The compound that will react most readily with gaseous bromine has the formula
A.
$${C_3}{H_6}$$
B.
$${C_2}{H_2}$$
C.
$${C_4}{H_{10}}$$
D.
$${C_2}{H_4}$$
Answer :
$${C_4}{H_{10}}$$
Solution :
In gaseous state, $$B{r_2}$$ forms free radicals and saturated hydrocarbons are more prone to have free radical substitutions. As $${C_4}{H_{10}}$$ reacts most readily with gaseous bromine via free radical mechanism as shown below :
$${C_4}{H_{10}} + B{r_2} \to {C_4}{H_9}Br + HBr$$
Therefore, option (C) is correct.
Releted MCQ Question on Organic Chemistry >> Hydrocarbons (Alkane, Alkene and Alkyne)