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ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/3/a/i/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / 3 / a / i / Solution by

Codex  0 2026-10-20

The products are hexanenitrile, $C H_{3} (C H_{2})_{4} C \equiv N$ , and ethyl triphenylmethyl ether, $P h_{3} C - O - C H_{2} C H_{3}$ . 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

C N^{-}

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

C l^{-}

. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to

P h_{3} C^{+}

and

C l^{-}

. An ethanol oxygen lone pair then attacks the carbocation, giving

P h_{3} C - 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.

Figure 1.
Original electron-pair arrow schemes for cyanide substitution and triphenylmethyl ether formation
. **Details**.

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