Hydrocarbons 2 Question 5
5. Which one of the following alkenes when treated with $\mathrm{HCl}$ yields majorly an anti Markownikov 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) $\mathrm{F}{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}{2}$
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Answer:
Correct Answer: 5. (b)
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-Markownikov’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 $\mathrm{HCl}$.
Similarly,
