Hydrocarbons Question 48
Question: 3-Phenylpropene on reaction with HBr gives (as a major product) [AIIMS 2005]
Options:
A) $ C_6H_5CH_2CH(Br)CH_3 $
B) $ C_6H_5CH(Br)CH_2CH_3 $
C) $ C_6H_5CH_2CH_2CH_2Br $
D) $ C_6H_5CH(Br)CH=CH_2 $
Show Answer
Answer:
Correct Answer: B
Solution:
According to Markownikoff’s rule, the negative part of the unsymmetrical reagent adds to less hydrogenated (more substituted) carbon atom of the double bond.
$C_6H_5CH = CH - CH_3 + HBr $
$ \overset{\overset{Br}{\mathop{|}}}{\mathop{C_6H_5CHC{H _{_2}}CH_3}} $
BETA
