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

Vedclass · Accepted Answer

$CH_3-CH_2-CH_2-Br$

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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:
$CH_3-CH(Br)-CH_3$ $\xrightarrow{KOH/C_2H_5OH} A$ $\xrightarrow{(C_6H_5CO)_2O_2, HBr} B$

Similar Questions

$HCl$ does not show the peroxide effect because the reaction mechanism involves the following steps:
$CH_3-CH=CH_2 + \dot{X} \to CH_3-\dot{C}H-CH_2X$ $(I)$
$CH_3-\dot{C}H-CH_2X + HX \to CH_3-CH_2-CH_2X + \dot{X}$ $(II)$
Why does $HCl$ fail to undergo this reaction?

Select the incorrect option based on the given reactions:
Reaction $1$: Methylenecyclohexane + $dil. H_2SO_4 \rightarrow (P)$ (major product)
Reaction $2$: $1-$Methylcyclohexanol + $dil. H_2SO_4 \rightarrow (Q)$ (major product)

What is the product formed in the given reaction?

Baeyer's reagent is used for the detection of

Ethylene is prepared by the dehydration of

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