### Reactions of Organohalides with Nucleophiles / Bases

Two characteristics of the nucleophilic reactant must be kept in mind. These are nucleophilicity and basicity. Nucleophilicity generally refers to the facility with which a nucleophile bonds to an electrophilic carbon atom. In the present case, we can estimate it by the rate of substitution at the halogen-bearing carbon atom of methyl bromide, CH3-Br. For the four nucleophiles examined here the nucleophilicity in methanol solution increases in the order: CH3CO2(–) < HO(–) < CN(–) < CH3S(–)

Basicity on the other hand refers to the proton affinity of the nucleophile. This can be estimated from the pKa's of the corresponding acids, remembering that strong acids have weak conjugate bases and vice versa. The four nucleophiles used in this problems have conjugate acids with pKa's ranging from 4.8 to 15.7. Acetic acid is the strongest acid (4.8), methanethiol (7.3) and hydrogen cyanide (9.1) are next, and water (15.7) is the weakest acid.

Select the nucleophilic reagent of interest from the following menu.

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### Acetate Anion

Of the four nucleophiles used in this problem, acetate is both the weakest nucleophile and the weakest base.
With 1° halides such as B, C & I substitution by an SN2 mechanism is likely. The rate of substitution for B will be enhanced by its benzylic character, and for I will be reduced due to substituents at the μ carbon.
The 3° halides G & J will not undergo a SN2 reaction with a nucleophile as weak as acetate. Since acetate is not a strong base, if an an E2 elimination takes place it will be slow. Ethanol is a polar solvent, but not normally sufficient to permit SN1 or E1 reactions.
The 2° halides A, F & H are all different. A is an allylic halide so both SN2 and SN1 reactions will be enhanced. In ethanol the former mode of reaction should predominate. F is a very hindered halide, due to the dimethyl substitution at each β carbon atom. It is unlikely to react at all. H is an unexceptional 2° halide, and may give slow substitution by the SN2 path. The necessary axial orientation of the nucleophile in that case would be a negative factor. Acetate is probably too weak a base to cause an E2 elimination.
The vinyl and aryl halides D & E do not undergo substitution or elimination reactions with the reagents used here.

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### Methanethiolate Anion

Of the four nucleophiles used in this problem, methanethiolate is the strongest nucleophile, and at best a modest base (pKa = 7.3).
With 1° halides such as B, C & I substitution by an SN2 mechanism should be rapid, especially for benzyl bromide and methyl iodide.
The 3° halides G & J are unlikely to give a SN2 reaction even with a strong nucleophile like thiolate. Thiolate is not a strong base, but it could induce a slow E2 elimination reaction. Ethanol is a polar solvent, but not normally sufficient to permit SN1 or E1 reactions.
The 2° halides A & H should give SN2 substitution with this strong nucleophile. The remaining compound F is severely hindered, and could only undergo substitution very slowly. It cannot give elimination products without rearranging.
The vinyl and aryl halides D & E do not undergo substitution or elimination reactions with the reagents used here.

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### Cyanide Anion

Of the four nucleophiles used in this problem, cyanide is relatively good nucleophile, and a moderate base (pKa = 9.1).
With 1° halides such as B, C & I substitution by an SN2 mechanism should be favored, especially for benzyl bromide and methyl iodide. The rate of substitution for I will be reduced due to substituents at the β-carbon, but substitution should still be the predominate course.
The 3° halides G & J are unlikely to give a SN2 reaction, but cyanide is a sufficiently strong base to cause E2 elimination reactions.
The 2° halides A, F & H are all different. A is an allylic halide so both SN2 and SN1 reactions will be enhanced. Cyanide is a fairly strong nucleophile, so the former path should be favored. H will not have this allylic enhancement, so a mixture of substitution (SN2) and elimination (E2) is expected. The remaining compound F is severely hindered, and could only undergo substitution very slowly. It cannot give elimination products without rearranging.
The vinyl and aryl halides D & E do not undergo substitution or elimination reactions with the reagents used here.

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### Hydroxide Anion

Of the four nucleophiles used in this problem, hydroxide is a relatively good nucleophile, and the strongest base (pKa = 15.7). This should lead to more E2 elimination reactions than in the other cases.
With 1° halides such as B, C & I substitution by an SN2 mechanism should be favored, especially for benzyl bromide and methyl iodide, which cannot give elimination products. The rate of substitution for I will be reduced due to substituents at the β-carbon, and the strongly basic hydroxide anion should produce a mixture of substitution and elimination products.
The 3° halides G & J will not give a SN2 reaction with hydroxide, but this strong base will instead cause rapid E2 elimination reactions.
2° Alkyl halides generally give E2 reactions with this strong base, assuming there are β-hydrogens. This is the case for A & H; however, A is an allylic halide and will show enhanced SN2 activity. Consequently a mixture of elimination and substitution products may result. Compound F has no β-hydrogens and is severely hindered. Substitution would be very slow if at all. Consequently, no reaction is expected, unless a skeletal rearrangement occurs.
The vinyl and aryl halides D & E do not undergo substitution or elimination reactions with the reagents used here.

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