Predicting SN2, E2, SN1, and E1 Reactions For each of the following reactions, predict the products and the mechanisms by which the products are formed. (a) CH3CH2CH2CH2Br + (CH3)3CO^-Na^+ → (b) (c) Analyze We follow the six-step approach outlined above and use the decision tree in Figure 27-13 to

Question

Predicting [latex]S_N2,[/latex] E2, [latex]S_N1,[/latex] and E1 Reactions

For each of the following reactions, predict the products and the mechanisms by which the products are formed.

(a) [latex]CH_3CH_2CH_2CH_2Br + (CH_3)_3CO^-Na^+ \xrightarrow{(CH_3)_3COH}[/latex]

(b) [latex]\begin{matrix} \ \underset{|}{Br} \quad\quad\quad\quad\quad\quad \underset{||}O \quad\quad\quad\quad\quad\quad \ CH_3CHCH_3 + CH_3C-O^-Na^+ \xrightarrow{ ext{acetone}} \end{matrix}[/latex] (c) [latex]\begin{matrix} \underset{|}{Br}\quad\quad\quad \ CH_3CH_2CHCH_2CH_3 \xrightarrow[{80}^{\circ}C]{H_2O} \end{matrix}[/latex]

Analyze
We follow the six-step approach outlined above and use the decision tree in Figure 27-13 to guide our thinking.

Accepted Answer

(a) We identify the electrophile as [latex]CH_3CH_3CH_2Br[/latex], which is a primary haloalkane. The nucleophile is [latex](CH_3)_3CO^-[/latex], 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 [latex]S_N2[/latex] will yield [latex]CH_3CH_2CH_3CH_2OC(CH_3)_3[/latex], 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 [latex]S_N2[/latex].

(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 [latex]S_N2[/latex] or E2 but rather [latex]S_N1[/latex] and E1. [latex]S_N1[/latex] and E1 always occur together. Secondary carbocations are relatively stable, especially in a polar protic solvent, such as water. We expect both [latex]S_N1[/latex] and E1 products. Because [latex]H_2O[/latex] is a very weak base and a fair nucleophile, substitution will dominate over elimination.

[latex]\begin{matrix} \underset{|}{Br} \ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \underset{|}{OH} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \ CH_3CH_2CHCH_2CH_3 + H_2O \xrightarrow{H_2O} CH_3CH_2CHCH_2CH_3 ext{ or } CH_3CH=CHCH_2CH_3 + HBr \ \ \end{matrix}[/latex][latex]\quad ext{Electrophile} \quad\quad\quad ext{Weak base} \quad\quad\quad\quad\quad\quad S_N1 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad E1 \quad \ \quad \ ext{(secondary)} \quad\quad ext{(weak nucleophile)} \quad\quad\quad ext{(major)} \quad\quad\quad\quad\quad\quad\quad\quad ext{(minor)}[/latex]

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 [latex]S_N1[/latex] and E1 products, including using arrows to show the movement of electrons. Also in (c), (E) and (Z) isomers of 2-pentene are possible.

[latex]\begin{matrix} \ H\ _{_\diagdown }\quad\quad\quad _{_\diagup }H \C=C\ \quad\quad H_3C\ ^{^\diagup }\quad \quad \ \ ^{^\diagdown }CH_2CH_3 \ \ ext{(Z)-2-pentene}\end{matrix}[/latex] [latex]\begin{matrix} H_3C\ _{_\diagdown }\quad\quad\quad _{_\diagup }H \quad \C=C\ \quad\quad\quad H\ ^{^\diagup }\quad \quad \ \ ^{^\diagdown }CH_2CH_3 \ \ ext{(E)-2-pentene}\end{matrix}[/latex]

Question 27.5: Predicting SN2, E2, SN1, and E1 Reactions For each of the fo...

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