Question 1

Which statements are True? A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of Br_2 are consumed per mole of an amide B. Hoffmann bromamide reaction is not given by alkyl amides. C. P

Accepted Answer

The balanced chemical equation for Hoffmann bromamide degradation is: R-CONH _2 + Br _2 + 4 NaOH R-NH _2 + Na _2 CO _3 + 2 NaBr + 2 H _2 O Statement A is false because 1 mole of Br _2 and 4 moles of NaOH are consumed per mole of amide. Statement B is false because primary alkyl amides readily undergo this reaction. Statement C is true as the primary product of this reaction is a primary amine. Statement D is false because the reaction is strictly given by primary amides ( R-CONH _2), not seconda

Question 2

Consider the following organic reaction sequence. Choose the final product (X) from the following (consider the major product in all intermediate reactions)

Accepted Answer

The starting material is propanoic acid (CH_3CH_2COOH). In step (i), propanoic acid reacts with NH_3 upon heating to form propanamide. CH_3CH_2COOH NH_3, CH_3CH_2CONH_2 In step (ii), propanamide undergoes Hoffmann bromamide degradation with NaOH and Br_2 to form ethylamine. CH_3CH_2CONH_2 NaOH, Br_2 CH_3CH_2NH_2 In step (iii), ethylamine reacts with nitrous acid (HNO_2) to form an unstable diazonium salt, which hydrolyzes to give ethanol. CH_3CH_2NH_2 HNO_2, H_2O CH_3CH_2OH In step (iv), ethanol

Question 3

The number of compounds from the following which can undergo reaction with Br_2/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel pht

Accepted Answer

The reaction of an amide with Br _2/ KOH (alcoholic) is the Hoffmann bromamide degradation reaction. This reaction is specific to primary amides (R- CONH _2) and produces primary amines (R- NH _2) with one less carbon atom. Among the given compounds, the primary amides are: 1. Benzamide ( C _6 H _5 CONH _2) 2. 2-Phenylacetamide ( C _6 H _5 CH _2 CONH _2) 3. Acetamide ( CH _3 CONH _2) 4. Cyclohexanecarboxamide ( C _6 H _ 11 CONH _2) The other two compounds (N-ethylcyclohexanecarboxamide and N-met

Question 4

Consider the following sequence of reactions. The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol^ -1 H:1, C:12, N:14)

Accepted Answer

The given starting material is diazoaminobenzene (C_6H_5-N=N-NH-C_6H_5). When diazoaminobenzene is treated with an acid catalyst (HCl) and excess aniline, and warmed to 40-45^ C , it undergoes a rearrangement reaction (or dissociation followed by coupling) to form p-aminoazobenzene. The reaction sequence is as follows: 1. Diazoaminobenzene reacts with HCl to form benzenediazonium chloride and aniline. 2. The benzenediazonium ion then couples with the excess aniline at the para position to yield

Question 5

Given below are two statements : Statement I: Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine. Statement II: Nitration of aniline with HNO_3/H_2SO_4 a

Accepted Answer

Statement I describes the Hoffmann bromamide degradation reaction. When benzamide (C_6H_5CONH_2) is heated with bromine in an ethanolic solution of sodium hydroxide, it produces aniline (C_6H_5NH_2), not benzylamine. Thus, Statement I is false. Statement II describes the nitration of aniline. Direct nitration of aniline with HNO_3/H_2SO_4 yields a mixture of p-nitroaniline (51\%), m-nitroaniline (47\%), and o-nitroaniline (2\%). The significant formation of the meta derivative is due to the prot

Question 6

Identify the incorrect statements. Choose the correct answer from the options given below:

Accepted Answer

Statement A: In benzylamine (C_6H_5CH_2NH_2), the lone pair of electrons on the nitrogen atom is localized and not involved in resonance. In aniline (C_6H_5NH_2), the lone pair is delocalized over the benzene ring. Hence, benzylamine is a stronger base than aniline. Statement A is correct. Statement B: Gabriel phthalimide synthesis involves the nucleophilic substitution (S_N2) of an alkyl halide by the phthalimide anion. Aryl halides do not undergo nucleophilic substitution easily due to the par

Question 7

The strongest conjugate acid will result from:

Accepted Answer

The strength of a conjugate acid is inversely proportional to the basic strength of its corresponding base. Thus, the weakest base will form the strongest conjugate acid. The basicity of anilines depends on the availability of the lone pair on the nitrogen atom. Electron-withdrawing groups decrease basicity by withdrawing electron density, while electron-donating groups increase basicity. In 4-nitroaniline, the - NO _2 group is a strong electron-withdrawing group due to its -R and -I effects. Th

Question 8

Product C of the following reaction sequence will be

Accepted Answer

Aniline on reaction with bromine water gives 2,4,6-tribromoaniline (A) as the major product due to the strong activating nature of the - NH _2 group. Reaction of (A) with NaNO _2/ HCl at 273-278 K (diazotization) yields 2,4,6-tribromobenzenediazonium chloride (B). Treatment of (B) with HBF _4 forms 2,4,6-tribromobenzenediazonium fluoroborate. Heating this intermediate with NaNO _2 and Cu replaces the diazonium group with a nitro group, forming 2,4,6-tribromonitrobenzene (C). To assign the IUPAC

Question 9

Consider the following reactions sequence When the product (P) is subjected to Carius analysis using AgNO_3, 1.0 g of the product (P) will produce _______ g of the precipitate of AgBr. (Nearest Intege

Accepted Answer

The given reactant is p-nitrotoluene. Step (i): Reduction with Sn/HCl followed by OH^- converts the -NO_2 group to an -NH_2 group, yielding p-toluidine. Step (ii): Reaction with acetic anhydride, (CH_3CO)_2O, protects the amino group by forming N-(4-methylphenyl)acetamide. Step (iii): Bromination with Br_2/AlBr_3 directs the incoming bromine to the ortho position relative to the strongly activating -NHCOCH_3 group (the para position is blocked by the methyl group). This yields 2-bromo-4-methylac

Question 10

Match List - I with List - II. List - I Mixture of Compounds List - II Reagent used to distinguish A. Diethyl amine + Ethyl amine I. Bromine water B. Acetaldehyde + Acetone II. CHCl_3 + KOH, C. Ethano

Accepted Answer

Diethyl amine is a secondary amine and ethyl amine is a primary amine. Primary amines give the carbylamine test with CHCl_3 and KOH on heating, while secondary amines do not. Thus, A matches with II. Acetaldehyde is an aldehyde and acetone is a ketone. Aldehydes reduce ammoniacal silver nitrate (Tollens' reagent) to form a silver mirror, while ketones do not. Thus, B matches with IV. Phenol reacts with neutral FeCl_3 to form a violet-colored complex, whereas ethanol does not react. Thus, C match

Question 11

Consider the three aromatic molecules (P, Q and R) whose structures have been given below : order regarding the reactivity of these compounds with Ph-N NCl^ (-) _ (+) under optimum but slightly acidic

Accepted Answer

The reaction of the given aromatic compounds with benzenediazonium chloride (Ph-N_2^+ Cl^-) is an azo coupling reaction, which is a type of electrophilic aromatic substitution. The electrophile, the diazonium ion (Ph-N_2^+), is a weak electrophile and requires a strongly activated aromatic ring to undergo coupling. The reactivity of the ring depends on its electron density, which is enhanced by the +M (mesomeric) effect of the -NMe_2 group. 1. In molecule P (N,N-dimethylaniline), the -NMe_2 grou

Question 12

Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is \_\_\_\_. Aniline, N-Methylaniline, Methanamine, N, N-Dimethylmethanamine, N-Methyl

Accepted Answer

Hinsberg's reagent (benzenesulfonyl chloride) classifies amines by their reaction products: Primary amines (RNH₂) form alkali-soluble sulfonamides (soluble in NaOH). Secondary amines (R₂NH) form alkali-insoluble sulfonamides (precipitate in NaOH). Tertiary amines (R₃N) do not react. Analyzing the given amines: Aniline (primary), Methanamine (primary), Phenylmethanamine (primary), Allyl amine (primary), Isopropyl amine (primary) → alkali-soluble. N,N-Dimethylmethanamine (tertiary), N,N-Dimethylan

Question 13

Consider the following reactions giving major product. Identify the correct reaction.

Accepted Answer

In reaction (1), Gabriel phthalimide synthesis cannot be used to prepare aniline because aryl halides do not undergo nucleophilic substitution (S_N2) with potassium phthalimide due to partial double bond character of the C-X bond. In reaction (2), the reduction of isocyanide (R-NC) with Sn/HCl yields a secondary amine (R-NHCH_3), not a primary amine (R-CH_2NH_2). In reaction (3), propanamide reacts with Br_2 and KOH (Hoffmann Bromamide Degradation) to give ethanamine. This is a correct reaction.

Question 14

Given below are two statements: Statement I : Griss-Ilosvay test is used for the detection of nitrite ion, which involves the use of sulphanilic acid and -naphthylamine reagent. Statement II : In the

Accepted Answer

The Griss-Ilosvay test for nitrite ion detection involves adding sulphanilic acid and -naphthylamine reagent. Acidified nitrite ions diazotize sulphanilic acid, forming a diazonium salt that undergoes azo coupling with -naphthylamine to produce a characteristic red azo-dye. Both statements accurately describe this test and its mechanism, making both statements true.

Question 15

Given below are two statements: Statement I : The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce : R-CN hydrolyses under acidic medium to pro

Accepted Answer

Statement I: The dipole moment of alkyl cyanides (R-CN) is indeed greater than alkyl isocyanides (R-NC) because the C N bond is more polar than the N^+ C^- bond where formal charges oppose the electronegativity difference. However, the acidic hydrolysis of R-NC produces a primary amine (R-NH_2) and formic acid (HCOOH), not a carboxylic acid (R-COOH). Thus, Statement I is false. Statement II: R-CN on acidic hydrolysis gives R-COOH. Treatment with SOCl_2 gives R-COCl. Addition of NH_3 gives R-CONH

Question 16

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3 g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The

Accepted Answer

Molar mass of aniline (C_6H_5NH_2) = 93 g/mol. Moles of aniline = 9.3/93 = 0.1 mol. From the reaction, theoretical yield of acetanilide = 0.1 mol. Molar mass of acetanilide (C_6H_5NHCOCH_3) = 135 g/mol. Theoretical mass = 0.1 × 135 = 13.5 g. Percentage yield = (11/13.5) × 100 = 81.5%

Question 17

The correct stability order of the following diazonium salts is

Accepted Answer

The stability of benzenediazonium salts is governed by the electronic effects of the substituents present on the benzene ring. The diazonium group -N_2^+ is strongly electron-withdrawing. Electron-donating groups (EDG) stabilize the diazonium cation by dispersing the positive charge through resonance or inductive effects, whereas electron-withdrawing groups (EWG) destabilize it by increasing the positive charge density. In (A), the -OCH_3 group at the para position is a strong electron-donating

Question 18

A student has been given a compound " x " of molecular formula- C _ 6 H _ 7 ~N . ' x ' is sparingly soluble in water. However, on addition of dilute mineral acid, ' x ' becomes soluble in water. ' x '

Accepted Answer

Compound X has molecular formula C_6H_7N, is sparingly soluble in water but soluble in dilute mineral acid → primary aromatic amine (aniline). X with CHCl _3 + KOH (alc.) gives Y with unpleasant smell → carbylamine reaction, confirms primary amine. X with benzenesulphonyl chloride gives Z which is soluble in alkali → Hinsberg test confirms primary amine. Z is N-phenylbenzenesulfonamide. Structure of Z: C _6 H _5 -NH-SO _2 -C _6 H _5

Question 19

Given below are two statements: Statement I: can be synthesized from using simpler reagents in the order i) Acidic KMnO _ 4 , ii) Ammonia, iii) Bromine and alkali Statement II: can be converted into u

Accepted Answer

Statement I: Statement II: Therefore, both statements are true.

Question 20

Consider the following sequence of reactions. Assuming that the reaction proceeds to completion, then 137 mg of 4 -nitrotoluene will produce \_\_\_\_ mg of B. (Given molar mass in g mol ^ -1 H : 1, C

Accepted Answer

Starting with 137 mg of 4-nitrotoluene (C₇H₉NO₂, MW = 137). Step 1: Reduction with Sn/HCl converts NO₂ to NH₂, producing 4-aminotoluene (A). Step 2: Acetylation with (CH₃CO)₂O produces the acetamide intermediate, then bromination with Br₂/AcOH adds bromine to the para position (ortho-para directing due to acetamide group), yielding 4-bromo-N-(4-methylphenyl)acetamide (B) with molecular formula C₉H₁₀BrNO and MW = 228 g/mol. Moles of starting material = 137 mg ÷ 137 g/mol = 1 mmol. Assuming comple

Amines — Previous Year Questions & Solutions