Q: Consider the following four compounds I, II, III, and IV. Choose the correct statement(s).

The correct basic strength order is (A) K b : IV > II > ∣ > III ; IV is strongest base due to SIR effect. III is weakest base due to - M group of three nitro groups present at Ortho and Para positions. Il is stronger than I since III is tertiary and I primary aromatic amine. (B) IV is found to be 40 , 000 times more basic than III . While I and II differ very little in basic strength. (C,D) Due to SIR effect in IV both - NO 2 and - N CH 3 2 will be

Question 1

Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option. List-I List-II (P) (1) (Q) (2) (R) (3) (S) (4) (5)

Accepted Answer

For the reaction in (P), 2-bromo-5-nitrobenzaldehyde oxime reacts with aqueous NaOH. The strong base deprotonates the oxime group. The resulting oximate anion undergoes intramolecular nucleophilic aromatic substitution (S_NAr), attacking the ortho-carbon and displacing the bromide ion to form a 1,2-benzisoxazole intermediate. Since it is derived from an aldoxime, this intermediate has a hydrogen atom at the C3 position. In the presence of NaOH, it undergoes base-catalyzed Kemp elimination (ring

Question 2

In the following reaction sequence, P , Q , S and T are the major products. The correct statement(s) about P , Q , S and T is(are)

Accepted Answer

In the first reaction sequence: Step 1: Aniline reacts with acetic anhydride in the presence of pyridine to undergo acetylation, forming acetanilide. Step 2: Acetanilide is treated with a nitrating mixture (conc. HNO _3 and conc. H _2 SO _4) at 288 K . The - NHCOCH _3 group is moderately activating and ortho/para directing. Due to steric hindrance, the major product P is p-nitroacetanilide. Step 3: P is hydrolyzed with H _3 O ^+ to yield p-nitroaniline. Step 4: p-nitroaniline is diazotized using

Question 3

Passage: Consider the following reaction sequence in which J , K , L and M are the major products. The volume of 1 M aqueous H _2 SO _4 required to completely neutralize the ammonia evolved from 5.72

Accepted Answer

The reaction sequence proceeds as follows: 1. Friedel-Crafts Acylation: Reaction of m-xylene (1,3-dimethylbenzene) with chloroacetyl chloride ( ClCH _2 COCl ) in the presence of anhydrous AlCl _3 occurs at the less sterically hindered activated position (position 4) to give 1-(2,4-dimethylphenyl)-2-chloroethan-1-one. 2. Finkelstein Reaction: Heating with NaI replaces the chloride with iodide, yielding 1-(2,4-dimethylphenyl)-2-iodoethan-1-one. 3. Williamson Ether Synthesis: Reaction with sodium 3

Question 4

The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the

Accepted Answer

Answer option index: 1

Question 5

Match the compounds in List-I with the appropriate observations in List-II and choose the correct option. LIST - I LIST - II (P) (1) Reaction with phenyl diazonium salt gives #ff0 dye. (Q) (2) Reactio

Accepted Answer

(P) Gives FeCl _3 test because of presence of phenolic group and ninhydrin test. (Q) Amino acid obtained gives (+) Ninhydrin test but not phthalein dye test. (R) Reaction with phenyl diazonium salt gives yellow dye. (S) Compound will form hydrazone derivative with aldehydic group of glucose.

Question 6

For the reaction sequence given below, the correct statement(s) is(are)

Accepted Answer

Answer option index: 2

Question 7

List-I contains various reaction sequences and List-II contains different phenolic compounds. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

Accepted Answer

Answer option index: 2

Question 8

The total number of sp 2 hybridised carbon atoms in the major product P (a non-heterocyclic compound) of the following reaction is

Accepted Answer

Lithium aluminium hydride is a strong reducing agent. It reduces cyanide to primary amine with no loss of carbon. Primary amine can undergo nucleophilic addition reaction with acetophenone to produce imine derivative. The carbon with one pi bond are  sp 2  hybridised carbon. All the carbon in benzene ring are  sp 2  hybridised. Hence, totally 28  sp 2  carbon are present in the molecule.

Question 9

Considering the following reaction sequence, the correct option(s) is(are)

Accepted Answer

First step of the reaction involves reduction of nitro group to primary amine this can be done using Sn in presence of HCl or H 2 /Pd, but aromatic nitro compounds on reaction with LiAlH 4 produces azo products.  Second step of the reaction involves diazotization, which involves the use of HNO 2  (or) NaNO 2  in HCl as the reagent ,Para toluene diazonium chloride formed is formed which on hydrolysis produces para-cresol. And Para toluene diazonium chloride

Question 10

Match the compounds in LIST- I with the observations in LIST- II , and choose the correct option.   LIST- I   LIST- II I Aniline P Sodium fusion extract of the compound on boiling with

Accepted Answer

I Since it contains both carbon and nitrogen so its sodium fusion extract with boiling  FeSO 4 , followed by acidification with conc.  H 2 SO 4  gives Prussian blue colour. Fe 2 + + 6 CN - &#8594; Fe CN 6 4 - 3 Fe CN 6 - 4 + 4 Fe 3 + &#8594; xH 2 O Fe 4 Fe CN 6 3 . xH 2 O &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#16

Question 11

Correct option(s) for the following sequence of reactions is(are)

Accepted Answer

Answer option index: 3

Question 12

Consider the following four compounds I, II, III, and IV. Choose the correct statement(s).

Accepted Answer

The correct basic strength order is (A)  K b : IV &#62; II &#62; &#8739; &#62; III ; IV is strongest base due to SIR effect. III is weakest base due to - M group of three nitro groups present at Ortho and Para positions. Il is stronger than I since III is tertiary and I primary aromatic amine. (B)  IV is found to be 40 , 000  times more basic than III . While I  and  II differ very little in basic strength. (C,D) Due to SIR effect in IV both - NO 2 and - N CH 3 2 will be

Question 13

Scheme 1 and 2 describe the conversion of P to Q and R to S , respectively. Scheme 3 describes the synthesis of T from Q and S . The total number of B r atoms in a molecule of T is Scheme 1 : Scheme 2

Accepted Answer

Answer option index: 0

Amines — Previous Year Questions & Solutions