Friedel Crafts

Answer: C. Friedel-Crafts reactions require a Lewis acid catalyst — anhydrous AlCl₃ is the classic choice (NCERT Class 11 Chemistry, Chapter 9, Part 2, page 322). The catalyst must be anhydrous because water decomposes AlCl₃.

Why A is wrong: A is wrong because aqueous HCl is a Brønsted acid and cannot generate the required carbocation electrophile from R–Cl. Friedel-Crafts specifically needs a Lewis acid.

Why B is wrong: B is wrong because Pd/C is a hydrogenation catalyst (used for reduction reactions), not an electrophilic aromatic substitution catalyst.

Why D is wrong: D is wrong because concentrated H₂SO₄ is used for sulfonation of benzene, not for Friedel-Crafts reactions. It cannot activate alkyl halides.

Answer: A. The nitro group (–NO₂) is a strong meta-directing deactivator that makes the ring too electron-poor for the electrophile to attack. Friedel-Crafts reactions require a sufficiently nucleophilic aromatic ring (NCERT Class 11 Chemistry, Chapter 9, Part 2).

Why B is wrong: B is wrong because –OH is a strong activating group (o,p-director). Phenol undergoes electrophilic substitution easily. However, –OH can coordinate with AlCl₃, so in practice modified conditions may be needed — but the reaction does NOT 'fail' as it does with –NO₂.

Why C is wrong: C is wrong because –CH₃ is an activating group (electron-donating via hyperconjugation). Toluene readily undergoes Friedel-Crafts reactions — in fact, it reacts faster than benzene.

Why D is wrong: D is wrong because –Cl is a weak deactivator but still allows Friedel-Crafts reactions to proceed. Chlorobenzene undergoes alkylation/acylation, though more slowly than benzene.

Answer: B. The acyl group (–COR) is an electron-withdrawing group that deactivates the aromatic ring toward further electrophilic attack. This prevents polysubstitution — a key advantage of acylation over alkylation.

Why A is wrong: A is wrong because the acylium ion (RC≡O⁺) is actually resonance-stabilised and quite stable — that stability is why it does NOT rearrange. The stopping of the reaction is due to product deactivation, not reagent instability.

Why C is wrong: C is wrong because AlCl₃ is regenerated during the reaction mechanism (it is a catalyst, used in stoichiometric amounts for acylation due to complexation, but it is not 'consumed' in the sense of preventing further reaction).

Why D is wrong: D is wrong because acylation does not require higher temperatures that would decompose the product. The reaction is typically done at room temperature or mild heating. The stopping point is electronic, not thermal.

Answer: B. The primary carbocation (CH₃CH₂CH₂⁺) formed from n-propyl chloride rearranges via a 1,2-hydride shift to the more stable secondary carbocation (CH₃CH⁺CH₃). This isopropyl cation attacks benzene to give isopropylbenzene (cumene). Carbocation rearrangement is a hallmark problem of Friedel-Crafts alkylation.

Why A is wrong: A is wrong because it assumes no rearrangement occurs. The primary carbocation is less stable than the secondary carbocation, so a 1,2-hydride shift gives the isopropyl cation — the product is isopropylbenzene, not n-propylbenzene.

Why C is wrong: C is wrong because propanal would be a product of oxidation or ozonolysis, not Friedel-Crafts alkylation. No aldehyde is formed when an alkyl halide reacts with benzene and AlCl₃.

Why D is wrong: D is wrong because biphenyl formation requires a coupling reaction (e.g., Wurtz-Fittig or Ullmann), not Friedel-Crafts alkylation with an alkyl halide.

Answer: B. Acylation with propanoyl chloride gives phenyl ethyl ketone (no rearrangement, because the acylium ion is resonance-stabilised). Subsequent Clemmensen reduction (Zn-Hg/conc. HCl) reduces the C=O to CH₂, yielding n-propylbenzene. This two-step route avoids the carbocation rearrangement inherent in direct alkylation.

Why A is wrong: A is wrong because the primary carbocation from n-propyl chloride rearranges to a secondary carbocation (1,2-hydride shift), giving isopropylbenzene — not the desired straight-chain product.

Why C is wrong: C is wrong because isopropyl chloride gives a secondary carbocation directly, which attaches as an isopropyl group — this is branched, not the straight-chain propyl group required.

Why D is wrong: D is wrong because propan-1-ol with H₂SO₄ also generates a primary carbocation that rearranges to secondary. The same rearrangement problem as option A occurs regardless of the leaving group source.

Answer: D. The methyl group is an electron-donating group (hyperconjugation + inductive effect) that activates the aromatic ring. Toluene is therefore more reactive than benzene, so it undergoes further alkylation to give xylenes and trimethylbenzenes — the classic polyalkylation problem of Friedel-Crafts alkylation.

Why A is wrong: A is wrong because the issue is not catalyst selectivity — it is the electronic activation of the product ring. Even a perfectly selective catalyst cannot prevent polyalkylation when the product is inherently more reactive than the starting material.

Why B is wrong: B is wrong because polyalkylation occurs even at low temperatures. The problem is thermodynamic/electronic (product more reactive than reactant), not kinetic (temperature-dependent).

Why C is wrong: C is wrong because CH₃Cl does not disproportionate. The methyl carbocation (or methyl-AlCl₃ complex) directly attacks the ring. Multiple substitutions occur because the product ring is more nucleophilic, not because the reagent changes.

Answer: A. The lone pair on the –NH₂ nitrogen acts as a Lewis base and forms a coordinate bond with the Lewis acid AlCl₃. This AlCl₃-amine complex is stable, effectively removing the catalyst from the reaction mixture. Without active catalyst, the electrophile is never generated.

Why B is wrong: B is wrong because –NH₂ is actually a strong activating group (lone pair donation into the ring). It makes the ring MORE electron-rich. The failure is due to catalyst poisoning by the amine, not ring deactivation.

Why C is wrong: C is wrong because –NH₂ is a small group with minimal steric demands. It does not physically block ring positions. Even large substituents rarely prevent Friedel-Crafts by steric hindrance alone — the issue here is purely electronic (catalyst coordination).

Why D is wrong: D is wrong because there is no polymerisation concern with aromatic amines under Friedel-Crafts conditions. The reaction simply does not proceed because the Lewis acid catalyst is sequestered by the amine nitrogen.

Answer: C. C is correct. Step 1: acylation puts an acetyl group on the ring, giving acetophenone, C₆H₅COCH₃. Step 2: Clemmensen reduction converts the C=O into CH₂ and keeps every carbon, so –COCH₃ becomes –CH₂CH₃ and the product is ethylbenzene. Step 3: the ethyl group activates the ring and directs ortho/para; steric hindrance next to the ethyl group makes para the major site, giving 4-ethylacetophenone. Watch the carbon count in step 2: reduction removes the oxygen, not a carbon, so the product is ethylbenzene, not toluene.

Why A is wrong: A is wrong because the ethyl group is an ortho/para director. It donates electron density to the ring, which stabilises attack at the ortho and para positions; meta substitution is major only when the existing group withdraws electrons (–NO₂, –COOH, –COCH₃).

Why B is wrong: B is wrong because although ethyl directs to both ortho and para, the ortho position sits next to the ethyl group, and the bulky acylium electrophile is hindered there. The para position is unhindered, so the para isomer is the major product.

Why D is wrong: D is wrong because ethylbenzene is MORE reactive than benzene towards electrophiles: the alkyl group activates the ring. The ring would be deactivated, and Friedel–Crafts would fail, only if a strongly electron-withdrawing group were present.