Question

Give structures of the organic products of the following reactions and account for their formation: (a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow}$ (b) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{NaI}$ acetone (c) $\mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{NaI} \stackrel{\text { acetone. }}{\longrightarrow}$ (a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$. $\mathrm{Br}^{-}$is a better leaving group than $\mathrm{Cl}^{-}$. (b) $\mathrm{CH}_3 \mathrm{CHICH}_3$. Equilibrium is shifted to the right because $\mathrm{Nal}$ is soluble in acetone, while $\mathrm{NaBr}$ is not and precipitates. (c) $\mathrm{ICH}_2 \mathrm{CH}=\mathrm{CH}_2 . \mathrm{Nal}$ is soluble in acetone and $\mathrm{NaCl}$ is insoluble.

    Give structures of the organic products of the following reactions and account for their formation:
(a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow}$
(b) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{NaI}$ acetone
(c) $\mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{NaI} \stackrel{\text { acetone. }}{\longrightarrow}$

(a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$. $\mathrm{Br}^{-}$is a better leaving group than $\mathrm{Cl}^{-}$.
(b) $\mathrm{CH}_3 \mathrm{CHICH}_3$. Equilibrium is shifted to the right because $\mathrm{Nal}$ is soluble in acetone, while $\mathrm{NaBr}$ is not and precipitates.
(c) $\mathrm{ICH}_2 \mathrm{CH}=\mathrm{CH}_2 . \mathrm{Nal}$ is soluble in acetone and $\mathrm{NaCl}$ is insoluble.
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Schaum's Outline of Organic Chemistry
Schaum's Outline of Organic Chemistry
George Hademenos,… 3rd Edition
Chapter 7, Problem 41 ↓

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The cyanide ion ($\mathrm{CN}^-$) will attack the carbon atom bonded to the bromine because bromine is a better leaving group compared to chlorine. This is a nucleophilic substitution reaction (SN2 mechanism). - The product formed is $\mathrm{ClCH}_2 \mathrm{CH}_2  Show more…

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Give structures of the organic products of the following reactions and account for their formation: (a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{NaCN} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow}$ (b) $\mathrm{CH}_3 \mathrm{CHBrCH}_3+\mathrm{NaI}$ acetone (c) $\mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{NaI} \stackrel{\text { acetone. }}{\longrightarrow}$ (a) $\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$. $\mathrm{Br}^{-}$is a better leaving group than $\mathrm{Cl}^{-}$. (b) $\mathrm{CH}_3 \mathrm{CHICH}_3$. Equilibrium is shifted to the right because $\mathrm{Nal}$ is soluble in acetone, while $\mathrm{NaBr}$ is not and precipitates. (c) $\mathrm{ICH}_2 \mathrm{CH}=\mathrm{CH}_2 . \mathrm{Nal}$ is soluble in acetone and $\mathrm{NaCl}$ is insoluble.
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