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Acetophenone (C6H5COCH3) is subjected to the following sequence of reactions: (i) Cl2 with anhydrous AlCl3, (ii) Zn-Hg with concentrated HCl, (iii) HNO3 with H2SO4. Which of the following correctly describes the major product Z?

1. A benzene ring bearing an ethyl group at C-1, a chloro group at C-3, and a nitro group at C-4
2. A benzene ring bearing an acetyl group at C-1, a chloro group at C-2, and a nitro group at C-4
3. A benzene ring bearing an ethyl group at C-1, a chloro group at C-4, and a nitro group at C-2
4. A benzene ring bearing an ethyl group at C-1, a chloro group at C-3, and a nitro group at C-2

Correct answer: A benzene ring bearing an ethyl group at C-1, a chloro group at C-3, and a nitro group at C-4

Solution

Step (i): Cl2/AlCl3 is an electrophilic aromatic substitution (Friedel-Crafts). The acetyl group (COCH3) is an electron-withdrawing meta-director, so Cl enters at the meta position giving 3-chloroacetophenone. Step (ii): Zn-Hg/conc. HCl is the Clemmensen reduction which converts C=O to CH2, so COCH3 becomes C2H5, giving 3-chloroethylbenzene (ethyl at C-1, Cl at C-3). Step (iii): Nitration. Ethyl group at C-1 is o/p director; Cl at C-3 is also an o/p director (relative to its own position, directing to C-2, C-4, C-6). Both groups cooperatively direct the incoming NO2 to position C-4 (para to ethyl, ortho to Cl). Major product is 1-ethyl-3-chloro-4-nitrobenzene.

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