= Solution
\b[The products are <hexanenitrile>, $\mathrm{CH_3(CH_2)_4C\equiv N}$, and <ethyl triphenylmethyl ether>, $\mathrm{Ph_3C-O-CH_2CH_3}$.] The nitrile carbon comes from <cyanide>, so replacing chloride in the five-carbon chain gives a six-carbon product.
For the primary chloride, the carbon-end <lone pair> of $\mathrm{CN^-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted <nucleophilic substitution> step forms the C–C bond and releases $\mathrm{Cl^-}$. Attack through the carbon end gives the nitrile, not an isocyanide.
For the triphenylmethyl chloride, the C–Cl bond first undergoes <heterolysis> to $\mathrm{Ph_3C^+}$ and $\mathrm{Cl^-}$. An <ethanol> <oxygen> <lone pair> then attacks the <carbocation>, giving $\mathrm{Ph_3C-O^+(H)Et}$. Finally <pyridine> accepts that O–H <proton>, with the O–H bond pair returning to <oxygen>, giving the neutral <ether> and pyridinium chloride.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-substitution-mechanisms.png]
{title=Original <electron-pair arrow> schemes for cyanide substitution and triphenylmethyl ether formation}
Back to article page
