Question

In the given reaction two products are expected.
${CH}_3-CH={CH}_2+HBr⟶{CH}_{3}C{H}_2{CH}_{2}Br\left(A\right)+{CH}_3-C\underset{\stackrel{|}{Br}}{H}-C{H}_3\left(B\right)$

• A. The carbocation ${CH}_3-C\underset{|}{H}{-CH}_3$ is formed which is more stable
• B. the carbocation ${CH}_3-C{H}_2{-\stackrel{+}{C}H}_3$ is formed which is more stable
• C. Both carbon are equally stable but the nucleophile attacks on central atom
• D. ${CH}_3-C\underset{|}{H}{-CH}_3$ can easily gives proton to Br${}^{+}$

Answer 1

The correct option is A The carbocation ${CH}_3-C\underset{|}{H}{-CH}_3$ is formed which is more stable

As Secondary carbocation is more stab;le than that of primry carbocation the carbocation ${CH}_3-C\underset{|}{H}{-CH}_3$ is formed which is more stable

Hence, the correct option is $A$