JEE PYQ: Hydrocarbons Question 30
Question 30 - 2019 (12 Apr Shift 1)
The major product of the following addition reaction is:
$\text{H}_3\text{C}-\text{CH}=\text{CH}_2 \xrightarrow{\text{Cl}_2/\text{H}_2\text{O}}$
(1) $\text{H}_3\text{C}-\text{CH}(-\text{OH})-\text{CH}_2(-\text{Cl})$
(2) $\text{H}_3\text{C}-\text{CH}(-\text{Cl})-\text{CH}_2(-\text{OH})$ (Markovnikov)
(3) Propylene oxide (epoxide)
(4) 2,3-Dimethyloxirane
Show Answer
Answer: (2)
Solution
$\text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{Cl}_2/\text{H}_2\text{O}}$ $\text{CH}_3-\text{CH}(-\text{Cl}^+)-\text{CH}_2$ then $\text{H}_2\text{O}$ attacks the more substituted carbon (Markovnikov) giving $\text{CH}_3-\text{CH}(-\text{OH})-\text{CH}_2\text{Cl}$.
BETA
