Which one of the following halide can be used in the Friedel-Crafts reaction?

The correct option is Isopropyl chloride.

To understand why isopropyl chloride is the correct choice, let's look at the mechanism and requirements of the Friedel-Crafts alkylation reaction:

1. Mechanism of Friedel-Crafts Reaction:
In a Friedel-Crafts alkylation, an alkyl halide reacts with a Lewis acid catalyst (such as $\mathrm{Al}{\mathrm{Cl}}_3$) to generate a carbocation electrophile ($R^{+}$). This carbocation then undergoes electrophilic aromatic substitution with an aromatic ring.

2. Requirement for Carbocation Formation:
For the reaction to proceed, the carbon-halogen bond (C-X) in the halide must be easily cleaved to form a relatively stable carbocation. The ease of carbocation formation depends on the hybridization of the carbon bonded to the halogen:

- In Bromobenzene and Chlorobenzene, the halogen is attached to an ${\mathrm{sp}}^2$ hybridized carbon of the benzene ring. Due to resonance, the lone pairs on the halogen atom are delocalized into the aromatic ring, giving the C-X bond a partial double-bond character. This makes the bond extremely strong and difficult to break. Consequently, aryl carbocations do not form under Friedel-Crafts conditions.

- In Chloroethene (vinyl chloride), the chlorine is attached to an ${\mathrm{sp}}^2$ hybridized carbon of a double bond. Similar resonance occurs, imparting a partial double-bond character to the C-Cl bond and making vinyl carbocations highly unstable and difficult to form.

- In Isopropyl chloride ($({\mathrm{CH}}_3{)}_2\mathrm{CH}-\mathrm{Cl}$), the chlorine atom is bonded to an ${\mathrm{sp}}^3$ hybridized carbon. There is no resonance stabilization of the C-Cl bond. When it reacts with a Lewis acid like $\mathrm{Al}{\mathrm{Cl}}_3$, the C-Cl bond easily cleaves to yield a stable secondary isopropyl carbocation ($({\mathrm{CH}}_3{)}_2{\mathrm{CH}}^{+}$). This carbocation acts as a powerful electrophile, allowing the Friedel-Crafts alkylation to proceed successfully.