The final product B of the below reaction seq | vedclass

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(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ gives propene $(A)$ as the major product.$CH_3-CH(Br)-CH_3 \xrightarrow{alc

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$CH_3-CH_2-CH_2-Br$

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(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ gives propene $(A)$ as the major product.$CH_3-CH(Br)-CH_3 \xrightarrow{alc. KOH} CH_3-CH=CH_2 (A) + HBr$Step $2$: Addition of $HBr$ to propene in the presence of peroxide $( (C_6H_5CO)_2O_2 )$ follows the Anti-Markownikoff rule (Kharasch effect).$CH_3-CH=CH_2 + HBr \xrightarrow{peroxide} CH_3-CH_2-CH_2-Br (B)$Thus,the final product $B$ is $1$-bromopropane.Hence,option $C$ is correct.

The final product $B$ of the below reaction sequence is:
$CH_3-CH(Br)-CH_3$ $\xrightarrow{KOH/C_2H_5OH} A$ $\xrightarrow{(C_6H_5CO)_2O_2, HBr} B$

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The final product $B$ of the below reaction sequence is:$CH_3-CH(Br)-CH_3$ $\xrightarrow{KOH/C_2H_5OH} A$ $\xrightarrow{(C_6H_5CO)_2O_2, HBr} B$