Br-CH_2-(CH_2)_2-CH_2-Br + CH_3NH_2 to The pr | vedclass

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Question

(A) The reaction is an intramolecular nucleophilic substitution $(S_N2)$ reaction.$1$. The lone pair on the nitrogen atom of $CH_3NH_2$ attacks one of

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$N$-methylpyrrolidine

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(A) The reaction is an intramolecular nucleophilic substitution $(S_N2)$ reaction.$1$. The lone pair on the nitrogen atom of $CH_3NH_2$ attacks one of the terminal carbon atoms of $Br-(CH_2)_4-Br$,displacing a $Br^-$ ion.$2$. This forms a secondary amine intermediate: $CH_3-NH_2^+-CH_2-(CH_2)_2-CH_2-Br$.$3$. $A$ proton is lost from the nitrogen to form $CH_3-NH-CH_2-(CH_2)_2-CH_2-Br$.$4$. The lone pair on the nitrogen then attacks the other terminal carbon atom,leading to ring closure (cyclization) and the loss of the second $Br^-$ ion.$5$. The final product is $N$-methylpyrrolidine,which is a five-membered ring containing nitrogen with a methyl group attached to the nitrogen atom.

$Br-CH_2-(CH_2)_2-CH_2-Br + CH_3NH_2 \to$ The product of the reaction is:

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