Question 27.5: Predicting SN2, E2, SN1, and E1 Reactions. For each of the f......
Predicting S_{N}2, E2, S_{N}1, and E1 Reactions
For each of the following reactions, predict the products and the mechanisms by which the products are formed.
(a) CH_{3}CH_{2}CH_{2}CH_{2}Br + (CH_{3})_{3}CO^{−}Na^{+} \xrightarrow[]{(CH_{3})_{3}COH}
(b) CH_{3}\overset{\begin{matrix} Br \\ \mid \end{matrix} }{C} HCH_{3} + CH_{3}\overset{\begin{matrix} O \\ \parallel \end{matrix} }{C}— O^{−}Na^{+} \xrightarrow[]{acetone}
(c) CH_{3}CH_{2}\overset{\begin{matrix} Br \\ \mid \end{matrix} }{C}HCH_{2}CH_{3} \xrightarrow[80 °C]{H_{2}O}
Analyze
We follow the six-step approach outlined above and use the decision tree in Figure 27-13 to guide our thinking.
Solve
(a) We identify the electrophile as CH_{3}CH_{2}CH_{2}CH_{2}Br, which is a primary haloalkane. The nucleophile is (CH_{3})_{3}CO^{−}, a sterically hindered strong base.
Although a strong base is usually a strong nucleophile, the bulkiness of this base disfavors substitution, and we expect that elimination by E2 will provide the major product.
Substitution by S_{N}2 will yield CH_{3}CH_{2}CH_{2}CH_{2}OC(CH_{3})_{3}, but the substitution product will be the minor product.
(b) The electrophile is a secondary haloalkane. The electron pair donor is a weak base (the conjugate base of a weak acid) and, because it is negatively charged, it is a reasonably good nucleophile, especially in a polar aprotic solvent. We expect that the main reaction will be S_{N}2.
(c) Molecules of the solvent also serve as the electron pair donor. The reaction involves a secondary haloalkane with a weak base/weak nucleophile in a polar protic solvent. These conditions do not favor S_{N}2 or E2 but rather S_{N}1 and E1. S_{N}1 and E1 always occur together. Secondary carbocations are relatively stable, especially in a polar protic solvent, such as water. We expect both S_{N}1 and E1 products. Because H_{2}O is a very weak base and a fair nucleophile, substitution will dominate over elimination.
\underset{\begin{matrix} \text{Electrophile} \\ (\text{secondary)} \end{matrix} }{CH_{3}CH_{2}\overset{\begin{matrix} Br \\ \mid \end{matrix} }{C}HCH_{2}CH_{3}} + \underset{\begin{matrix} \text{Weak base} \\ \text{(weak nucleophile)} \end{matrix} }{H_{2}O} \xrightarrow[]{H_{2}O} \underset{\begin{matrix} S_{N}1 \\ \text{(major)} \end{matrix} }{CH_{3}CH_{2}\overset{\begin{matrix} OH \\ \mid \end{matrix} }{C}HCH_{2}CH_{3}} or \underset{\begin{matrix} E1 \\ \text{(minor)} \end{matrix} }{CH_{3}CH \xlongequal[]{}CHCH_{2}CH_{3}} + HBr
Assess
The mechanism for the reaction in (c) was not explicitly shown. We must make sure we are able to show the steps involved in forming the S_{N}1 and E1 products, including using arrows to show the movement of electrons.
Also in (c), (E) and (Z) isomers of pent-2-ene are possible.
