1,4-dichlorohexane (1 mole) + NaI (1 mole) | vedclass

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Question

(B) The reaction is a Finkelstein reaction,which proceeds via an $S_N2$ mechanism. In $1,4$-dichlorohexane,there are two chlorine atoms: one at the $C

Vedclass · Accepted Answer

$I-CH_2-CH_2-CH_2-CH(Cl)-CH_2-CH_3$

Vedclass · Answer

(B) The reaction is a Finkelstein reaction,which proceeds via an $S_N2$ mechanism. In $1,4$-dichlorohexane,there are two chlorine atoms: one at the $C1$ position (primary,$1^{\circ}$) and one at the $C4$ position (secondary,$2^{\circ}$). $S_N2$ reactions are faster at less sterically hindered positions. Therefore,the primary $(1^{\circ})$ chloride is more reactive towards $S_N2$ substitution than the secondary $(2^{\circ})$ chloride. With $1 \ mole$ of $NaI$,the substitution occurs selectively at the $C1$ position. The product is $1$-iodo-$4$-chlorohexane,which is $I-CH_2-CH_2-CH_2-CH(Cl)-CH_2-CH_3$.

$1,4$-dichlorohexane $(1 \ mole)$ + $NaI$ $(1 \ mole)$ $\xrightarrow{Acetone}$ Product of the reaction is

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