Question

Select the best conditions for the reactions. CH3OH NaOCH3, CH3OH H+, NaOCH3, CH3OH NaOCH3, DMSO H+, NaOCH3, DMSO CH3OH NaOCH3, CH3OH cat. H+, NaOCH3, CH3OH NaOCH3, DMSO cat. H+, NaOCH3, DMSO

Answer 1

(a) NaOCH3, DMSO
(b) CH3OH

Answer 2

Both products shown are ethers, so both reactions are substitution reactions. The reaction conditions and the structure of the alkyl halide determine the substitution mechanism, SN1 or SN2. Unimolecular substitution occurs through the formation of a carbocation and is favored by a poor nucleophile and a protic polar solvent. A bimolecular mechanism involves a transition state formed when a nucleophile attacks the carbon bonded to the halide, so SN2 reactions are favored by a strong nucleophile and an aprotic polar solvent. Since primary carbocations are very unstable, primary alkyl halides cannot undergo SN1 substitution, only SN2. Secondary alkyl halides can undergo SN1 and SN2 reactions. In tertiary alkyl halides, steric hindrance makes the nucleophilic attack on the tertiary carbon too difficult; therefore, tertiary alkyl halides undergo only SN1 reactions. (a) 1-Bromo-3-methylbutane is a primary alkyl bromide; therefore, the substitution follows an SN2 mechanism. The combination of a strong base/good nucleophile and an aprotic polar solvent would provide the best conditions for the SN2 reaction. Acidic conditions would deactivate a base, so we can eliminate the choices with H . Methanol is a weak nucleophile and so it does not promote an SN2 reaction. The two remaining choices both have the same strong nucleophile, methoxide, but different solvents. Since SN2 reactions are favored by an aprotic polar solvent, DMSO (dimethyl sulfoxide) is a better choice than methanol. (b) 2-Bromo-2-methylbutane is a tertiary alkyl bromide; therefore, the substitution follows an SN1 mechanism. In the SN1 reaction, the nucleophile cannot be a strong base because that would favor an elimination reaction. Therefore, we can exclude methoxide in methanol and methoxide in DMSO. Acid will protonate methoxide, converting it to methanol, but only a catalytic (i.e., trace) amount of acid is added, so methoxide will still be prevalent. The best choice here is methanol. It is a weak nucleophile that favors SN1 reactions, and it is also a polar protic solvent that can stabilize both the carbocation (via nonbonding electrons on the oxygen) and the leaving Br