Question

The rate of reaction of t-butyl bromide and NaOH in water depends on the concentration of

• A. Both t-butyl bromide \& NaOH
• B. NaOH
• C. Independent of concentration
• D. t-butyl bromide

Hint:

\textbf{S$_N$1/E1 reactions:} Rate = $k$[Alkyl Halide]. First order. Favored for tertiary>secondary>primary alkyl halides, and by polar protic solvents. Rate is independent of nucleophile/base concentration (if it's involved after the RDS).
\textbf{S$_N$2 reactions:} Rate = $k$[Alkyl Halide][Nucleophile]. Second order. Favored for primary>secondary alkyl halides. Sterically hindered for tertiary. Favored by polar aprotic solvents and strong nucleophiles.
\textbf{E2 reactions:} Rate = $k$[Alkyl Halide][Base]. Second order. Favored by strong bases, high temperatures. Can occur for primary, secondary, tertiary.
t-Butyl bromide is a tertiary alkyl halide. In water (polar protic solvent), it primarily undergoes S$_N$1 and/or E1 reactions. Both S$_N$1 and E1 have the same rate-determining step: formation of the carbocation, which depends only on the concentration of the alkyl halide.

Answer 1

The Correct Option is D

t-Butyl bromide is a tertiary alkyl halide ((CH$_3$)$_3$C-Br). NaOH is a strong nucleophile (OH$^-$) and a strong base. The reaction of a tertiary alkyl halide with a nucleophile/base can proceed via S$_N$1 (substitution, nucleophilic, unimolecular) or E1 (elimination, unimolecular) mechanisms, or S$_N$2/E2 if conditions are very specific (which is less common for tertiary substrates due to steric hindrance). In water, which is a polar protic solvent, S$_N$1 and E1 pathways are favored for tertiary substrates. S$_N$1/E1 Mechanism: The first step in both S$_N$1 and E1 mechanisms is the formation of a carbocation, which is the rate-determining step (slow step). (CH$_3$)$_3$C-Br $\xrightarrow{\text{slow}}$ (CH$_3$)$_3$C$^+$ + Br$^-$ This step involves only the alkyl halide (t-butyl bromide). The rate of this step, and thus the overall rate of the S$_N$1/E1 reaction, depends only on the concentration of the alkyl halide. Rate = $k$[t-butyl bromide] The carbocation ((CH$_3$)$_3$C$^+$) then reacts rapidly with the nucleophile (OH$^-$ or H$_2$O for S$_N$1) or a base abstracts a proton (OH$^-$ or H$_2$O for E1). \begin{itemize} \item S$_N$1 product: (CH$_3$)$_3$C-OH (t-butyl alcohol) \item E1 product: (CH$_3$)$_2$C=CH$_2$ (isobutylene or 2-methylpropene) \end{itemize} Since the concentration of NaOH (or OH$^-_~{\text{aq}}$) does not appear in the rate equation for the rate-determining step of S$_N$1/E1, the rate of reaction depends only on the concentration of t-butyl bromide. S$_N$2/E2 Mechanism: S$_N$2 reactions are highly disfavored for tertiary alkyl halides due to steric hindrance at the reaction center. E2 reactions are possible for tertiary alkyl halides with strong, bulky bases, especially in non-polar solvents or at high temperatures. However, NaOH in water (polar protic solvent) would favor S$_N$1/E1 for a tertiary substrate. If E2 were to occur, its rate would be: Rate = $k$[t-butyl bromide][NaOH]. Given the substrate (tertiary alkyl halide) and the solvent (water, polar protic), the S$_N$1/E1 pathways are dominant. Therefore, the rate of the reaction depends on the concentration of t-butyl bromide only. \[ \boxed{\text{t-butyl bromide}} \]