Which of the following will not give S_N1.

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

To understand why this compound does not undergo the $S_{\text{N}}1$ mechanism, let us analyze the reactivity of each given option towards nucleophilic substitution:

1. CH₂ = CH – CH₂ – Cl (Allyl chloride):
When the chloride ion leaves, it forms the allyl carbocation:
${\text{CH}}_2=\text{CH}-{\text{CH}}_2^{+}$
This carbocation is highly stable because the positive charge is delocalized over the adjacent carbon-carbon double bond via resonance:
${\text{CH}}_2=\text{CH}-{\text{CH}}_2^{+}\leftrightarrow {\text{CH}}_2^{+}-\text{CH}={\text{CH}}_2$
Therefore, it readily undergoes $S_{\text{N}}1$ reaction.

2. Benzyl chloride (shown in the first image, C₆H₅ – CH₂ – Cl):
Ionization of benzyl chloride yields the benzyl carbocation:
${\text{C}}_6{\text{H}}_5-{\text{CH}}_2^{+}$
This carbocation is exceptionally stable due to the resonance delocalization of the positive charge throughout the aromatic benzene ring. Hence, it undergoes $S_{\text{N}}1$ reactions very easily.

3. tert-Butyl chloride (shown in the second image, (CH₃)₃C – Cl):
Upon loss of chloride, it forms a tertiary carbocation:
${\left({\text{CH}}_3\right)}_3{\text{C}}^{+}$
A tertiary carbocation is stabilized by nine hyperconjugative hydrogens (from three methyl groups) and strong electron-donating inductive effects (+I). Thus, it undergoes $S_{\text{N}}1$ reactions very rapidly.

4. CH₃ – CH = CH – Cl (1-chloroprop-1-ene / Vinylic halide):
In this compound, the chlorine atom is directly attached to an sp² hybridized carbon atom (vinylic carbon). The lone pairs on the chlorine atom undergo resonance with the double bond:
${\text{CH}}_3-\text{CH}=\text{CH}-\ddot{\text{Cl}}\leftrightarrow {\text{CH}}_3-{\text{CH}}^{-}-\text{CH}={\text{Cl}}^{+}$
This resonance imparts a partial double-bond character to the C – Cl bond, making it shorter, stronger, and extremely difficult to cleave. Furthermore, if the C – Cl bond were to break, it would produce a vinylic carbocation:
${\text{CH}}_3-\text{CH}={\text{CH}}^{+}$
This vinylic carbocation is highly unstable because the positive charge resides on a highly electronegative sp² hybridized carbon. Due to both the strength of the C – Cl bond and the instability of the resulting carbocation, this compound does not undergo $S_{\text{N}}1$ substitution reactions.