Vitamin D, classified as a fat-soluble vitamin, behaves differently from water-soluble vitamins such as vitamin C and the B vitamins. The primary characteristic of fat-soluble vitamins is their ability to be stored in the body's fatty tissues and liver. This means that when you consume vitamin D, it is absorbed through the intestines, transported via the bloodstream, and then stored in the fat cells and liver for later use. This characteristic makes Assertion (A) true, as it correctly states that Vitamin D is a fat-soluble vitamin.


The confusion arises with Reason (R), which incorrectly states that Vitamin D is excreted from the body in urine. This is not accurate for fat-soluble vitamins. Unlike water-soluble vitamins, which are excreted through urine when in excess, fat-soluble vitamins like Vitamin D are stored in the body. They do not get excreted through the urinary system; instead, they remain in fat and liver tissues for future utilization. This makes Reason (R) false.


Step 2:
Thus, based on the explanation above, the correct answer is option (3). Quick Tip: It’s important to remember the distinction between fat-soluble and water-soluble vitamins. Fat-soluble vitamins like Vitamin D, A, E, and K are stored in the liver and fatty tissues, while water-soluble vitamins are not stored in the body and are usually excreted through urine when in excess.

The Gabriel Phthalimide synthesis is a well-known method for synthesizing primary amines, but it is not effective for preparing aromatic amines. The reason for this is that aryl halides do not undergo nucleophilic substitution reactions with the phthalimide anion. This is due to the poor leaving group ability of the halides in aryl compounds, which hinders the substitution process. However, this issue is not the primary reason why aromatic primary amines cannot be synthesized using this method.


Step 2:
Therefore, while both Assertion (A) and Reason (R) are true, Reason (R) does not correctly explain the statement in Assertion (A). Quick Tip: The Gabriel synthesis is an effective method for preparing primary amines from alkyl halides, but it does not work well with aryl halides due to their poor reactivity in nucleophilic substitution reactions.

Copper (Cu) does not release hydrogen gas when it reacts with dilute mineral acids because it is a less reactive metal. Due to its low reactivity, copper is unable to displace hydrogen from the acids. While copper has a positive electrode potential, which indicates that it is less prone to losing electrons and participating in such reactions, this property does not directly explain why it does not release hydrogen gas.


Step 2:
Therefore, Assertion (A) is incorrect, while Reason (R) is true, but they are not logically linked to each other. Quick Tip: Metals such as copper, with a positive electrode potential, are less reactive and do not readily produce hydrogen gas when reacting with dilute acids.

In the case of a first-order reaction, the half-life is constant and does not depend on the initial concentration of the reactant. This means the half-life remains the same, regardless of any changes in the concentration of the reactant over time. Therefore, Assertion (A) is incorrect, while Reason (R) is correct.


Step 2:
Thus, Assertion (A) is false, and Reason (R) is true. Quick Tip: In first-order reactions, the half-life is unaffected by the concentration of reactants, which is a distinctive characteristic of these reactions.

Scurvy is a condition that results from a lack of Vitamin C, also known as ascorbic acid. Vitamin C plays a vital role in the synthesis of collagen, a key protein that supports the structure of connective tissues in the body.


Step 2:
Therefore, the correct answer is option (3). Quick Tip: Vitamin C (ascorbic acid) is essential for the formation of collagen and the maintenance of connective tissues. A deficiency in this vitamin leads to scurvy.

In nucleic acids such as DNA and RNA, nucleotides are linked together by phosphodiester bonds. These bonds occur when a phosphate group attaches the 3' carbon of one sugar molecule to the 5' carbon of the next sugar molecule in the chain.


Step 2:
Therefore, the correct answer is option (4). Quick Tip: Phosphodiester bonds form the structural backbone of nucleic acids, linking the sugar and phosphate groups of consecutive nucleotides.

Protein denaturation occurs when proteins lose their original structure due to external influences such as changes in temperature or pH levels. The processes of egg white coagulation and milk curdling are both examples of protein denaturation.


Step 2:
Thus, the correct answer is option (4). Quick Tip: Proteins can denature when exposed to heat, acid, or mechanical forces, resulting in the disruption of their functional shape.

The Reimer-Tiemann reaction is employed to convert phenols into salicylic acid. In this process, phenol is treated with chloroform and a base, resulting in the formation of salicylic acid.


Step 2:
Therefore, the correct answer is option (1). Quick Tip: The Reimer-Tiemann reaction consists of the chlorination of phenol followed by hydrolysis, producing salicylic acid, which is a key precursor for aspirin.

When a secondary alcohol is oxidized with chromic anhydride (CrO3), it is converted into a ketone. Secondary alcohols do not undergo oxidation to form aldehydes or carboxylic acids.


Step 2:
Therefore, the correct answer is option (2). Quick Tip: Chromic anhydride is a powerful oxidizing agent that specifically oxidizes secondary alcohols to ketones, without affecting aldehydes or carboxylic acids.

The compound contains both a methyl group and a chlorine atom attached to a benzene ring. The appropriate IUPAC name is 1-Chloro-4-Methylbenzene, where the numbering begins from the position of the chlorine atom and continues to the position of the methyl group.


Step 2:
Therefore, the correct answer is option (3). Quick Tip: For aromatic compounds with multiple substituents, the substituents should be numbered in such a way that the locants are as low as possible, with the substituents listed in alphabetical order.

Diamagnetic substances are those that do not contain any unpaired electrons. Among the given options, \( [Ni(CN)_4]^{2-} \) is diamagnetic because the \( Ni^{2+} \) ion in this complex has no unpaired electrons.


Step 2:
Therefore, the correct answer is option (1). Quick Tip: Diamagnetic species have all electrons paired and are repelled by magnetic fields, in contrast to paramagnetic species, which possess unpaired electrons.

Linkage isomerism arises when a ligand can coordinate to a metal ion in two distinct ways. In this case, \( NO_2 \) and \( ONO \) are two different forms of the same ligand, resulting in linkage isomerism.


Step 2:
Hence, the correct answer is option (2). Quick Tip: Linkage isomerism occurs when a ligand binds to a metal ion through different atoms, creating isomers that have the same molecular formula but distinct structural arrangements.

The electron configuration \( [Ar] 3d^{10} 4s^1 \) is characteristic of Copper (Cu), which has an atomic number of 29.


Step 2:
Therefore, the correct answer is option (1). Quick Tip: Copper (Cu) follows the electron configuration \( [Ar] 3d^{10} 4s^1 \), which deviates from the usual pattern due to the enhanced stability provided by a fully filled 3d subshell.

Molecularity of a reaction refers to the number of reactant molecules that participate in an elementary reaction. It is an essential concept used to understand the reaction mechanism.


Step 2:
Therefore, the correct answer is option (3). Quick Tip: Molecularity represents the number of reactant molecules that collide in an elementary step to initiate a reaction.

Molecularity refers to the number of reacting species in an elementary reaction, and it cannot be zero. A reaction cannot take place without any reacting molecules.


Step 2:
Therefore, the correct answer is option (3). Quick Tip: Molecularity is always a positive integer, indicating the number of molecules involved in an elementary reaction step.

In the reduction of \( MnO_4^- \) to \( MnO_2 \), the manganese ion undergoes a change in oxidation state from +7 to +4, which requires 5 moles of electrons. Hence, the charge required is 5 F (Faradays).


Step 2:
Thus, the correct answer is option (3). Quick Tip: The number of Faradays required for a redox reaction corresponds to the total number of electrons involved in the oxidation or reduction of the species.

Step 1: Understanding the \( S_N1 \) Mechanism.

The \( S_N1 \) reaction follows a two-step process:

1. Formation of a carbocation (the rate-determining step).

2. Nucleophilic attack on the carbocation.


The rate of the \( S_N1 \) reaction is influenced by the stability of the carbocation. Tertiary carbocations are more stable due to greater alkyl group substitution, which provides enhanced electron donation through hyperconjugation and inductive effects. This increased stability facilitates faster carbocation formation, thereby accelerating the reaction rate.

When comparing two halogen compounds, the one with the tertiary halogenated carbon (more substituted carbon) will form a more stable carbocation and therefore undergo the \( S_N1 \) reaction at a faster rate. Quick Tip: Keep in mind that in \( S_N1 \) reactions, the stability of the carbocation intermediate is key. More substituted carbocations are typically more stable and form more quickly, speeding up the overall reaction.

Step 1: Understanding the \( S_N2 \) Mechanism.

The \( S_N2 \) mechanism occurs in a single step where the nucleophile simultaneously attacks the electrophilic carbon and displaces the leaving group. This process is highly sensitive to steric hindrance; thus, carbons with fewer substituents are more reactive because they offer easier access to the nucleophile.


2-Bromo-2-methylbutane: Being a tertiary bromide, this compound is highly hindered, making \( S_N2 \) reactions very slow.

2-Bromopentane: As a secondary bromide, it is less hindered than the tertiary one but still more so than a primary bromide, placing it in the middle in terms of reactivity.

1-Bromopentane: This is a primary bromide with minimal steric hindrance, which promotes the fastest \( S_N2 \) reaction.


Thus, the order of increasing reactivity towards \( S_N2 \) displacement is: \[ 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane. \] Quick Tip: For \( S_N2 \) reactions, remember that less steric hindrance at the reaction center results in a faster reaction. Primary halides are generally more reactive than secondary and tertiary halides.

Step 1: Understanding Second-Order Reactions.
For a second-order reaction, the rate law is expressed as: \[ Rate = k [A]^2 \]
where \( [A] \) represents the concentration of the reactant, and \( k \) is the rate constant.


Step 2: Effect of Concentration Change.
(i) If the concentration of \( A \) is doubled, the rate will increase by a factor of \( 2^2 = 4 \) because the rate is proportional to the square of the concentration. \[ New Rate = k (2[A])^2 = 4k [A]^2 \]

(ii) If the concentration of \( A \) is halved, the rate will decrease by a factor of \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \). \[ New Rate = k \left(\frac{1}{2}[A]\right)^2 = \frac{1}{4} k [A]^2 \]

Thus, doubling the concentration increases the rate by a factor of 4, while halving the concentration reduces the rate to one-quarter. Quick Tip: For second-order reactions, the rate is proportional to the square of the concentration. Doubling the concentration increases the rate by a factor of 4, and halving it decreases the rate by a factor of 4.

Step 1: Identifying Compound (A).

When \( FeCr_2O_4 \) is fused with \( Na_2CO_3 \) in the presence of air, it produces a yellow solution of sodium chromate (\( Na_2CrO_4 \)).

Thus, \( A = Na_2CrO_4 \).


Step 2: Identifying Compound (B).

Upon acidifying \( Na_2CrO_4 \), it is converted to potassium dichromate (\( K_2Cr_2O_7 \)), which is orange in color.

Thus, \( B = K_2Cr_2O_7 \).


Step 3: Identifying Compound (C).

When \( K_2Cr_2O_7 \) reacts with KCl, it forms orange-colored \( Cr_2O_7^{2-} \) ions. Hence, \( C \) is an orange-colored chromium compound.

Thus, \( C = K_2Cr_2O_7 \).



Step 4: Identifying Compound (D).

An acidified solution of \( C \) oxidizes sodium sulfite (\( Na_2SO_3 \)) to sodium sulfate (\( Na_2SO_4 \)). This is a typical redox reaction where chromium compounds act as oxidizing agents.

Thus, \( D = Na_2SO_4 \).


So, the identities are:
\( A = Na_2CrO_4 \)
\( B = K_2Cr_2O_7 \)
\( C = K_2Cr_2O_7 \)
\( D = Na_2SO_4 \)
Quick Tip: Chromates and dichromates are powerful oxidizing agents. Chromates appear yellow in alkaline conditions, while dichromates appear orange in acidic conditions.

Step 1: Understanding Inner and Outer Orbital Complexes.
Inner orbital complexes involve the use of \( d \)-orbitals from the inner shell (typically the \( 3d \)-orbitals in transition metals). This type of hybridization leads to low-spin complexes, where ligands coordinate using these inner \( d \)-orbitals.

Outer orbital complexes involve the use of \( d \)-orbitals from the outer shell (typically \( 4d \)-orbitals for transition metals in higher oxidation states). This type of hybridization leads to high-spin complexes, where ligands coordinate using these outer \( d \)-orbitals.


Step 2: Identifying the Hybridization in \( [Co(NH_3)_6]^{3+} \).
Cobalt in the \( +3 \) oxidation state has an electronic configuration of \( 3d^6 \), and the complex \( [Co(NH_3)_6]^{3+} \) uses inner \( 3d \)-orbitals for bonding with the ligands. The \( 3d^2sp^3 \) hybridization results in an inner orbital complex.


Step 3: Identifying the Hybridization in \( [Ni(NH_3)_6]^{2+} \).
Nickel in the \( +2 \) oxidation state has an electronic configuration of \( 3d^8 \), and the complex \( [Ni(NH_3)_6]^{2+} \) uses outer \( 4d \)-orbitals for bonding. The \( sp^3d^2 \) hybridization results in an outer orbital complex.

Thus, \( [Co(NH_3)_6]^{3+} \) is an inner orbital complex, while \( [Ni(NH_3)_6]^{2+} \) is an outer orbital complex. Quick Tip: For inner orbital complexes, low-spin configurations are typical with \( d^2sp^3 \) hybridization. For outer orbital complexes, high-spin configurations are typical with \( sp^3d^2 \) hybridization.

For a zero-order reaction, the integrated rate law is given by: \[ [A] = [A_0] - kt \]
where:

\( [A] \) represents the concentration of A at time \( t \),

\( [A_0] \) is the initial concentration,

\( k \) is the rate constant, and

\( t \) is the time elapsed.


Given values are:

\( [A_0] = 0.10 \, M \),
\( [A] = 0.075 \, M \),

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


Substituting these values into the rate law:

\[ 0.075 = 0.10 - (0.0030)(t) \]

Rearranging to solve for \( t \):
\[ 0.0030t = 0.10 - 0.075 = 0.025 \] \[ t = \frac{0.025}{0.0030} = 8.33 \, seconds \]

Thus, it will take approximately \( 8.33 \, seconds \) for the concentration of A to decrease from 0.10 M to 0.075 M. Quick Tip: In zero-order reactions, concentration decreases at a constant rate over time. The time for a specific concentration change can be determined using the integrated rate law.

In a zero-order reaction, the rate of reaction remains constant and is equal to the rate constant \( k \). The rate of product formation is directly proportional to this rate constant. In this case, the rate constant is \( k = 2.5 \times 10^{-4} \, mol L^{-1} \, s^{-1} \).

For the decomposition of \( NH_3 \) on the platinum surface, the products \( N_2 \) and \( H_2 \) are produced in a 1:3 molar ratio. As a result, the rates of production of both \( N_2 \) and \( H_2 \) will be equal to the rate constant \( k \).

Thus, the rates of production of \( N_2 \) and \( H_2 \) are both: \[ Rate of production of N_2 = Rate of production of H_2 = 2.5 \times 10^{-4} \, mol L^{-1} \, s^{-1}. \] Quick Tip: In zero-order reactions, the rate of product formation remains constant and is equal to the rate constant, irrespective of the reactant concentration.

Osmotic pressure (\( \pi \)) is given by the formula: \[ \pi = \frac{nRT}{V} \]
where:

\( n \) is the number of moles of solute,

\( R \) is the gas constant,

\( T \) is the temperature,

\( V \) is the volume of the solution.


Since both solutions have the same osmotic pressure, the number of moles of solute in both solutions must be equal. For urea, the number of moles \( n_1 \) is calculated as: \[ n_1 = \frac{15 \, g}{60 \, g/mol} = 0.25 \, mol. \]

Let the mass of glucose be \( m_2 \). The number of moles of glucose \( n_2 \) is: \[ n_2 = \frac{m_2}{180 \, g/mol}. \]

Since the osmotic pressures are identical, the number of moles of urea must equal the number of moles of glucose: \[ 0.25 = \frac{m_2}{180}. \]

Solving for \( m_2 \): \[ m_2 = 0.25 \times 180 = 45 \, g. \]

Therefore, the mass of glucose present in one litre of its solution is \( 45 \, g \). Quick Tip: When two solutions have the same osmotic pressure, the number of moles of solute is inversely proportional to the molar mass of the solute.

First, we need to calculate the molar conductivity at infinite dilution, \( \Lambda_m^0 \), for acetic acid, which is the sum of the conductivity contributions from the ions of acetic acid and sodium acetate: \[ \Lambda_m^0 (Acetic acid) = \Lambda_m^0 (HC) - \Lambda_m^0 (NaCl) = 426 - 126 = 300 \, \Omega^{-1} \, cm^2 \, mol^{-1}. \]

Next, we can calculate the degree of dissociation \( \alpha \) by comparing the observed molar conductivity of acetic acid to its expected conductivity at infinite dilution: \[ \alpha = \frac{\Lambda_m}{\Lambda_m^0}. \]
Substituting the given values: \[ \alpha = \frac{48.1}{300} \approx 0.16. \]

Thus, the degree of dissociation \( \alpha \) is 0.16, and the molar conductivity at infinite dilution for acetic acid is 300 \( \Omega^{-1} \, cm^2 \, mol^{-1} \). Quick Tip: The degree of dissociation \( \alpha \) represents the proportion of the total molecules that dissociate into ions in a solution.

Chromium in the \( Cr^{2+} \) state has an electronic configuration of \( [Ar] 3d^4 \), which is relatively unstable and easily oxidizes to the more stable \( Cr^{3+} \) configuration, \( [Ar] 3d^5 \). This instability makes \( Cr^{2+} \) a strong reducing agent because it readily loses electrons.

In contrast, \( Mn^{3+} \) has an electronic configuration of \( [Ar] 3d^4 \), which is also unstable. It tends to gain electrons to achieve the more stable \( Mn^{2+} \) configuration, \( [Ar] 3d^5 \). Hence, \( Mn^{3+} \) acts as a strong oxidizing agent, readily accepting electrons.


Step 2:
The differences in the electronic configurations of \( Cr^{2+} \) and \( Mn^{3+} \) explain why \( Cr^{2+} \) is a strong reducing agent and \( Mn^{3+} \) is a strong oxidizing agent. Quick Tip: Elements with unstable electronic configurations in certain oxidation states (like \( Cr^{2+} \) or \( Mn^{3+} \)) are more likely to either gain or lose electrons to reach more stable configurations.

Lanthanide contraction refers to the gradual reduction in the ionic radius of the lanthanide series as the atomic number increases, due to the ineffective shielding of the 4f-electrons. Two significant consequences of lanthanide contraction are:

1. Decreasing ionic radii: As the atomic number increases from \( La^{3+} \) to \( Lu^{3+} \), the size of the ions decreases. This results in a higher effective nuclear charge acting on the electrons, causing the ionic radii to shrink.

2. Similarity between 3d and 4d/5d elements: Lanthanide contraction creates similarities in the properties of the lanthanide elements and the d-block elements in the same period. For example, \( Zr \) (a 4d element) and \( Hf \) (a 5d element) exhibit very similar ionic radii due to this contraction.


Step 2:
Thus, lanthanide contraction leads to smaller ionic radii and greater similarities in the properties of lanthanides and d-block elements. Quick Tip: The lanthanide contraction accounts for many of the similarities in the properties of d-block and f-block elements, especially concerning ionic radii and chemical behavior.

The element in the 3d series with the lowest enthalpy of atomisation is copper (Cu). Copper has a completely filled \( 3d^{10} \) configuration, which contributes to its high stability. This stable electron configuration reduces the energy required to break the metallic bonds during atomisation, resulting in a lower enthalpy of atomisation.

In comparison, elements like zinc (Zn) and manganese (Mn) have less stable electron configurations, requiring more energy to break the metallic bonds during atomisation.


Step 2:
Therefore, copper (Cu) has the lowest enthalpy of atomisation due to the stability provided by its filled \( 3d^{10} \) configuration. Quick Tip: Elements with stable electron configurations, such as copper’s \( 3d^{10} \), have lower enthalpies of atomisation because less energy is needed to break the bonds.

(a) \( [Fe(en)_2Cl_2]^+ \)

In this complex:

The metal ion is \( Fe^{3+} \), representing iron in the +3 oxidation state.

\( en \) stands for ethylenediamine, a bidentate ligand.

Chlorine (Cl) acts as a monodentate ligand.

The overall charge of the complex is +1.


The IUPAC name for this complex is: \[ bis(ethylenediamine)(chlorido)iron(III)^+ \]

(b) \( [Co(NH_3)_4(H_2O)Br]SO_4 \)


In this complex:

The metal ion is \( Co^{3+} \), cobalt in the +3 oxidation state.

\( NH_3 \) is ammonia, a monodentate ligand.

\( H_2O \) is water, also a monodentate ligand.

\( Br \) is bromide, a monodentate ligand.

The anion \( SO_4^{2-} \) balances the charge.


The IUPAC name for this complex is: \[ tetraammin(aqua)bromidocobalt(III) sulfate \]

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

In this complex:

The metal ion is \( Ni^{2+} \), nickel in the +2 oxidation state.

\( CN^- \) is cyanide, a monodentate ligand.

The overall charge of the complex is -2.


The IUPAC name for this complex is: \[ tetra(cyanido)nickelate(II)^{2-} \] Quick Tip: When naming coordination complexes, follow these steps:
1. Name the ligands first, in alphabetical order, using prefixes like "di-" or "tri-" to indicate the number of identical ligands.
2. The metal comes after the ligands, with its oxidation state in Roman numerals in parentheses.
3. For anionic complexes, the metal name ends in "-ate" (e.g., nickel becomes "nickelate").
4. Use common names for neutral ligands (e.g., ammonia for \( NH_3 \), water for \( H_2O \)).

(a) Wolff-Kishner Reduction
The Wolff-Kishner reduction is a method used to reduce a carbonyl group (C=O) to a methylene group (CH\(_2\)) by employing hydrazine (H\(_2\)NNH\(_2\)) in the presence of a strong base such as potassium hydroxide (KOH), with heating. This reaction effectively removes the oxygen atom from the carbonyl group, converting it into a hydrocarbon.

Reaction: \[ R-CO-R' + H_2NNH_2 \xrightarrow{KOH, heat} R-CH_2R' \]
This process reduces aldehydes or ketones to alkanes.


(b) Etard Reaction
The Etard reaction involves the oxidation of toluene (methylbenzene) to benzaldehyde using chromium-based reagents such as chromium trioxide (CrO\(_3\)) or pyridinium chlorochromate (PCC). In this reaction, an aldehyde group is added to the methyl group of toluene.

Reaction: \[ C_6H_5CH_3 + CrO_3 \rightarrow C_6H_5CHO \]
Here, toluene is oxidized to benzaldehyde.


(c) Cannizzaro Reaction
The Cannizzaro reaction is a redox disproportionation reaction that occurs with aldehydes lacking an alpha hydrogen. In the presence of a strong base, one molecule of the aldehyde is reduced to an alcohol, while another is oxidized to a carboxylate anion.

Reaction: \[ 2 R-CHO \xrightarrow{strong base} R-CH_2OH + R-COO^- \]
In this reaction, two molecules of aldehyde participate: one is reduced to alcohol, and the other is oxidized to a carboxylate. Quick Tip: In the Wolff-Kishner reduction, remember that hydrazine and strong base under heat are used to reduce carbonyl compounds, whereas in the Cannizzaro reaction, aldehydes without alpha hydrogens undergo disproportionation.

(a) Reaction Sequence:

1. CH\(_3\)COOH \( \xrightarrow{SOCl_2} \) CH\(_3\)COCl:

Acetic acid (CH\(_3\)COOH) reacts with thionyl chloride (SOCl\(_2\)) to form acetyl chloride (CH\(_3\)COCl), which is an acyl chloride.


2. CH\(_3\)COCl \( \xrightarrow{H_2, Pd-BaSO_4} \) CH\(_3\)CH\(_2\)NH\(_2\):

The acetyl chloride undergoes a reduction reaction using hydrogen (H\(_2\)) in the presence of a palladium catalyst on barium sulfate (Pd-BaSO\(_4\)) to form ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)).


3. CH\(_3\)CH\(_2\)NH\(_2\) \( \xrightarrow{H_2N-NH_2} \) C:

The ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)) reacts with hydrazine (H\(_2\)N-NH\(_2\)) to form a product C, which is likely a hydrazone or another nitrogen-containing compound.



(b) Reaction Sequence:
1. CH\(_3\)CN \( \xrightarrow{1. DIBAL-H, 2. H}_2O \) A:

Methyl cyanide (CH\(_3\)CN) reacts with DIBAL-H (diisobutylaluminum hydride) to reduce the nitrile group (CN) to an aldehyde (CH\(_3\)CHO), which is compound A.


2. A \( \xrightarrow{NaOH, heat} \) B:

The aldehyde A undergoes an aldol condensation in the presence of sodium hydroxide (NaOH) and heat, forming a product B, which is likely a β-hydroxy aldehyde or ketone.


3. B \( \xrightarrow{C} \) C:

The compound B undergoes further reactions, likely a dehydration or other transformation to form the final product C.
Quick Tip: In this reaction sequence, DIBAL-H is used to reduce nitriles to aldehydes, and aldol condensation with NaOH under heat leads to the formation of β-hydroxy aldehydes or ketones, which can undergo further transformations to form the final product.

(i) Enantiomers:

Enantiomers are pairs of molecules that are mirror images of each other but cannot be superimposed. These molecules have the same physical properties but differ in how they interact with plane-polarized light and with other chiral substances. For instance, the right- and left-handed forms of a chiral molecule are enantiomers.


(ii) Racemic mixture:

A racemic mixture consists of equal amounts of two enantiomers of a chiral compound. In such a mixture, the optical activities of the enantiomers cancel each other out, leading to no overall optical rotation.
Quick Tip: Enantiomers can be separated using chiral chromatography, while racemic mixtures can be resolved into their individual enantiomers through methods like recrystallization or enzymatic processes.

Chlorobenzene is resistant to nucleophilic substitution due to the electron-withdrawing effect of the chlorine atom, which decreases the electron density on the benzene ring. This makes the carbon attached to the chlorine less electrophilic, which reduces its ability to undergo nucleophilic attack. Additionally, the resonance stabilization of the chlorine atom with the aromatic ring further makes the carbon-chlorine bond stronger, thus making nucleophilic substitution less favorable. Quick Tip: The electron-withdrawing effect of the chlorine atom and the resonance stabilization in chlorobenzene make it resistant to nucleophilic substitution reactions, unlike alkyl halides where such substitution is easier.

When D-glucose reacts with hydroxylamine (\( H_2N - OH \)), it forms an oxime. The aldehyde group of D-glucose reacts with hydroxylamine, resulting in the formation of a stable oxime with the general structure \( R - C = N - OH \), where R represents the rest of the glucose molecule. In this reaction, the glucose aldehyde undergoes a nucleophilic addition to yield the corresponding oxime. Quick Tip: Hydroxylamine reacts with aldehydes and ketones to produce oximes, which serve as valuable intermediates in organic synthesis.

Amino acids exhibit amphoteric behavior because they contain both acidic and basic functional groups. The amino group (\( -NH_2 \)) can function as a base by accepting protons, while the carboxyl group (\( -COOH \)) can donate protons, acting as an acid. This dual functionality allows amino acids to behave as both acids and bases, depending on the pH of the solution, illustrating their amphoteric nature. Quick Tip: Amino acids can exist in various ionic forms (zwitterions) based on the pH, where the amino group is protonated (\( NH_3^+ \)) and the carboxyl group is deprotonated (\( COO^- \)).

Vitamin C (ascorbic acid) cannot be stored in the body because it is water-soluble, and the body lacks the ability to store water-soluble vitamins. As a result, any excess vitamin C is excreted in the urine. To maintain sufficient levels of vitamin C, it is crucial to include it regularly in the diet since the body cannot store it for extended periods. Quick Tip: Since vitamin C is not stored in the body, it needs to be consumed regularly. Good sources of vitamin C include citrus fruits, tomatoes, and green leafy vegetables.

Step 1: Understanding the Factors Influencing Basicity

The basicity of amines is mainly determined by the electron density on the nitrogen atom. The more electron-donating the substituents, the more basic the amine will be, as they enhance the availability of the nitrogen's lone pair for protonation. On the other hand, electron-withdrawing groups decrease the electron density on nitrogen, reducing its basicity.

Electron-donating groups (like \(–NH_2\) and \(–CH_3\)) increase the basicity by raising the electron density on nitrogen.

Electron-withdrawing groups (such as \(–NO_2\) and \(–CN\)) decrease the basicity as they pull electron density away from nitrogen, making the lone pair less available to accept a proton.



Step 2: Analyzing the Substituents

Now, let’s analyze the substituents attached to the nitrogen atoms in the given compounds:


\(–NH_2\) is an amine group, an electron-donating group. It will increase the electron density on nitrogen, making the amine more basic.

\(–NO_2\) is a strong electron-withdrawing group. It pulls electron density away from nitrogen, reducing the availability of the lone pair and lowering the basicity.

\(–CH_3\) is also an electron-donating group, but weaker than \(–NH_2\). Therefore, it increases basicity, but not as strongly as \(–NH_2\).



Step 3: Arranging in Order of Basicity

Now, let’s rank the compounds based on the strength of their electron-donating or electron-withdrawing effects:


\(–NO_2\) (electron-withdrawing) will have the lowest basicity.

\(–NO_2\) (when positioned para) will be less basic than \(–NH_2\) due to the electron-withdrawing nature of \(–NO_2\).

\(–NH_2\) will be the most basic because it is a strong electron-donating group.

\(–CH_3\) is a mild electron-donating group, so it will be more basic than \(–NO_2\), but less basic than \(–NH_2\).


Therefore, the increasing order of basicity is: \[ NO_2 < NO_2 < CH_3 < NH_2 \] Quick Tip: The basicity of amines is determined by the electron-donating or electron-withdrawing effect of the substituents attached to the nitrogen. Electron-donating groups increase basicity, while electron-withdrawing groups decrease it.

Step 1: Understanding Basicity

The basicity of amines refers to the ability of the nitrogen atom to donate its lone pair of electrons to a proton. The higher the availability of the nitrogen’s lone pair for protonation, the greater the basicity.

In aniline (C₆H₅NH₂), the nitrogen’s lone pair can participate in resonance with the benzene ring, reducing its availability for protonation. This lowers the electron density on the nitrogen, making aniline a weaker base compared to methylamine (CH₃NH₂), where the methyl group is an electron-donating group. The methyl group increases the electron density on nitrogen, making the lone pair more available for protonation and enhancing the basicity.


Step 2: Resonance Effect in Aniline
In aniline, the nitrogen’s lone pair can be delocalized into the aromatic ring through resonance, forming a structure where the lone pair is less available for protonation. This resonance effect decreases the basicity of aniline when compared to methylamine.


Step 3: Methylamine and the Inductive Effect
Methylamine, in contrast, has a methyl group (–CH₃) attached to nitrogen. The methyl group donates electrons via the inductive effect, increasing the electron density on nitrogen and making its lone pair more available for protonation, which increases the basicity.

As a result, methylamine has a greater basicity than aniline, leading to a higher pKₐ for methylamine compared to aniline. Quick Tip: The basicity of aromatic amines like aniline is reduced due to resonance with the aromatic ring, while aliphatic amines like methylamine are more basic due to the electron-donating inductive effect of the alkyl group.

Step 1: Understanding the Effect of Alkyl Groups on Basicity
The basicity of amines in aqueous solutions increases with the number of alkyl groups attached to the nitrogen atom. Alkyl groups are electron-donating groups, and they increase the electron density on nitrogen, making the lone pair more available to accept a proton.


Trimethylamine \((CH_3)_3N\) has three methyl groups attached to nitrogen, making it highly electron-donating and thus the most basic.

Dimethylamine \((CH_3)_2\)NH has two methyl groups, so it is also basic, but not as basic as trimethylamine.

Methylamine \((CH_3NH_2)\) has one methyl group, which donates less electron density than two or three groups.

Ammonia \((NH_3)\) has no alkyl groups, so it is the least basic.



Step 2: Arranging in Order of Basicity
The increasing order of basicity is:
\[\((NH_3)\) < \((CH_3NH_2)\) < \((CH_3)_2\)NH < \((CH_3)_3N\)\] Quick Tip: The basicity of amines increases as the number of electron-donating alkyl groups attached to nitrogen increases. Alkyl groups push electron density towards the nitrogen, enhancing the availability of its lone pair for protonation.

Step 1: The Ammonolysis Process
Ammonolysis of alkyl halides involves the reaction of an alkyl halide with ammonia (\( NH_3 \)), where the nucleophilic nitrogen from ammonia attacks the electrophilic carbon in the alkyl halide. This displaces the halide ion and forms a primary amine.

However, this reaction has a significant limitation: the product amine can further react with excess alkyl halide, resulting in the formation of secondary and tertiary amines.


Step 2: Multiple Reactions
The alkyl groups attached to the nitrogen of the amines formed during ammonolysis can cause the reaction to proceed further, leading to secondary and tertiary amines. This results in a mixture of amines, rather than a pure primary amine. For example, if methylamine (\( CH_3NH_2 \)) is formed in the first step, it can react with another molecule of alkyl halide to form dimethylamine (\( CH_3NHCH_3 \)), and further to trimethylamine (\( CH_3N(CH_3)_2 \)).


Step 3: Alternative Methods
To obtain pure primary amines, alternative methods such as reductive amination or Gabriel synthesis are preferred. These methods avoid the formation of secondary and tertiary amines. Quick Tip: Ammonolysis of alkyl halides often leads to a mixture of primary, secondary, and tertiary amines. To avoid this, alternative synthetic methods, like Gabriel synthesis, can be used to obtain pure primary amines.

Reverse osmosis is a process where solvent moves from a solution with a higher concentration to one with a lower concentration through a semipermeable membrane. However, this occurs only when external pressure is applied that exceeds the osmotic pressure, reversing the natural osmotic flow.

A common example of a semipermeable membrane (SPM) used in reverse osmosis is cellulose acetate. Quick Tip: Reverse osmosis is the process of driving solvent through a semipermeable membrane from a more concentrated solution to a less concentrated one by applying pressure greater than the osmotic pressure. It is widely used for water purification.

When red blood cells (RBCs) are placed in a 0.5% NaCl solution, which is hypotonic compared to the intracellular fluid, water moves into the RBCs by osmosis. This results in the cells swelling as water enters. If too much water accumulates, the RBCs may rupture, a process called hemolysis. Quick Tip: Hypotonic solutions cause water to move into RBCs by osmosis, leading to swelling and potentially causing the cells to burst (hemolysis).

Osmotic pressure is directly proportional to the number of solute particles in solution. Since KCl dissociates into two ions (K\(^+\) and Cl\(^-\)) in solution, a 1 M KCl solution will have a higher osmotic pressure than a 1 M urea solution, which does not dissociate into ions. Therefore, the 1 M KCl solution will have a higher osmotic pressure. Quick Tip: Osmotic pressure depends on the number of particles in solution. Ionic compounds like KCl dissociate into more particles, increasing osmotic pressure compared to non-ionic compounds like urea.

Osmotic pressure is a colligative property because it depends only on the number of solute particles in a solution, not on their nature or identity. The greater the number of solute particles, the higher the osmotic pressure, regardless of whether the solute is ionic or non-ionic. Quick Tip: Osmotic pressure is a colligative property because it depends on the concentration of solute particles in a solution, not their chemical nature.

Compound \( A \) has the molecular formula \( C_2H_6O \), indicating that it is ethanol.

When ethanol is oxidized by \( Cr_2O_7 \), it forms acetic acid (\( CH_3COOH \)), which is compound \( B \).

When acetic acid reacts with iodine and NaOH, it produces a yellow precipitate of iodoform (\( CHI_3 \)), which is compound \( C \).

When ethanol is heated with concentrated \( H_2SO_4 \) at 413 K, it undergoes dehydration to form ethene (\( C_2H_4 \)), which is compound \( D \).

When ethene reacts with excess HI, it forms iodoethane (\( C_2H_5I \)), which is compound \( E \).


The chemical equations involved are as follows:


1. \( C_2H_5OH + Cr_2O_7^{2-} \to CH_3COOH \)

2. \( CH_3COOH + I_2 + NaOH \to CHI_3 \) (yellow precipitate)

3. \( C_2H_5OH \xrightarrow{H_2SO_4, 413K} C_2H_4 + H_2O \)

4. \( C_2H_4 + I_2 \to C_2H_5I \)
Quick Tip: The iodoform test (formation of a yellow precipitate) is used to detect compounds containing the structure \( CH_3CO \) or \( CH_3CH \), such as aldehydes and alcohols.

1. Phenol treated with conc. \( HNO_3 \):
Phenol reacts with concentrated nitric acid (\( HNO_3 \)) to form a mixture of nitrophenols. The reaction is as follows:
\[ C_6H_5OH + HNO_3 \to 2,4,6 - Nitrophenol + H_2O \]

2. Propene treated with \( B_2H_6 \) followed by oxidation by \( H_2O_2 / OH^- \):
The hydroboration-oxidation reaction of propene results in the formation of an alcohol:
\[ C_3H_6 + B_2H_6 \to C_3H_7B \quad (Hydroboration) \]
Followed by oxidation:
\[ C_3H_7B + H_2O_2 / OH^- \to C_3H_7OH \quad (Propan-2-ol) \]

3. Sodium t-butoxide treated with \( CH_3Cl \):
Sodium t-butoxide undergoes nucleophilic substitution with methyl chloride to form t-butyl methyl ether:
\[ (CH_3)_3CO^- + CH_3Cl \to (CH_3)_3COCH_3 \] Quick Tip: The hydroboration-oxidation of alkenes adds water in a syn fashion, which is different from the anti-addition in acid-catalyzed hydration.

A straightforward chemical test to differentiate between butan-1-ol and butan-2-ol is the oxidation test using potassium dichromate (\( K_2Cr_2O_7 \)) in an acidic medium. Butan-1-ol will be oxidized to butanoic acid, while butan-2-ol will be oxidized to acetone.


Butan-1-ol: Upon oxidation, it forms a carboxylic acid, butanoic acid.

Butan-2-ol: Upon oxidation, it forms a ketone, acetone.


Another test involves the reaction with iodine and NaOH:

Butan-1-ol forms a yellow precipitate of iodoform when treated with iodine and NaOH.

Butan-2-ol also forms a yellow precipitate of iodoform, but the intensity of the reaction can be used for differentiation.
Quick Tip: The oxidation test and iodoform test can help distinguish between primary and secondary alcohols. Primary alcohols are oxidized to carboxylic acids, while secondary alcohols form ketones.

To arrange these compounds in increasing order of acid strength, we consider the ability of these compounds to donate a proton. The stronger the ability to donate a proton (H\(^+\)), the stronger the acid.

1. Ethanol (CH\(_3\)CH\(_2\)OH): Ethanol has a hydroxyl group (-OH) but it is a very weak acid compared to water or phenol.
2. Water (H\(_2\)O): Water is neutral but can act as a weak acid, donating a proton to form \( OH^- \).
3. Phenol (C\(_6\)H\(_5\)OH): Phenol is a stronger acid than both ethanol and water due to the resonance stabilization of the phenoxide ion (C\(_6\)H\(_5\)O\(^-\)) after proton donation.

Thus, the increasing order of acid strength is: \[ Ethanol < Water < Phenol \] Quick Tip: The acidity of alcohols and phenols increases with the ability to stabilize the conjugate base. Phenols are more acidic due to the resonance stabilization of the phenoxide ion.

The IUPAC name of the compound \( CH_3 = CH - CH = CH - CHO \) is pent-2,4-enal.

The compound consists of a 5-carbon chain with two double bonds at positions 2 and 4.

The aldehyde group (\( -CHO \)) is located at the end of the chain, and the suffix "-al" is used to indicate the aldehyde functional group.


Thus, the IUPAC name is pent-2,4-enal. Quick Tip: When naming compounds with multiple functional groups, the highest-priority functional group is given the lowest possible number, and double bonds are numbered to ensure the lowest positions.

A simple chemical test to differentiate between propanal (an aldehyde) and propanone (a ketone) is the Tollens' test.


Propanal (Aldehyde): When treated with Tollens' reagent (ammoniacal silver nitrate), aldehydes reduce silver ions to metallic silver, creating a silver mirror on the inner surface of the test tube.

Propanone (Ketone): Ketones do not react with Tollens' reagent and, therefore, do not form a silver mirror.


Thus, the Tollens' test effectively distinguishes between an aldehyde and a ketone. Quick Tip: The Tollens' test is a traditional method to identify aldehydes. Aldehydes reduce silver ions to form a silver mirror, whereas ketones do not react.

1. Toluene to benzoic acid:
Toluene can be oxidized to benzoic acid by treating it with potassium permanganate (KMnO\(_4\)).
\[ C_6H_5CH_3 + KMnO_4 \to C_6H_5COOH \]
This is a typical oxidation reaction where the methyl group (-CH\(_3\)) is oxidized to a carboxyl group (-COOH).

2. Ethanol to propan-2-ol:
Ethanol can be converted to propan-2-ol by using a suitable catalyst for a reduction reaction. For example, it can be oxidized to acetaldehyde (CH\(_3\)CHO), which is then reduced to propan-2-ol using a reducing agent like lithium aluminum hydride (LiAlH\(_4\)).
\[ C_2H_5OH \xrightarrow{[O]} C_2H_4O \xrightarrow{LiAlH_4} C_3H_7OH \]
Alternatively, it can be directly hydrated from propene.

3. Propanal to 2-hydroxy propanoic acid:
Propanal can be converted to 2-hydroxy propanoic acid by the hydroxylation reaction, specifically by adding a hydroxyl group to the alpha-carbon of the aldehyde. A suitable reagent for this transformation is hydroxylamine or a mild oxidizing agent.
\[ C_3H_7CHO + OH^- \to C_3H_6(OH)COOH \] Quick Tip: The oxidation of toluene with potassium permanganate is a strong method to obtain benzoic acid. Hydration of propene and reduction of aldehydes are common transformations in organic synthesis.

(a) \[ \ce{O} + \ce{HO-NH2} \xrightarrow{\ce{H+}} \ce{NH2OH} \]

The reaction involves the formation of hydroxylamine (\(\ce{NH2OH}\)) from an aldehyde (\(\ce{O}\)) and hydroxylamine (\(\ce{HO-NH2}\)) in the presence of an acid catalyst (\(\ce{H+}\)).


(b) \[ \ce{CH2=CH2} \xrightarrow{(i) O_3 (ii) Zn-H_2O} \ce{2HCHO} \xrightarrow{\Delta} No further change \]

The reaction involves ozonolysis of ethene (\(\ce{CH2=CH2}\)) followed by reductive workup to form formaldehyde (\(\ce{HCHO}\)). Heating (\(\Delta\)) does not change the product further.

(c) \[ \ce{OH} \xrightarrow{\ce{SOCl2}} \ce{Cl} \]

The reaction involves the conversion of an alcohol (\(\ce{OH}\)) to an alkyl chloride (\(\ce{Cl}\)) using thionyl chloride (\(\ce{SOCl2}\)) under heating (\(\Delta\)).



(d) \[ \ce{CHO} \xrightarrow{\ce{NaCN/HCl}} \ce{COOH} \]

The reaction involves the conversion of an aldehyde (\(\ce{CHO}\)) to a carboxylic acid (\(\ce{COOH}\)) using sodium cyanide (\(\ce{NaCN}\)) and hydrochloric acid (\(\ce{HCl}\)).

(e) \[ \ce{Cl2C6H5} \xrightarrow{\ce{CH3}} \ce{Cl2C6H4CH3} \]

The reaction involves the methylation of chlorobenzene (\(\ce{Cl2C6H5}\)) to form a methylated chlorobenzene derivative (\(\ce{Cl2C6H4CH3}\)). Quick Tip: \textbf{Quick Tips for Organic Synthesis:} \textbf{(a)} Aldehydes react with hydroxylamine (\(\ce{HO-NH2}\)) in the presence of an acid to form oximes, which are useful intermediates in organic synthesis. \textbf{(b)} Ozonolysis cleaves double bonds in alkenes, producing carbonyl compounds (aldehydes or ketones) depending on the structure of the starting alkene. \textbf{(c)} Thionyl chloride (\(\ce{SOCl2}\)) is a common reagent for converting alcohols to alkyl chlorides, with gaseous byproducts (\(\ce{SO2}\) and \(\ce{HCl}\)) that are easy to remove. \textbf{(d)} Aldehydes can be converted to carboxylic acids using sodium cyanide (\(\ce{NaCN}\)) and hydrochloric acid (\(\ce{HCl}\)) via cyanohydrin formation and hydrolysis. \textbf{(e)} Electrophilic aromatic substitution reactions, such as methylation, introduce substituents like \(\ce{CH3}\) onto the benzene ring, commonly seen in aromatic compounds.

The standard Gibbs free energy change (\( \Delta G^\circ \)) is related to the standard electrode potential change (\( \Delta E^\circ \)) using the following equation: \[ \Delta G^\circ = -nF \Delta E^\circ \]
Where:

\( n \) is the number of moles of electrons transferred in the reaction (which is 3 for this reaction).

\( F \) is the Faraday constant (\( 96500 \, C mol^{-1} \)).

\( \Delta E^\circ \) is the cell potential, calculated as:
\[ \Delta E^\circ = E^\circ_{cathode} - E^\circ_{anode} = 1.5 - (-2.87) = 4.37 \, V. \]

Now, calculating \( \Delta G^\circ \): \[ \Delta G^\circ = -3 \times 96500 \times 4.37 = -1.26 \times 10^6 \, J/mol. \]

Since \( \Delta G^\circ \) is negative, the reaction is spontaneous. Quick Tip: A negative value for \( \Delta G^\circ \) indicates that the reaction is spontaneous. The magnitude of \( \Delta G^\circ \) is directly related to the cell potential, which can be determined using the Nernst equation.

The reaction involves the reduction of \( Ag_2S \) to silver, and the oxidation of aluminium to aluminium ion. The cell potential for this reaction is the difference between the reduction potential of silver and aluminium: \[ \Delta E^\circ = E^\circ_{cathode} - E^\circ_{anode} = (-0.71) - (-1.66) = 0.95 \, V. \]
Since the cell potential is positive, the reaction is spontaneous, meaning that tarnish can be removed by using an aluminium pan containing NaCl. Quick Tip: In electrochemical reactions, the substance with the more positive reduction potential (cathode) will be reduced, and the one with the more negative reduction potential (anode) will be oxidized.

Cell potential is the difference in the electric potential between two electrodes in an electrochemical cell. It is a measure of the ability of a redox reaction to occur spontaneously. The cell potential is typically measured in volts (V) and is denoted by \( E_{cell} \). The higher the cell potential, the greater the driving force for the reaction. Quick Tip: The cell potential is calculated as the difference between the reduction potentials of the cathode and anode, and it indicates the direction of electron flow in an electrochemical cell.

A fuel cell is an electrochemical cell that converts the chemical energy of a fuel (such as hydrogen or methanol) into electrical energy through a redox reaction. The reactants are continuously supplied to the cell, and the products are continuously removed. The most common example is the hydrogen fuel cell, where hydrogen reacts with oxygen to produce water and electricity. Quick Tip: Fuel cells are environmentally friendly because their only byproduct is usually water, making them an important alternative to traditional power sources.

The emf of the cell can be determined using the Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
Where:

\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \).

\( n = 2 \) (the number of electrons transferred).


The standard cell potential is: \[ E^\circ_{cell} = (-0.76) - (-0.40) = -0.36 \, V. \]

Substituting values into the Nernst equation: \[ E_{cell} = -0.36 - \frac{0.0592}{2} \log \left( \frac{0.01}{0.1} \right) \] \[ E_{cell} = -0.36 - \frac{0.0592}{2} \log (0.1) \] \[ E_{cell} = -0.36 - \frac{0.0592}{2} \times (-1) = -0.36 + 0.0296 = -0.33 \, V. \]

Thus, the emf of the cell at 25°C is \( -0.33 \, V \). Quick Tip: The Nernst equation is used to calculate the emf of an electrochemical cell under non-standard conditions by considering the ion concentrations.