Previous Year NEET Question- Hydrocarbons L-4

Question: The compound that will react most readily with gaseous bromine has the formula#

A) $\mathrm{C}_3 \mathrm{H}_6$

B) $\mathrm{C}_2 \mathrm{H}_2$

C) $\mathrm{C}4 \mathrm{H}{10}$

D) $\mathrm{C}_2 \mathrm{H}_4$

Answer: $\mathrm{C}4 \mathrm{H}{10}$#

Sol:#

In gaseous state, $\mathrm{Br}_2$ can undergo homolytic cleavage to form free radicals and saturated hydrocarbons are more prone to free radical substitutions. As $\mathrm{C}4 \mathrm{H}{10}$ reacts most readily with gaseous bromine via free radical mechanism as shown below: $$ \mathrm{C}4 \mathrm{H}{10}+\mathrm{Br}_2 \longrightarrow \mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}+\mathrm{HBr} $$

Therefore, option (c) is correct.