What are A, B and C in the following reaction | vedclass

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Question

(A) The reaction sequence is as follows:$1.$ Addition of $HBr$ to ethene $(CH_2=CH_2)$ follows Markovnikov's rule to give ethyl bromide $(A)$:$CH_2=CH

Vedclass · Accepted Answer

$A: CH_3-CH_2-Br, B: CH_3-CH_2-CN, C: CH_3-CH_2-COOH$

Vedclass · Answer

(A) The reaction sequence is as follows:$1.$ Addition of $HBr$ to ethene $(CH_2=CH_2)$ follows Markovnikov's rule to give ethyl bromide $(A)$:$CH_2=CH_2 + HBr \rightarrow CH_3-CH_2-Br$$2.$ Nucleophilic substitution of ethyl bromide with $KCN$ gives ethyl cyanide $(B)$:$CH_3-CH_2-Br + KCN \rightarrow CH_3-CH_2-CN + KBr$$3.$ Acidic hydrolysis of ethyl cyanide gives propanoic acid $(C)$:$CH_3-CH_2-CN \xrightarrow{H^{+}/H_2O} CH_3-CH_2-COOH$Therefore,the correct option is $A$.

What are $A, B$ and $C$ in the following reaction sequence?
$CH_2=CH_2$ $\xrightarrow{HBr} A$ $\xrightarrow{KCN} B$ $\xrightarrow{H^{+} / H_2O} C$

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What are $A, B$ and $C$ in the following reaction sequence?$CH_2=CH_2$ $\xrightarrow{HBr} A$ $\xrightarrow{KCN} B$ $\xrightarrow{H^{+} / H_2O} C$