HO-CH_2CH_2CH_2CH_2-Br xrightarrow{KOH} text{ | vedclass

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Question

(B) The reaction involves an intramolecular nucleophilic substitution $(S_N2)$ reaction.$1$. The base $KOH$ deprotonates the hydroxyl group $(-OH)$ to

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Tetrahydrofuran

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(B) The reaction involves an intramolecular nucleophilic substitution $(S_N2)$ reaction.$1$. The base $KOH$ deprotonates the hydroxyl group $(-OH)$ to form an alkoxide ion: $HO-CH_2CH_2CH_2CH_2-Br + OH^- \rightarrow ^-O-CH_2CH_2CH_2CH_2-Br + H_2O$.$2$. The negatively charged oxygen atom acts as an internal nucleophile and attacks the carbon atom attached to the bromine atom,displacing the bromide ion.$3$. This cyclization leads to the formation of a five-membered cyclic ether,which is Tetrahydrofuran.

$HO-CH_2CH_2CH_2CH_2-Br \xrightarrow{KOH} \text{Product}$. The product is:

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