The reaction of CH_3-CH=CH-C_6H_4-OH with HBr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ follows Markovnikov's rule.In this reaction,the proton $(H^+)$ adds to the carbon atom of the do

Vedclass · Accepted Answer

$CH_3-CH_2-CHBr-C_6H_4-OH$

Vedclass · Answer

(B) The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ follows Markovnikov's rule.In this reaction,the proton $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom.The carbocation formed at the benzylic position is more stable due to resonance with the benzene ring.Therefore,the product is $CH_3-CH_2-CHBr-C_6H_4-OH$.

The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ gives:

Similar Questions

Which of the following statements is correct for the reactivity in $S_N2$ reaction?

What is formed when toluene reacts with chlorine in the presence of light?

Which one of the following is likely to give a precipitate with $AgNO_3$ solution?

Arrange the following compounds in order of decreasing rate of hydrolysis for $S_N1$ reaction:
$(i)$ $C_6H_5CH_2Br$
(ii) $p-CH_3-C_6H_4-CH_2Br$
(iii) $p-CH_3CH_2-C_6H_4-CH_2Br$
(iv) $p-(CH_3)_2CH-C_6H_4-CH_2Br$

Before using chloroform as an anesthetic,its sample is tested by ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module