CBSE Class 12 Chemistry Organic Compounds Containing Nitrogen Worksheet

Question: The hybridisation and geometry of amines are sp3 and pyramidal respectively due to the presence of
a) divalent N-atom
b) trivalent N-atom
c) monovalent N-aom
d) tetravalent N-atom

Answer: b

Question: Which of the following reactions is appropriate for converting acetamide to methanamine?
a) Carbylamine reaction
b) Hofmann bromamide reaction
c) Stephen’s reaction
d) Gabriel’s phthalimide synthesis

Answer: b

Question: If one H-atom of ammonia is replaced by alkyl group, the amine, thus obtained is known as
a) secondary amine
b) primary amine
c) tertiary amine
d) quaternary amine

Answer: b

Question: Which of the following is an aromatic amine?
a) Aniline
b) N-methyl aniline
c) 2-phenyl ethanamine
d) None of the above

Answer: a

Question: Which of the following can be prepared using Gabriel phthalimide synthesis?
a) Primary aromatic amines
b) Secondary amines
c) Primary aliphatic amines
d) Tertiary amines

Answer: c

Question: Ethylamine (C₂H₅HN₂)    can be obtained from N-ethylphthalimide on treatment with
a) NaBH₄
b) NH₂NH₂
c) H₂O 
d) CaH₂

Answer: b

Question: Amines behave as a nucleophile because of
a) the presence of unshared pair of electrons on N-atom
b) the absence of unshared pair of electrons on N-atom
c) the vacant p-orbital of N-atom
d) All of the above

Answer: a

Question: What is the correct order of reactivity of halides with amines?
a) RCl > RBr > RI
b) RI > RBr > RCl
c) RCl > RI > RBr
d) RI > RCl > RBr

Answer: b

Question: Amines on treatment with acids yield salt because they are
a) basic in nature
b) acidic in nature
c) amphoteric in nature
d) None of these

Answer: a

Question: What is the bond angle of C—N—E (E = C or H) in case of trimethylamine?
a) 109.5°
b) More than 109.5°
c) 108°
d) 90°

Answer: c

Question: The correct order of the basic strength of methyl substituted amines in aqueous solution is
a) (CH₃)₃ N > CH₃NH₂ > (CH₃)₂NH
b) (CH₃)₃ N > (CH₃)₂NH > CH₃NH₂
c) CH₃NH₂ > (CH₃)₂NH > (CH₃)₂N
d) (CH₃)₂ NH> CH₃NH₂ > (CH₃)₃N

Answer: d

Question: The chemical formula of Hinsberg’s reagent is
a) HNO₂
b) NaOH+ CaO
c) C₆H₅SO₂Cl   
d) CH₃CONH₂

Answer: c

Question: Aniline is more stable than anilinium ion because
a) it has more resonating structures.
b) it has less resonating structures.
c) it has more p-bonding.
d) it has less p-bonding

Answer: a

Question: The carbylamine reaction is given by
a) (C₂H₅)₃N  
b) (C₂H₅)₂NH   
c) C₂H₅NH₂
d) C₃H₇NHC₂H₅

Answer: c

Question: —NH₂ group in aniline is activating group and, hence reaction occurs at
a) para position
b) ortho position
c) meta position
d) Both (a) and (b)

Answer: d

Question: Which of the following can form H-bond?
a) NH₃
b) R—CH₃
c) R—O—R
d) R—Br

Answer: a

Question: Aniline does not undergo Friedel-Crafts reaction due to
a) less reactivity of aniline
b) salt formation with AlCl₃
c) electron accepting effect of—NH₂ group present in aniline
d) None of the above

Answer: b

Question: The reagent used to form aryl fluoride from arene diazonium chloride is
a) HF
b) KF
c) HBF₄
d) None of these

Answer: c

Question: Which of the following amine will form stable diazonium salt at 273-278 K?
a) C₂H₅NH₂
b) C₆H₅NH₂
c) C₆H₅CH₂NH₂
d) C₆H₅N(CH₃)₂

Answer: b

Question: Name the product(s) formed during the reaction of primary aliphatic amines with nitrous acid at room temperature?
a) R NO₂
b) ROH
c) Both (a) and
b) d) None of these

Answer: b

Question: Which of the following compound is water insoluble and stable at room temperature?
a) Benzene diazonium chloride
b) Benzene diazonium fluoroborate
c) Both (a) and (b)
d) None of the above

Answer: b

Question: Coupling reaction is an example of
a) nucleophilic addition reaction.
b) nucleophilic substitution reaction.
c) electrophilic substitution reaction.
d) electrophilic addition reaction.

Answer: c

CBSE Class 12 Chemistry Organic Compounds containing Nitrogen. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

CLASS XII CHAPTER – ORGANIC COMPOUNDS CONTAINING NITROGEN ONE MARK QUESTIONS

1.

Why do amines react as nucleophiles? (2007)

2.

Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt. (2008)

3.

Give the IUPAC name of H2N-CH2-CH2CH=CH2 (2010)

4.

Why is an alkylamine more basic than ammonia? (2011)

TWO MARK QUESTIONS

1.

Effect the following conversions:

a) Aniline to p-nitro aniline

b) Benzyl bromide to 2-Phenyl ethanamine

c) Acetaldehyde to ethyl amine

d) Nitro Benzene to Benzene

e) Methyl cyanide to acetone

2.

Account for the following:

a) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

b) Amines are more basic than alcohols of comparable molecular masses.(2011)

THREE MARK QUESTIONS

1.

Illustrate the following reactions giving a chemical equation in each case:

a) Carbylamine reaction

b) Coupling reaction

c) Gabriel-Phthalimide synthesis (2011)

Give the structures of A , B and C in the following reactions:

a) CH3CH2I A B C

b) C6H5NO2 A BC

FIVE MARK QUESTIONS

1. Effect the following conversions:

a) Aniline to p-nitro aniline

b) Benzyl bromide to 2-Phenyl ethanamine

c) Acetaldehyde to ethyl amine

d) Nitro Benzene to Benzene

e) Methyl cyanide to acetone

VALUE BASED QUESTION

1. Mrinal wanted to establish a chemical plant for the production of ethanamine. He is

in favour of its production through ammonolysis of alkyl halides. But his friend

suggested him to employ Gabriel phthalimide synthesis process.

a) Is the suggestion given by Mrinal’s friend correct? What are the values shown

by Mrinal’s friend?

b) What do you mean by ammonolysis? Explain ammonolysis of alkyl halides.

c) Why Mrinal’s friend suggested him to employ Gabriel phthalimide synthesis

method?

Please click the link below to download CBSE Class 12 Chemistry Organic Compounds containing Nitrogen

One Mark Questions

Question 1. Why do amines react as nucleophiles?
Answer: Amines possess a non-bonding lone pair of electrons on the nitrogen atom. This makes them electron-rich species that can easily donate their electron pair to electron-deficient centers, allowing them to act as nucleophiles.
In simple words: Amines have an extra pair of electrons on their nitrogen atom that they can share with other atoms that need electrons.

Exam Tip: To get full marks, always mention the presence of the "unshared/lone pair of electrons on the nitrogen atom".

Question 2. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt. 
Answer: When an aqueous solution of benzenediazonium chloride is warmed with potassium iodide (KI), the diazonium group is replaced by the iodide ion to yield iodobenzene. \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{KI} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{I} + \text{N}_2 + \text{KCl} \]
In simple words: Warming benzenediazonium chloride with potassium iodide replaces the nitrogen-based diazonium group with iodine, creating iodobenzene.

Exam Tip: Be sure to write the complete balanced equation, including side products like nitrogen gas and potassium chloride.

Question 3. Give the IUPAC name of H2N-CH2-CH2CH=CH2 
Answer: The IUPAC name of the compound \( \text{H}_2\text{N-CH}_2\text{-CH}_2\text{-CH=CH}_2 \) is but-3-en-1-amine.
In simple words: The carbon chain is numbered starting from the end with the amine group. It has four carbons, a double bond at the third carbon, and an amine at the first carbon.

Exam Tip: The amine group has higher priority than the double bond in IUPAC nomenclature, so the carbon chain must be numbered starting from the carbon attached to the nitrogen atom.

Question 4. Why is an alkylamine more basic than ammonia? 
Answer: Alkyl groups are electron-releasing in nature due to their +I (inductive) effect. This electron displacement increases the electron density on the nitrogen atom in alkylamines compared to ammonia, making the lone pair more available for protonation. Furthermore, the alkyl group disperses the positive charge of the resulting alkylammonium cation, stabilizing it.
In simple words: Alkyl chains push electrons towards the nitrogen atom. This makes the nitrogen in alkylamines more electron-rich and better at capturing protons than the nitrogen in ammonia.

Exam Tip: Always support your explanation by mentioning both the "+I effect of the alkyl group" and the "stabilization of the conjugate acid".

Two Mark Questions

Question 1. Effect the following conversions:
(a) Aniline to p-nitro aniline
(b) Benzyl bromide to 2-Phenyl ethanamine
(c) Acetaldehyde to ethyl amine
(d) Nitro Benzene to Benzene
(e) Methyl cyanide to acetone
Answer:
(a) Aniline to p-nitro aniline: Direct nitration of aniline is not possible because it oxidizes the ring and protonates the amino group. Thus, aniline is first protected by acetylation with acetic anhydride in pyridine to form acetanilide, which is then nitrated using a mixture of concentrated nitric and sulfuric acids. Finally, acidic or basic hydrolysis yields p-nitroaniline. \[ \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{(\text{CH}_3\text{CO})_2\text{O, Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4} p\text{-O}_2\text{NC}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}_3\text{O}^+} p\text{-O}_2\text{NC}_6\text{H}_4\text{NH}_2 \]
(b) Benzyl bromide to 2-Phenyl ethanamine: Benzyl bromide is treated with alcoholic potassium cyanide (KCN) to produce benzyl cyanide. This nitrile is then reduced with lithium aluminium hydride (\( \text{LiAlH}_4 \)) or hydrogen in the presence of nickel to yield 2-phenyl ethanamine. \[ \text{C}_6\text{H}_5\text{CH}_2\text{Br} \xrightarrow{\text{KCN (alc.)}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2 \]
(c) Acetaldehyde to ethyl amine: Acetaldehyde is first reacted with hydroxylamine (\( \text{NH}_2\text{OH} \)) to produce acetaldoxime, which is subsequently reduced using sodium in ethanol or \( \text{LiAlH}_4 \) to obtain ethyl amine. \[ \text{CH}_3\text{CHO} \xrightarrow{\text{NH}_2\text{OH}} \text{CH}_3\text{CH=NOH} \xrightarrow{\text{Na / C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \]
(d) Nitro Benzene to Benzene: Nitrobenzene is reduced to aniline using iron scrap and hydrochloric acid. Aniline is then converted to benzenediazonium chloride through diazotisation at 273 - 278 K. This diazonium salt is reduced to benzene using hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) in water. \[ \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl, } 273\text{-}278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2 + \text{H}_2\text{O}} \text{C}_6\text{H}_6 \]
(e) Methyl cyanide to acetone: Methyl cyanide is reacted with methylmagnesium bromide in dry ether. The addition product obtained is then subjected to acid hydrolysis to give acetone. \[ \text{CH}_3\text{CN} + \text{CH}_3\text{MgBr} \xrightarrow{\text{dry ether}} \text{CH}_3\text{-C(CH}_3\text{)=N-MgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COCH}_3 \]
In simple words: These reactions show how to protect sensitive groups, add carbons using cyanides, reduce functional groups, and use organometallic reagents to build complex molecules.

Exam Tip: Clearly write down all intermediate reagents and reaction conditions over the conversion arrows to ensure you score full marks.

Question 2. Account for the following:
(a) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(b) Amines are more basic than alcohols of comparable molecular masses. (2011)
Answer:
(a) Stability of diazonium salts: Aromatic diazonium salts are stabilized by resonance because the positive charge on the diazonium group can be delocalized over the aromatic benzene ring. On the other hand, aliphatic diazonium salts do not have any resonance stabilization and rapidly decompose to release nitrogen gas and highly reactive carbocations, even at low temperatures.
(b) Basicity of amines vs alcohols: Nitrogen is less electronegative than oxygen. Consequently, nitrogen holds its lone pair of electrons less tightly and can donate them to a proton much more easily than the oxygen atom in alcohols, making amines more basic than alcohols of similar molecular mass.
In simple words: Benzene rings help spread out the positive charge on aromatic diazonium salts, keeping them stable. For basicity, nitrogen is less electronegative than oxygen, so it shares its electron pair more readily.

Exam Tip: Use key terms like "resonance stabilization," "charge delocalization," and "difference in electronegativity" to build strong theoretical answers.

Three Mark Questions

Question 1. Illustrate the following reactions giving a chemical equation in each case:
(a) Carbylamine reaction
(b) Coupling reaction
(c) Gabriel-Phthalimide synthesis (2011)
Answer:
(a) Carbylamine reaction: Aliphatic and aromatic primary amines, when heated with chloroform and ethanolic potassium hydroxide, form extremely foul-smelling isocyanides (carbylamines). This reaction is used as a test to identify primary amines. \[ \text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{R-NC (Isocyanide)} + 3\text{KCl} + 3\text{H}_2\text{O} \] Example using aniline: \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \]
(b) Coupling reaction: Arene diazonium salts react with highly reactive aromatic compounds like phenols or aromatic amines to form brightly colored azo compounds. The coupling usually takes place at the para-position of the ring. Example: Benzenediazonium chloride couples with phenol in a weakly basic medium to yield p-hydroxyazobenzene (an orange dye). \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-, \text{ pH } 9\text{-}10} p\text{-HOC}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 + \text{Cl}^- + \text{H}_2\text{O} \]
(c) Gabriel-Phthalimide synthesis: This method is used to prepare pure primary aliphatic amines. Phthalimide is treated with ethanolic KOH to yield potassium phthalimide, which is heated with an alkyl halide to produce N-alkylphthalimide. Subsequent alkaline hydrolysis of the N-alkylphthalimide yields a primary amine. \[ \text{C}_6\text{H}_4(\text{CO})_2\text{NH} \xrightarrow{\text{KOH (eth.)}} \text{C}_6\text{H}_4(\text{CO})_2\text{N}^-\text{K}^+ \xrightarrow{\text{R-X}} \text{C}_6\text{H}_4(\text{CO})_2\text{N-R} \xrightarrow{2\text{NaOH (aq)}} \text{R-NH}_2 + \text{C}_6\text{H}_4(\text{COONa})_2 \]
In simple words: Primary amines produce a terrible smell when heated with chloroform and base (Carbylamine test). Diazonium salts react with phenols to form colorful dyes (Coupling reaction). Phthalimide is used to selectively build pure primary amines (Gabriel synthesis).

Exam Tip: Remember that secondary and tertiary amines do not undergo Gabriel-Phthalimide synthesis or the Carbylamine reaction. Highlighting these details shows a strong understanding to the examiner.

Question 2. Give the structures of A, B and C in the following reactions:
(a) \( \text{CH}_3\text{CH}_2\text{I} \xrightarrow{\text{NaCN}} A \xrightarrow[\text{hydrolysis}]{\text{OH}^-, \text{ Partial}} B \xrightarrow{\text{NaOH} + \text{Br}_2} C \)
(b) \( \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} A \xrightarrow[\text{273 K}]{\text{HNO}_2} B \xrightarrow{\text{C}_6\text{H}_5\text{OH}} C \)
Answer:
(a) Reaction sequence (a): - Ethyl iodide reacts with NaCN via nucleophilic substitution to produce propanenitrile (A). - Partial hydrolysis of propanenitrile in the presence of hydroxide ions produces propanamide (B). - Propanamide undergoes Hofmann bromamide degradation with NaOH and bromine to yield ethanamine (C). \[ \text{CH}_3\text{CH}_2\text{I} \xrightarrow{\text{NaCN}} \text{CH}_3\text{CH}_2\text{CN (A)} \xrightarrow[\text{hydrolysis}]{\text{OH}^-, \text{ Partial}} \text{CH}_3\text{CH}_2\text{CONH}_2 \text{ (B)} \xrightarrow{\text{NaOH} + \text{Br}_2} \text{CH}_3\text{CH}_2\text{NH}_2 \text{ (C)} \] The structures are: - **A:** \( \text{CH}_3\text{CH}_2\text{CN} \) - **B:** \( \text{CH}_3\text{CH}_2\text{CONH}_2 \) - **C:** \( \text{CH}_3\text{CH}_2\text{NH}_2 \)
(b) Reaction sequence (b): - Nitrobenzene is reduced by Fe/HCl to aniline (A). - Aniline reacts with nitrous acid at 273 K to form benzenediazonium chloride (B). - Benzenediazonium chloride undergoes a coupling reaction with phenol to yield p-hydroxyazobenzene (C). \[ \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \text{ (A)} \xrightarrow[\text{273 K}]{\text{HNO}_2} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \text{ (B)} \xrightarrow{\text{C}_6\text{H}_5\text{OH}} p\text{-HOC}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 \text{ (C)} \] The structures are: - **A:** \( \text{C}_6\text{H}_5\text{NH}_2 \) - **B:** \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \) - **C:** \( p\text{-HOC}_6\text{H}_4\text{-N=N-C}_6\text{H}_5 \)
In simple words: In the first scheme, nitrile is formed, partially hydrolyzed to an amide, and then degraded to an amine. In the second, nitrobenzene is reduced, diazotized, and coupled with phenol to form a colored dye.

Exam Tip: Be careful with nitrile hydrolysis - partial hydrolysis yields an amide (\( \text{-CONH}_2 \)), whereas complete acidic hydrolysis yields a carboxylic acid (\( \text{-COOH} \)).

Five Mark Questions

Question 1. Effect the following conversions:
(a) Aniline to p-nitro aniline
(b) Benzyl bromide to 2-Phenyl ethanamine
(c) Acetaldehyde to ethyl amine
(d) Nitro Benzene to Benzene
(e) Methyl cyanide to acetone
Answer:
(a) Aniline to p-nitro aniline: Direct nitration of aniline is not possible because it oxidizes the ring and protonates the amino group. Thus, aniline is first protected by acetylation with acetic anhydride in pyridine to form acetanilide, which is then nitrated using a mixture of concentrated nitric and sulfuric acids. Finally, acidic or basic hydrolysis yields p-nitroaniline. \[ \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{(\text{CH}_3\text{CO})_2\text{O, Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4} p\text{-O}_2\text{NC}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}_3\text{O}^+} p\text{-O}_2\text{NC}_6\text{H}_4\text{NH}_2 \]
(b) Benzyl bromide to 2-Phenyl ethanamine: Benzyl bromide is treated with alcoholic potassium cyanide (KCN) to produce benzyl cyanide. This nitrile is then reduced with lithium aluminium hydride (\( \text{LiAlH}_4 \)) or hydrogen in the presence of nickel to yield 2-phenyl ethanamine. \[ \text{C}_6\text{H}_5\text{CH}_2\text{Br} \xrightarrow{\text{KCN (alc.)}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2 \]
(c) Acetaldehyde to ethyl amine: Acetaldehyde is first reacted with hydroxylamine (\( \text{NH}_2\text{OH} \)) to produce acetaldoxime, which is subsequently reduced using sodium in ethanol or \( \text{LiAlH}_4 \) to obtain ethyl amine. \[ \text{CH}_3\text{CHO} \xrightarrow{\text{NH}_2\text{OH}} \text{CH}_3\text{CH=NOH} \xrightarrow{\text{Na / C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \]
(d) Nitro Benzene to Benzene: Nitrobenzene is reduced to aniline using iron scrap and hydrochloric acid. Aniline is then converted to benzenediazonium chloride through diazotisation at 273 - 278 K. This diazonium salt is reduced to benzene using hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) in water. \[ \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl, } 273\text{-}278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2 + \text{H}_2\text{O}} \text{C}_6\text{H}_6 \]
(e) Methyl cyanide to acetone: Methyl cyanide is reacted with methylmagnesium bromide in dry ether. The addition product obtained is then subjected to acid hydrolysis to give acetone. \[ \text{CH}_3\text{CN} + \text{CH}_3\text{MgBr} \xrightarrow{\text{dry ether}} \text{CH}_3\text{-C(CH}_3\text{)=N-MgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COCH}_3 \text{ (Acetone)} \]
In simple words: These multi-step conversions show how we can build, alter, and reduce carbon skeletons by protecting active sites, introducing nitriles, or utilizing Grignard reagents.

Exam Tip: Practice step-by-step mechanisms for common reactions (like the Grignard addition and diazotisation) as they are frequently evaluated in 5-mark conversion questions.

Value Based Question

Question 1. Mrinal wanted to establish a chemical plant for the production of ethanamine. He is in favour of its production through ammonolysis of alkyl halides. But his friend suggested him to employ Gabriel phthalimide synthesis process.
(a) Is the suggestion given by Mrinal’s friend correct? What are the values shown by Mrinal’s friend?
(b) What do you mean by ammonolysis? Explain ammonolysis of alkyl halides.
(c) Why Mrinal’s friend suggested him to employ Gabriel phthalimide synthesis method?
Answer:
(a) Yes, the suggestion given by Mrinal's friend is correct. The values shown by his friend include helpfulness, sharing of scientific knowledge, concern for the economic feasibility of Mrinal's plant, and a highly analytical, problem-solving approach.
(b) Ammonolysis refers to the cleavage of the carbon-halogen (C-X) bond in alkyl halides by ammonia. When an alkyl halide is heated with an ethanolic solution of ammonia at 373 K in a sealed tube, the halogen atom is replaced by the amino group, forming a primary amine. Since the resulting primary amine still has nucleophilic nitrogen, it continues to react with excess alkyl halide, leading to a mixture of secondary and tertiary amines, and eventually a quaternary ammonium salt. \[ \text{NH}_3 \xrightarrow{\text{R-X, } - \text{HX}} \text{R-NH}_2 \xrightarrow{\text{R-X, } - \text{HX}} \text{R}_2\text{NH} \xrightarrow{\text{R-X, } - \text{HX}} \text{R}_3\text{N} \xrightarrow{\text{R-X}} \text{R}_4\text{N}^+\text{X}^- \]
(c) Gabriel phthalimide synthesis is preferred for industrial production because it yields exclusively pure primary aliphatic amines without any contamination from secondary or tertiary amines or quaternary salts. In contrast, ammonolysis of alkyl halides yields a mixture of products that are extremely difficult to separate, which significantly lowers the overall yield and purity of the desired ethanamine. Therefore, Gabriel phthalimide synthesis is a much better choice for Mrinal's plant.
In simple words: Mrinal's friend is right because ammonolysis creates a messy mixture of different amines that are hard to separate. Gabriel phthalimide synthesis produces only the pure primary amine, which is ideal for a chemical plant.

Exam Tip: In value-based questions, make sure to explicitly list distinct values like "scientific temperament," "helpfulness," or "economic awareness" to score complete marks on the value-oriented sub-part.