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Chapter 7: Problem 54

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. A compound (A) reacts with phenyl hydrazine to give a crystalline yellow precipitate. When (A) is added to a solution of rosaniline-hydrochloride decolourized by \(\mathrm{SO}_{2}\) gas, the solution remains colourless. (A) is reduced with LiAlH to give (B). (B) is a (a) phenol (b) primary alcohol (c) aromatic polyhydric alcohol (d) secondary alcohol

Short Answer

Expert verified
Answer: (d) secondary alcohol

Step by step solution

01

Reaction of compound (A) with phenyl hydrazine

It is given that compound (A) reacts with phenyl hydrazine to form a crystalline yellow precipitate. This reaction is characteristic of carbonyl groups in aldehydes and ketones. The formation of phenylhydrazone (yellow precipitate) indicates that compound (A) contains a carbonyl group.
02

Reaction of compound (A) with rosaniline-hydrochloride solution decolorized by SO2 gas

The exercise tells us that when compound (A) is added to the decolorized rosaniline-hydrochloride solution (decolorized by SO2 gas), the solution remains colorless. Such behavior is exhibited by ketones that do not have an alpha-hydrogen. Aldehydes with alpha-hydrogens would cause the solution to regain its color due to the reaction between SO2, aldehyde, and rosaniline. Hence, we can conclude that compound (A) is a ketone with no alpha-hydrogens.
03

Reduction of compound (A) by LiAlH4 to give compound (B)

It is mentioned that compound (A) is reduced with LiAlH4 to give compound (B). LiAlH4 is a strong reducing agent that is known to reduce carbonyl groups in aldehydes and ketones, converting them to alcohols. Since compound (A) is a ketone with no alpha-hydrogens, its reduction would give a secondary alcohol. Hence, compound (B) is a secondary alcohol. Thus, the correct answer to this multiple-choice question is: (d) secondary alcohol

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