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Step 1: Evaluate the synthesis of chlorobenzene from each starting material.
(A) Aniline (C H NH ): Aniline can be easily converted to chlorobenzene via the Sandmeyer reaction. First, aniline is treated with NaNO /HCl at 0-5 C to form benzene diazonium chloride. This is then treated with CuCl/HCl to yield chlorobenzene. This is a standard and easy preparation.
(B) Benzene (C H ): Chlorobenzene is prepared commercially by the direct halogenation of benzene. This is an electrophilic aromatic substitution reaction carried out by reacting benzene with chlorine in the presence of a Lewis acid catalyst like FeCl or AlCl . This is an easy and direct method.
(C) Phenol (C H OH): The C-O bond in phenol is very strong and difficult to break. This is because the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring, giving the C-O bond a partial double bond character. Simple reagents used to convert aliphatic alcohols to alkyl chlorides (like HCl, SOCl , PCl ) do not work effectively with phenol to produce chlorobenzene. The reaction is not easy and requires harsh conditions with very low yield.
(D) Benzene diazonium chloride (C H N Cl ): This is an excellent starting material for preparing chlorobenzene. As mentioned in part (A), treatment with cuprous chloride (CuCl) in HCl (Sandmeyer reaction) or with copper powder in HCl (Gattermann reaction) easily converts it to chlorobenzene.
Step 2: Final Answer.
Due to the partial double bond character of the C-O bond, it is very difficult to prepare chlorobenzene from phenol. Therefore, this is the compound from which chlorobenzene cannot be prepared easily.

Step 1: Understanding Racemization in S 1 Reactions.
The S 1 reaction mechanism proceeds through a carbocation intermediate. If the carbocation is formed at a chiral center, the intermediate will be planar (sp hybridized) and achiral. The incoming nucleophile can then attack this planar carbocation from either the front side or the back side with equal probability. This leads to the formation of both possible enantiomers in equal amounts. A 50:50 mixture of two enantiomers is called a racemic mixture. Therefore, for a racemic mixture to be produced, the starting alkyl halide must be chiral.
Step 2: Analyzing the Chirality of the given Alkyl Halides.
A molecule is chiral if it has a carbon atom bonded to four different groups. Let's examine each option.
(A) Tertiary butyl bromide ((CH ) C-Br): The carbon atom bonded to bromine is also bonded to three identical methyl groups. It is not a chiral center. The product, tert-butanol, is also achiral.
(B) 2-bromobutane (CH -CH(Br)-CH -CH ): The second carbon atom is bonded to four different groups: a hydrogen atom (-H), a bromine atom (-Br), a methyl group (-CH ), and an ethyl group (-CH CH ). Therefore, 2-bromobutane is chiral. Its hydrolysis via the S 1 mechanism will form a planar secondary carbocation, which upon attack by H O will produce a racemic mixture of (R)-butan-2-ol and (S)-butan-2-ol.
(C) Isopropyl bromide ((CH ) CH-Br): The carbon atom bonded to bromine is also bonded to two identical methyl groups. It is not a chiral center.
(D) Methyl bromide (CH Br): The carbon atom is bonded to three identical hydrogen atoms. It is not a chiral center. (Also, methyl halides primarily undergo S 2 reactions).
Step 3: Final Answer.
Only 2-bromobutane is a chiral alkyl halide among the options. Therefore, its hydrolysis via an S 1 mechanism will produce a racemic mixture.