See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Friedel-Crafts alkylation requires the generation of a carbocation (or carbocation-like electrophile) from an alkyl halide in the presence of a Lewis acid (AlCl3). The carbocation then attacks the benzene ring. Step 1 - Evaluate each option for its ability to form a carbocation with AlCl3: (a) (CH3)3CCl (tert-butyl chloride): AlCl3 readily abstracts Cl- to form a stable tertiary carbocation (CH3)3C+. This carbocation attacks benzene to give tert-butylbenzene. So this WILL give a Friedel-Crafts product. (b) CH2=CHCH2Cl (allyl chloride): AlCl3 abstracts Cl- to form an allylic carbocation CH2=CH-CH2+ (resonance-stabilized). This carbocation can attack benzene. So this WILL give a Friedel-Crafts product. (c) CH3CH2Cl (ethyl chloride): AlCl3 abstracts Cl- to form CH3CH2+ (ethyl carbocation). This carbocation attacks benzene to give ethylbenzene. So this WILL give a Friedel-Crafts product. (d) CH2=CHCl (vinyl chloride): The carbon bearing the Cl is sp2-hybridized. The C-Cl bond has partial double bond character due to resonance between the lone pairs on Cl and the pi system. This makes the C-Cl bond much stronger and harder to cleave. AlCl3 cannot readily abstract Cl- to form a vinyl carbocation (CH2=CH+) because vinyl carbocations are extremely unstable (the positive charge is on an sp2 carbon, which is less able to stabilize positive charge compared to sp3 carbons). Therefore, no carbocation is formed and no Friedel-Crafts alkylation occurs. Step 2 - Why other options fail to be the answer: Options (a), (b), and (c) all readily form carbocations (tertiary, allylic, and primary/secondary respectively) that are stable enough to carry out electrophilic aromatic substitution. Step 3 - Conclusion: Vinyl chloride (CH2=CHCl) does not undergo Friedel-Crafts alkylation because it cannot form a stable vinyl carbocation under these conditions. Therefore, the correct answer is D.