Out of the following which will not undergo S_N^1 reaction pathway:

The correct answer is Option (3) → 1-Bromo-3,3-dimethylbutane

An S_N1 reaction goes through a carbocation formed in the slow first step, so its rate depends on how stable that carbocation is. Tertiary, allylic and benzylic halides react readily; primary alkyl halides do not follow the S_N1 pathway, because a primary carbocation is too unstable to form.

• Option (1) 1-Chloro-1,1,1-triphenylmethane — Undergoes S_N1. \( \ce{(C6H5)3C-Cl} \) gives the triphenylmethyl cation, whose charge is spread over three rings by resonance.

• Option (2) 2-Iodo-2-methylpropane — Undergoes S_N1. It is a tertiary halide and gives the stable tert-butyl cation.

• Option (3) 1-Bromo-3,3-dimethylbutane — Does not undergo S_N1. \( \ce{BrCH2CH2C(CH3)3} \) has Br on a \( \ce{CH2} \) group, so ionisation would give a primary carbocation.

• Option (4) 3-Bromoprop-1-ene — Undergoes S_N1. \( \ce{CH2=CH-CH2Br} \) is an allylic halide; the allyl cation is stabilised by resonance.

NCERT (2026-27 edition):
Substitution Nucleophilic Unimolecular (S_N1) – “S_N1 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. … Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo S_N1 reaction very fast because of the high stability of 3° carbocations. … For the same reasons, allylic and benzylic halides show high reactivity towards the S_N1 reaction.”