Cl-CH_2-C(CH_3)_2-CH_2-CH_2-Cl + I^- xrightar | vedclass

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Question

(B) The reaction involves a nucleophilic substitution $(S_N2)$ mechanism using $I^-$ as the nucleophile in a polar aprotic solvent $(DMF)$.$S_N2$ reac

Vedclass · Accepted Answer

$Cl-CH_2-C(CH_3)_2-CH_2-CH_2-I$

Vedclass · Answer

(B) The reaction involves a nucleophilic substitution $(S_N2)$ mechanism using $I^-$ as the nucleophile in a polar aprotic solvent $(DMF)$.$S_N2$ reactions are highly sensitive to steric hindrance.In the substrate $Cl-CH_2-C(CH_3)_2-CH_2-CH_2-Cl$,there are two primary chloride sites.The site at the end of the chain $(-CH_2-CH_2-Cl)$ is less sterically hindered compared to the site near the bulky tert-butyl group $(-CH_2-C(CH_3)_2-)$.Therefore,the nucleophile $I^-$ preferentially attacks the less hindered primary carbon atom,resulting in the formation of $Cl-CH_2-C(CH_3)_2-CH_2-CH_2-I$ as the major product.

$Cl-CH_2-C(CH_3)_2-CH_2-CH_2-Cl + I^- \xrightarrow[DMF]{} \text{product}$; The major product of this reaction is:

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