Question

Hydrocarbon (A) on monobrination forms an alkyl bromide which by Wurtz reaction is converted to a gaseous hydrocarbon containing less than four carbon atoms. (A) is:

• A. $C{H}_3-C{H}_3$
• B. $C{H}_2=C{H}_2$
• C. $CH\equiv CH$
• D. $C{H}_4$

Answer 1

The correct option is D $C{H}_4$

The given reaction takes place as follows:
$\underset{\left(A\right)}{C{H}_4}\underset{\text{Step 1}}{\overset{B{r}_2/hv}{\to }}C{H}_{3}Br\underset{\text{Wurtz reaction (Step 2)}}{\overset{\text{Na/dry ether}}{\to }}\underset{\left(A\right)}{C{H}_3-C{H}_3}$
Step I: Alkyl halide is formed by free radical halogenation of alkanes in the presence of UV-light.
Step II: The formed alkyl halide reacts with sodium in presence of dry ether to form alkane containing double number of carbon atoms present in the alkyl halide.
This reaction is known as Wurtz reaction.
From the above mechanism, it is concluded that option (d) is correct as in all other cases the hydrocarbon formed in step 2 will contain more than four carbon atoms.