The Assertion (A) is true because all naturally occurring α-amino acids, except for glycine, are optically active due to the presence of a chiral carbon, which leads to non-superimposable mirror images (enantiomers). Glycine does not have a chiral carbon, making it optically inactive.

The Reason (R) is also true in stating that most naturally occurring amino acids have the L-configuration. However, the L-configuration does not directly explain the optical activity of amino acids. Optical activity arises from the chiral nature of the molecule, not from its specific configuration. Therefore, Reason (R) does not serve as the correct explanation for the optical activity stated in Assertion (A).


Final Answer: (2) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Quick Tip: When studying amino acids, remember that their L-configuration is common, but optical activity depends on the presence of a chiral center.

The Assertion (A) is true because ethanol has a higher boiling point compared to methoxyethane. The reason behind this difference in boiling points lies in the presence of **intramolecular hydrogen bonding** in ethanol. In ethanol, the hydroxyl group (-OH) can form hydrogen bonds with other ethanol molecules, requiring more energy to separate the molecules, thus raising its boiling point. In contrast, methoxyethane does not form hydrogen bonds, so its boiling point is lower.


Thus, Reason (R) correctly explains why ethanol has a higher boiling point than methoxyethane.


Final Answer: (1) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
Quick Tip: Hydrogen bonding significantly impacts the boiling point of alcohols. Ethanol’s higher boiling point is due to intramolecular hydrogen bonding.

The boiling points of alkyl halides increase with the size of the halogen. Heavier halogens such as iodine (I) have stronger van der Waals forces due to their larger size and greater number of electrons. These stronger intermolecular forces require more energy to overcome, leading to a higher boiling point. Hence, the order of boiling points is: RI \(>\) RBr \(>\) RCl \(>\) RF.

The Reason (R) is also true as alkyl halides have higher boiling points than hydrocarbons of similar molecular mass because the halogens introduce stronger intermolecular forces. Thus, the assertion and reason are both true, and the reason provides the correct explanation for the boiling point order.

Final Answer: (1) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
Quick Tip: Remember that the size and number of electrons in the halogen atoms affect the boiling points of alkyl halides.

Homoleptic complexes are coordination compounds in which all ligands attached to the central metal atom or ion are the same. In the given complexes [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2, both have six water molecules coordinated to the central metal ion (Cr or Fe), making them homoleptic complexes. The ligands in both complexes are the same, which explains why they are homoleptic.

Thus, the Reason (R) correctly explains the Assertion (A) because homoleptic complexes specifically involve identical ligands.

Final Answer: (1) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
Quick Tip: Homoleptic complexes always have identical ligands attached to the central metal atom or ion, making them distinct from heteroleptic complexes.

When an unripe mango is placed in a concentrated salt solution, water moves out of the mango cells due to osmosis. Osmosis occurs when water moves from an area of higher concentration to an area of lower concentration. The concentrated salt solution causes water to move from the mango cells to the surrounding solution, causing the mango to shrivel.
Quick Tip: When working with osmosis, remember that water moves from a region of lower solute concentration to higher solute concentration.

Glucose is an aldose, which means it contains an aldehyde group. When glucose is heated with hydriodic acid (HI), it undergoes a reduction reaction to form n-hexane. Glucose exists in two cyclic forms: pyranose (six-membered ring) and furanose (five-membered ring). Schiff's test is used to detect aldehydes, and since glucose contains an aldehyde group, it gives a positive Schiff's test.
Quick Tip: Schiff's test is a common test for detecting aldehydes; glucose, being an aldose, gives a positive result.

LiAlH\(_4\) (Lithium aluminum hydride) is a powerful reducing agent that can reduce amides like propanamide into their corresponding amines, such as propanamine. The other reagents listed, such as excess H\(_2\) or Br\(_2\) in aqueous NaOH, do not reduce amides to amines effectively. Lithium aluminum hydride is the best choice for this reduction reaction.
Quick Tip: For reducing amides to amines, remember that LiAlH\(_4\) is the most effective reagent.

When propyl magnesium bromide (C\(_3\)H\(_7\)MgBr) reacts with CO\(_2\) in a Grignard reaction, it forms a carboxylic acid. The reaction leads to the formation of propionic acid (C\(_3\)H\(_6\)COOH). This is a typical reaction of Grignard reagents with CO\(_2\), where the nucleophilic carbon in the Grignard reagent attacks the electrophilic carbon in CO\(_2\).
Quick Tip: When using Grignard reagents, treat them with CO\(_2\) to form carboxylic acids.

The acid strength is based on the ability of the compound to donate a proton (H\(^+\)). Water (H\(_2\)O) is a stronger acid than phenol (C\(_6\)H\(_5\)OH) and alcohol (ROH) because it has a more stable hydronium ion (H\(_3\)O\(^+\)). Phenol is stronger than alcohol due to the resonance stabilization of its conjugate base, the phenoxide ion, which stabilizes the negative charge better than the alkoxide ion from alcohols. Thus, the correct order of acid strength is H\(_2\)O \(>\) C\(_6\)H\(_5\)OH \(>\) ROH.
Quick Tip: For comparing acid strengths, remember that the conjugate base stability is key. Resonance stabilization in phenol makes it a stronger acid than alcohol.

In nucleophilic bimolecular substitution reactions (SN2), the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group.

This results in a simultaneous inversion of configuration at the carbon center.

The mechanism involves a single concerted step, and no intermediate carbocation is formed.

Therefore, the correct answer is inversion of configuration.
Quick Tip: In SN2 reactions, remember that the configuration inverts as the nucleophile attacks from the opposite side of the leaving group.

The boiling point of a compound depends on factors like molecular weight and intermolecular forces.

In this case, CH\(_3\)C\(\equiv\)CH (iii) is the smallest molecule with the lowest boiling point.

CH\(_3\)CH\(_2\)Br (i) has a higher boiling point than CH\(_3\)C\(\equiv\)CH due to the presence of the halogen and its molecular size.

The longest chain compound, CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)Br (ii), has the highest boiling point due to increased van der Waals forces.

Hence, the correct order is (iii) \(<\) (i) \(<\) (ii).
Quick Tip: For comparing boiling points, larger molecules with longer chains generally have higher boiling points due to increased intermolecular forces.

The complex [Pt(NH\(_3\))\(_4\)Cl\(_2\)]\(^{2+}\) contains platinum in the +4 oxidation state, as indicated by the Roman numeral "IV".

The "di" prefix indicates two chloride ligands, and "ammines" are ammonia ligands.

Therefore, the correct IUPAC name is Diamminedichloridoplatinum (IV).
Quick Tip: Remember that the oxidation state of platinum in a complex is determined by the charges of the ligands and the overall charge of the complex.

In an acidic medium, potassium permanganate (KMnO\(_4\)) acts as a strong oxidizing agent and oxidizes sulfite ions (SO\(_3^{2-}\)) to sulfate ions (SO\(_4^{2-}\)).

Thus, the correct product of this oxidation reaction is SO\(_4^{2-}\), which corresponds to option (D).
Quick Tip: KMnO\(_4\) is a strong oxidizing agent that converts sulfite (SO\(_3^{2-}\)) to sulfate (SO\(_4^{2-}\)) in acidic conditions.

For Cr\(^{3+}\) ion, the electronic configuration is [Ar] 3d\(^3\). The magnetic moment is given by the formula: \[ \mu = \sqrt{n(n+2)} \, B.M. \]
where \(n\) is the number of unpaired electrons. For Cr\(^{3+}\), \(n = 3\), so: \[ \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \, B.M. \]
Thus, the correct answer is 3.47 B.M. for spin-only magnetic moment, considering the given choices.
Quick Tip: For transition metal ions, the magnetic moment can be calculated using the number of unpaired electrons in the ion's electronic configuration.

The cell potential can be calculated using the standard electrode potentials of the two half-reactions.
The cell potential \(E_{cell}\) is given by: \[ E_{cell} = E_{cathode} - E_{anode} \]
For this cell, the Sn\(^{4+}\)/Sn\(^{2+}\) couple will be reduced (cathode) and the Cr\(^{3+}\)/Cr couple will be oxidized (anode).
Thus, \[ E_{cell} = (+0.15 \, V) - (-0.74 \, V) = 1.19 \, V. \]
Hence, the correct answer is +1.19 V.
Quick Tip: The cell potential is the difference between the reduction potentials of the two half-cells.

In the case of association, the solute particles associate with each other to form larger complexes. This leads to an increase in the apparent molar mass because the number of particles in the solution decreases, which affects colligative properties.
Quick Tip: For associated solutes, the apparent molar mass increases because the number of solute particles decreases in solution.

For reaction (a): CH\(_3\)CH\(_2\)Cl reacts with NaCN to form CH\(_3\)CH\(_2\)CN (ethyl cyanide), and upon reduction with H\(_2\) and Ni, it gives ethylamine (C\(_6\)H\(_5\)NH\(_2\)). Thus, A is CH\(_3\)CH\(_2\)CN, and B is ethylamine.

For reaction (b): Aniline (C\(_6\)H\(_5\)NH\(_2\)) reacts with NaNO\(_2\) in the presence of HCl at low temperatures (0-5°C) to form diazonium salt (C\(_6\)H\(_5\)N\(_2^+\). This is A. When treated with H\(^+\), the diazonium salt is converted into aniline.
Quick Tip: In organic reactions, diazonium salts are often formed by reacting aromatic amines with nitrous acid, and can be reduced to amines in subsequent steps.

In this reaction sequence:
- FeCr\(_2\)O\(_4\) fused with Na\(_2\)CO\(_3\) in air forms sodium chromate (Na\(_2\)CrO\(_4\)), which is compound (A).
- Acidification of sodium chromate (Na\(_2\)CrO\(_4\)) forms chromium trioxide (CrO\(_3\)), which is compound (B).
- Compound (B) reacts with KCl to form potassium chromyl chloride (CrCl\(_3\)), which is compound (C).
- An acidified solution of potassium chromyl chloride (C) oxidizes iodide to iodine (I\(_2\)), which is compound (D).
Quick Tip: The oxidation states of chromium change in different reaction steps. Sodium chromate and chromyl chloride are important intermediates in this reaction.

Benzaldehyde is more reactive in nucleophilic addition reactions than propanal. This is because the phenyl group in benzaldehyde withdraws electron density from the carbonyl group, making the carbonyl carbon more electrophilic and, therefore, more susceptible to attack by nucleophiles. On the other hand, propanal has an alkyl group (which is electron-donating) attached to the carbonyl group, making it slightly less reactive than benzaldehyde in nucleophilic addition reactions.
Quick Tip: In carbonyl compounds, electron-withdrawing groups increase reactivity by stabilizing the negative charge on the oxygen atom, making the carbonyl carbon more electrophilic.

(a) Cooking is faster in a pressure cooker because the pressure inside the cooker is higher, which increases the boiling point of water. The higher boiling point means food cooks at a higher temperature, speeding up the cooking process. In an open pan, the boiling point is lower, and the cooking time is longer.


(b) The decrease in volume after mixing liquids X and Y indicates that the interaction between the molecules of X and Y is stronger than the interactions between the molecules of each individual liquid. This results in negative deviation from Raoult’s law, where the vapor pressure of the solution is lower than expected. In this case, the solution shows a temperature decrease, as the decrease in volume leads to stronger intermolecular forces between the molecules, lowering the vapor pressure and cooling the solution.
Quick Tip: When liquids mix and show negative deviation from Raoult's law, the intermolecular forces between the components are stronger than those within each liquid.

An azeotrope is a mixture of two or more liquids that distills at a constant boiling point and has the same composition in the vapor phase as in the liquid phase. The mixture behaves like a single substance when boiled.

In the case of negative deviation from Raoult’s law, the azeotrope formed exhibits a boiling point that is lower than the boiling points of the pure components. The intermolecular attractions between the different molecules are stronger than those between similar molecules, leading to a reduced vapor pressure.

An example of an azeotrope formed due to negative deviation is the mixture of water and hydrochloric acid (HCl), which forms an azeotrope with a boiling point of 108°C (lower than the boiling point of pure water).
Quick Tip: In a negative deviation azeotrope, the mixture distills at a constant boiling point lower than the boiling points of the individual components.

(a) \(\alpha\)-amino acids have both an amino group (-NH\(_2\)) and a carboxyl group (-COOH) in their structure. This allows for stronger intermolecular hydrogen bonding between molecules, increasing the melting point. In contrast, carboxylic acids only have one carboxyl group, leading to weaker intermolecular interactions and a lower melting point.


(b) Amino acids show amphoteric behavior because they contain both a basic amino group (-NH\(_2\)) and an acidic carboxyl group (-COOH). The amino group can accept a proton, acting as a base, while the carboxyl group can donate a proton, acting as an acid. This dual capability allows amino acids to react with both acids and bases.
Quick Tip: Amphoteric compounds, like amino acids, can act as both acids and bases depending on the pH of the solution.

Osmotic pressure is given by the formula: \[ \Pi = \dfrac{nRT}{V} \]
where \(\Pi\) is the osmotic pressure, \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume of the solution. Since the osmotic pressures of the urea and glucose solutions are equal, the moles of solute will be the same.

For urea, \[ n = \dfrac{15 \, g}{60 \, g/mol} = 0.25 \, mol \]

For glucose, the number of moles is: \[ n = 0.25 \, mol \quad and \quad mass of glucose = 0.25 \, mol \times 180 \, g/mol = 45 \, g \]

Thus, the mass of glucose required to produce the same osmotic pressure as the urea solution is 45 g.
Quick Tip: Osmotic pressure is directly proportional to the number of moles of solute, so for isotonic solutions, the number of moles of urea and glucose must be the same.

First, calculate the standard cell potential (\(E^\circ_{cell}\)) using the equation: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 \, V - 0.77 \, V = 0.03 \, V \]

Next, use the Nernst equation to calculate \(\Delta_r G^\circ\) using the formula: \[ \Delta_r G^\circ = -nFE^\circ_{cell} \]
For this reaction, \(n = 1\) (since one electron is involved in the reaction). \[ \Delta_r G^\circ = -1 \times 96500 \times 0.03 = -2895 \, J/mol \]

Now, calculate the equilibrium constant \(K_c\) using the relationship: \[ \Delta_r G^\circ = -RT \ln K_c \]
Rearrange to solve for \(K_c\): \[ \ln K_c = -\dfrac{\Delta_r G^\circ}{RT} \] \[ K_c = e^{-\frac{-2895}{8.314 \times 298}} = e^{1.16} \approx 3.19 \]
Therefore, \(\Delta_r G^\circ = -2895 \, J/mol\) and log \(K_c = \log 3.19 = 0.51\).
Quick Tip: To calculate \(\Delta_r G^\circ\) and \(K_c\), use the standard cell potential and apply the Nernst equation and the relationship between \(\Delta_r G^\circ\) and \(K_c\).

The pK\(_b\) values are inversely related to the basicity of the compounds. In general, electron-withdrawing groups (such as NO\(_2\) in p-nitroaniline) reduce the basicity, while electron-donating groups (such as -CH\(_3\) in p-methylaniline) increase it. Therefore, the order of decreasing pK\(_b\) (i.e., increasing basicity) is: \[ p-methylaniline > Aniline > p-nitroaniline \] Quick Tip: Basicity decreases with the presence of electron-withdrawing groups and increases with electron-donating groups.

(i) Diazonium salts of aromatic amines are more stable than those of aliphatic amines because the aryl group can stabilize the positive charge on the nitrogen atom through resonance. In contrast, the alkyl group in aliphatic amines does not provide this stabilization, making the diazonium salt less stable.

(ii) Methylamine reacts with FeCl\(_3\) in water to form a complex that precipitates hydrated ferric oxide. This occurs because FeCl\(_3\) is an electron-deficient Lewis acid and reacts with the lone pair of electrons on the nitrogen atom of methylamine, forming a complex that leads to precipitation.
Quick Tip: Diazonium salts of aromatic amines are stabilized by resonance, making them more stable than those of aliphatic amines.

(a) CH\(_2\)CH\(_2\)Cl reacts with Br\(_2\) under heat to give 1,2-dibromoethane as the major product.

(b) Toluene (C\(_6\)H\(_5\)CH\(_3\)) reacts with HBr under heat to form benzyl bromide (C\(_6\)H\(_5\)CH\(_2\)Br).

(c) The reaction of HO-CH\(_2\)C with HCl under heat will result in the formation of chloroethanol (C\(_2\)H\(_5\)Cl).
Quick Tip: In electrophilic aromatic substitution reactions, the methyl group on toluene activates the benzene ring, making it more reactive towards electrophiles such as HBr.

(a) To convert chlorobenzene to biphenyl, use the Wurtz-Fittig reaction, where chlorobenzene is reacted with sodium in the presence of dry ether.

(b) Propene can be converted to 1-iodopropane by reacting it with HI (Hydroiodic acid) in the presence of heat.

(c) To convert 2-bromobutane to but-2-ene, perform an elimination reaction using a strong base such as NaOH to remove a hydrogen atom and a bromine atom, leading to the formation of the alkene.
Quick Tip: The Wurtz-Fittig reaction is a useful method for coupling aromatic compounds, while elimination reactions with strong bases are key for producing alkenes.

(a) Copper has an exceptionally positive \(E^{\circ}\) value for the Cu\(^{2+}\) / Cu half-reaction because copper has a completely filled 3d-subshell in its +2 oxidation state, which makes it very stable and thus has a high tendency to be reduced.

(b) Zn\(^{2+}\) is a strong reducing agent in its +2 oxidation state because its \(d^{10}\) configuration is stable, and it has a strong tendency to donate electrons to reduce other species.

(c) Zn\(^{2+}\) salts are colourless because the Zn\(^{2+}\) ion has a completely filled 3d subshell (d\(^{10}\) configuration) and no unpaired electrons to absorb visible light.
Quick Tip: Transition metals with a completely filled d-subshell or with no unpaired electrons in their lower oxidation states tend to form colourless salts.

For a first-order reaction, we can use the Arrhenius equation in the form: \[ \ln \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right) \]
where \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1\) and \(T_2\), respectively, \(E_a\) is the activation energy, and \(R\) is the gas constant.


From the given data, we know that the reaction is 50% complete at both temperatures, so the rate constants are proportional to the time taken for the reaction to reach 50%. Thus, \[ \dfrac{k_2}{k_1} = \dfrac{t_1}{t_2} = \dfrac{20}{5} = 4 \]
Now, substituting the values in the equation: \[ \ln(4) = \dfrac{E_a}{8.314} \left( \dfrac{1}{300} - \dfrac{1}{350} \right) \] \[ 0.602 = \dfrac{E_a}{8.314} \left( \dfrac{1}{300} - \dfrac{1}{350} \right) \]
Simplifying the right-hand side: \[ \left( \dfrac{1}{300} - \dfrac{1}{350} \right) = \dfrac{50}{300 \times 350} = 4.76 \times 10^{-5} \]
Now, solving for \(E_a\): \[ E_a = \dfrac{0.602 \times 8.314}{4.76 \times 10^{-5}} = 105.6 \, kJ/mol \]

Thus, the activation energy is \(105.6 \, kJ/mol\).
Quick Tip: To calculate the activation energy, use the Arrhenius equation and apply the known temperatures and reaction times for the first-order reaction.

(i) When D-glucose reacts with hydroxylamine (NH\(_2\)OH), it forms a hydroxamic acid derivative: \[ C_6H_{12}O_6 + NH_2OH \longrightarrow C_6H_{11}NO_5 + H_2O \]

(ii) When D-glucose reacts with acetic anhydride, it forms glucose pentaacetate: \[ C_6H_{12}O_6 + (CH_3CO)_2O \longrightarrow C_6H_7O_6C_2H_3O_2 + CH_3COOH \] Quick Tip: When glucose reacts with acetic anhydride, it undergoes esterification to form glucose pentaacetate.

Vitamin C (ascorbic acid) is a water-soluble vitamin, and the body cannot store it for long periods. It is excreted in urine when consumed in excess. Unlike fat-soluble vitamins (A, D, E, K), which are stored in the liver and adipose tissue, water-soluble vitamins like vitamin C need to be consumed regularly in the diet to maintain proper health.
Quick Tip: Vitamin C is water-soluble and is not stored in the body. It must be obtained regularly through diet or supplements.

(a)

(i) When phenol reacts with Br\(_2\)/CS\(_2\), it undergoes bromination at the ortho and para positions, resulting in the formation of 2,4,6-tribromophenol.

(ii) When phenol reacts with concentrated HNO\(_3\), it undergoes nitration at the ortho and para positions to form 2,4,6-trinitrophenol (picric acid).
Quick Tip: In electrophilic substitution reactions with phenol, the OH group makes the benzene ring more reactive at the ortho and para positions.

The rate-determining step is the slowest step in a multi-step reaction. It controls the overall rate of the reaction as the entire reaction cannot proceed faster than its slowest step. The rate constant of this step determines the rate of the entire reaction.
Quick Tip: In a multi-step reaction, identify the slowest step (rate-determining step) to determine the overall reaction rate.

(i) Ethanol can be converted to But-2-enal by first oxidizing ethanol to acetaldehyde (CH\(_3\)CHO), followed by a reaction with a strong base (like NaOEt) to form the enolate ion, which undergoes an aldol condensation. The product is But-2-enal.

(ii) Propanoic acid can be reduced to ethane using a strong reducing agent like LiAlH\(_4\) (Lithium Aluminium Hydride), which reduces the carboxyl group to an alkane.
Quick Tip: In organic chemistry, oxidation and reduction reactions are commonly used to convert between alcohols, aldehydes, and carboxylic acids.

(A) The organic compound (A) is phenyl acetate (C\(_8\)H\(_8\)O\(_2\)).

(B) The carboxylic acid (B) formed is acetic acid (CH\(_3\)COOH).

(C) The alcohol (C) formed is ethanol (CH\(_3\)CH\(_2\)OH).


The reactions involved:
1. Hydrolysis of phenyl acetate with dilute sulfuric acid: \[ C_8H_8O_2 + H_2O \xrightarrow{H_2SO_4} CH_3COOH + CH_3CH_2OH \]

2. Oxidation of ethanol (C) with chromic acid to acetic acid (B): \[ CH_3CH_2OH \xrightarrow{Cr_2O_7^{2-}, \, H^+} CH_3COOH \]

3. Dehydration of ethanol (C) gives But-1-ene: \[ CH_3CH_2OH \xrightarrow{H_2SO_4, \, heat} CH_2=CHCH_3 \] Quick Tip: In organic chemistry, hydrolysis reactions often involve breaking an ester bond to form a carboxylic acid and alcohol. Oxidation reactions can be used to convert alcohols into carboxylic acids, and dehydration reactions can produce alkenes.

(i) The electrolysis of aqueous NaCl involves the liberation of Cl\(_2\) gas at the anode instead of O\(_2\). This is because chlorine ions (Cl\(^-\)) are easier to oxidize than water molecules, even though the theoretical standard electrode potential for the oxidation of water to oxygen (O\(_2\)) is more positive than that for the oxidation of chloride ions. In aqueous solution, chloride ions are preferentially oxidized at the anode due to overpotentials and the specific conditions of electrolysis.

(ii) The conductivity of CH\(_3\)COOH decreases on dilution because the dissociation of acetic acid (CH\(_3\)COOH) into ions is limited. As the solution is diluted, the number of ions in solution decreases, leading to a decrease in the overall conductivity. The extent of dissociation of CH\(_3\)COOH is low, and dilution reduces the concentration of the dissociated ions.
Quick Tip: In electrolysis, the substance that gets oxidized depends on its concentration and the overpotential, not just the theoretical electrode potentials.