Which of the following will not give S_N1.

The correct option is CH₃ – CH = CH – Cl.

To determine which of the given compounds will not undergo an S_N1 (substitution nucleophilic unimolecular) reaction, we must analyze the stability of the carbocation intermediate formed in the first step of the reaction, as well as the strength of the carbon-chlorine (C–Cl) bond.

1. Mechanism of S_N1 Reaction:
An S_N1 reaction proceeds via the formation of a carbocation intermediate. The rate of the reaction is directly proportional to the stability of this carbocation. If a compound cannot form a stable carbocation or has a very strong C–Cl bond that is difficult to break, it will not undergo an S_N1 reaction.

2. Analysis of the Options:
Let us evaluate each option:

• CH₂ = CH – CH₂ – Cl (Allyl chloride): Upon ionization, it forms the allyl carbocation:

  ${\mathrm{CH}}_2=\mathrm{CH}-{\mathrm{CH}}_2^{+}$

  This carbocation is highly stabilized by resonance. Therefore, it undergoes the S_N1 reaction readily.
• Benzyl chloride (shown in the first image, C₆H₅–CH₂–Cl): Upon ionization, it forms the benzyl carbocation:

  $C_6{H}_5-{\mathrm{CH}}_2^{+}$

  This carbocation is extensively stabilized by delocalization of the positive charge over the aromatic benzene ring (resonance stabilization). Hence, it readily undergoes the S_N1 reaction.
• CH₃ – CH = CH – Cl (1-chloroprop-1-ene, a vinylic halide): In this compound, the chlorine atom is directly attached to an ${\mathrm{sp}}^2$-hybridized carbon atom. The lone pair of electrons on the chlorine atom is in conjugation with the double bond:

  ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}\leftrightarrow {\mathrm{CH}}_3-{\mathrm{CH}}^{-}-\mathrm{CH}={\mathrm{Cl}}^{+}$

  This resonance resonance-imparts a partial double-bond character to the C–Cl bond, making it extremely strong and difficult to break. Furthermore, the corresponding vinylic carbocation (${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}^{+}$) is highly unstable because the positive charge resides on an ${\mathrm{sp}}^2$-hybridized carbon, which is highly electronegative. Therefore, this compound will not undergo an S_N1 reaction.
• tert-Butyl chloride (shown in the second image, (CH₃)₃C–Cl): Upon ionization, it forms a tertiary carbocation:

  $({\mathrm{CH}}_3{)}_3{C}^{+}$

  This carbocation is highly stable due to the strong inductive effect (+I) and hyperconjugation from the nine hydrogen atoms of the three methyl groups. Thus, it undergoes the S_N1 reaction very easily.

Therefore, CH₃ – CH = CH – Cl will not give S_N1 reaction.