 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/2/b/ii/solution

Draw a curly arrow from the nitrogen lone pair to methyl carbon, and a second from the C-I bond to I:

HOC H_{2} C H_{2} NM e_{2} + MeI ⟶ [HOC H_{2} C H_{2} NM e_{3}]^{+} I^{-} .

This is concerted SN2 quaternization. The product is choline iodide; nitrogen has four carbon substituents and positive charge, while the alcohol remains OH.

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