Step 1: Understanding the Concept:
$S_{N}1$ reactions proceed via a two-step mechanism involving a planar, $sp^2$ hybridized carbocation intermediate. If the starting alkyl halide has a chiral center, the planar intermediate loses the original stereochemistry, allowing nucleophilic attack from both faces, yielding a racemic mixture.
Step 2: Key Formula or Approach:
Identify which of the given alkyl halides possesses a chiral carbon (an asymmetric carbon atom bonded to four different groups). Only optically active substrates can yield a noticeably racemic mixture upon $S_N1$ hydrolysis.
Step 3: Detailed Explanation:
1. Tertiary butyl bromide: The central carbon is attached to three identical methyl groups and one bromine. It is achiral. The product (tert-butanol) is also achiral. It cannot form a racemic mixture.
2. 2-bromobutane: The $C2$ carbon is attached to four distinct groups: a Hydrogen atom, a Methyl group ($CH_3$), an Ethyl group ($C_2H_5$), and a Bromine atom. It is a chiral center. Upon ionizing to form the planar carbocation ($CH_{3}-CH^{+}-C_{2}H_{5}$), water can attack equally from the top or bottom face, leading to equal amounts of the $(R)$ and $(S)$ enantiomers of 2-butanol (a racemic mixture). (Correct)
3. Isopropyl bromide: The central carbon is attached to two identical methyl groups. Achiral.
4. Methyl bromide: Achiral (and it typically undergoes $S_{N}2$ exclusively anyway).
Step 4: Final Answer:
2-bromobutane is the optically active halide that produces a racemic mixture.