\(_3\)) is the weakest as it lacks the +I effect of alkyl groups.


Thus, the increasing order of basic strength is: NH\(_3\) \textless (CH\(_3\))\(_3\)N \textless CH\(_3\)NH\(_2\) \textless (CH\(_3\))\(_2\)NH
Quick Tip: The order of basicity of amines is different in the gaseous phase and aqueous solution. \textbf{Gaseous phase (only +I effect matters):} 3° \textgreater 2° \textgreater 1° \textgreater NH\(_3\) \textbf{Aqueous solution (all 3 effects matter):} For methyl groups: 2° \textgreater 1° \textgreater 3° \textgreater NH\(_3\) For ethyl groups: 2° \textgreater 3° \textgreater 1° \textgreater NH\(_3\)

Step 1: Understanding the Concept:

Ammonolysis is the reaction of an alkyl halide with ammonia (a nucleophile) in a nucleophilic substitution reaction to form a primary amine. The question asks why this method is problematic for synthesis.


Step 2: Detailed Explanation:

1. Initial Reaction: The reaction starts with ammonia attacking the alkyl halide (R-X) to form a primary amine salt, which is then deprotonated by excess ammonia to give the primary amine (R-NH\(_2\)).
\[ R-X + NH_3 \rightarrow R-NH_3^+X^- \xrightarrow{NH_3} R-NH_2 + NH_4^+X^- \]
2. Problem of Over-alkylation: The primary amine (R-NH\(_2\)) formed is also a nucleophile, often even more nucleophilic than ammonia itself. Therefore, it can react with another molecule of the alkyl halide.

3. Formation of a Mixture: This leads to a series of subsequent reactions:

- The primary amine reacts with R-X to form a secondary amine (R\(_2\)NH).

\[ R-NH_2 + R-X \rightarrow R_2NH_2^+X^- \xrightarrow{NH_3} R_2NH + NH_4^+X^- \]
- The secondary amine reacts with R-X to form a tertiary amine (R\(_3\)N).

\[ R_2NH + R-X \rightarrow R_3NH^+X^- \xrightarrow{NH_3} R_3N + NH_4^+X^- \]
- The tertiary amine reacts with R-X to form a quaternary ammonium salt (R\(_4\)N\(^+\)X\(^-\)).

\[ R_3N + R-X \rightarrow R_4N^+X^- \]
4. Separation Difficulty: As a result, the final product is a mixture of primary, secondary, and tertiary amines, plus the quaternary ammonium salt. Separating these components is often difficult due to their similar boiling points, making this method impractical for the synthesis of a pure primary amine.

(Note: A pure primary amine can be obtained if a large excess of ammonia is used, but the yield is often low.)
Quick Tip: To synthesize a pure primary amine without the problem of over-alkylation, a better method is the \textbf{Gabriel Phthalimide Synthesis}. This method specifically prevents the formation of secondary and tertiary amines.

Step 1: Understanding the Concept:

We need to apply the IUPAC rules for naming an organic compound containing two functional groups: an aldehyde and a double bond.


Step 2: Detailed Explanation:

1. Identify the Principal Functional Group: The compound contains an aldehyde group (-CHO) and a double bond (C=C). The aldehyde group has higher priority, so it determines the suffix of the name. The suffix for an aldehyde is "-al".

2. Identify the Parent Chain: The longest carbon chain containing the principal functional group has four carbon atoms. The root word for four carbons is "but".

3. Number the Parent Chain: Numbering starts from the carbon of the principal functional group. So, the aldehyde carbon is C-1.

\[ \chemfig{CH_3-[4]CH=[3]CH-[2]CHO} \]
(The numbers indicate position: 4-3-2-1)

4. Locate the Substituents/Multiple Bonds: The double bond starts at carbon C-2. The infix for a double bond is "-en-". Its position is indicated as "2-en".

5. Combine the Parts: The full name is constructed as: (Root word) + (Position of double bond) + (Suffix for double bond) + (Suffix for aldehyde).

But + 2-en + al = But-2-enal.
Quick Tip: When a chain contains both a multiple bond and a functional group, the functional group gets priority in numbering, unless a specific rule dictates otherwise. For aldehydes, the numbering always starts from the -CHO carbon as C1.

Step 1: Understanding the Concept:

Propanal (CH\(_3\)CH\(_2\)CHO) is an aldehyde, while propanone (CH\(_3\)COCH\(_3\)) is a ketone. Aldehydes are easily oxidized, whereas ketones are resistant to oxidation by mild oxidizing agents. This difference in reactivity forms the basis for several chemical tests.


Step 2: Detailed Explanation:

Test: Tollen's Test

Reagent: Tollen's reagent, which is an ammoniacal silver nitrate solution, [Ag(NH\(_3\))\(_2\)]\(^+\)OH\(^-\).

Procedure: Add a few drops of each sample (propanal and propanone) to separate test tubes containing freshly prepared Tollen's reagent. Warm the test tubes gently in a water bath.

Observation:

- With Propanal (Aldehyde): A bright silver mirror is formed on the inner walls of the test tube, or a black precipitate of silver is formed. This is because the aldehyde is oxidized to the carboxylate anion, and the Ag\(^+\) ions in the reagent are reduced to metallic silver (Ag).

- With Propanone (Ketone): No reaction occurs. The solution remains clear.


Chemical Equation (for Propanal):
\[ CH_3CH_2CHO + 2[Ag(NH_3)_2]^+(aq) + 3OH^-(aq) \rightarrow CH_3CH_2COO^-(aq) + 2Ag(s) + 4NH_3(aq) + 2H_2O(l) \] Quick Tip: Other tests that can distinguish aldehydes from ketones include Fehling's test (aldehydes give a red-brown precipitate of Cu\(_2\)O) and Benedict's test. Tollen's test is generally the most reliable for both aliphatic and aromatic aldehydes.

Step 1: Understanding the Concept:

This conversion involves the oxidation of the methyl group (-CH\(_3\)) of toluene to a carboxylic acid group (-COOH). This requires a strong oxidizing agent.


Step 2: Reaction Scheme:

The reaction is a two-step process:

1. Oxidation: Toluene is heated with a strong oxidizing agent, such as potassium permanganate (KMnO\(_4\)) in a basic medium (e.g., KOH). This oxidizes the methyl group to form the potassium salt of benzoic acid (potassium benzoate).

2. Acidification: The resulting solution is acidified with a dilute acid (like HCl or H\(_2\)SO\(_4\)) to protonate the benzoate salt and precipitate the free benzoic acid.


Chemical Equation:
\[ \chemfig{C_6H_5-CH_3} \xrightarrow[2. H_3O^+]{1. KMnO_4, KOH, \Delta} \chemfig{C_6H_5-COOH} \] Quick Tip: Any alkylbenzene with at least one benzylic hydrogen (a hydrogen on the carbon directly attached to the ring) will be oxidized by strong oxidizing agents like KMnO\(_4\) or K\(_2\)Cr\(_2\)O\(_7\) to benzoic acid, regardless of the length of the alkyl chain. For example, ethylbenzene and propylbenzene also give benzoic acid.

Step 1: Understanding the Concept:

This is a step-up reaction where the carbon chain is increased by one carbon (from 2 carbons in ethanol to 3 in propan-2-ol). Such conversions often involve Grignard reagents.


Step 2: Reaction Scheme:

Step 1: Oxidation of Ethanol to Ethanal

Ethanol (a primary alcohol) is oxidized to ethanal (an aldehyde) using a mild oxidizing agent like Pyridinium chlorochromate (PCC) to prevent over-oxidation to carboxylic acid.
\[ CH_3CH_2OH \xrightarrow{PCC} CH_3CHO \]
Step 2: Grignard Reaction

Ethanal is treated with methyl magnesium bromide (CH\(_3\)MgBr), a Grignard reagent. The nucleophilic methyl group of the Grignard reagent attacks the electrophilic carbonyl carbon of ethanal to form an alkoxide adduct.
\[ CH_3CHO + CH_3MgBr \rightarrow CH_3CH(OMgBr)CH_3 \]
Step 3: Hydrolysis

The adduct is hydrolyzed with dilute acid (H\(_3\)O\(^+\)) to produce the final product, propan-2-ol (a secondary alcohol).
\[ CH_3CH(OMgBr)CH_3 \xrightarrow{H_3O^+} CH_3CH(OH)CH_3 + Mg(OH)Br \] Quick Tip: Grignard reagents are extremely useful for forming new carbon-carbon bonds. Remember these general patterns: Formaldehyde + Grignard reagent \(\rightarrow\) Primary alcohol Other Aldehydes + Grignard reagent \(\rightarrow\) Secondary alcohol Ketones + Grignard reagent \(\rightarrow\) Tertiary alcohol

Step 1: Understanding the Concept:

This conversion requires adding a carboxyl group (-COOH) and a hydroxyl group (-OH) to the original molecule. This is a characteristic two-step synthesis via a cyanohydrin intermediate. It is also a step-up reaction, adding one carbon to the chain.

*(Note: There appears to be a typo in the question. Starting with propanal (3 carbons) and adding a nitrile group (1 carbon) will result in a 4-carbon acid. The product should be 2-hydroxybutanoic acid. The solution below shows this intended conversion. To obtain 2-hydroxypropanoic acid, one would need to start with ethanal.)*


Step 2: Reaction Scheme (Propanal \(\rightarrow\) 2-hydroxybutanoic acid):

Step 1: Cyanohydrin Formation

Propanal (CH\(_3\)CH\(_2\)CHO) is treated with hydrogen cyanide (HCN). The cyanide ion (CN\(^-\)) acts as a nucleophile and attacks the carbonyl carbon, followed by protonation of the oxygen atom. This forms propanal cyanohydrin.
\[ CH_3CH_2CHO + HCN \rightarrow CH_3CH_2CH(OH)CN \]
The product is 2-hydroxybutanenitrile.


Step 2: Hydrolysis of the Nitrile

The nitrile group (-C\(\equiv\)N) of the cyanohydrin is then subjected to complete acid-catalyzed hydrolysis (boiling with dilute acid, H\(_3\)O\(^+\)/\(\Delta\)). This converts the nitrile group into a carboxylic acid group (-COOH).
\[ CH_3CH_2CH(OH)CN + 2H_2O \xrightarrow{H^+, \Delta} CH_3CH_2CH(OH)COOH + NH_3 \]
The final product is 2-hydroxybutanoic acid.
Quick Tip: The cyanohydrin reaction followed by hydrolysis is a powerful tool in organic synthesis. It's a step-up reaction that adds one carbon to the chain and simultaneously introduces both a hydroxyl and a carboxylic acid group at the position of the original carbonyl group.

Step 1: Understanding the Concept:

This reaction shows a ketone (cyclopentanone) reacting with hydroxylamine (HO-NH\(_2\)) in the presence of an acid catalyst (H\(^+\)). This is a condensation reaction, a characteristic reaction of aldehydes and ketones with ammonia derivatives.


Step 2: Reaction Mechanism and Equation:

The reaction involves the nucleophilic addition of hydroxylamine to the carbonyl carbon, followed by the elimination of a water molecule. The C=O double bond is replaced by a C=N-OH double bond. The product formed is called an oxime.
Quick Tip: Reactions of carbonyl compounds with ammonia derivatives (Z-NH\(_2\)) generally follow the same pattern: the C=O group is converted to a C=N-Z group, with the loss of a water molecule.

Step 1: Understanding the Concept:

This is a reductive ozonolysis reaction. The reaction cleaves a carbon-carbon double bond (C=C) and replaces it with two carbon-oxygen double bonds (C=O). We are given the product and need to determine the starting alkene.


Step 2: Working Backwards:

The product is two molecules of cyclohexanone. Ozonolysis cleaves a double bond. To find the starting material, we can reverse the process. Take the two carbonyl groups of the product molecules and join them together with a double bond, removing the oxygen atoms.

Starting with two molecules of cyclohexanone
To reverse the reaction, remove the two oxygen atoms and form a double bond between the two carbonyl carbons:
The starting material is an alkene where two cyclohexyl rings are joined by a double bond. This molecule is called Bicyclohexylidene.
Quick Tip: To find the alkene from the products of ozonolysis, place the two product molecules side-by-side with their carbonyl groups facing each other. Then, erase the oxygen atoms and draw a double bond between the two carbonyl carbons.

Step 1: Understanding the Concept:

The starting material is a dicarboxylic acid (cyclohexane-1,2-dicarboxylic acid). It is being treated with thionyl chloride (SOCl\(_2\)) and heat. SOCl\(_2\) is a reagent used to convert carboxylic acids to acid chlorides. However, when two carboxylic acid groups are close to each other (e.g., in a 1,2- or 1,3-dicarboxylic acid), heating can cause an intramolecular reaction to form a cyclic anhydride.


Step 2: Reaction Mechanism and Equation:

First, thionyl chloride can convert one or both -COOH groups into -COCl groups. However, the presence of heat (\(\Delta\)) with two adjacent carboxylic acid groups strongly favors dehydration to form a stable five or six-membered cyclic anhydride. One molecule of water is eliminated from the two carboxyl groups. SOCl\(_2\) is an excellent dehydrating agent, facilitating this process.

The product is a cyclic anhydride.
Quick Tip: When you see a dicarboxylic acid that can form a 5 or 6-membered ring (like succinic acid, glutaric acid, or phthalic acid) being heated, especially with a dehydrating agent like SOCl\(_2\) or P\(_2\)O\(_5\), think of cyclic anhydride formation.

Step 1: Understanding the Concept:

The starting material has both an aldehyde (-CHO) and a carboxylic acid (-COOH) functional group. The reagent is NaCN/HCl, which is a source of hydrogen cyanide (HCN). HCN reacts with carbonyl compounds (aldehydes and ketones) to form cyanohydrins. We need to determine which functional group will react.


Step 2: Reaction Mechanism and Equation:

- Aldehydes are generally more reactive towards nucleophilic addition than carboxylic acids. The carbonyl carbon of an aldehyde is more electrophilic.

- The cyanide ion (CN\(^-\)), a strong nucleophile, will preferentially attack the aldehyde's carbonyl carbon.

- The carboxylic acid group does not react under these conditions.

The reaction is a nucleophilic addition to the aldehyde.
Quick Tip: Remember the reactivity order of carbonyl compounds towards nucleophiles: Aldehydes \textgreater Ketones. Carboxylic acids and their derivatives (esters, amides) are generally less reactive towards nucleophilic addition because the carbonyl carbon is less electrophilic due to resonance.

Step 1: Understanding the Concept:

This conversion shows the introduction of an acetyl group (-COCH\(_3\)) onto a benzene ring. This is a classic example of an electrophilic aromatic substitution reaction.


Step 2: Reaction Name and Reagents:

The specific reaction is the Friedel-Crafts Acylation.

Reagents:

- Acylating Agent: An acid chloride (like acetyl chloride, CH\(_3\)COCl) or an acid anhydride (like acetic anhydride, (CH\(_3\)CO)\(_2\)O).

- Catalyst: A strong Lewis acid, typically anhydrous aluminum chloride (AlCl\(_3\)).

The Lewis acid catalyst reacts with the acylating agent to generate a highly electrophilic species, the acylium ion ([CH\(_3\)C=O]\(^+\)), which then attacks the electron-rich benzene ring.


Chemical Equation:
\[ \chemfig{C_6H_6} + \chemfig{CH_3-C(=O)-Cl} \xrightarrow{Anhyd. AlCl_3} \chemfig{C_6H_5-C(=O)-CH_3} + HCl \]
The product is acetophenone.
Quick Tip: Distinguish between Friedel-Crafts Acylation and Alkylation. Acylation uses an acyl halide (RCOCl) and is generally a more reliable reaction, free from carbocation rearrangements and poly-substitution that can plague alkylation reactions.

Step 1: Understanding the Concept:

We need to calculate the standard Gibbs free energy change (\(\Delta_r G^\circ\)) for a redox reaction using the standard cell potential (E\(^\circ_{cell}\)). The sign of \(\Delta_r G^\circ\) will then tell us if the reaction is spontaneous.


Step 2: Key Formula or Approach:

1. The standard cell potential is calculated as: \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \).

2. The standard Gibbs free energy change is related to the cell potential by: \( \Delta_r G^\circ = -nFE^\circ_{cell} \).

3. Spontaneity condition: If \(\Delta_r G^\circ \textless 0\), the reaction is spontaneous. If \(\Delta_r G^\circ \textgreater 0\), the reaction is non-spontaneous.


Step 3: Detailed Explanation:

1. Identify Half-Reactions and Electrodes:

The overall reaction is: Au(s) + Ca\(^{2+}\)(1M) \(\rightarrow\) Au\(^{3+}\)(1M) + Ca(s)

- Oxidation (Anode): Gold is oxidized from Au(s) to Au\(^{3+}\).
\[ Au(s) \rightarrow Au^{3+}(aq) + 3e^- \]
- Reduction (Cathode): Calcium ion is reduced from Ca\(^{2+}\) to Ca(s).
\[ Ca^{2+}(aq) + 2e^- \rightarrow Ca(s) \]

2. Calculate Standard Cell Potential (E\(^\circ_{cell}\)):
\(E^\circ_{cathode}\) = Standard reduction potential of Ca\(^{2+}\)/Ca = -2.87 V
\(E^\circ_{anode}\) = Standard reduction potential of Au\(^{3+}\)/Au = +1.5 V
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-2.87 \, V) - (+1.5 \, V) = -4.37 \, V \]

3. Determine 'n' (moles of electrons transferred):

To balance the overall reaction, we need to find the least common multiple of electrons in the half-reactions (3e\(^-\) for Au, 2e\(^-\) for Ca). The LCM is 6.

Multiply oxidation half-reaction by 2: \( 2Au \rightarrow 2Au^{3+} + 6e^- \)

Multiply reduction half-reaction by 3: \( 3Ca^{2+} + 6e^- \rightarrow 3Ca \)

The number of moles of electrons transferred in the balanced reaction is \( n = 6 \).


4. Calculate Standard Gibbs Energy (\(\Delta_r G^\circ\)):
\( F \) (Faraday constant) = 96500 C mol\(^{-1}\)
\[ \Delta_r G^\circ = -nFE^\circ_{cell} \] \[ \Delta_r G^\circ = -(6) \times (96500 \, C mol^{-1}) \times (-4.37 \, V) \] \[ \Delta_r G^\circ = 2530950 \, J mol^{-1} \]
Converting to kJ mol\(^{-1}\): \[ \Delta_r G^\circ = +2530.95 \, kJ mol^{-1} \]

5. Predict Spontaneity:

Since the calculated value of \( \Delta_r G^\circ \) is positive (+2530.95 kJ mol\(^{-1}\)), the reaction is non-spontaneous in the forward direction under standard conditions.
Quick Tip: A quick way to check for spontaneity without calculation is to look at the E\(^\circ_{cell}\). If E\(^\circ_{cell}\) is positive, the reaction is spontaneous (\(\Delta G^\circ\) will be negative). If E\(^\circ_{cell}\) is negative, the reaction is non-spontaneous (\(\Delta G^\circ\) will be positive). Here, E\(^\circ_{cell}\) is -4.37 V, immediately indicating a non-spontaneous reaction.

Step 1: Understanding the Concept:

This problem describes setting up an electrochemical cell (a galvanic cell) where the tarnished silver (Ag\(_2\)S) and the aluminum pan act as electrodes. For the tarnish to be removed, the Ag\(_2\)S must be reduced back to Ag, and the aluminum must be oxidized. This process will be spontaneous if the overall cell potential (E\(^\circ_{cell}\)) is positive.


Step 2: Key Formula or Approach:

We need to set up the half-reactions for the desired process and calculate the E\(^\circ_{cell}\).

Desired Process:

- Reduction (Cathode): Removal of tarnish means Ag\(_2\)S is reduced to Ag.

\[ Ag_2S(s) + 2e^- \rightarrow 2Ag(s) + S^{2-}(aq); \quad E^\circ = -0.71 \, V \]
- Oxidation (Anode): The aluminum pan is oxidized to Al\(^{3+}\).

\[ Al(s) \rightarrow Al^{3+}(aq) + 3e^-; \quad E^\circ_{ox} = -E^\circ_{red} = -(-1.66 \, V) = +1.66 \, V \]
The overall cell potential is calculated as: \( E^\circ_{cell} = E^\circ_{cathode} + E^\circ_{anode\_oxidation} \) or \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode\_reduction} \).


Step 3: Detailed Explanation:

Let's calculate the E\(^\circ_{cell}\) for the proposed galvanic cell.

Cathode (Reduction): Ag\(_2\)S/Ag, \( E^\circ_{cathode} = -0.71 \, V \)

Anode (Oxidation): Al/Al\(^{3+}\), \( E^\circ_{anode} = -1.66 \, V \) (this is the reduction potential)

Using the formula \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \): \[ E^\circ_{cell} = (-0.71 \, V) - (-1.66 \, V) \] \[ E^\circ_{cell} = -0.71 + 1.66 = +0.95 \, V \]
Conclusion:

Since the calculated E\(^\circ_{cell}\) is positive (+0.95 V), the corresponding Gibbs free energy change (\(\Delta G^\circ = -nFE^\circ_{cell}\)) will be negative. This indicates that the reaction is spontaneous. Therefore, placing the tarnished silverware in an aluminum pan with an electrolyte will create a galvanic cell that spontaneously reduces the Ag\(_2\)S back to shiny Ag, thus removing the tarnish. The aluminum pan acts as a sacrificial anode and gets oxidized. The NaCl solution simply acts as a salt bridge, allowing ions to flow and complete the circuit.
Quick Tip: In any galvanic cell, the half-reaction with the more negative (or less positive) standard reduction potential will act as the anode (oxidation), and the one with the more positive (or less negative) potential will act as the cathode (reduction). Here, Al (-1.66 V) is much more negative than Ag\(_2\)S (-0.71 V), so Al will be oxidized and Ag\(_2\)S will be reduced.

Step 1: Understanding the Concept:

Cell potential is the fundamental property that quantifies the energy difference between the two half-cells in an electrochemical cell.


Step 2: Detailed Explanation:

- In a galvanic (voltaic) cell, spontaneous redox reactions occur, generating an electric current. This current flows because there is a difference in electrical potential between the two electrodes.

- This potential difference is called the cell potential or electromotive force (EMF).

- It is determined by the difference in the reduction potentials of the cathode and the anode: \( E_{cell} = E_{cathode} - E_{anode} \).

- The cell potential represents the maximum amount of work that can be done by the cell per unit of charge that passes through it (\( W_{max} = -nFE_{cell} \)).

- When measured under standard conditions (1 M concentration for solutes, 1 atm pressure for gases, 25 \(^\circ\)C), it is called the standard cell potential (E\(^\circ_{cell}\)).
Quick Tip: Think of cell potential as analogous to the pressure difference in a water pipe that causes water to flow, or the voltage in a simple circuit. A larger cell potential means a stronger driving force for the redox reaction.

Step 1: Understanding the Concept:

A fuel cell is an electrochemical device designed for continuous energy production from an external fuel source.


Step 2: Detailed Explanation:

- Working Principle: A fuel cell operates like a battery but does not run down or need recharging. It produces electricity and heat as long as fuel and an oxidant are supplied.

- Components: It typically consists of an anode, a cathode, and an electrolyte that allows ions to pass between them.

- Example (Hydrogen-Oxygen Fuel Cell): This is the most common type of fuel cell.
- Fuel: Hydrogen gas (H\(_2\)) is supplied to the anode.
- Oxidant: Oxygen gas (O\(_2\)) is supplied to the cathode.
- Reactions (in acidic or basic medium):
- At the Anode (Oxidation): \( 2H_2 \rightarrow 4H^+ + 4e^- \)
- At the Cathode (Reduction): \( O_2 + 4H^+ + 4e^- \rightarrow 2H_2O \)
- Overall Reaction: \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
- Advantages: Fuel cells are highly efficient and produce very low emissions. In the case of the H\(_2\)-O\(_2\) fuel cell, the only byproduct is water, making them an environmentally friendly energy source. They were used in the Apollo space missions.
Quick Tip: The key difference between a battery and a fuel cell is that a battery stores chemical energy, while a fuel cell converts externally supplied chemical energy. A battery is a closed system, whereas a fuel cell is an open system.

Step 1: Understanding the Concept:

We need to calculate the electromotive force (EMF) or cell potential (E\(_{cell}\)) of a galvanic cell under non-standard conditions (concentrations are not 1M). This requires the use of the Nernst equation.


Step 2: Key Formula or Approach:

1. Standard Cell Potential (E\(^\circ_{cell}\)): First, calculate the standard EMF.

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
2. Nernst Equation: Then, apply the Nernst equation to account for the non-standard concentrations.

\[ E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q \]
At 25 \(^\circ\)C (298 K), this simplifies to:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where \( Q \) is the reaction quotient.


Step 3: Detailed Explanation:

1. Identify Anode, Cathode, and Overall Reaction:

- Anode (Oxidation): The electrode with the more negative E\(^\circ\) is the anode. E\(^\circ_{Zn^{2+}/Zn}\) (-0.76 V) is more negative than E\(^\circ_{Cd^{2+}/Cd}\) (-0.40 V). So, Zn is the anode.
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \]
- Cathode (Reduction): Cd is the cathode.
\[ Cd^{2+}(aq) + 2e^- \rightarrow Cd(s) \]
- Overall Reaction:
\[ Zn(s) + Cd^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cd(s) \]

2. Calculate E\(^\circ_{cell}\):
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.40 \, V) - (-0.76 \, V) = 0.36 \, V \]

3. Apply the Nernst Equation:

- Number of electrons transferred, \( n = 2 \).
- Reaction Quotient, \( Q = \frac{[Products]}{[Reactants]} = \frac{[Zn^{2+}]}{[Cd^{2+}]} \). The concentrations of pure solids (Zn and Cd) are taken as 1.
- Given concentrations: [Zn\(^{2+}\)] = 0.1 M, [Cd\(^{2+}\)] = 0.01 M. \[ Q = \frac{0.1}{0.01} = 10 \]
- Now, substitute the values into the Nernst equation: \[ E_{cell} = 0.36 - \frac{0.0591}{2} \log(10) \]
Given that log(10) = 1: \[ E_{cell} = 0.36 - \frac{0.0591}{2} \times 1 \] \[ E_{cell} = 0.36 - 0.02955 \] \[ E_{cell} = 0.33045 \, V \]

Step 4: Final Answer:

The emf of the cell at 25 \(^\circ\)C is approximately 0.3305 V.
Quick Tip: To easily identify the anode and cathode from E\(^\circ\) values, remember "LEO says GER" (Lose Electrons Oxidation, Gain Electrons Reduction) and that oxidation occurs at the anode. The half-reaction with the MORE NEGATIVE reduction potential is more likely to be oxidized, hence it's the anode.

Step 1: Identify Compound 'A'

- Molecular formula C\(_2\)H\(_6\)O corresponds to two possible isomers: Ethanol (CH\(_3\)CH\(_2\)OH) and Dimethyl ether (CH\(_3\)OCH\(_3\)).
- The problem states that 'A' is oxidized by CrO\(_3\). Ethers are resistant to oxidation, while primary alcohols are oxidized. Therefore, 'A' must be Ethanol.
\[ \textbf{A = CH\(_3\)CH\(_2\)OH (Ethanol)} \]

Step 2: Identify Compound 'B'

- 'A' (Ethanol, a primary alcohol) is oxidized by CrO\(_3\) (a mild oxidizing agent) to form 'B'. The mild oxidation of a primary alcohol yields an aldehyde.
\[ \chemfig{CH_3CH_2OH} \xrightarrow{CrO_3} \chemfig{CH_3CHO} \]
- So, 'B' is Ethanal (Acetaldehyde).
\[ \textbf{B = CH_3CHO (Ethanal)} \]

Step 3: Identify Compound 'C'

- Compound 'B' (Ethanal) is warmed with iodine and NaOH. This is the Iodoform Test.
- The iodoform test gives a positive result (yellow precipitate of iodoform) for compounds containing a methyl ketone group (CH\(_3\)C=O) or groups that can be oxidized to it (like ethanol). Ethanal has a CH\(_3\)C=O group.
\[ \chemfig{CH_3CHO} + 3I_2 + 4NaOH \rightarrow \chemfig{CHI_3 \downarrow} + HCOONa + 3NaI + 3H_2O \]
- The yellow precipitate 'C' is Iodoform.
\[ \textbf{C = CHI\(_3\) (Iodoform)} \]

Step 4: Identify Compound 'D'

- Compound 'A' (Ethanol) is heated with concentrated H\(_2\)SO\(_4\) at 413 K (140 \(^\circ\)C).
- This is the acid-catalyzed intermolecular dehydration of an alcohol to form an ether. Two molecules of ethanol combine to form diethyl ether.
\[ 2\chemfig{CH_3CH_2OH} \xrightarrow[413 K]{conc. H_2SO_4} \chemfig{CH_3CH_2-O-CH_2CH_3} + H_2O \]
- So, 'D' is Diethyl ether.
\[ \textbf{D = CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\) (Diethyl ether)} \]

Step 5: Identify Compound 'E'

- Compound 'D' (Diethyl ether) reacts with excess HI.
- This is the cleavage of an ether by a hydrohalic acid. With excess HI and heat, both alkyl groups are converted to alkyl iodides.
\[ \chemfig{CH_3CH_2-O-CH_2CH_3} + 2HI \xrightarrow{\Delta} 2\chemfig{CH_3CH_2I} + H_2O \]
- So, 'E' is Iodoethane.
\[ \textbf{E = CH\(_3\)CH\(_2\)I (Iodoethane)} \] Quick Tip: For these types of "road-map" problems, start by deducing the structure of the initial compound from its molecular formula and the first reaction. Then, systematically follow the reaction sequence, using your knowledge of named reactions and standard transformations (oxidation, dehydration, iodoform test, etc.).

Step 1: Understanding the Concept:

This is the nitration of phenol. Phenol is a highly activated aromatic ring due to the strong electron-donating nature of the -OH group. Treatment with a strong nitrating mixture (concentrated nitric acid, usually with concentrated sulfuric acid as a catalyst) leads to extensive substitution.


Step 2: Chemical Equation:

The -OH group is an ortho-, para- directing group. Due to the high reactivity of phenol, nitration with concentrated HNO\(_3\) (in the presence of conc. H\(_2\)SO\(_4\)) results in the substitution of nitro groups at all available ortho and para positions.

The product is 2,4,6-trinitrophenol, which is commonly known as Picric acid.
Quick Tip: Note the difference in nitration of phenol with dilute vs. concentrated nitric acid. \textbf{Dilute HNO\(_3\)}: Gives a mixture of ortho-nitrophenol and para-nitrophenol (monosubstitution). \textbf{Concentrated HNO\(_3\)}: Gives 2,4,6-trinitrophenol (polysubstitution).

Step 1: Understanding the Concept:

This two-step reaction is the Hydroboration-Oxidation of an alkene (propene). It is a method for the hydration of alkenes to produce alcohols.


Step 2: Chemical Equation and Regiochemistry:

Step 1: Hydroboration. Propene reacts with diborane (B\(_2\)H\(_6\), which exists as BH\(_3\)). The boron atom adds to the less substituted carbon of the double bond, and the hydrogen atom adds to the more substituted carbon. This is an anti-Markovnikov addition. This happens three times to form a trialkylborane intermediate.
\[ 3\chemfig{CH_3-CH=CH_2} + BH_3 \rightarrow (\chemfig{CH_3-CH_2-CH_2})_3B \]
Step 2: Oxidation. The trialkylborane intermediate is then oxidized using hydrogen peroxide (H\(_2\)O\(_2\)) in a basic medium (OH\(^-\)). The boron group is replaced by a hydroxyl (-OH) group with retention of stereochemistry.
\[ (\chemfig{CH_3-CH_2-CH_2})_3B + 3H_2O_2 + 3OH^- \rightarrow 3\chemfig{CH_3-CH_2-CH_2-OH} + B(OH)_4^- \]
The final product is the anti-Markovnikov alcohol.

Overall Reaction: Propene gives Propan-1-ol.
Quick Tip: Hydroboration-oxidation is the go-to reaction for anti-Markovnikov hydration of an alkene (adding -OH to the less substituted carbon). This contrasts with acid-catalyzed hydration, which follows Markovnikov's rule (adding -OH to the more substituted carbon).

Step 1: Understanding the Concept:

This reaction involves a sodium alkoxide (sodium t-butoxide) and an alkyl halide (methyl chloride). This is the Williamson Ether Synthesis.


Step 2: Chemical Equation and Mechanism:

The reaction proceeds via an S\(_N\)2 mechanism. The alkoxide ion ((CH\(_3\))\(_3\)CO\(^-\)) acts as a nucleophile and attacks the methyl chloride. The chloride ion is displaced, and an ether is formed.

In this specific case, the alkyl halide is primary (methyl chloride), which is unhindered and ideal for S\(_N\)2. The alkoxide is tertiary and bulky, but since it's the nucleophile and the halide is primary, the substitution reaction proceeds efficiently to form the ether.
\[ \chemfig{(CH_3)_3C-O^-Na^+} + \chemfig{CH_3-Cl} \rightarrow \chemfig{(CH_3)_3C-O-CH_3} + NaCl \]
The product is methyl tert-butyl ether (MTBE).
Quick Tip: For a successful Williamson ether synthesis, it is best to use a primary alkyl halide and any alkoxide (primary, secondary, or tertiary). If a secondary or tertiary alkyl halide is used with a strong, bulky base like t-butoxide, the elimination (E2) reaction will dominate over substitution, forming an alkene instead of an ether.

Step 1: Understanding the Concept:

Butan-1-ol is a primary (1\(^\circ\)) alcohol, and butan-2-ol is a secondary (2\(^\circ\)) alcohol. They can be distinguished based on the differing rates at which they react with certain reagents, particularly in reactions that proceed via a carbocation intermediate.


Step 2: Detailed Explanation:

Test: Lucas Test

Reagent: Lucas reagent, which is a solution of anhydrous zinc chloride (ZnCl\(_2\)) in concentrated hydrochloric acid (HCl).

Principle: The reaction involves the conversion of the alcohol to the corresponding alkyl chloride, which is insoluble in the reagent and appears as a cloudy precipitate or turbidity. The reaction proceeds via an S\(_N\)1 mechanism, and its rate depends on the stability of the carbocation formed (3\(^\circ\) \textgreater 2\(^\circ\) \textgreater 1\(^\circ\)).

Procedure: Add a few drops of each alcohol to separate test tubes containing the Lucas reagent at room temperature. Shake the tubes and observe.

Observation:

- With Butan-1-ol (Primary alcohol): No turbidity appears at room temperature. The solution remains clear. (Turbidity may appear only upon heating).

- With Butan-2-ol (Secondary alcohol): Turbidity appears slowly, typically within 5-10 minutes.

(For comparison, a tertiary alcohol like t-butanol would give immediate turbidity).

This difference in the time taken for turbidity to appear allows for the distinction between the primary and secondary alcohols.
Quick Tip: Remember the Lucas test results for different alcohols at room temperature: \textbf{3\(^\circ\) Alcohol:} Immediate turbidity. \textbf{2\(^\circ\) Alcohol:} Turbidity in 5-10 minutes. \textbf{1\(^\circ\) Alcohol:} No turbidity (or only on heating).

Step 1: Understanding the Concept:

The acidic strength of these compounds depends on the stability of the conjugate base formed after donating a proton (H\(^+\)). The more stable the conjugate base, the stronger the acid.


Step 2: Detailed Explanation:

Let's analyze the conjugate base of each compound:

1. Ethanol (CH\(_3\)CH\(_2\)OH):
- Acidic reaction: \( CH_3CH_2OH \rightleftharpoons CH_3CH_2O^- + H^+ \)
- The conjugate base is the ethoxide ion (CH\(_3\)CH\(_2\)O\(^-\)).
- The ethyl group (-CH\(_2\)CH\(_3\)) is an electron-donating group (+I effect). It pushes electron density onto the oxygen atom, which already has a negative charge. This intensifies the negative charge and destabilizes the ethoxide ion, making it a very strong base. Therefore, ethanol is a very weak acid.


2. Water (H\(_2\)O):
- Acidic reaction: \( H_2O \rightleftharpoons OH^- + H^+ \)
- The conjugate base is the hydroxide ion (OH\(^-\)).
- There are no electron-donating or withdrawing groups attached to the oxygen, so the hydroxide ion is more stable than the ethoxide ion. Thus, water is a stronger acid than ethanol.


3. Phenol (C\(_6\)H\(_5\)OH):
- Acidic reaction: \( C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+ \)
- The conjugate base is the phenoxide ion (C\(_6\)H\(_5\)O\(^-\)).
- The phenoxide ion is highly stabilized because the negative charge on the oxygen atom is delocalized over the benzene ring through resonance. The negative charge is spread across the ortho and para positions of the ring. This extensive delocalization makes the phenoxide ion much more stable than the hydroxide ion.
- Because its conjugate base is very stable, phenol is the strongest acid among the three.


Conclusion:

Based on the stability of the conjugate bases (phenoxide \textgreater hydroxide \textgreater ethoxide), the increasing order of acidic strength is:

ethanol \textless water \textless phenol
Quick Tip: A simple rule of thumb: Alcohols are generally weaker acids than water, and phenols are significantly more acidic than both water and alcohols due to the resonance stabilization of the phenoxide ion.