Reaction Mechanisms

In an SN2 reaction, the transition state involves the central carbon atom being bonded to five different entities: the nucleophile, the leaving group, and the two substituents that are attached to the carbon. This results in the carbon being tetra coordinated in the transition state, as it is simultaneously interacting with four groups in a partially bonded manner. Thus, the correct answer is (3) Tetra coordinated.

The reaction sequence starts with aniline reacting with acetic anhydride in the presence of pyridine.
Acetic anhydride is an acetylating agent, and pyridine acts as a base to facilitate the reaction by removing the byproduct.
This reaction will acetylate the amino group ( ) of aniline to form an amide ( ), which is acetanilide.
The amino group is a strong activating and ortho/para-directing group in electrophilic aromatic substitution.
By converting it to an amide, we make it moderately activating and still ortho/para-directing, but less so.
This step is done to control the subsequent bromination.
Thus, X is acetanilide.
The structure of X (acetanilide) is: In the second step, acetanilide reacts with in acetic acid ( ).
Bromine is an electrophile, and acetanilide will undergo electrophilic aromatic substitution.
The amide group ( ) is an ortho/para-directing group.
However, due to steric hindrance at the ortho positions (adjacent to the bulky amide group) and the slightly higher activation of the para position, the major product will be the para-substituted bromoacetanilide.
The structure of Y (para-bromoacetanilide) is: Comparing these structures with the options, option (D) correctly shows X as acetanilide and Y as para-bromoacetanilide.

The starting material is a benzenediazonium salt, .
Let's assume X is Cl or HSO typically.
Step 1: Product (referred to as X in options, let's call it P1).
Reaction of diazonium salts with ethanol (C H OH) can lead to reduction of the diazonium group to -H (forming benzene) or substitution by -OC H (forming phenetole, an ether).
The reduction to benzene is favored, especially if a reducing agent like H PO is also present or if the ethanol acts as the reducing agent (it gets oxidized to ethanal).
So, P1 (X in the options) is Benzene ( ).
Step 2: P1 ( ) Y This reaction of benzene with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous AlCl (often with CuCl as a co-catalyst) is the Gattermann-Koch reaction.
It is used to introduce an aldehyde group (-CHO) onto the benzene ring.
(The HCl is formally regenerated or used in forming the electrophile [HCO] ).
Product Y is Benzaldehyde ( ).
So, X (P1) is Benzene and Y is Benzaldehyde.
This corresponds to the structures shown in option (1).
Let's check other possibilities if P1 were Phenol: If reacts with warm water or dilute acid, it forms phenol ( ).
Ethanol can sometimes lead to ether formation too.
If P1 was Phenol: Phenol Y This reaction (Gattermann-Koch) on phenol is not standard for simple formylation.
Phenols are highly activated.
Formylation of phenol usually occurs via Reimer-Tiemann (CHCl /NaOH) or Gattermann aldehyde synthesis (using HCN/HCl then hydrolysis, not CO/HCl).
If Gattermann-Koch were applied to phenol, it would likely give p-hydroxybenzaldehyde or a mixture.
Options (3) and (4) suggest P1 is Phenol.
Option (3) makes Y Salicylic acid (o-Hydroxybenzoic acid).
Option (4) makes Y Salicylaldehyde (o-Hydroxybenzaldehyde).
Salicylic acid is COOH, Gattermann-Koch gives CHO.
Salicylaldehyde is formylation at ortho.
The most standard interpretations are: - Diazonium salt + Ethanol Benzene (reduction).
- Benzene + CO/HCl/AlCl (Gattermann-Koch) Benzaldehyde.
This makes X=Benzene, Y=Benzaldehyde.

Step 1: Diazotization of Aniline Aniline ( ) reacts with sodium nitrite ( ) and hydrochloric acid ( ) at low temperatures (273 - 278 K), forming benzene diazonium chloride ( ). Step 2: Sandmeyer Reaction - Cyanation Benzene diazonium chloride reacts with copper(I) cyanide ( ) and potassium cyanide ( ), replacing the diazonium group with cyano (-CN) group, forming benzonitrile ( ). Step 3: Hydrolysis to Benzoic Acid Benzonitrile undergoes acidic hydrolysis using (aqueous acid), converting the cyano (-CN) group into a carboxyl (-COOH) group, forming benzoic acid ( ). Conclusion Thus, the correct answer is:

Benzyl amine can be synthesized via the reaction in option (4), which involves the reduction of the corresponding nitrile to the amine using LiAlH₄.

Step 1: Diazotization of Aniline Aniline ( ) reacts with sodium nitrite ( ) and hydrochloric acid ( ) at low temperatures (273 - 278 K), forming benzene diazonium chloride ( ). Step 2: Sandmeyer Reaction - Cyanation Benzene diazonium chloride reacts with copper(I) cyanide ( ) and potassium cyanide ( ), replacing the diazonium group with cyano (-CN) group, forming benzonitrile ( ). Step 3: Hydrolysis to Benzoic Acid Benzonitrile undergoes acidic hydrolysis using (aqueous acid), converting the cyano (-CN) group into a carboxyl (-COOH) group, forming benzoic acid ( ). Conclusion Thus, the correct answer is:

In the SN1 mechanism (reaction I), the C--Cl bond breaks in the slow step, forming a carbocation. In reaction II (aromatic nucleophilic substitution), the presence of strong electron-withdrawing groups facilitates faster bond cleavage.

Statement-I is correct: CH NH is more basic than NH (due to +I effect), while C H NH is less basic due to resonance.
However, in aqueous medium, basicity does not follow the order given in II due to solvation effects; hence Statement-II is incorrect.

The starting material is a benzenediazonium salt, .
Let's assume X is Cl or HSO typically.
Step 1: Product (referred to as X in options, let's call it P1).
Reaction of diazonium salts with ethanol (C H OH) can lead to reduction of the diazonium group to -H (forming benzene) or substitution by -OC H (forming phenetole, an ether).
The reduction to benzene is favored, especially if a reducing agent like H PO is also present or if the ethanol acts as the reducing agent (it gets oxidized to ethanal).
So, P1 (X in the options) is Benzene ( ).
Step 2: P1 ( ) Y This reaction of benzene with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of anhydrous AlCl (often with CuCl as a co-catalyst) is the Gattermann-Koch reaction.
It is used to introduce an aldehyde group (-CHO) onto the benzene ring.
(The HCl is formally regenerated or used in forming the electrophile [HCO] ).
Product Y is Benzaldehyde ( ).
So, X (P1) is Benzene and Y is Benzaldehyde.
This corresponds to the structures shown in option (1).
Let's check other possibilities if P1 were Phenol: If reacts with warm water or dilute acid, it forms phenol ( ).
Ethanol can sometimes lead to ether formation too.
If P1 was Phenol: Phenol Y This reaction (Gattermann-Koch) on phenol is not standard for simple formylation.
Phenols are highly activated.
Formylation of phenol usually occurs via Reimer-Tiemann (CHCl /NaOH) or Gattermann aldehyde synthesis (using HCN/HCl then hydrolysis, not CO/HCl).
If Gattermann-Koch were applied to phenol, it would likely give p-hydroxybenzaldehyde or a mixture.
Options (3) and (4) suggest P1 is Phenol.
Option (3) makes Y Salicylic acid (o-Hydroxybenzoic acid).
Option (4) makes Y Salicylaldehyde (o-Hydroxybenzaldehyde).
Salicylic acid is COOH, Gattermann-Koch gives CHO.
Salicylaldehyde is formylation at ortho.
The most standard interpretations are: - Diazonium salt + Ethanol Benzene (reduction).
- Benzene + CO/HCl/AlCl (Gattermann-Koch) Benzaldehyde.
This makes X=Benzene, Y=Benzaldehyde.