Question 27.3: Recognizing SN1 and SN2 Reactions and Predicting the Product......

Analyze
To decide whether a reaction of this type will take place, we first identify the electrophile, nucleophile, and leaving group. Keep in mind that the order of reactivity for S_{N}2 is methyl > primary > secondary > tertiary and that the order is the opposite for S_{N}1.

Solve
(a) The nucleophile is the CN^{−} ion, the electrophile is the chloropropane, and the leaving group is the Cl^{−} ion. The CN^{−} ion is a much stronger base than the Cl^{−} ion, and so the equilibrium constant for the reaction should be large. The chloropropane is a primary haloalkane, so the likely mechanism is S_{N}2 and the product will be CH_{3}CH_{2}CH_{2}CN.
(b) The nucleophile is the Br^{−} ion, the electrophile is the ethanol, and the potential leaving group is the OH^{−} ion. The OH^{−} ion is a much stronger base than the Br^{−} ion, so the equilibrium constant for the reaction is much less than one. No reaction is expected.
(c) The nucleophile is the CH_{3}OH molecule, the electrophile is the t-butyl chloride, and the leaving group is the Cl^{−} ion. In this case we have to know the relative basicities of CH_{3}OH  and  Cl^{−} to decide in which direction the reaction will proceed. If we assume that the basicities of methanol and chloride ion are about the same (both are very weak bases), we expect an equilibrium to be established. However, the fact that we are using a large excess of methanol (methanol is the solvent) shifts the equilibrium toward the product (both are very weak bases). The product is (CH_{3})_{3}COCH_{3}, and the likely  mechanism is S_{N}1 since the electrophile is a tertiary haloalkane.

Assess
In Section 27-3 we will see that other products, not just substitution products, are possible in some of these reactions.