The major product 'B' in the below mentioned | vedclass

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(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ leads to the formation of propene $(A)$ via an elimination reaction.$CH_3-C

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$2-$Bromopropane

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(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ leads to the formation of propene $(A)$ via an elimination reaction.$CH_3-CH(Br)-CH_3 + 	ext{Alc. KOH} \xrightarrow{\Delta} CH_3-CH=CH_2 (A) + KBr + H_2O$Step $2$: The addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule, where the negative part of the addendum $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.$CH_3-CH=CH_2 + HBr \xrightarrow{	ext{Peroxide absent}} CH_3-CH(Br)-CH_3 (B)$Thus, the major product '$B$' is $2$-bromopropane.

The major product '$B$' in the below mentioned reaction is:
$CH_3-CH(Br)-CH_3 \xrightarrow[\Delta]{\text{Alc. KOH}} A \xrightarrow[\text{Peroxide absent}]{\text{HBr}} B$

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The major product '$B$' in the below mentioned reaction is:$CH_3-CH(Br)-CH_3 \xrightarrow[\Delta]{\text{Alc. KOH}} A \xrightarrow[\text{Peroxide absent}]{\text{HBr}} B$