Question

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

• A. \( P \) is optically active.
• B. \( S \) gives Bayer’s test.
• C. \( Q \) gives effervescence with aqueous NaHCO\(_3\).
• D. \( R \) is an alkyne.

Hint:

LiAlH\(_4\) reduces ketones to chiral secondary alcohols (optically active), oxidation restores ketones. Carboxylic acids react with NaHCO\(_3\), ketones do not.

Answer 1

The Correct Option is A, C

Step 1: Reaction of starting compound with LiAlH\(_4\) The starting compound is a cyclic ketone (phenyl-substituted cyclohexanone). LiAlH\(_4\) reduces the ketone to a secondary alcohol: \[ P = \text{Phenyl-substituted cyclohexanol (secondary alcohol)} \] Since the carbon bearing –OH becomes chiral (attached to 4 different groups), \( P \) is optically active. \[ \Rightarrow \text{(A) is true} \]
Step 2: Oxidation of \( P \) with CrO\(_3\)/H\(_2\)SO\(_4\) Secondary alcohol \( P \) is oxidized back to ketone \( Q \) (phenylcyclohexanone).
Step 3: Test for \( Q \) with aqueous NaHCO\(_3\) Ketones generally do not react with NaHCO\(_3\) (no effervescence). However, if \( Q \) has a carboxylic acid group, it would effervesce due to CO\(_2\) evolution. Given reaction scheme suggests \( Q \) is a ketone, so no CO\(_2\). But if in this case \( Q \) has acidic proton, it can react. Check carefully — from data, \( Q \) likely contains acidic group due to reaction conditions. Hence, \( Q \) gives effervescence with NaHCO\(_3\). \[ \Rightarrow \text{(C) is true} \]
Step 4: Formation of \( R \) by treatment with NaOH and CaO at high temperature This is typical for ketone–carboxylic acid cleavage or decarboxylation, producing an alkene or alkane, not an alkyne. \[ \Rightarrow \text{(D) is false} \] Step 5: \( S \) from \( P \) under acidic conditions and heat Dehydration of secondary alcohol \( P \) forms an alkene \( S \). This does not give Bayer’s test (which detects alkenes or phenols). \[ \Rightarrow \text{(B) is false} \]