Question

Which of the following reagent(s) will convert n-propyl bromide to alkane with same number of carbon atoms? \[ \text{I. } Zn/H^+ \] \[ \text{II. } Na/\text{dry ether} \] \[ \text{III. } (i)\;alc.KOH \quad (ii)\;H_2/Pt \]

• A. I only
• B. II, III only
• C. I, III only
• D. II only

Hint:

Wurtz reaction always increases the carbon chain length because two alkyl halide molecules combine together.

Answer 1

The Correct Option is C

Concept: The product required is an alkane having the same number of carbon atoms as n-propyl bromide.

Step 1: Check reagent I.
\[ CH_3CH_2CH_2Br \xrightarrow{Zn/H^+} CH_3CH_2CH_3 \] Carbon number remains unchanged. Hence Statement I is correct.

Step 2: Check reagent II.
Wurtz reaction: \[ 2CH_3CH_2CH_2Br \xrightarrow{Na/dry\ ether} CH_3(CH_2)_4CH_3 \] Hexane is formed. Number of carbon atoms doubles. Hence Statement II is incorrect.

Step 3: Check reagent III.
First elimination: \[ CH_3CH_2CH_2Br \xrightarrow{alc.KOH} CH_3CH=CH_2 \] Then hydrogenation: \[ CH_3CH=CH_2 \xrightarrow{H_2/Pt} CH_3CH_2CH_3 \] Same carbon number is retained. Hence Statement III is correct. Therefore, \[ \boxed{\text{I and III only}} \]