Identify Z in the following series of reactio | vedclass

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Question

(A) Step $1$: Dehydrohalogenation of $1$-iodopropane with alcoholic $KOH$ gives propene $(X)$: $CH_{3}CH_{2}CH_{2}I + KOH(alc.) \xrightarrow{\Delta} C

Vedclass · Accepted Answer

$CH_{3}CH_{2}CH_{2}CN$

Vedclass · Answer

(A) Step $1$: Dehydrohalogenation of $1$-iodopropane with alcoholic $KOH$ gives propene $(X)$: $CH_{3}CH_{2}CH_{2}I + KOH(alc.) \xrightarrow{\Delta} CH_{3}CH=CH_{2} (X) + KI + H_{2}O$ Step $2$: Anti-Markovnikov addition of $HBr$ to propene in the presence of peroxide gives $1$-bromopropane $(Y)$: $CH_{3}CH=CH_{2} + HBr \xrightarrow{	ext{peroxide}} CH_{3}CH_{2}CH_{2}Br (Y)$ Step $3$: Nucleophilic substitution of $1$-bromopropane with $KCN$ gives butanenitrile $(Z)$: $CH_{3}CH_{2}CH_{2}Br + KCN \xrightarrow{	ext{alcohol}, \Delta} CH_{3}CH_{2}CH_{2}CN (Z) + KBr$ Thus,$Z$ is $CH_{3}CH_{2}CH_{2}CN$.

Identify $Z$ in the following series of reactions:

Similar Questions

The increasing order of the reactivity of the following halides for the $S_N1$ reaction is:
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$(II)$ $CH_3-CH_2-CH_2-Cl$
$(III)$ $p-CH_3O-C_6H_4-CH_2Cl$

The rate of $S_N1$ reaction will be faster for which of the following bromides?

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What is the correct order of reactivity of halides for the $S_{N}2$ reaction?

$tert$-Butylbromide reacts with $aq.$ $NaOH$ by $S_N1$ mechanism while $n$-butylbromide reacts by $S_N2$ mechanism. Why?

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