CH_3 - CH = CH_2 xrightarrow[(2) CH_3CO_2T]{( | vedclass

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(C) The reaction is a hydroboration-oxidation (or reduction) sequence using $BD_3$ and $CH_3CO_2T$.Step $1$: Hydroboration of propene with $BD_3$ in $

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$CH_3 - CHD - CH_2T$

Vedclass · Answer

(C) The reaction is a hydroboration-oxidation (or reduction) sequence using $BD_3$ and $CH_3CO_2T$.Step $1$: Hydroboration of propene with $BD_3$ in $THF$ follows anti-Markovnikov addition,where $D$ adds to the more substituted carbon and $B$ adds to the less substituted carbon,forming $(CH_3-CH(D)-CH_2)_3B$.Step $2$: Treatment with $CH_3CO_2T$ (deuterated acetic acid) replaces the boron atom with a tritium $(T)$ atom via protodeboronation.Therefore,the final product $(A)$ is $CH_3-CHD-CH_2T$.

$CH_3 - CH = CH_2 \xrightarrow[(2) CH_3CO_2T]{(1) THF : BD_3} (A)$; Product $(A)$ of the above reaction is

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