Hydrocarbons - Result Question 4
5. Which one of the following alkenes when treated with HCl yields majorly an anti-Markovnikov product?
(2019 Main, 8 April II)
(a) $\mathrm{Cl}-\mathrm{CH}=\mathrm{CH}_2$
(b) $\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}=\mathrm{CH}_2$
(c) $\mathrm{CH}_3 \mathrm{O}-\mathrm{CH}=\mathrm{CH}_2$
(d) $\mathrm{F}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2$
Show Answer
Answer:
Correct Answer: 5 . (d)
Solution:
Attachment of electron donating group $(+R$ or $+I)$ with $s p^2$-carbon of an unsymmetrical alkene supports Markownikov’s addition rule through electrophilic addition pathway. But, attachment of electron-withdrawing group ( $-R$ or $-I$ ) for the same will follow anti-Markovnikov’s pathway (even in absence of organic peroxide which favours free radical addition) through electrophilic addition pathway. The product formed by given alkenes when treated with HCl.
BETA
