Question

Draw a structural formula for the product of each \(S_{N} 2\) reaction. Where configuration of the starting material is given, show the configuration of the product. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+} \underset{\text { ethanol }}{\longrightarrow}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}+\mathrm{CH}_{3} \mathrm{I} \underset{\text { acetone }}{\longrightarrow}\) (c) C C N = C = N (d) CC1CCC(Cl)CC1 (e) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Li}^{+} \underset{\substack{\text { diethyl } \\ \text { ether }}}{\longrightarrow}\) (f) NCC1CCCC1 (g) C C C C C C C C C C C C C (h) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{Na}^{+} \mathrm{CN}^{-} \underset{\text { acetone }}{\longrightarrow}\)

Short answer

Answer: The product of this \(S_{N}2\) reaction is \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{O}\mathrm{CH}_{2}\mathrm{CH}_{3}\).

Step-by-step solution

Identify nucleophile and leaving group

In this reaction, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is the nucleophile and \(\mathrm{Cl}^{-}\) is the leaving group.

Draw the product

Replace the leaving group with the nucleophile, resulting in the product: \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{O}\mathrm{CH}_{2} \mathrm{CH}_{3}\). (b)

Identify nucleophile and leaving group

In this reaction, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\) is the nucleophile and \(\mathrm{I}^{-}\) is the leaving group.

Draw the product

Replace the leaving group with the nucleophile, resulting in the product: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \mathrm{CH}_{3}\). (c)

Identify nucleophile and leaving group

This reaction does not have a nucleophile and leaving group mentioned. The exercise needs more information to determine the product. (d)

Identify nucleophile and leaving group

This reaction does not have a nucleophile and leaving group mentioned. The exercise needs more information to determine the product. (e)

Identify nucleophile and leaving group

In this reaction, \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-}\) is the nucleophile and \(\mathrm{Cl}^{-}\) is the leaving group.

Draw the product

Replace the leaving group with the nucleophile, resulting in the product: \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{CH}_{2} \mathrm{C}\equiv\mathrm{C}\mathrm{CH}_{3}\). (f)

Identify nucleophile and leaving group

This reaction does not have a nucleophile and leaving group mentioned. The exercise needs more information to determine the product. (g)

Identify nucleophile and leaving group

This reaction does not have a nucleophile and leaving group mentioned. The exercise needs more information to determine the product. (h)

Identify nucleophile and leaving group

In this reaction, \(\mathrm{CN}^{-}\) is the nucleophile and \(\mathrm{Br}^{-}\) is the leaving group.

Draw the product

Replace the leaving group with the nucleophile, resulting in the product: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}\mathrm{CN}\).