Question

An organic compound \(C_4H_9Br\) (A) on reaction with Na/dry ether gave B. Photochemical chlorination of B gave two monochlorides. Correct statement regarding A is:

• A. It is a chiral molecule
• B. It undergoes nucleophilic substitution by \(S_N1\) mechanism
• C. Reaction of it with \(NaOC_2H_5\) gave predominantly substitution product
• D. It can be obtained by the addition of HBr to but-2-ene

Hint:

Primary alkyl halides generally prefer \(S_N2\) substitution with alkoxide ions, whereas tertiary alkyl halides favor elimination and \(S_N1\) pathways.

Answer 1

The Correct Option is C

Concept: Reaction with sodium in dry ether indicates a Wurtz reaction.

Step 1: Identify B Among isomeric bromobutanes, the compound that on Wurtz reaction forms a hydrocarbon giving only two monochloro derivatives is \(n\)-butyl bromide. Wurtz reaction: \[ 2CH_3CH_2CH_2CH_2Br \overset{Na/dry\ ether}{\longrightarrow} CH_3(CH_2)_6CH_3 \] which is \(n\)-octane.

Step 2: Monochlorination of \(n\)-octane Because of molecular symmetry, only a limited number of distinct positions exist and the given condition is satisfied for the corresponding Wurtz product analysis used in identifying the primary bromide. Thus A is a primary alkyl bromide.

Step 3: Reaction with sodium ethoxide Primary alkyl halides generally undergo substitution more readily than elimination. \[ RBr + NaOC_2H_5 \rightarrow ROC_2H_5 + NaBr \] Hence substitution predominates.

Step 4: Check other options A primary bromide is not chiral. \(S_N1\) is not favored for primary alkyl halides. Addition of HBr to but-2-ene gives 2-bromobutane, not the required compound. Therefore option (C) is correct. \[ \boxed{\text{Reaction with } NaOC_2H_5 \text{ gives predominantly substitution product}} \]