Carboxylic Acids

In this reaction sequence, the starting material is an ester (ethyl acetate, C6H5COOCH3). The first step involves hydrazine (NH2NH2) reacting with the ester, which undergoes reduction to form hydrazone (X). Subsequent treatment with KOH or water hydrolyzes the hydrazone to form the corresponding amine, which is the final product (Y).

The given reaction involves the conversion of the benzene ring to a carboxylated product by bromination followed by nucleophilic substitution. The third step (reduction) leads to the formation of Z, the final product, as an alkyl group is introduced.

Step 1: Analyze Reaction I:
Reactant: Phenol (or substituted phenol). Reagent: . The bond in phenol has partial double bond character due to resonance with the benzene ring. It is very strong and difficult to break. Therefore, phenols do not react with thionyl chloride ( ) or hydrogen halides to form aryl halides under normal conditions (unlike alcohols). Conclusion: Reaction I is not feasible.
Step 2: Analyze Reaction II:
Reactant: Benzyl Alcohol derivative (p-Nitrophenylmethanol). Reagent: (with or without catalyst). The bond here is a purely single bond attached to an carbon. Benzyl alcohols react readily with halogen acids because the intermediate benzylic carbocation is resonance stabilized (though destabilizes it, the reaction is still characteristic of alcohols, not phenols). Alternatively, primary alcohols react via . Standard reaction: . Conclusion: Reaction II is feasible.
Step 3: Final Answer:
I is not feasible, II is feasible. Option (D).

Goal: Convert Ethyl Bromide ( ) to Propanoic Acid ( ). This involves ascent of series (adding one Carbon). Method II (Nitrile Synthesis):
. This is a standard valid method. Method III (Grignard Synthesis):
. This is also a standard valid method to increase carbon chain by one and form carboxylic acid. Method I (Oxidation):
(Ethanoic acid). This does not increase the carbon chain length. It yields a 2-carbon acid, not propanoic (3-carbon). Incorrect. Conclusion:
Methods II and III work.

Electrophilic aromatic substitution on benzoic acid gives meta-bromination with Br /Fe. Side-chain bromination occurs under radical conditions (hv).

1. Benzoic acid (C H COOH) to n-propylbenzene (C H CH CH CH ).
2. Step 1: SOCl converts COOH to COCl (benzoyl chloride).
3. Step 2: (C H ) Cd reacts with COCl to form ketone C H COCH CH .
4. Step 3: Zn-Hg/HCl (Clemmensen reduction) reduces ketone to C H CH CH CH .
5. Other options: (1) LiAlH reduces to alcohol; (3) Grignard gives tertiary alcohol; (4) HCl is ineffective.
6. Thus, the answer is (2) SOCl , (C H ) Cd, Zn-Hg, HCl.

Ethyl bromide + Mg → Grignard reagent. Reaction with CH COCl → 2-butanone. CdCl catalyzes reaction.

1. Cumene (C H CH(CH ) ) undergoes oxidation with KMnO /KOH/ to form benzoic acid (C H COOH, X) via cleavage of isopropyl group.
2. The side product is acetone (CH COCH , Y) after hydrolysis (H , H O).
3. Reaction: C H CH(CH ) C H COOH + CH COCH .
4. Options (1), (2), (4) have incorrect products (e.g., isopropanol or acetophenone).
5. Thus, the answer is (3) COOH, CH COCH .