CH_3-CH_2-CH_2-Br xrightarrow{text{alk. KOH}} | vedclass

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Question

(C) $CH_3-CH_2-CH_2-Br \xrightarrow{	ext{alk. KOH}} CH_3-CH=CH_2$ $(A)$.$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^{-}]{(i) BD_3/THF} CH_3-CH(D)-CH_2O

Vedclass · Accepted Answer

$CH_3-CH(D)-CH_2OH$

Vedclass · Answer

(C) $CH_3-CH_2-CH_2-Br \xrightarrow{	ext{alk. KOH}} CH_3-CH=CH_2$ $(A)$.$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^{-}]{(i) BD_3/THF} CH_3-CH(D)-CH_2OH$ $(B)$.Hydroboration-oxidation involves syn-addition.$D$ from $BD_3$ adds to the less substituted carbon $(C1)$ and $OH$ from $H_2O_2$ adds to the more substituted carbon $(C2)$ is incorrect; actually,in hydroboration-oxidation,$B$ (from $BD_3$) adds to the less substituted carbon $(C1)$ and $H$ (or $D$) adds to the more substituted carbon $(C2)$ following Anti-Markovnikov regioselectivity.Thus,the product is $CH_3-CH(D)-CH_2OH$.

$CH_3-CH_2-CH_2-Br$ $\xrightarrow{\text{alk. KOH}} A$ $\xrightarrow[(ii) H_2O_2/OH^{-}]{(i) BD_3/THF} B$; $B$ is:

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