Ph-CH_2-CH(D)-CH_3 xrightarrow{Br_2/hv} Produ | vedclass

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

Question

(A) The reaction is a free radical bromination.The most stable radical is formed at the benzylic position $(Ph-\dot{C}H-CH(D)-CH_3)$ due to resonance

Vedclass · Accepted Answer

Diastereomers

Vedclass · Answer

(A) The reaction is a free radical bromination.The most stable radical is formed at the benzylic position $(Ph-\dot{C}H-CH(D)-CH_3)$ due to resonance stabilization with the phenyl ring.Bromination at this position creates a new chiral center.Since the starting molecule already contains a chiral center at the carbon atom bonded to deuterium $(D)$,the formation of a second chiral center results in a pair of diastereomers.

$Ph-CH_2-CH(D)-CH_3 \xrightarrow{Br_2/hv}$
Product of the above reaction will be:

Similar Questions

The octane number $100$ is assigned to .............

Which of the following does not react with chlorine in the dark?

The angle strain in cyclopropane is:

In the case of the homologous series of alkanes,which one of the following statements is incorrect?

Which product is obtained on heating sodium acetate with sodalime?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$Ph-CH_2-CH(D)-CH_3 \xrightarrow{Br_2/hv}$ Product of the above reaction will be: