Consider the reaction of isopropyl iodide with various nucleophiles. For each pair, predict which will give the larger substitution/elimination ratio. (a) I or \(\mathrm{Cl}\) (b) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) or \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) or \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

Short Answer

Expert verified
(a) Cl will give the larger substitution/elimination ratio, as it is a stronger nucleophile than I. (b) \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) will give the larger substitution/elimination ratio, as it is a stronger nucleophile and base compared to \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\). (c) \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) will give the larger substitution/elimination ratio, as it is a better nucleophile but a weaker base than \(\mathrm{CH}_{3} \mathrm{O}^{-}\).

Step by step solution

01

(a) Compare I and Cl

(For this pair, we are comparing two halide ions: iodide (I) and chloride (Cl). In terms of nucleophilicity, iodide is a weaker nucleophile than chloride due to its larger size and lower charge density. This means that iodide will favor elimination over substitution more than chloride. Therefore, Cl will give the larger substitution/elimination ratio.)
02

(b) Compare \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) and \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\)

(In this pair, we are comparing two similar molecules containing nitrogen (N) and phosphorus (P) with three methyl (CH3) groups attached. The key difference between these two molecules is that nitrogen is more electronegative and smaller than phosphorus. As a result, \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) is both a stronger nucleophile and a stronger base compared to \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\). This makes \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) more likely to favor substitution over elimination. Therefore, \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) will give the larger substitution/elimination ratio.)
03

(c) Compare \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) and \(\mathrm{CH}_{3} \mathrm{O}^{-}\)

(Here we are comparing two ions: the thiolate ion (\(\mathrm{CH}_{3} \mathrm{~S}^{-}\)) and the alkoxide ion (\(\mathrm{CH}_{3} \mathrm{O}^{-}\)). The sulfur in the thiolate ion is larger and less electronegative than the oxygen in the alkoxide ion. This difference makes the thiolate ion a better nucleophile but a weaker base than the alkoxide ion. Since a stronger nucleophile favors substitution over elimination, \(\mathrm{CH}_{3} \mathrm{~S}^{-}\) will give the larger substitution/elimination ratio.)

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Most popular questions from this chapter

Which compound of each of the following sets would you expect to give the higher yield of substitution product under conditions for bimolecular reaction? (a) ethyl bromide or \(\beta\) -phenylethy 1 bromide; (b) \(\alpha\) -phenylethy 1 bromide or \(\beta\) -phenylethyl bromide; (c) isobutyl bromide or \(\mathrm{n}\) -butyl bromide; (d) isobuty1 bromide or tert-buty1 bromide.

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