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

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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

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Diastereomers

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(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:

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