# Why is the nucleophilic substitution of OH- with 1-bromopentane slower than with 2-bromo-2-methylbutane?

Which statement about the reaction of a hydroxide ion with the organic reagent is correct?

A. 1-bromopentane predominantly follows an $$\mathrm{S_N1}$$ mechanism.

B. 2-bromo-2-methylbutane predominantly follows an $$\mathrm{S_N2}$$ mechanism.

C. Reaction with 1-bromopentane occurs at a slower rate than with 1-chloropentane.

D. Reaction with 1-bromopentane occurs at a slower rate than with 2-bromo-2-methylbutane.

I got the answer correct (D), but only by eliminating the other three options. However, why is D true? Does this mean that $$\mathrm{S_N1}$$ reactions are always faster than $$\mathrm{S_N2}$$ reactions when they are reacting with the same nucleophile?

I'd really appreciate some help.

• There is no SN1 in D. The tertiary bromide in the presence of hydroxide will give E2 elimination. Whether the SN2 rxn. on the primary bromide is slower than the E2 is at issue. – user55119 May 21 '19 at 0:55