RBSE Solutions Class 12 Chemistry Chapter 13 Organic Compounds with Functional Group-C

Rajasthan Board Rbse Class 12 Chemistry Chapter 13 Organic Compounds With Functional Group-Containing Nitrogen

Rbse Class 12 Chemistry Chapter 13 Text Book Questions

Rbse Class 12 Chemistry Chapter 13 Multiple Choice Questions

Question 1. Which of the following is most basic?
(a) CH3NH2
(b) (CH3)2NH
(c) (CH3)3N
(d) C6H5NH2
Answer: (b) (CH3)2NH
In simple words: Dimethylamine is the strongest base among the options. It has two methyl groups that push electrons towards the nitrogen, making the nitrogen more ready to share its electrons with a proton.

🎯 Exam Tip: Remember that alkyl groups are electron-donating. More alkyl groups around the nitrogen atom (up to secondary amines) generally increase basicity in solution due to stabilization of the conjugate acid and the inductive effect, but steric hindrance can reduce basicity in tertiary amines in aqueous solutions.

Question 2. Hinsberg's reagent is
(a) Benzene sulphonyl chloride
Answer: (a) Benzene sulphonyl chloride
In simple words: Hinsberg's reagent is benzene sulphonyl chloride. This special chemical is used to find out if an amine is primary, secondary, or tertiary.

🎯 Exam Tip: Know the chemical formula for Hinsberg's reagent: \( \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \). It is key for the Hinsberg test, which helps classify amines.

Question 3. C3H9N does not show
(a) Primary amine
(b) Quaternary salt
(c) Tertiary amine
(d) Secondary amine
Answer: (b) Quaternary salt
In simple words: The chemical formula C3H9N can be a primary, secondary, or tertiary amine. However, it cannot be a quaternary salt because that needs four alkyl groups attached to nitrogen, which this formula doesn't allow.

🎯 Exam Tip: Practice drawing the possible structures for a given molecular formula to understand which functional groups or classes of compounds it can represent.

Question 4. The hybridisation of N atom in alkyl amine is
(a) sp²
(b) sp³
(c) sp
(d) sp³d
Answer: (b) sp³
In simple words: In an alkylamine, the nitrogen atom always has \( \text{sp}^3 \) hybridization. This means it forms three bonds and has one lone pair of electrons, making its overall shape like a pyramid.

🎯 Exam Tip: Remember that nitrogen in most amines has a lone pair of electrons, which, along with the three bonding pairs, results in \( \text{sp}^3 \) hybridization and a trigonal pyramidal geometry.

Question 5. The molecular formula of a compound having mustard oil smell is
(a) RCN
(b) RNC
(c) RNCO
(d) RNCS
Answer: (d) RNCS
In simple words: Compounds that have a characteristic mustard oil smell are called alkyl isothiocyanates. Their general chemical formula is RNCS.

🎯 Exam Tip: Relate functional groups to their common names and characteristic properties, such as specific odors, to easily identify them in MCQs.

Question 6. The formula for chloropicrin is
(a) C(NO2)CI3
(b) CCI(NO2)3
(c) C(NO2)2CI2
(d) None of these
Answer: (a) C(NO2)CI3
In simple words: Chloropicrin is a chemical with the formula \( \text{C(NO}_2)\text{Cl}_3 \). It is also known as trichloronitromethane.

🎯 Exam Tip: Familiarize yourself with the chemical names and formulas of common or named organic compounds, especially those with unique properties or uses.

Question 7. Nitration of benzene gives nitrobenzene when HNO3 and H2SO4 take part in the reaction. Here HNO3 works as
(a) as a base
(b) as an acid
(c) a reducing agent
(d) as a catalyst
Answer: (a) as a base
In simple words: In the nitration of benzene, nitric acid acts like a base. It takes a proton from sulfuric acid, which then helps create the nitronium ion for the reaction.

🎯 Exam Tip: For nitration reactions, remember that concentrated sulfuric acid protonates nitric acid, making nitric acid act as a base to generate the electrophilic nitronium ion (\( \text{NO}_2^+ \)).

Question 9. The formula of acetonitrile is
(a) CH3CN
(b) CH3COCN
(c) CH3CH2CN
(d) CNCH2COOH
Answer: (a) CH3CN
In simple words: Acetonitrile has the chemical formula \( \text{CH}_3\text{CN} \). It is a simple organic compound that is commonly used as a solvent.

🎯 Exam Tip: Learn the common names and corresponding molecular formulas for important organic compounds, especially those with one or two carbon atoms.

Question 10. The major product of the reaction of methylamine with Tilden reagent is
(a) CH3OH
(b) CH3CHO
(c) CH3CI
(d) CH3COOH
Answer: (a) CH3OH
In simple words: When methylamine reacts with Tilden reagent (nitrosyl chloride), one important product that can be formed is methyl alcohol. This reaction often produces several chemicals depending on the exact conditions.

🎯 Exam Tip: Be aware that reactions of primary amines with nitrous acid (or Tilden reagent) can lead to a mixture of products including alcohols and alkyl halides, sometimes with rearrangements.

Rbse Class 12 Chemistry Chapter 13 Very Short Answer Type Questions

Question 1. What is the reason that aromatic diazonium salt is more stable than aliphatic diazonium salt?
Answer: Aromatic diazonium salts are more stable than aliphatic diazonium salts. This is because the aromatic ring can share electrons with the diazonium group through a process called resonance. This resonance helps to spread out the charge, which makes the molecule more stable overall.
In simple words: Aromatic diazonium salts are stronger because their ring structure helps stabilize the salt by spreading electrons around. Aliphatic salts do not have this same stabilizing effect.

🎯 Exam Tip: Always mention 'resonance stabilization' when explaining the higher stability of aromatic compounds compared to their aliphatic counterparts, as this is a key concept.

Question 3. Identify X and Y in the following reactions: RR - CO NH2 Br2/NaOH \( \times \) CHCl3/KOH Y
Answer: In the given reactions, X is an alkylamine (R-\( \text{NH}_2 \)) and Y is an alkyl isocyanide (RNC). The first reaction shown is the Hoffmann bromamide degradation, which converts an amide (R-\( \text{CONH}_2 \)) into a primary amine (R-\( \text{NH}_2 \)). The second reaction is the carbylamine reaction, which converts a primary amine into an isocyanide.
In simple words: For the reactions shown, X is an alkylamine (R-\( \text{NH}_2 \)) and Y is an alkyl isocyanide (RNC). The first step makes an amine from an amide, and the second step makes an isocyanide from that amine.

🎯 Exam Tip: Recognize the reagents and products of common named reactions like Hoffmann bromamide degradation and carbylamine reaction, as they frequently appear in identification questions.

Question 4. Identify A and B in the following reactions. C6H5 N2 Cl HOH A Br 2 B
Answer: In the first reaction step, benzene diazonium chloride (\( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \)) reacts with water (\( \text{HOH} \)) under heating (\( \Delta \)) to form phenol (\( \text{A} \)). Next, phenol (\( \text{A} \)) reacts with bromine (\( \text{Br}_2 \)) in an aqueous solution (\( 3\text{Br}_2(\text{aq}) \)) to yield 2,4,6-tribromophenol (\( \text{B} \)). This is an electrophilic substitution reaction where bromine atoms replace hydrogen atoms on the phenol ring.
In simple words: First, benzene diazonium chloride reacts with water and heat to become phenol, which is A. Then, this phenol (A) reacts with bromine water to form 2,4,6-tribromophenol, which is B.

🎯 Exam Tip: Remember that diazonium salts are highly reactive and can be converted to many different functional groups, with hydrolysis to phenol being a key transformation. Also, recall that phenol undergoes easy bromination.

Question 5. Dimethylamine is a stronger base than methylamine. Give reason.
Answer: Dimethylamine is a stronger base than methylamine. This is because dimethylamine has two methyl groups, while methylamine has only one. Methyl groups are electron-donating, meaning they push electrons towards the nitrogen atom. This increases the electron density on the nitrogen, making it more available to accept a proton, which in turn makes the compound a stronger base. The additional methyl group in dimethylamine provides a greater inductive (+I) effect compared to methylamine, enhancing its basicity.
In simple words: Dimethylamine is a stronger base than methylamine because it has two methyl groups that push more electrons onto the nitrogen. This makes the nitrogen better at taking a proton, so it acts like a stronger base.

🎯 Exam Tip: When comparing the basicity of amines, consider the inductive effect of alkyl groups (+I effect) and resonance effects. Alkyl groups increase basicity by stabilizing the conjugate acid.

Question 7. Write the equation of the Mendius reduction.
Answer: The Mendius reduction is a process where alkyl cyanides are reduced to primary amines using nascent hydrogen, typically generated from sodium and ethanol. This reaction is important for synthesizing amines. \[ \text{CH}_3\text{C}\equiv\text{N} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \] (Methyl cyanide is reduced to Ethylamine.)
In simple words: The Mendius reduction changes a methyl cyanide into an ethylamine. We use sodium and ethanol to add hydrogen atoms to the methyl cyanide, making it ethylamine.

🎯 Exam Tip: Remember that the Mendius reduction specifically converts nitriles (cyanides) to primary amines by using reducing agents like \( \text{Na/C}_2\text{H}_5\text{OH} \) or \( \text{LiAlH}_4 \).

Question 8. Write the equation for the conversion of aniline to phenyl isocyanide.
Answer: The conversion of aniline to phenyl isocyanide is known as the carbylamine reaction (or isocyanide test). This reaction involves heating aniline with chloroform and an alcoholic solution of potassium hydroxide. Phenyl isocyanide has a very unpleasant smell. \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{heat}} \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \] (Aniline reacts with chloroform and potassium hydroxide to form phenyl isocyanide, potassium chloride, and water.)
In simple words: To change aniline into phenyl isocyanide, you heat it with chloroform and potassium hydroxide. This reaction, called the carbylamine reaction, also produces potassium chloride and water.

🎯 Exam Tip: The carbylamine reaction is a distinctive test for primary amines, characterized by the formation of foul-smelling isocyanides.

Question 9. Write the reaction to obtain ethanol from Ethylamine.
Answer: Ethanol can be obtained from ethylamine through a reaction with nitrous acid (\( \text{HNO}_2 \)), which is typically prepared in situ (in the reaction mixture) from sodium nitrite (\( \text{NaNO}_2 \)) and hydrochloric acid (\( \text{HCl} \)). This is a common method to convert primary amines into alcohols. \[ \text{C}_2\text{H}_5\text{NH}_2 + \text{HNO}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}} \text{C}_2\text{H}_5\text{OH} + \text{N}_2 + \text{H}_2\text{O} \] (Ethylamine reacts with nitrous acid to produce ethyl alcohol, nitrogen gas, and water.)
In simple words: To get ethanol from ethylamine, you react ethylamine with nitrous acid. This makes ethyl alcohol, nitrogen gas, and water. It's a way to turn primary amines into alcohols.

🎯 Exam Tip: Primary aliphatic amines react with nitrous acid to give highly unstable diazonium salts, which immediately decompose to form alcohols with the evolution of nitrogen gas.

Question 10. Give the structural formula of urea and write its IUPAC name.
Answer: The structural formula of urea is \( \text{NH}_2-\text{CO}-\text{NH}_2 \). Its IUPAC name is carbonyl diamide. Urea is a common organic compound that plays a significant role in biological systems and agriculture as a fertilizer.
In simple words: Urea's structure has a carbonyl group (\( \text{CO} \)) connected to two amino groups (\( \text{NH}_2 \)). Its full chemical name is carbonyl diamide.

🎯 Exam Tip: Familiarize yourself with the structures and IUPAC names of simple, important organic compounds like urea, which bridges inorganic and organic chemistry.

Question 11. Write the reaction for the reduction of nitrobenzene in the presence of Zn + NH4CI.
Answer: The reduction of nitrobenzene in a neutral medium, typically achieved using a combination of zinc and ammonium chloride (\( \text{Zn} + \text{NH}_4\text{Cl} \)), yields N-phenylhydroxylamine. This is a selective reduction method. \[ \text{C}_6\text{H}_5\text{NO}_2 + 4[\text{H}] \xrightarrow{\text{Zn/NH}_4\text{Cl}} \text{C}_6\text{H}_5\text{NHOH} + \text{H}_2\text{O} \] (Nitrobenzene reacts with four hydrogen atoms in the presence of zinc and ammonium chloride to form N-phenylhydroxylamine and water.)
In simple words: If you reduce nitrobenzene using zinc and ammonium chloride, you get N-phenylhydroxylamine. This reaction adds hydrogen atoms to the nitrobenzene molecule in a controlled way.

🎯 Exam Tip: The type of product obtained from nitrobenzene reduction heavily depends on the pH of the medium. Neutral reduction conditions often lead to hydroxylamine derivatives.

Question 12. Complete the following reaction NH4CNO \( \rightarrow \) K?
Answer: The reaction \( \text{NH}_4\text{CNO} \rightarrow \text{K} \) represents the rearrangement of ammonium cyanate (\( \text{NH}_4\text{CNO} \)) to urea (\( \text{NH}_2\text{CONH}_2 \)). This is famously known as Wöhler's synthesis, a historical reaction that proved organic compounds could be synthesized from inorganic ones. \[ \text{NH}_4\text{CNO} \xrightarrow{\text{Rearrangement}} \text{NH}_2\text{CONH}_2 \] (Ammonium cyanate undergoes rearrangement to form urea.)
In simple words: The reaction shows ammonium cyanate changing into urea. This is a special type of chemical change where atoms move around inside the molecule to make a new compound.

🎯 Exam Tip: Wöhler's synthesis is a landmark reaction in organic chemistry; remember its reactants (ammonium cyanate) and product (urea).

Question 13. Give one equation to show the basic nature of ethanolamine.
Answer: Ethanolamine, being an amine, exhibits basic properties by accepting a proton from water. This reaction forms an ethylammonium hydroxide (or ethanolammonium hydroxide). \[ \text{C}_2\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{NH}_3^+\text{OH}^- \] (Ethylamine reacts with water to form ethylammonium hydroxide.) This equation shows that ethylamine acts as a base by producing hydroxide ions in solution.
In simple words: Ethanolamine is basic because it can take a hydrogen atom from water. When it does this, it forms ethylammonium hydroxide, which proves its basic nature.

🎯 Exam Tip: All amines are basic due to the lone pair of electrons on the nitrogen atom, which can readily accept a proton from an acid or water.

Question 14. The boiling point of primary amine is greater than tertiary amine. Why?
Answer: Primary amines have a higher boiling point than tertiary amines because primary amines can form stronger intermolecular hydrogen bonds with each other. This is due to the presence of two hydrogen atoms directly attached to the nitrogen, allowing for extensive hydrogen bonding between molecules. Tertiary amines, however, lack any hydrogen atoms directly bonded to the nitrogen, so they cannot form hydrogen bonds with other tertiary amine molecules. This leads to weaker intermolecular forces in tertiary amines and, consequently, lower boiling points.
In simple words: Primary amines boil at a higher temperature than tertiary amines. This is because primary amines can make many strong connections called hydrogen bonds between their molecules. Tertiary amines cannot make these connections because they don't have hydrogen atoms attached to their nitrogen.

🎯 Exam Tip: Hydrogen bonding capability significantly impacts physical properties like boiling point. More opportunities for hydrogen bonding (like in primary amines) lead to higher boiling points.

Rbse Class 12 Chemistry Chapter 13 Short Answer Type Questions

Question 1. What is the biuret test for urea? Explain with the chemical equation.
Answer: The biuret test is used to detect the presence of urea. When two molecules of urea are heated to about 155°C, they react together to release ammonia (\( \text{NH}_3 \)) and form biuret, which is a white crystalline solid. \[ \text{2NH}_2\text{CONH}_2 \xrightarrow{155^\circ\text{C}} \text{NH}_2\text{CONHCONH}_2 + \text{NH}_3 \] (Two urea molecules combine to form biuret and ammonia.) To perform the test, an alkaline solution of copper sulfate is added to the biuret. The formation of a characteristic violet color confirms the presence of biuret, and thus, indirectly, urea. This color change indicates the presence of peptide bonds in the biuret molecule.
In simple words: The biuret test checks for urea. First, you heat two urea molecules, which releases ammonia and forms a white solid called biuret. Then, you add an alkaline copper sulfate solution to this biuret. If it turns violet, it means urea was present.

🎯 Exam Tip: Remember both parts of the biuret test: the initial heating of urea to form biuret, and the subsequent color reaction with alkaline copper sulfate solution.

Question 2. Give the reaction of urea with:
1. Formaldehyde
2. Hydrazine
3. Malonic acid
Answer: Urea reacts differently with formaldehyde, hydrazine, and malonic acid:
1. **Reaction with Formaldehyde:** Urea reacts with formaldehyde to form dimethylolurea. This is an important step in the production of urea-formaldehyde resins. \[ \text{NH}_2-\text{CO}-\text{NH}_2 + 2\text{HCHO} \rightarrow \text{HOCH}_2-\text{NH}-\text{CO}-\text{NH}-\text{CH}_2\text{OH} \] (Urea reacts with two molecules of formaldehyde to form dimethylolurea.)
2. **Reaction with Hydrazine:** Urea reacts with hydrazine to form semicarbazide. This reaction involves the removal of an ammonia molecule. \[ \text{NH}_2\text{CONH}_2 + \text{NH}_2\text{NH}_2 \rightarrow \text{H}_2\text{N}-\text{CO}-\text{NHNH}_2 + \text{NH}_3 \] (Urea reacts with hydrazine to form semicarbazide and ammonia.) Semicarbazide is used in organic synthesis.
3. **Reaction with Malonic Acid:** Urea reacts with malonic acid in the presence of a dehydrating agent like phosphorus oxychloride (\( \text{POCl}_3 \)) to form barbituric acid (also called malonylurea). This condensation reaction is significant in the synthesis of barbiturates. \[ \text{NH}_2\text{CONH}_2 + \text{CH}_2(\text{COOH})_2 \xrightarrow{\text{POCl}_3} \text{Malonyl Urea} + 2\text{H}_2\text{O} \] (Urea reacts with malonic acid to produce malonyl urea and water.)
In simple words: Urea and formaldehyde combine to make dimethylolurea. Urea and hydrazine combine to make semicarbazide, releasing ammonia. Urea and malonic acid react, with the help of \( \text{POCl}_3 \), to form malonyl urea (barbituric acid) and water.

🎯 Exam Tip: Be familiar with the various condensation reactions of urea, particularly with aldehydes, hydrazine, and dicarboxylic acids, as they are crucial for synthesizing many important organic compounds.

Question 3. Complete the following reactions: R-X + KCN\( \rightarrow \)?
Answer: When an alkyl halide (R-X) reacts with potassium cyanide (KCN), the main product formed is an alkyl cyanide (R-CN). KCN is an ionic compound, so it provides a cyanide ion (\( \text{CN}^- \)), which is an ambident nucleophile (meaning it can attack through either its carbon or nitrogen atom). However, the carbon atom in cyanide is a stronger nucleophile, causing the alkyl halide to predominantly form an alkyl cyanide. If silver cyanide (AgCN) were used instead, which is covalent, the attack would primarily occur through the nitrogen atom, leading to an alkyl isocyanide (R-NC). \[ \text{R-X} + \text{KCN} \rightarrow \text{R-CN} + \text{KX} \] (An alkyl halide reacts with potassium cyanide to yield an alkyl cyanide and a potassium halide.)
In simple words: When an alkyl halide reacts with potassium cyanide, it mostly forms an alkyl cyanide. This is because the carbon part of the cyanide likes to attack the alkyl group more. If you used silver cyanide, you would mostly get an alkyl isocyanide instead.

🎯 Exam Tip: Remember the distinction between KCN (ionic, C-attack) and AgCN (covalent, N-attack) in reactions with alkyl halides to produce cyanides versus isocyanides. This is a common point of confusion.

Question 4. Write the balanced chemical equation for the reduction of nitrobenzene in
1. Basic medium
2. Neutral medium
Answer: The reduction of nitrobenzene depends on the reaction medium, leading to different products:
1. **In Basic medium:** When nitrobenzene is reduced in a basic medium (e.g., with \( \text{Zn/NaOH} \) or \( \text{CH}_3\text{OH} \)), it forms azobenzene. This transformation involves an 8-electron reduction. \[ 2\text{C}_6\text{H}_5\text{NO}_2 + 8[\text{H}] \xrightarrow{\text{Zn/NaOH or CH}_3\text{OH}} \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_5 + 4\text{H}_2\text{O} \] (Two molecules of nitrobenzene are reduced to azobenzene and water in a basic environment.)
2. **In Neutral medium:** In a neutral medium (e.g., with \( \text{Zn/NH}_4\text{Cl} \)), nitrobenzene is reduced to N-phenylhydroxylamine. This is a partial reduction. \[ \text{C}_6\text{H}_5\text{NO}_2 + 4[\text{H}] \xrightarrow{\text{Zn/NH}_4\text{Cl}} \text{C}_6\text{H}_5\text{NHOH} + \text{H}_2\text{O} \] (Nitrobenzene is reduced to N-phenylhydroxylamine and water in a neutral environment.)
In simple words: Nitrobenzene changes differently when it's reduced, depending on if the liquid is basic or neutral. In a basic liquid, it can form azobenzene, while in a neutral liquid, it forms N-phenylhydroxylamine.

🎯 Exam Tip: Pay close attention to the reaction conditions (especially the pH of the medium) when considering the reduction of nitro compounds, as they dictate the final product.

Question 5. Write the aliphatic amines in increasing order of their basicity and give a short note on basicity.
Answer: Amines are basic because the nitrogen atom has a lone pair of electrons that it can donate. Aliphatic amines are generally stronger bases than ammonia. This is due to the electron-donating inductive (+I) effect of the alkyl groups, which increases the electron density on the nitrogen, making it more willing to donate its lone pair. In contrast, aromatic amines are weaker bases than ammonia because the aromatic ring withdraws electrons from the nitrogen through resonance, making the lone pair less available for donation. The increasing order of basicity for aliphatic amines in the gaseous phase is: \[ \text{NH}_3 < 1^\circ \text{ Amines} < 2^\circ \text{ Amines} < 3^\circ \text{ Amines} \] (Ammonia is the least basic, followed by primary, then secondary, and tertiary amines are the most basic in the gaseous state.)
In simple words: Amines are basic because their nitrogen atom has extra electrons to give away. Aliphatic amines are more basic than ammonia because alkyl groups push electrons to the nitrogen. Aromatic amines are less basic than ammonia because the ring pulls electrons away. In the gas state, tertiary amines are the most basic, then secondary, then primary, and ammonia is the least basic.

🎯 Exam Tip: Remember to consider both inductive and resonance effects when discussing amine basicity. Also, note that the order of basicity in aqueous solution can differ from the gaseous phase due to solvation effects.

Question 6. Write the structures of A, B and C in the following reactions:
(i) \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{CuCN/KCN}} \text{A} \xrightarrow{\text{H}_2\text{O/H}^+} \text{B} \)
(ii) \( \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3/\Delta} \text{A} \xrightarrow{\text{NaOBr}} \text{B} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C} \)
(iii) \( \text{CH}_3\text{Br} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{LiAlH}_4} \text{B} \xrightarrow{\text{HNO}_2} \text{C} \)
Answer: Here are the structures for A, B, and C in each reaction:
(i) **Reaction:** \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{CuCN/KCN}} \text{A} \xrightarrow{\text{H}_2\text{O/H}^+} \text{B} \) * **A (Benzenenitrile):** \( \text{C}_6\text{H}_5\text{CN} \) (Benzene diazonium chloride reacts with cuprous cyanide to form benzenenitrile.) * **B (Benzoic Acid):** \( \text{C}_6\text{H}_5\text{COOH} \) (Benzenenitrile is then hydrolyzed with water and acid to yield benzoic acid.)
In simple words: First, \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \) changes into benzenenitrile (A) with \( \text{CuCN} \). Then, benzenenitrile (A) changes into benzoic acid (B) with water and acid.
(ii) **Reaction:** \( \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3/\Delta} \text{A} \xrightarrow{\text{NaOBr}} \text{B} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C} \) * **A (Acetamide):** \( \text{CH}_3\text{CONH}_2 \) (Acetic acid reacts with ammonia and heat to form acetamide.) * **B (Methylamine):** \( \text{CH}_3\text{NH}_2 \) (Acetamide undergoes Hoffmann bromamide degradation with sodium hypobromite to produce methylamine.) * **C (Methyl Alcohol):** \( \text{CH}_3\text{OH} \) (Methylamine reacts with nitrous acid (from \( \text{NaNO}_2/\text{HCl} \)) to form methyl alcohol.)
In simple words: Acetic acid becomes acetamide (A) with ammonia and heat. Acetamide (A) becomes methylamine (B) with \( \text{NaOBr} \). Finally, methylamine (B) becomes methyl alcohol (C) with \( \text{NaNO}_2/\text{HCl} \).
(iii) **Reaction:** \( \text{CH}_3\text{Br} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{LiAlH}_4} \text{B} \xrightarrow{\text{HNO}_2} \text{C} \) * **A (Methyl cyanide):** \( \text{CH}_3\text{CN} \) (Methyl bromide reacts with potassium cyanide to form methyl cyanide.) * **B (Ethylamine):** \( \text{CH}_3\text{CH}_2\text{NH}_2 \) (Methyl cyanide is reduced by lithium aluminum hydride to produce ethylamine.) * **C (Ethyl Alcohol):** \( \text{CH}_3\text{CH}_2\text{OH} \) (Ethylamine reacts with nitrous acid to form ethyl alcohol.)
In simple words: Methyl bromide becomes methyl cyanide (A) with \( \text{KCN} \). Methyl cyanide (A) becomes ethylamine (B) with \( \text{LiAlH}_4 \). Lastly, ethylamine (B) becomes ethyl alcohol (C) with \( \text{HNO}_2 \).

🎯 Exam Tip: For multi-step synthesis problems, break down the reaction into individual steps and identify the reagent for each transformation. Knowing common reactions for functional group interconversions is crucial.