Maharashtra Board Class 12 Chemistry Chapter 13 Amine Solutions

Amine Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 13 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 13 Exercise Solutions

Question 1. Choose the most correct option.

Question i. The hybridisation of nitrogen in primary amine is ............
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³
In simple words: Primary amines have nitrogen with three single bonds and one lone pair, which results in sp³ hybridization for optimal electron pair repulsion.

🎯 Exam Tip: Recognize that the lone pair on nitrogen contributes to its hybridization state, leading to a tetrahedral electron geometry and pyramidal molecular geometry.

Question ii. Isobutylamine is an example of ............
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine
In simple words: Amines are classified by the number of alkyl groups attached to the nitrogen atom; isobutylamine has two alkyl groups attached to nitrogen, making it a secondary amine.

🎯 Exam Tip: Pay attention to the number of carbon atoms directly bonded to the nitrogen to correctly classify an amine as primary, secondary, or tertiary.

Question iii. Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine
In simple words: Straight-chain primary amines, like n-Butylamine, exhibit stronger intermolecular hydrogen bonding and less steric hindrance compared to branched or more substituted amines, resulting in higher boiling points.

🎯 Exam Tip: Boiling points of amines are influenced by the extent of hydrogen bonding (primary > secondary > tertiary) and molecular branching (less branching generally leads to higher boiling points).

Question iv. Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine
In simple words: Dimethylamine typically exhibits the highest basic strength among simple methylamines in aqueous solution due to a favorable balance between the electron-donating inductive effect of alkyl groups and the stabilization of its conjugate acid through solvation.

🎯 Exam Tip: When comparing basicity in aqueous solutions, consider the combined effects of inductive electron donation, solvation of the conjugate acid, and steric hindrance around the nitrogen atom.

Question v. Which type of amine does produce N2 when treated with HNO2?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine
In simple words: Primary aliphatic amines react with nitrous acid to form unstable diazonium salts, which rapidly decompose to release nitrogen gas, distinguishing them from secondary and tertiary amines.

🎯 Exam Tip: The evolution of N2 gas upon reaction with nitrous acid (from NaNO2/HCl) is a characteristic test for primary aliphatic amines.

Question vi. Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine
In simple words: The carbylamine test, also known as the isocyanide test, is a specific chemical test for primary amines (both aliphatic and aromatic) which produce intensely foul-smelling isocyanides when heated with chloroform and a base.

🎯 Exam Tip: The production of an extremely offensive odor is the definitive positive result for the carbylamine test, confirming the presence of a primary amine.

Question vii. Which one of the following compounds does not react with acetyl chloride?
a. CH3-CH2-NH2
b. (CH3-CH2)2NH
c. (CH3-CH2)3N
d. C6H5-NH2
Answer:
c. (CH3-CH2)3N
In simple words: Acetyl chloride reacts with amines via an acylation reaction, which requires at least one hydrogen atom attached to the nitrogen; tertiary amines lack this hydrogen, hence they do not react.

🎯 Exam Tip: For an acylation reaction to occur, a hydrogen atom on the amine nitrogen is necessary, distinguishing primary and secondary amines from tertiary amines.

Question viii. Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine
In simple words: Primary amines react with Hinsberg's reagent to form N-alkylbenzenesulfonamides, which contain an acidic hydrogen on the nitrogen and therefore dissolve in aqueous NaOH solution.

🎯 Exam Tip: The solubility of the sulfonamide product in aqueous NaOH is a key diagnostic feature of the Hinsberg test, allowing differentiation between primary, secondary, and tertiary amines.

Question ix. Identify 'B' in the following reactions
CH3-C≡N \( \xrightarrow{Na/C_2H_5OH} \) A \( \xrightarrow{NaNO_2/dilHCl} \) B
a. CH3-CH2-NH2
b. CH3-CH2-NO2
c. CH3-CH2N2Cl
d. CH3-CH2-OH
Answer:
d. CH3-CH2-OH
In simple words: Methyl cyanide is first reduced to ethylamine (A) using sodium in ethanol. This primary amine (ethylamine) then reacts with nitrous acid to form an unstable diazonium salt, which decomposes to yield ethyl alcohol (B).

🎯 Exam Tip: Remember that reduction of nitriles produces primary amines, and subsequent reaction of primary aliphatic amines with nitrous acid typically leads to alcohols, with the evolution of nitrogen gas.

Question x. Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene
In simple words: An azo linkage is characterized by a -N=N- group connecting two organic moieties; p-hydroxyazobenzene contains this specific linkage.

🎯 Exam Tip: Recognize the -N=N- functional group as the defining feature of an azo compound, often associated with colored dyes.

2. Answer In One Sentence.

Question i. Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:
\(H_3C-\text{C}_6\text{H}_4-\text{SO}_2\text{Cl} + \text{HN}(\text{C}_2\text{H}_5)_2 \xrightarrow{-\text{HCl}} H_3C-\text{C}_6\text{H}_4-\text{SO}_2\text{N}(\text{C}_2\text{H}_5)_2 \)
p-toluene sulphonyl chloride Diethyl amine N,N-diethyl-p-toluene sulphon amide
In simple words: p-Toluenesulfonyl chloride reacts with diethylamine, a secondary amine, to form N,N-diethyl-p-toluenesulfonamide, releasing hydrochloric acid as a byproduct.

🎯 Exam Tip: For the Hinsberg reaction with a secondary amine, ensure the product is an N,N-disubstituted sulfonamide, and remember that HCl is eliminated.

Question ii. How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.
In simple words: To transform ethanamine (a primary amine) into N,N-dimethyl ethanamine (a tertiary amine), two methyl groups need to be successively introduced, each requiring one mole of methyl bromide.

🎯 Exam Tip: Each alkylation step on the nitrogen atom (from primary to secondary, and then to tertiary) consumes one equivalent of the alkyl halide.

Question iii. Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH3 - CH2 - CONH2) produces ethanamine by Hofmann bromamide degradation reaction.
In simple words: The Hofmann bromamide degradation reaction converts an amide into a primary amine with one fewer carbon atom; thus, to obtain ethanamine (two carbons), the starting amide must be propanamide (three carbons).

🎯 Exam Tip: Remember that the Hofmann bromamide degradation is a 'step-down' reaction, always yielding an amine with one carbon less than the original amide.

Question iv. Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH3.
In simple words: In the gaseous phase, the basicity of aliphatic amines increases with the number of electron-donating alkyl groups attached to nitrogen, as this enhances the availability of the lone pair for protonation.

🎯 Exam Tip: For gaseous phase basicity, inductive effects are dominant; more alkyl groups mean stronger electron donation and higher basicity.

Question v. Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.
In simple words: Alkyl groups are electron-donating, and when present in primary aliphatic amines, they increase the electron density on the nitrogen, making its lone pair more readily available for donation compared to ammonia.

🎯 Exam Tip: The inductive effect (+I effect) of alkyl groups directly contributes to the enhanced basicity of amines relative to ammonia.

Question vi. Predict the product of the following reaction. Nitrobenzene \( \xrightarrow{Sn/Conc.\ HCl} \)?
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह अभिक्रिया नाइट्रोबेंजीन (एक बेंजीन रिंग जिस पर एक नाइट्रो -NO2 समूह जुड़ा है) की कमी को दर्शाती है। टिन और सांद्रित हाइड्रोक्लोरिक एसिड (Sn/Conc. HCl) एक प्रबल अपचायक एजेंट के रूप में कार्य करते हैं, जो नाइट्रो समूह को एक एमिनो -NH2 समूह में परिवर्तित करता है। उत्पाद एनिलिन होगा, जिसमें बेंजीन रिंग पर एक एमिनो समूह जुड़ा होता है।
The product is aniline/ \( \text{C}_6\text{H}_5-\text{NH}_2 \).
In simple words: Nitrobenzene, upon reaction with tin and concentrated hydrochloric acid, undergoes reduction where the nitro group is converted into an amino group, yielding aniline.

🎯 Exam Tip: Remember that Sn/Conc. HCl is a common reagent used for the reduction of nitro aromatic compounds to their corresponding anilines.

Question vii. Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.
In simple words: Benzylamine, structured as a phenyl group attached to a -CH2-NH2 unit, is systematically named as a derivative of methanamine (CH3NH2) with a phenyl substituent.

🎯 Exam Tip: For IUPAC nomenclature, treat the phenyl-CH2- group as a substituent (benzyl or phenylmethyl) on the main amine chain, or name the entire structure as a substituted alkane-amine.

Question viii. Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine
In simple words: The boiling points increase from tertiary to secondary to primary amines due to increasing capability for hydrogen bonding: trimethylamine (tertiary) has no N-H bonds, ethylmethyl amine (secondary) has one, and n-propylamine (primary) has two, allowing for stronger intermolecular attraction.

🎯 Exam Tip: The ability to form hydrogen bonds (number of N-H hydrogens) is the primary factor determining the relative boiling points among isomeric amines of similar molecular mass.

Question ix. Write the balanced chemical equations for the action of dil H2SO4 on diethylamine.
Answer:
\( 2(\text{C}_2\text{H}_5)_2 \text{NH} + \text{H}_2\text{SO}_4 \to [(\text{C}_2\text{H}_5)_2 \text{NH}_2]_2 \text{SO}_4 \)
Diethyl amine diethylammonium sulphate
In simple words: Diethylamine, being a basic amine, reacts with dilute sulfuric acid to form its corresponding salt, diethylammonium sulfate, demonstrating the acid-base nature of amines.

🎯 Exam Tip: Amines act as bases and react with acids to form ammonium salts; ensure the reaction is stoichiometrically balanced, especially for polyprotic acids like H2SO4.

Question x. Arrange the following amines in the increasing order of their pK♭ values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (pKb 3.34), aniline (pKb 9.13) 4-nitroaniline (pKb 12.99)
In simple words: The increasing order of pKb values reflects decreasing basic strength: cyclohexylamine is a strong aliphatic base, aniline is a weaker aromatic base due to resonance, and 4-nitroaniline is the weakest because of the additional electron-withdrawing nitro group.

🎯 Exam Tip: A higher pKb value indicates a weaker base. Electron-donating groups increase basicity (lower pKb), while electron-withdrawing groups decrease basicity (higher pKb).

3. Answer The Following

Question i. Identify A and B in the following reactions.
\( \text{C}_6\text{H}_5\text{CH}_2\text{Br} \xrightarrow{\text{alco. KCN}} \text{A} \xrightarrow{\text{Na/ethanol}} \text{B}. \)
Answer:
\( \text{C}_6\text{H}_5\text{CH}_2\text{Br} \xrightarrow{\text{alc.KCN}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} + \text{KBr} \)
phenyl acetonitrile (A)

\( \text{C}_6\text{H}_5\text{CH}_2\text{CN} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{C}_6\text{H}_5\text{CH}_2-\text{CH}_2-\text{NH}_2 \)
β-phenyl ethyl amine (B)
A = C6H5CH2CN phenyl acetonitrile
B = C6H5CH2-CH2-NH2 β-phenyl ethyl amine
In simple words: Benzyl bromide (C6H5CH2Br) reacts with alcoholic KCN to form phenyl acetonitrile (A), which then undergoes reduction using sodium and ethanol to yield β-phenyl ethyl amine (B).

🎯 Exam Tip: The reaction with KCN introduces a new carbon atom (nitrile formation), and subsequent reduction of the nitrile functional group is a common method for synthesizing primary amines with an extended carbon chain.

Question ii. Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K♭ or pK♭ value. According to Lowry-Bronsted theory the basic nature of amines is explained by the following equilibrium equation.
\( \overset{\bullet\bullet}{\text{N}} + \text{H}_2\text{O} \rightleftharpoons \overset{+}{\text{N}}\text{H} + \text{OH}^- \)
Amine Conjugate acid
In this equilibrium amine accepts H+, hence an amine is a Lowry-Bronsted base.
According to Lewis theory, the species which donates a pair of electrons is called a base.
The nitrogen atom in amines has a lone pair of electrons, which can be donated to suitable acceptor like proton H+.
The aqueous solutions of amines are basic in nature due to release of free OH¯ ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :
\( \text{R} - \text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{RNH}_3^+ + \text{OH}^- \)
Since OH¯ is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH¯.
Here, K♭ value is smaller and pK♭ value is larger.
Hence amines are weak bases.
In simple words: Amines are basic because the nitrogen atom possesses a lone pair of electrons, enabling them to accept a proton (Brønsted-Lowry base) or donate an electron pair (Lewis base), which results in the formation of hydroxide ions in aqueous solutions.

🎯 Exam Tip: When describing amine basicity, refer to both the Brønsted-Lowry (proton acceptor) and Lewis (electron pair donor) definitions, and explain how factors like inductive effects and resonance influence the availability of the nitrogen's lone pair.

Question iii. What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 - 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह अभिक्रिया एनिलिन के डायज़ोटीकरण को दर्शाती है। एनिलिन (एक बेंजीन रिंग जिस पर एक NH2 समूह जुड़ा है) सोडियम नाइट्राइट (NaNO2) और हाइड्रोक्लोरिक एसिड (2HCl) के साथ 273-278 K के ठंडे तापमान पर अभिक्रिया करता है। इस प्रक्रिया से बेंजीन डायज़ोनियम क्लोराइड (एक बेंजीन रिंग जिस पर एक N2+Cl- समूह जुड़ा है), सोडियम क्लोराइड (NaCl) और पानी (2H2O) का निर्माण होता है।
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278 \text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \)
Aniline Benzene diazonium chloride
In simple words: Diazotisation is the process where primary aromatic amines react with nitrous acid at low temperatures (273-278 K) to form stable arenediazonium salts.

🎯 Exam Tip: Always specify the low temperature condition (0-5°C or 273-278 K) for diazotisation, as arenediazonium salts are unstable at higher temperatures.

Question iv. Write reaction to convert acetic acid into methylamine.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह संश्लेषण एसिटिक एसिड से मेथिलैमीन में परिवर्तन को दर्शाता है, जिसमें कई चरण शामिल हैं। पहले, एसिटिक एसिड (CH3COOH) अमोनिया के साथ अभिक्रिया करके अमोनियम एसीटेट (CH3COONH4) बनाता है, जो फिर निर्जलीकरण (-H2O) से एसीटामाइड (CH3CONH2) में बदल जाता है। अंत में, एसीटामाइड का हॉफमैन ब्रोमामाइड निम्नीकरण (Br2/4KOH) से मेथिलैमीन (CH3NH2) का उत्पादन होता है, साथ में 2KBr, K2CO3 और H2O सह-उत्पाद के रूप में बनते हैं।
\( \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, -\text{H}_2\text{O}} \text{CH}_3\text{COONH}_4 \xrightarrow{\Delta, -\text{H}_2\text{O}} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{4KOH}} \text{CH}_3\text{NH}_2 \)
Acetic acid Ammonium acetate Acetamide methyl amine
In simple words: Acetic acid is first converted into acetamide, usually via ammonium acetate, which then undergoes the Hofmann bromamide degradation reaction to yield methylamine, effectively reducing the carbon chain by one atom.

🎯 Exam Tip: To convert a carboxylic acid to an amine with one less carbon, the sequence of forming the amide followed by Hofmann degradation is a standard and important synthetic route.

Question v. Write a short note on coupling reactions.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह खंड डायज़ोनियम लवणों की विभिन्न अभिक्रियाओं को दर्शाता है। पहले भाग (1) में, एक प्राथमिक सुगंधित ऐमीन (Ar-NH2) नाइट्रस एसिड (NaNO2/2HCl) के साथ 273-278 K पर अभिक्रिया करके डायज़ोनियम क्लोराइड (Ar-N2+Cl-) बनाता है, जो फिर क्यूप्रस क्लोराइड (CuCl) के साथ अभिक्रिया करके एरिल क्लोराइड (ArCl) और नाइट्रोजन गैस देता है। दूसरे भाग (2) में, डायज़ोनियम क्लोराइड (Ar-N2+Cl-) H3PO2 और H2O के साथ अभिक्रिया करके एक एरीन (ArH) और नाइट्रोजन गैस देता है। अंतिम पंक्ति "Reactions involving retention of diazo group : (Coupling reactions) :" बताती है कि इसके बाद युग्मन अभिक्रियाओं का विवरण आना चाहिए, हालांकि उनके उदाहरण यहां नहीं दिए गए हैं।
(1) \( \text{Ar}-\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{2HCl}} \text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cuprous chloride}} \text{ArCl} + \text{N}_2 \uparrow \)
(2) \( \text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2/\text{H}_2\text{O}} \text{ArH} + \text{N}_2 + \text{H}_3\text{PO}_3 + \text{HCl} \)
Reactions involving retention of diazo group : (Coupling reactions) :
In simple words: Coupling reactions are characteristic reactions of arenediazonium salts where they react with highly activated aromatic compounds (like phenols or aromatic amines) to form brightly colored azo compounds, retaining the -N=N- diazo group.

🎯 Exam Tip: Coupling reactions are crucial for the synthesis of azo dyes and are highly sensitive to pH, often requiring slightly acidic or alkaline conditions for optimal yield.

Question vi. Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

ℹ️ चित्र व्याख्या (Diagram Explanation): गैब्रियल थैलिमाइड संश्लेषण प्राथमिक ऐमीन बनाने की एक विधि है। पहले चरण में, थैलिमाइड अल्कोहलिक KOH के साथ अभिक्रिया करके पोटेशियम थैलिमाइड बनाता है। दूसरे चरण में, पोटेशियम थैलिमाइड एथिल आयोडाइड के साथ अभिक्रिया करके N-एथिल थैलिमाइड बनाता है। अंत में, N-एथिल थैलिमाइड का जलीय NaOH के साथ जल-अपघटन करने पर एथिलैमीन और सोडियम थैलेट प्राप्त होता है।
\( \text{Phthalimide} \xrightarrow{\text{KOH (alc.), -H}_2\text{O}} \text{Potassium phthalimide} \xrightarrow{\text{C}_2\text{H}_5\text{I}, -\text{KI}} \text{N-Ethyl phthalimide} \xrightarrow{\text{NaOH (aq)}} \text{Sodium phthalate} + \text{C}_2\text{H}_5-\text{NH}_2 \)
In simple words: Gabriel phthalimide synthesis is a valuable method for preparing pure primary amines by reacting phthalimide with an alcoholic base, then with an alkyl halide, followed by hydrolysis of the resulting N-alkylphthalimide.

🎯 Exam Tip: This synthesis is advantageous because it exclusively yields primary amines, avoiding the formation of secondary and tertiary amines often seen in other alkylation methods.

Question vii. Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.
Secondary and tertiary amines do not give this test.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह कार्बिलामाइन अभिक्रिया को दर्शाता है, जो प्राथमिक ऐमीनों का पता लगाने के लिए एक परीक्षण है। एक प्राथमिक ऐमीन (R-NH2), क्लोरोफॉर्म (CHCl3) और अल्कोहलिक पोटेशियम हाइड्रॉक्साइड (3KOH) को गर्म करने पर एक अल्काइल आइसोसायनाइड (RNC) बनता है, जिसमें अत्यंत अप्रिय गंध होती है, साथ ही पोटेशियम क्लोराइड (3KCl) और पानी (3H2O) भी बनता है। उदाहरण के तौर पर, एथिलैमीन (C2H5-NH2) इसी अभिक्रिया से एथिल आइसोसायनाइड (C2H5NC) बनाता है।
\( \text{R}-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{RNC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
Primary amine Alkyl isocyanide
\( \text{C}_2\text{H}_5-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{C}_2\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
Ethylamine Ethyl isocyanide
In simple words: The carbylamine reaction (isocyanide test) is a diagnostic test for primary amines, where they react with chloroform and an alcoholic base upon heating to produce foul-smelling isocyanides.

🎯 Exam Tip: The distinct, highly unpleasant odor of the isocyanide formed is the key indicator for a positive carbylamine test, confirming the presence of a primary amine.

Question viii. Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

ℹ️ चित्र व्याख्या (Diagram Explanation): यह मेथिलैमीन को एथिलैमीन में बदलने की मल्टी-स्टेप अभिक्रिया को दर्शाता है, जिसमें कार्बन श्रृंखला में वृद्धि होती है। मेथिलैमीन (CH3-NH2) पहले नाइट्रस एसिड (NaNO2/HCl) से मेथनॉल (CH3-OH) में बदलता है। मेथनॉल PCl5 से अभिक्रिया करके मेथिल क्लोराइड (CH3Cl) बनाता है। मेथिल क्लोराइड अल्कोहलिक KCN के साथ अभिक्रिया करके एसीटोनिट्राइल (CH3CN) बनाता है। अंत में, एसीटोनिट्राइल का रेने Ni उत्प्रेरक की उपस्थिति में H2 गैस (2H2(g)) द्वारा अपचयन होकर एथिलैमीन (CH3-CH2-NH2) प्राप्त होता है।
\( \text{CH}_3-\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{CH}_3-\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{alc.KCN}} \text{CH}_3\text{CN} \)
Methanamine Methanol Methyl chloride Acetonitrile
\( \text{CH}_3\text{CN} + 2\text{H}_2(\text{g}) \xrightarrow{\text{Raney Ni, R.T.}} \text{CH}_3-\text{CH}_2-\text{NH}_2 \)
Ethanamine
(2) Aniline into p-bromo aniline

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एनिलिन से p-ब्रोमोएनिलिन में रूपांतरण को दर्शाता है। पहले, एनिलिन (बेंजीन रिंग पर NH2) को एसिटिल क्लोराइड (CH3COCl) और एक बेस के साथ एसिटाइलेट करके एसीटानिलिड (बेंजीन रिंग पर NHCOCH3) बनाया जाता है। फिर, एसीटानिलिड का ब्रोमीन/एसिटिक एसिड (Br2/acetic acid) के साथ ब्रोमिनेशन किया जाता है, जिससे p-ब्रोमोएसीटानिलिड (पैरा स्थिति पर Br के साथ बेंजीन-NHCOCH3) प्राप्त होता है। अंत में, p-ब्रोमोएसीटानिलिड का अम्लीय या क्षारीय जल-अपघटन (H+ or OH- hydrolysis) करके एसिटिल समूह को हटा दिया जाता है, जिससे p-ब्रोमोएनिलिन (पैरा स्थिति पर Br के साथ बेंजीन-NH2) प्राप्त होता है।
\( \text{Aniline} \xrightarrow{\text{CH}_3\text{COCl, base}} \text{Acetanilide} \xrightarrow{\text{Br}_2/\text{acetic acid}} \text{p-bromoacetanilide} \xrightarrow{\text{H}^+\text{ or OH}^- \text{hydrolysis}} \text{p-bromoaniline} \)
In simple words: Methanamine can be converted to ethanamine through a multi-step synthesis involving conversion to an alcohol, then an alkyl halide, then a nitrile (to increase the carbon chain), and finally reduction to the amine. Aniline can be converted to p-bromoaniline by protecting the amino group via acetylation, followed by para-bromination, and then hydrolysis to deprotect.

🎯 Exam Tip: To increase the carbon chain for amine synthesis, nitrile formation and reduction are key. For controlled aromatic substitution, protecting groups on highly activating substituents are often essential.

Question ix. Complete the following reactions :
a. C6H5N2Cl + C2H5OH →
b. C6H5NH2 + Br2(aq) → ?
Answer:
(a)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह अभिक्रिया बेंजीन डायज़ोनियम क्लोराइड (C6H5N2+Cl-) और इथेनॉल (C2H5OH) के बीच की अभिक्रिया को दर्शाती है, जहाँ असामान्य रूप से एक जटिल एज़ो यौगिक बनता है जिसमें एक अल्कोहल समूह भी होता है। उत्पाद के रूप में C6H5-N=N-CH2-CH2-OH और हाइड्रोक्लोरिक एसिड (HCl) प्राप्त होते हैं।
\( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_2\text{H}_5\text{OH} \to \text{C}_6\text{H}_5 - \text{N} = \text{N} - \text{CH}_2-\text{CH}_2-\text{OH} + \text{HCl} \)
(b)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एनिलिन (C6H5NH2) के जलीय ब्रोमीन (3Br2(aq)) के साथ अभिक्रिया को दर्शाता है। एमिनो समूह (NH2) एक प्रबल सक्रियण समूह है, इसलिए यह ऑर्थो और पैरा स्थितियों पर ब्रोमिनेशन को निर्देशित करता है, जिसके परिणामस्वरूप 2,4,6-ट्राइब्रोमोएनिलिन का निर्माण होता है।
\( \text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2 (\text{aq}) \to \text{C}_6\text{H}_2\text{Br}_3\text{NH}_2 + 3\text{HBr} \)
2,4,6-Tribromoaniline
In simple words: Benzene diazonium chloride reacts with ethanol to form a complex azo compound with an alcohol group, while aniline reacts with aqueous bromine to undergo tribromination at the ortho and para positions due to the strong activating effect of the amino group.

🎯 Exam Tip: Be aware that highly activated aromatic rings like aniline undergo multiple substitutions with strong electrophiles unless the activating group is protected; diazonium salt reactions with ethanol can lead to reduction or, in complex cases, unusual coupling products.

Question x. Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C - X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylammonium bromide (a quaternary ammonium salt).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह अल्काइल हैलाइड्स के अमोनोलिसिस को दर्शाता है, एक ऐसी प्रक्रिया जो ऐमीनों के मिश्रण का उत्पादन करती है। मेथिल ब्रोमाइड (CH3-Br) अमोनिया (NH3) के साथ अभिक्रिया करके मेथिलैमीन (CH3-NH2) बनाता है। मेथिलैमीन फिर से मेथिल ब्रोमाइड के साथ अभिक्रिया करके डाइमेथिलैमीन ((CH3)2NH) बनाता है। डाइमेथिलैमीन पुनः मेथिल ब्रोमाइड के साथ अभिक्रिया करके ट्राइमेथिलैमीन ((CH3)3N) बनाता है। अंत में, ट्राइमेथिलैमीन मेथिल ब्रोमाइड के साथ अभिक्रिया करके टेट्रामेथिल अमोनियम ब्रोमाइड ((CH3)4N+Br-) नामक एक चतुष्कोणीय अमोनियम लवण बनाता है।
\( \text{CH}_3-\text{Br} + \text{NH}_3 \xrightarrow{\Delta} \text{CH}_3-\text{NH}_2 + \text{HBr} \)
methylbromide methylamine
\( \text{CH}_3-\text{NH}_2 + \text{CH}_3-\text{Br} \xrightarrow{\Delta} (\text{CH}_3)_2\text{NH} + \text{HBr} \)
methyl amine dimethyl amine
\( (\text{CH}_3)_2\text{NH} + \text{CH}_3-\text{Br} \xrightarrow{\Delta} (\text{CH}_3)_3\text{N} + \text{HBr} \)
dimethyl amine trimethyl amine
\( (\text{CH}_3)_3\text{N} + \text{CH}_3-\text{Br} \xrightarrow{\Delta} (\text{CH}_3)_4\text{N}^+\text{Br}^- \)
trimethyl amine tetramethyl ammonium bromide (quaternary ammonium salt)
The order of reactivity of alkyl halides with ammonia is R - I > R - Br > R - Cl.
In simple words: Ammonolysis of alkyl halides is the reaction of alkyl halides with ammonia, leading to a mixture of primary, secondary, and tertiary amines, and finally a quaternary ammonium salt, due to successive alkylation steps.

🎯 Exam Tip: A major limitation of ammonolysis for preparing pure primary amines is the formation of a mixture of products, which requires careful separation techniques.

Question xi. Write reaction to convert ethylamine into methylamine.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एथिलैमीन से मेथिलैमीन में परिवर्तन को दर्शाता है, जिसमें कार्बन परमाणुओं की संख्या कम होती है। एथिलैमीन (C2H5NH2) नाइट्रस एसिड (HONO, -N2) के साथ अभिक्रिया करके एथिल अल्कोहल (C2H5OH) बनाता है। एथिल अल्कोहल को पोटेशियम डाइक्रोमेट और सल्फ्यूरिक एसिड (K2Cr2O7/H2SO4, 2[O]) द्वारा एसिटिक एसिड (CH3COOH) में ऑक्सीकृत किया जाता है। एसिटिक एसिड अमोनिया (NH3) के साथ अभिक्रिया करके अमोनियम एसीटेट (CH3COONH4) बनाता है, जो निर्जलीकरण (-H2O) से एसीटामाइड (CH3CONH2) में बदलता है। अंत में, एसीटामाइड का हॉफमैन ब्रोमामाइड निम्नीकरण (Br2/4KOH) से मेथिलैमीन (CH3NH2) प्राप्त होता है।
\( \text{C}_2\text{H}_5\text{NH}_2 \xrightarrow{\text{HONO, -N}_2} \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4, 2[\text{O}]} \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, -\text{H}_2\text{O}} \text{CH}_3\text{COONH}_4 \xrightarrow{\Delta, -\text{H}_2\text{O}} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{4KOH}} \text{CH}_3\text{NH}_2 \)
Ethyl amine Ethyl alcohol Acetic acid Ammonium acetate Acetamide methyl amine
In simple words: To convert ethylamine to methylamine, the carbon chain is shortened by first converting ethylamine to ethanol, then oxidizing it to acetic acid, which is then converted to acetamide, and finally subjected to Hofmann bromamide degradation.

🎯 Exam Tip: This multi-step synthesis illustrates a common strategy to reduce the carbon count in an amine by converting it to an acid and then using Hofmann degradation.

4. Answer The Following.

Question i. Write the IUPAC names of the following amines :
a. CH3-CH2-N-CH2-CH2-CH3
CH3
b. CH3-C-CH2-CH2-NH2
CH3
CH3
c. CH3-CH-NH-CH2-CH3
CH3
Answer:

Compound	IUPAC name
(1) CH3-CH2-N-CH2-CH2-CH3 CH3	N-Ethyl, N-methyl propan-1-amine
(2) CH3-C-CH2-CH2-NH2 CH3 CH3	3, 3-Dimethylbutan-1-amine
(3) CH3-CH-NH-CH2-CH3 CH3	N-Ethyl propan-2-amine

In simple words: IUPAC nomenclature for amines involves identifying the longest carbon chain containing the amino group, naming it as an alkane-amine, and using 'N-' prefixes for substituents directly attached to the nitrogen.

🎯 Exam Tip: Always assign the lowest possible locants to the amino group and other substituents, ensuring correct identification of the parent alkane chain.

Question ii. What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).
(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.
Examples:
(i) CH3 - NH2 methylamine
(ii) CH3 - CH2 - NH2 ethylamine
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिस पर सीधे एक -NH2 समूह जुड़ा हुआ है। इसे एनिलिन कहते हैं।
Aniline
(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.
Examples:
(i) C2H5 - NH - CH3 ethylmethylamine
(ii) CH3 - NH - CH3 dimethylamine
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिस पर सीधे एक -NHCH3 समूह जुड़ा हुआ है। इसे N-मेथिलैनिलिन कहते हैं।
N-methylaniline
(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.
Examples:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक नाइट्रोजन परमाणु को दर्शाता है जिससे तीन मेथिल समूह (CH3) जुड़े हुए हैं। यह ट्राइमेथिलैमीन है।
CH3-N-CH3 trimethylamine
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक नाइट्रोजन परमाणु को दर्शाता है जिससे तीन एथिल समूह (C2H5) जुड़े हुए हैं। यह ट्राइएथिलैमीन है।
C2H5-N-C2H5 triethylamine
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिस पर सीधे एक -N(CH3)2 समूह जुड़ा हुआ है। इसे N,N-डाइमेथिलैनिलिन कहते हैं।
N, N-dimethylaniline
Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.
(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.
C2H5-NH-C2H5 Diethylamine.
CH3-N-CH3
CH3 Trimethylamine
(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.
CH3
C2H5-N-C2H5 Diethyl methylamine
CH3-NH-C2H5 Ethyl methylamine
In simple words: Amines are organic derivatives of ammonia, classified as primary, secondary, or tertiary based on how many hydrogen atoms of ammonia are replaced by alkyl or aryl groups. They can also be categorized as simple (symmetrical) if all attached groups are identical, or mixed (unsymmetrical) if the attached groups are different.

🎯 Exam Tip: The classification (primary, secondary, tertiary) is determined by the number of carbon atoms directly bonded to the nitrogen, while symmetrical/unsymmetrical depends on whether the alkyl/aryl groups are the same or different.

Question iii. Write IUPAC names of the following amines.
a. H2N-(CH2)6-NH2
b.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिसमें एक मेथिल समूह (CH3) ऑर्थो स्थिति पर और एक ऐमीन समूह (NH2) पैरा स्थिति पर है।
c.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिसमें दो ऐमीन समूह (NH2) पैरा स्थितियों पर जुड़े हुए हैं।
Answer:

Compound	IUPAC name
(1) H2N-(CH2)6 - NH2	Hexan-1, 6-diamine
(2) ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिसमें एक ऐमीन समूह (NH2) C1 पर, एक मेथिल समूह (CH3) C2 पर और दूसरा मेथिल समूह (CH3) C4 पर जुड़ा हुआ है।	2, 4-Dimethylaniline
(3) ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन रिंग को दर्शाता है जिसमें एक ऐमीन समूह (NH2) C1 पर और दूसरा ऐमीन समूह (NH2) C4 पर जुड़ा हुआ है।	Benzene-1, 4-diamine

In simple words: IUPAC names provide a systematic way to identify organic compounds by indicating the parent hydrocarbon, the main functional group (amine), and the positions of any substituents.

🎯 Exam Tip: For polyfunctional or cyclic amines, always prioritize the numbering to give the lowest possible locants to the amino groups, followed by other substituents.

Question iv. Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एसीटोनिट्राइल (CH3-C≡N) से एथिलैमीन (CH3-CH2-NH2) के निर्माण को दर्शाता है। एसीटोनिट्राइल का Na/C2H5OH (सोडियम इन इथेनॉल) द्वारा अपचयन करने पर एथिलैमीन प्राप्त होता है, जहाँ 4 हाइड्रोजन परमाणु ([H]) भाग लेते हैं।
\( \text{CH}_3-\text{C} \equiv \text{N} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3-\text{CH}_2-\text{NH}_2 \)
Acetonitrile Ethanamine
b. Ethanamine from nitroethane :

ℹ️ चित्र व्याख्या (Diagram Explanation): यह नाइट्रोइथेन (CH3-CH2-NO2) से एथिलैमीन (CH3-CH2-NH2) के निर्माण को दर्शाता है। नाइट्रोइथेन का Sn/conc. HCl (टिन और सांद्रित हाइड्रोक्लोरिक एसिड) द्वारा अपचयन करने पर एथिलैमीन और पानी (2H2O) प्राप्त होता है, जहाँ 6 हाइड्रोजन परमाणु ([H]) भाग लेते हैं।
\( \text{CH}_3-\text{CH}_2-\text{NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/conc. HCl}} \text{CH}_3-\text{CH}_2-\text{NH}_2 + 2\text{H}_2\text{O} \)
Nitroethane Ethanamine
c. Ethanamine from Propionamide :

ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रोपेनामाइड (CH3-CH2-CO-NH2) से एथिलैमीन (CH3-CH2-NH2) के निर्माण को दर्शाता है, जिसे हॉफमैन ब्रोमामाइड निम्नीकरण अभिक्रिया के माध्यम से किया जाता है। प्रोपेनामाइड ब्रोमीन (Br2) और पोटेशियम हाइड्रॉक्साइड (4KOH) के साथ अभिक्रिया करके एथिलैमीन, पोटेशियम कार्बोनेट (K2CO3), पोटेशियम ब्रोमाइड (2KBr) और पानी (2H2O) बनाता है।
\( \text{CH}_3-\text{CH}_2-\text{CO}-\text{NH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{CH}_3-\text{CH}_2-\text{NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}. \)
Propionamide Ethanamine
In simple words: Ethanamine can be synthesized by reducing acetonitrile with Na/ethanol, by reducing nitroethane with Sn/HCl, or by the Hofmann bromamide degradation reaction of propionamide.

🎯 Exam Tip: Each synthesis method starts with a different functional group (nitrile, nitro compound, amide) and employs specific reagents for reduction or degradation to form the target amine.

 

Question vii. Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg's reagent?
Answer: This reaction is useful for the distinction of primary, secondary and tertiary amines.
(i) Primary amine (like ethyl amine) is treated with Hinsberg's reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.
In simple words: Primary, secondary, and tertiary amines can be identified using Hinsberg's reagent. Primary amines form a sulphonamide that dissolves in KOH, while secondary amines form a sulphonamide that does not dissolve. Tertiary amines do not react at all.

🎯 Exam Tip: Understanding the solubility of sulphonamide products in KOH is crucial for distinguishing between primary and secondary amines in the Hinsberg test.

 

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.
\[ \text{C}_{6}\text{H}_{5}\text{-SO}_{2}\text{Cl} + \text{HN}(\text{C}_{2}\text{H}_{5})_{2} \xrightarrow{\text{- HCl}} \text{C}_{6}\text{H}_{5}\text{-SO}_{2}\text{N}(\text{C}_{2}\text{H}_{5})_{2} \]
(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.
\[ \text{C}_{6}\text{H}_{5}\text{-SO}_{2}\text{Cl} + (\text{C}_{2}\text{H}_{5})_{3}\text{N} \xrightarrow{\text{No reaction}} \]
\[ (\text{C}_{2}\text{H}_{5})_{3}\text{N} + \text{HCl} \rightarrow (\text{C}_{2}\text{H}_{5})_{3}\text{NHCl} \]
In simple words: Primary, secondary, and tertiary amines can be identified using Hinsberg's reagent. Primary amines form a sulphonamide that dissolves in KOH, while secondary amines form a sulphonamide that does not dissolve. Tertiary amines do not react at all.

🎯 Exam Tip: Understanding the solubility of sulphonamide products in KOH is crucial for distinguishing between primary and secondary amines in the Hinsberg test.

 

Question viii. Write reactions to bring about the following conversions :
(a) Aniline into p-nitroaniline
(b) Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम एनिलीन से p-नाइट्रोएनिलीन बनाने की प्रक्रिया को दर्शाता है। पहले एनिलीन की एसिटिक एनहाइड्राइड से अभिक्रिया कर एसिटानिलाइड बनाते हैं, फिर इसका HNO3/H2SO4 के साथ नाइट्रेशन करने पर p-नाइट्रोएसिटानिलाइड बनता है, और अंत में इसके जल-अपघटन से p-नाइट्रोएनिलीन प्राप्त होता है।
In simple words: To convert aniline to p-nitroaniline, aniline is first acetylated to protect the amine group, then nitrated at the para position, and finally deacetylated.

🎯 Exam Tip: Remember that direct nitration of aniline leads to multiple products and oxidation, so protecting the amine group via acetylation is a key step to achieve para-nitration. This is an important concept for targeted substitution.

 

(2) Aniline into sulphanilic acid.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम एनिलीन से सल्फानिलिक एसिड बनाने की प्रक्रिया को दर्शाता है। एनिलीन की सल्फ्यूरिक एसिड के साथ अभिक्रिया से एनिलियम हाइड्रोजन सल्फेट बनता है, जिसे उच्च तापमान पर गर्म करने पर सल्फानिलिक एसिड प्राप्त होता है।
In simple words: Aniline reacts with concentrated sulfuric acid to form an anilinium hydrogen sulfate salt, which on heating undergoes rearrangement to yield sulphanilic acid.

🎯 Exam Tip: The sulfonation of aniline is a special reaction where the -NH2 group is protected by forming an anilinium ion, preventing its oxidation and directing sulfonation to the para position.

 

Activity :
• Prepare a chart of azodyes, colours and its application.
• Prepare a list of names and structures of N-containing ingredients of diet.

 

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

 

Use your brain power! (Textbook Page No 282)

 

Question 1. Classify the following amines as simple/mixed; 1º, 2º, 3º and aliphatic or aromatic.
(C2H5)2NH, (CH3)3N, C2H5 - NH - CH3,
(C2H5)2NH, (CH3)3N, C2H5 - NH - CH3,
ℹ️ चित्र व्याख्या (Diagram Explanation): यह प्रश्न विभिन्न एमीन्स की संरचनाओं को वर्गीकृत करने के लिए है। दिखाए गए रेखाचित्र क्रमशः एन-मिथाइल प्रोपैन-1-एमीन (मिश्रित, द्वितीयक, एलिफैटिक), 3,3-डाइमिथाइल ब्यूटेन-1-एमीन (सरल, प्राथमिक, एलिफैटिक), और एन-एथिल प्रोपेन-2-एमीन (सरल, प्राथमिक, एलिफैटिक) को दर्शाते हैं।
In simple words: Amines are classified based on how many hydrogen atoms of ammonia are replaced by alkyl or aryl groups, leading to primary, secondary, and tertiary amines, which can be simple or mixed, and aliphatic or aromatic.

🎯 Exam Tip: Correctly identifying the number of alkyl/aryl groups attached to the nitrogen atom and whether the groups are the same or different is key for accurate classification and naming.

 

Answer:

Compound	Simple/mixed	Type	Aliphatic/Aromatic amine
(1) (C2H5)2 NH	Simple	Secondary (2°) amine	Aliphatic
(2) (CH3)3N	Mixed	Tertiary (3°) amine	Aliphatic
(3) C2H5-NH - CH3	Mixed	Secondary (2°) amine	Aliphatic
(4) C2H5-NH2	Simple	Primary (1°) amine	Aromatic
(5) CH3-CH-CH3 NH2	Simple	Primary (1°) amine	Aliphatic
(6) ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेंजीन रिंग को दर्शाता है जिस पर एक NH-C6H5 समूह जुड़ा है, जो एक द्वितीयक, एरोमैटिक एमीन है।	Simple	Secondary (2°) amine	Aromatic
(7) CH3-C-NH2 CH3	Mixed	Primary (1°) amine	Aliphatic
(8) ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेंजीन रिंग को दर्शाता है जिस पर एक N(CH3)2 समूह जुड़ा है, जो एक तृतीयक, एरोमैटिक एमीन है।	Mixed	Tertiary (3°) amine	Aromatic
(9) ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बेंजीन रिंग को दर्शाता है जिस पर एक नाइट्रोजन परमाणु जुड़ा है, जो एक तृतीयक, एरोमैटिक एमीन है।	Mixed	Tertiary (3°) amine	Aromatic

(A) Common Names : Rules
1. According to common naming system, the amines are named as alkylamines.
2. The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word 'amine'.
Example: CH3 - NH2 : methyl-amine
3. The simple (symmetrical) secondary and tertiary amines are written by adding prefix 'di- (forpresence of two alkyl groups) and 'tri'- (for presence of three alkyl groups) respectively to the name of alkyl groups.
Examples: (i) CH3 - NH - CH3 dimethylamine, (ii) (C2H5)3 N triethylamine
4. The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word 'amine'.
Example : CH3 - CH2 - NH - CH3 ethyhnethylamine
In simple words: Amines are classified based on how many hydrogen atoms of ammonia are replaced by alkyl or aryl groups, leading to primary, secondary, and tertiary amines, which can be simple or mixed, and aliphatic or aromatic.

🎯 Exam Tip: Correctly identifying the number of alkyl/aryl groups attached to the nitrogen atom and whether the groups are the same or different is key for accurate classification and naming.

 

(B) IUPAC names : Rules
1. According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
2. The name of the amine is obtained by replacing the suffix 'e' from parent alkane's name by 'amine'.
3. The position of the amino group is indicated by the lowest possible locant.
Example :
\( \text{CH}_{3}\text{-CH}_{2}\text{-CH}(\text{NH}_{2})\text{-CH}_{3} \) Butan-2-amine
4. In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
Example : C6H5NH - CH3 N - Methylethanamine
5. A complete name of amine is written as one word.

 

Try this..... (Textbook Page No 283)

 

Question 1. Draw possible structures of all the isomers of C4H11N. Write their common as well as IUPAC names.
Answer:

Compound	Common name	IUPAC name	Primary/Secondary/ Tertiary amines
(1) CH3-CH2-CH2-CH2-NH2	1-Aminobutane	Butan-1-amine	Primary
(2) CH3-CH2-CH(NH2)-CH3	2-Aminobutane	Butan-2-amine	Primary
(3) CH3-CH(CH3)-CH2-NH2	1-Amino-2-methyl propane	2-Methyl propan-1-amine	Primary
(4) CH3-C(CH3)2-NH2	2-Amino-2-methyl propane	2-Methylpropan-2-amine	Primary
(5) CH3-CH2-CH2-NH-CH3	N-Methylaminopropane	N-Methylpropan-1-amine	Secondary
(6) CH3CHNHCH3	N-Methyl-2-aminopropane	N-Methylethanamine	Secondary
(7) CH3CH2NHC2H5	N-Ethyl aminoethane	N-Ethylethanamine	Secondary
(8) CH3CH2N(CH3)2	N, N-Dimethylaminoethane	N,N-Dimethylethanamine	Tertiary

In simple words: C4H11N has several isomeric amine structures, including primary, secondary, and tertiary types, each with unique common and IUPAC names derived from their carbon skeleton and nitrogen substitution.

🎯 Exam Tip: When drawing isomers, systematically vary the position of the amino group and the branching of the carbon chain, then consider secondary and tertiary amine possibilities by distributing alkyl groups on nitrogen.

 

Use your brain power! (Textbook Page No 283)

 

Question 1. Write chemical equations for
(i) reaction of alc. NH with C2H5I.
Answer:
(1) \( \text{CH}_{3}\text{-CH}_{2}\text{-I} + \text{NH}_{3} \xrightarrow{\Delta} \text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2} + \text{HI} \)
(2) \( \text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2} + \text{CH}_{3}\text{-CH}_{2}\text{-I} \xrightarrow{\Delta} (\text{CH}_{3}\text{-CH}_{2})_{2}\text{NH} + \text{HI} \)
(3) \( (\text{CH}_{3}\text{-CH}_{2})_{2}\text{NH} + \text{CH}_{3}\text{-CH}_{2}\text{-I} \xrightarrow{\Delta} (\text{C}_{2}\text{H}_{5})_{3}\text{N} + \text{HI} \)
(4) \( (\text{CH}_{3}\text{-CH}_{2})_{3}\text{N} + \text{CH}_{3}\text{-CH}_{2}\text{-I} \xrightarrow{\Delta} (\text{CH}_{3}\text{-CH}_{2})_{4}\text{NI} \)
In simple words: Alkyl halides react with ammonia and subsequently with primary, secondary, and tertiary amines in a stepwise fashion to form a mixture of higher substituted amines and a quaternary ammonium salt.

🎯 Exam Tip: Ammonolysis reactions typically yield a mixture of amines, making it challenging to isolate a single product. Excess ammonia can favor the formation of primary amines.

 

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम बेंजाइल क्लोराइड के अमोनोलिसिस और उसके बाद मेथिल आयोडाइड के साथ दो बार अभिक्रिया को दर्शाता है। पहले बेंजाइल क्लोराइड अमोनिया से अभिक्रिया कर बेंजाइल एमीन बनाता है। फिर यह मेथिल आयोडाइड के साथ अभिक्रिया करके N-मिथाइलफेनिल मेथानमिन और फिर N,N-डाइमिथाइलफेनिल मेथानमिन बनाता है।
(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer: In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.
In simple words: Ammonolysis of benzyl chloride initially forms benzylamine. Subsequent reactions with methyl iodide lead to N-methylbenzylamine and then N,N-dimethylbenzylamine, as each hydrogen on the nitrogen is successively replaced.

🎯 Exam Tip: To obtain a primary amine exclusively from ammonolysis, use a large excess of ammonia to minimize over-alkylation. Otherwise, a mixture of primary, secondary, tertiary, and quaternary ammonium salts will form.

 

Problem 13.1: (Textbook Page No 285)

 

Question 1. Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution: Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.
\( \text{CH}_{3}\text{-Br} + \text{KCN} \rightarrow \text{CH}_{3}\text{-CN} + \text{KBr} \)
\( \text{CH}_{3}\text{-CN} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH (reduction)}} \text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2} \)
The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.
In simple words: Methyl bromide is converted to methyl cyanide via KCN, which is then reduced to ethylamine using Na/ethanol, increasing the carbon chain length by one atom.

🎯 Exam Tip: The KCN reaction followed by reduction is a standard method for "step-up" conversions, increasing the carbon chain length, which is a common requirement in organic synthesis problems.

 

Use your brain power! (Textbook Page No 285)

 

Question 1. Identify 'A' and 'B' in the following conversions.
(1) \( \text{CH}_{3}\text{-I} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH}} \text{B} \)
(2) \( \text{CH}_{3}\text{-Br} \xrightarrow{\text{AgNO}_{2}} \text{A} \xrightarrow{\text{Sn/HCl}} \text{B} \)
(3) \( \text{C}_{2}\text{H}_{5}\text{-I} \xrightarrow{\text{AgCN}} \text{A} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH}} \text{B} \)
Answer:
(1) \( \text{CH}_{3}\text{-I} \xrightarrow{\text{KCN}} \text{CH}_{3}\text{CN (Acetonitrile, A)} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH (4H)}} \text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2} \text{(Ethyl amine, B)} \)
(2) \( \text{CH}_{3}\text{-Br} \xrightarrow{\text{AgNO}_{2}} \text{CH}_{3}\text{-NO}_{2} \text{(Nitromethane, A)} \xrightarrow{\text{Sn/HCl (6H)}} \text{CH}_{3}\text{-NH}_{2} \text{(Methyl amine, B)} + 2\text{H}_{2}\text{O} \)
(3) \( \text{C}_{2}\text{H}_{5}\text{-I} \xrightarrow{\text{AgCN}} \text{C}_{2}\text{H}_{5}\text{NC} \text{(Ethyl isocyanide, A)} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH}} \text{CH}_{3}\text{-CH}_{2}\text{-NH}\text{-CH}_{3} \text{(Ethyl methyl amine, B)} \)
In simple words: This question identifies intermediates (A) and final products (B) in various multi-step reactions involving alkyl halides, nitriles, nitro compounds, and isocyanides, leading to different amines.

🎯 Exam Tip: Pay close attention to the reagents used (e.g., KCN vs. AgCN, Na/C2H5OH vs. Sn/HCl) as they dictate the type of intermediate and final product formed, especially in carbon chain elongation and reduction reactions.

 

Use your brain power! (Textbook Page No 286)

 

Question 1. Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine
\( \text{CH}_{3}\text{-Cl} + \text{KCN} \xrightarrow{\text{alc. } \Delta} \text{CH}_{3}\text{-CN} + \text{KCl} \)
\( \text{CH}_{3}\text{CN} + 4\text{[H]} \xrightarrow{\text{Na/C}_{2}\text{H}_{5}\text{OH}} \text{CH}_{3}\text{-CH}_{2}\text{-NH}_{2} \)
In simple words: Methyl chloride is converted to methyl cyanide using KCN, which is then reduced to ethylamine using a reducing agent like Na/C2H5OH, adding one carbon atom to the chain.

🎯 Exam Tip: This two-step process (nucleophilic substitution followed by reduction) is a fundamental method for increasing the carbon chain length and forming primary amines.

 

(2) Benzamide to aniline
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम बेंजामाइड से एनिलीन बनाने की प्रक्रिया को दर्शाता है, जिसे हॉफमैन ब्रोमामाइड डिग्रेडेशन रिएक्शन कहते हैं। बेंजामाइड की ब्रोमीन (Br2) और पोटेशियम हाइड्रोक्साइड (4KOH) के साथ अभिक्रिया से एनिलीन बनता है, साथ में पोटेशियम कार्बोनेट और पोटेशियम ब्रोमाइड उपोत्पाद के रूप में प्राप्त होते हैं।
In simple words: Benzamide is converted to aniline through the Hofmann bromamide degradation reaction, which involves treating the amide with bromine and potassium hydroxide, resulting in a primary amine with one less carbon atom.

🎯 Exam Tip: The Hofmann bromamide degradation reaction is a crucial 'step-down' reaction, as it reduces the number of carbon atoms in the chain by one while converting an amide to a primary amine.

 

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम 1,4-डाइक्लोरोब्यूटेन से हेक्सेन-1,6-डाइऐमीन बनाने की प्रक्रिया को दर्शाता है। 1,4-डाइक्लोरोब्यूटेन की पोटेशियम साइनाइड से अभिक्रिया से एडिपोनिट्राइल बनता है, जिसे निकल की उपस्थिति में हाइड्रोजन के साथ अपचयित करने पर हेक्सेन-1,6-डाइऐमीन प्राप्त होता है।
In simple words: 1,4-Dichlorobutane reacts with KCN to form adiponitrile, which is then reduced by hydrogen in the presence of nickel to yield hexane-1,6-diamine, effectively increasing the carbon chain length and adding two amine groups.

🎯 Exam Tip: This reaction is a 'step-up' synthesis involving two carbon atoms for each -CN group, and the reduction of dinitriles is a common method for preparing diamines.

 

(4) Benzamide to benzylamine
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम बेंजामाइड से बेंजाइलैमीन बनाने की प्रक्रिया को दर्शाता है। बेंजामाइड का लिथियम एल्यूमीनियम हाइड्राइड (LiAlH4) और ईथर का उपयोग करके अपचयन करने पर बेंजाइलैमीन प्राप्त होता है, साथ में पानी भी उपोत्पाद के रूप में बनता है।
In simple words: Benzamide is reduced to benzylamine using lithium aluminum hydride (LiAlH4) in ether, converting the carbonyl group into a methylene group and retaining the carbon count.

🎯 Exam Tip: LiAlH4 is a powerful reducing agent capable of converting amides directly into amines without loss of carbon atoms, unlike the Hofmann bromamide degradation.

 

Use your brain power! (Textbook Page No 287)

 

Question 1. Arrange the following :
(1) In decreasing order of the boiling point C2H5 - OH, C2H5 - NH2, (CH3)2 NH
(2) In increasing order of solubility in water: C2H5 - NH2, C3H7 - NH2, C6H5 - NH2
Answer:
(1) Decreasing order of the boiling point : C2H5 - OH > C2H5 - NH2 > (CH3)2 NH
(2) Increasing order of solubility in water : C6H5NH2 < C3H7 - NH2 < C2H5 - NH2
In simple words: Boiling points of compounds depend on intermolecular forces, with alcohols generally higher than amines due to stronger hydrogen bonding, and primary amines having stronger bonding than secondary amines. Solubility in water decreases with increasing nonpolar hydrocarbon chain length and the presence of bulky groups hindering hydrogen bonding.

🎯 Exam Tip: Remember that alcohols typically have higher boiling points than amines of similar molar mass due to the stronger electronegativity of oxygen, leading to more polarized O-H bonds and stronger hydrogen bonding. For solubility, shorter chain amines and alcohols are more soluble due to effective hydrogen bonding with water.

 

Use your brain power! (Textbook Page No 288)

 

Question 1. Refer to pk♭ values and answer which compound from the following pairs is the stronger base?
(1) CH3 - NH2 and (CH3)2 NH
(2) (C2H5)2 NH and (C2H5)3 N
(3) NH3 and (CH3)2 CH - NH2
Answer:
(1) CH3 -NH2 and (CH3)2 NH
(CH3)2 NH is a stronger base
(2) (C2H5)2 NH and (C2H5)3 N
(C2H5)2 NH is a stronger base
(3) NH3 and (CH3)2 CHNH2
(CH3)2 CHNH2 is a stronger base
In simple words: The basicity of amines is influenced by the electron-donating effect of alkyl groups, which stabilize the conjugate acid. Secondary amines are generally stronger bases than primary or tertiary amines in aqueous solution due to a balance of inductive effects and solvation.

🎯 Exam Tip: For basic strength in aqueous solution, remember the order for aliphatic amines: secondary > primary > tertiary (for methyl groups) or secondary > tertiary > primary (for ethyl groups), generally stronger than ammonia, due to a combination of inductive effect, solvation, and steric hindrance.

 

Use your brain power! (Textbook Page No 290)

 

Question 1. Arrange the following amines in decreasing order of their basic strength :
NH3, CH3 - NH2, (CH3)2 NH, C6H5NH2
Answer: Decreasing order of basic strength :
(CH3)2NH > CH3 -NH2 > NH3 > C6H5NH2
In simple words: The basic strength of amines is primarily determined by the availability of the lone pair on nitrogen. Alkyl groups increase electron density on nitrogen making it more basic, while aryl groups (like in aniline) withdraw electron density through resonance, making them less basic than ammonia.

🎯 Exam Tip: Remember that aniline (C6H5NH2) is significantly less basic than aliphatic amines and even ammonia due to the resonance stabilization of the lone pair on nitrogen with the benzene ring.

 

Use your brain power! (Textbook Page No 291)

 

Question 1.
(1) C2H5NH2 + C2H5 - I \( \xrightarrow{\text{excess } \Delta} \)?
(2) (C2H5)2NH + CH3 - I \( \xrightarrow{\text{excess } \Delta} \)?
(3) C6H5NH2 + CH3 - I \( \xrightarrow{\text{excess } \Delta} \)?
Answer:
(1) \( \text{C}_{2}\text{H}_{5}\text{NH}_{2} + \text{excess C}_{2}\text{H}_{5}\text{-I} \xrightarrow{\Delta} (\text{C}_{2}\text{H}_{5})_{4}\text{N}^{+}\text{I}^{-} \)
(2) \( (\text{C}_{2}\text{H}_{5})_{2}\text{NH} + \text{excess CH}_{3}\text{-I} \xrightarrow{\Delta} (\text{C}_{2}\text{H}_{5})_{2}\text{N}(\text{CH}_{3})_{2}^{+}\text{I}^{-} \)
(3) \( \text{C}_{6}\text{H}_{5}\text{NH}_{2} + \text{excess CH}_{3}\text{-I} \xrightarrow{\Delta} \text{C}_{6}\text{H}_{5}\text{N}(\text{CH}_{3})_{3}^{+}\text{I}^{-} \)
In simple words: When a primary or secondary amine reacts with an excess of alkyl iodide, it undergoes exhaustive alkylation to form a quaternary ammonium salt, as all available hydrogens on the nitrogen are replaced by alkyl groups.

🎯 Exam Tip: The "excess" keyword in alkylation reactions indicates that the reaction proceeds to the highest possible substitution, resulting in a quaternary ammonium salt rather than a primary, secondary, or tertiary amine.

 

Use your brain power! (Textbook Page No 291)

 

Question 1. Complete the following reaction :
\( \text{CH}_{3}\text{-CH}_{2}\text{-N(CH}_{3})_{3}\text{I}^{-} \xrightarrow{\text{Moist Ag}_{2}\text{O}} ? \)
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह हॉफमैन एलिमिनेशन रिएक्शन को दर्शाता है। एथिलट्राइमिथाइल अमोनियम आयोडाइड की नम सिल्वर ऑक्साइड (Ag2O/H2O) से अभिक्रिया कराने पर एथिलट्राइमिथाइल अमोनियम हाइड्रॉक्साइड बनता है। इसे गर्म करने पर एथिलीन, ट्राइमिथाइल एमीन और पानी बनते हैं।
In simple words: The quaternary ammonium iodide reacts with moist silver oxide to form a strong base (quaternary ammonium hydroxide), which upon heating undergoes Hofmann elimination to yield an alkene, a tertiary amine, and water.

🎯 Exam Tip: Hofmann elimination, where a quaternary ammonium hydroxide is heated, is a predictable reaction leading to the least substituted alkene, a tertiary amine, and water. This is distinct from other elimination reactions.

 

Use your brain power! (Textbook Page No 292)

 

(1) CH3-NH2 + Ph – CO – Cl→ ?
Answer:
(1) \( \text{CH}_{3}\text{-NH}_{2} + \text{Ph-CO-Cl} \xrightarrow{\text{Pyridine } \Delta} \text{CH}_{3}\text{-NH-CO-Ph} \)
(2) \( (\text{CH}_{3})_{3}\text{N} + \text{Ph-CO-Cl} \rightarrow \text{No reaction} \)
In simple words: Primary amines react with acid chlorides to form N-substituted amides, while tertiary amines do not react due to the absence of a hydrogen atom on the nitrogen.

🎯 Exam Tip: Acylation reactions (like with acid chlorides) require a hydrogen atom attached to the nitrogen for the reaction to occur, which is why tertiary amines, lacking such a hydrogen, do not react.

 

Use your brain power! (Textbook Page No 292)

 

Question 1. Write the carbylamine reaction by using aniline as starting material.
Answer:
\( \text{C}_{6}\text{H}_{5}\text{-NH}_{2} + \text{CHCl}_{3} + 3\text{KOH} \xrightarrow{\Delta} \text{C}_{6}\text{H}_{5}\text{NC} + 3\text{HCl} + 3\text{H}_{2}\text{O} \)
In simple words: Aniline, a primary aromatic amine, reacts with chloroform and alcoholic potassium hydroxide to form phenyl isocyanide, which has a characteristic unpleasant smell, serving as a test for primary amines.

🎯 Exam Tip: The carbylamine test (or isocyanide test) is a specific and highly sensitive test for primary amines (both aliphatic and aromatic) due to the formation of the foul-smelling isocyanide product.

 

Can you tell? (Textbook Page No 292)

 

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H - O - N = O
(2) Nitrous acid cannot be stored in bottle.
In simple words: Nitrous acid (\( \text{HNO}_{2} \)) is unstable and cannot be stored, so it must be generated in situ for reactions.

🎯 Exam Tip: Recognize that unstable reagents like nitrous acid are often generated in-situ (e.g., from NaNO2 and HCl) immediately before use in reactions like diazotization.

 

Use your brain power! (Textbook Page No 294)

 

Question 1. How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg's test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र प्राथमिक एमीन (मिथाइल एमीन) की बेंजीन सल्फोनील क्लोराइड से अभिक्रिया को दर्शाता है, जिसमें N-मिथाइल बेंजीन सल्फोनामाइड बनता है। यह उत्पाद क्षारीय माध्यम में घुलनशील होता है, जो प्राथमिक एमीन की पहचान है।
(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N - dimethylbenzene sulphonamide.
In simple words: The Hinsberg test distinguishes primary, secondary, and tertiary amines: primary amines form a sulphonamide soluble in KOH; secondary amines form a sulphonamide insoluble in KOH; tertiary amines do not react with the reagent.

🎯 Exam Tip: The key to the Hinsberg test is the solubility of the sulphonamide product in aqueous KOH. Only sulphonamides from primary amines have an acidic hydrogen and are soluble in alkali.

 

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्रमशः द्वितीयक एमीन (डाइमिथाइल एमीन) की बेंजीन सल्फोनील क्लोराइड से अभिक्रिया और तृतीयक एमीन (ट्राइमिथाइल एमीन) की अभिक्रिया न होने को दर्शाता है। द्वितीयक एमीन से बना N,N-डाइमिथाइल बेंजीन सल्फोनामाइड क्षारीय माध्यम में अघुलनशील होता है, जबकि तृतीयक एमीन कोई अभिक्रिया नहीं करता।
(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH
In simple words: The Hinsberg test distinguishes primary, secondary, and tertiary amines: primary amines form a sulphonamide soluble in KOH; secondary amines form a sulphonamide insoluble in KOH; tertiary amines do not react with the reagent.

🎯 Exam Tip: The key to the Hinsberg test is the solubility of the sulphonamide product in aqueous KOH. Only sulphonamides from primary amines have an acidic hydrogen and are soluble in alkali.

 

Problem 13.1: (Textbook Page No 295)

 

Question 1. Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution: NH2 - group in aniline is highly ring activating and o - /p - directing due to involvement of the lone pair of electrons on 'N' in resonance with the ring. As a result, on reaction with Br2 it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of - NH2 group. This is done by acetylation of aniline. The lone pair of 'N' in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एसिटानिलाइड के दो अनुनादी संरचनाओं को दर्शाता है। बाईं ओर की संरचना में, नाइट्रोजन परमाणु पर एक लोन पेयर है जो कार्बन-ऑक्सीजन डबल बॉन्ड के साथ अनुनाद में है, जबकि दाहिनी ओर की संरचना में नाइट्रोजन पर सकारात्मक चार्ज और कार्बन-नाइट्रोजन डबल बॉन्ड है।
Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original - NH2 group is regenerated. The protection of - NH2 group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.
In simple words: To selectively prepare p-bromoaniline from aniline, the highly activating amine group is first protected by acetylation, which reduces its activating effect and allows for controlled monobromination at the para position, followed by deacetylation to restore the amine group.

🎯 Exam Tip: The protection and deprotection strategy (acetylation and subsequent hydrolysis) is a crucial technique in organic synthesis to control regioselectivity and prevent over-reaction of highly reactive functional groups like the -NH2 group on an aromatic ring.

 

ℹ️ चित्र व्याख्या (Diagram Explanation): यह अनुक्रम एनिलीन से p-ब्रोमोएनिलीन बनाने की प्रक्रिया को दर्शाता है। पहले एनिलीन की एसिटाइल क्लोराइड से अभिक्रिया कर एसिटानिलाइड बनाते हैं, फिर इसका ब्रोमीन/एसिटिक एसिड से ब्रोमिनेशन करने पर p-ब्रोमोएसिटानिलाइड बनता है, और अंत में इसके जल-अपघटन से p-ब्रोमोएनिलीन प्राप्त होता है।
In simple words: To selectively prepare p-bromoaniline from aniline, the highly activating amine group is first protected by acetylation, which reduces its activating effect and allows for controlled monobromination at the para position, followed by deacetylation to restore the amine group.

🎯 Exam Tip: The protection and deprotection strategy (acetylation and subsequent hydrolysis) is a crucial technique in organic synthesis to control regioselectivity and prevent over-reaction of highly reactive functional groups like the -NH2 group on an aromatic ring.

 

Use your brain power! (Textbook Page No 296)

 

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel - Craft's reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr's reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एनिलीन के एक लुईस एसिड (\( \text{AlCl}_{3} \)) के साथ अभिक्रिया करके एक नमक बनाने को दर्शाता है, जहाँ एनिलीन का नाइट्रोजन परमाणु \( \text{AlCl}_{3} \) के साथ बंधन बनाता है, जिससे नाइट्रोजन पर सकारात्मक चार्ज आ जाता है।
In simple words: Aniline reacts with Lewis acids like aluminum chloride to form an anilinium salt, which converts the activating amine group into a strong deactivating group, thus preventing Friedel-Crafts reactions.

🎯 Exam Tip: The formation of a positively charged anilinium ion with Lewis acids (like AlCl3) is critical to understanding why aniline does not undergo Friedel-Crafts reactions, as the deactivating effect prevents electrophilic substitution.

 

Can you tell? (Textbook Page No 294)

 

(1) Do tertiary amines have 'H' bonded to 'N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक तृतीयक एमीन की सामान्य संरचना को दर्शाता है, जिसमें नाइट्रोजन परमाणु तीन अल्काइल (R) समूहों से जुड़ा होता है और उस पर कोई हाइड्रोजन परमाणु नहीं होता है।
(1) Tertiary amines R do not have 'H' bonded to 'N'.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.
In simple words: Tertiary amines lack a hydrogen atom bonded directly to the nitrogen, which is why they cannot react with reagents like benzene sulfonyl chloride that require the removal of an acidic hydrogen from the nitrogen.

🎯 Exam Tip: The absence of a hydrogen atom on the nitrogen is the defining structural feature that prevents tertiary amines from undergoing reactions that involve N-H bond cleavage, such as acylation or Hinsberg test (sulfonylation).

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