The reaction in which a nucleophile replaces an existing group (the leaving group) in a molecule is called a nucleophilic substitution reaction. Haloalkanes are the substrates in these reactions. In this type of reaction, a nucleophile reacts with a haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to the halogen. A substitution reaction takes place and the halogen atom, called the leaving group, departs as a halide ion. Since the substitution reaction is initiated by a nucleophile, it is called a nucleophilic substitution reaction.
It is one of the most useful classes of organic reactions of alkyl halides in which the halogen is bonded to an sp3 hybridised carbon.
What is the product formed when tert-butyl chloride is boiled with water?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → tert-Butyl alcohol
tert-Butyl chloride, \( \ce{(CH3)3C-Cl} \), is a tertiary haloalkane. Boiling water is a polar protic solvent and also acts as the nucleophile, so the substitution follows the SN1 path in two steps:
Step I (slow): the C–Cl bond breaks to give a tertiary carbocation and a chloride ion.
\( \displaystyle \ce{(CH3)3C-Cl} \rightleftharpoons \ce{(CH3)3C^+ + Cl^-} \)
Step II (fast): a water molecule attacks the carbocation and loses a proton.
\( \displaystyle \ce{(CH3)3C^+ + H2O -> (CH3)3C-OH + H^+} \)
\( \displaystyle \ce{(CH3)3C-Cl + H2O} \xrightarrow{\text{boil}} \ce{(CH3)3C-OH + HCl} \)
The tertiary carbocation is very stable, so the reaction goes readily and gives 2-methylpropan-2-ol, tert-butyl alcohol. The same carbocation and the same alcohol form whether the halide is the chloride or the bromide, and whether the nucleophile is water or \( \ce{OH^-} \).
Why the other options are incorrect
Option (1) 2-Methylpropane — Incorrect. This would need the Cl to be replaced by H, which is a reduction. Water attaches an –OH group, not a hydrogen.
Option (3) 2,2-Dimethylpropane — Incorrect. This would need a new C–C bond to a methyl group. Water brings no carbon to the molecule.
Option (4) No reaction — Incorrect. Tertiary haloalkanes undergo SN1 reaction very fast in a polar protic solvent such as water, because the 3° carbocation is so stable.
NCERT (2026-27 edition):
(b) Substitution nucleophilic unimolecular (SN1) – “SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.). The reaction between tert-butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics, i.e., the rate of reaction depends upon the concentration of only one reactant, which is tert-butyl bromide. … In case of alkyl halides, 3° alkyl halides undergo SN1 reaction very fast because of the high stability of 3° carbocations.”