The following chemical transformations can be achieved through a 4-st…

The correct answer is Option (2) → B, A, D, C

In a multi-step synthesis on a benzene ring, the order of the steps matters, because each group already on the ring directs where the next group goes. –Cl is ortho/para directing, while an acyl group (–COCH₃) is meta directing. The product has Cl and the ethyl group para to each other and no –OH group.

(B) Zn dust first: heating phenol with zinc dust removes the –OH group and gives benzene.

    \( \displaystyle \ce{C6H5OH + Zn -> C6H6 + ZnO} \)

(A) Cl₂/FeCl₃: electrophilic chlorination gives chlorobenzene.

    \( \displaystyle \ce{C6H6 + Cl2} \xrightarrow{\ce{FeCl3}} \ce{C6H5Cl + HCl} \)

(D) CH₃COCl/AlCl₃: Friedel–Crafts acylation. Because Cl is ortho/para directing, the acetyl group enters mainly at the para position, giving 4-chloroacetophenone.

    \( \displaystyle \ce{C6H5Cl + CH3COCl} \xrightarrow{\ce{AlCl3}} \ce{4-ClC6H4COCH3 + HCl} \)

(C) Zn(Hg)/HCl: Clemmensen reduction turns the C=O into \( \ce{CH2} \), giving 1-chloro-4-ethylbenzene.

    \( \displaystyle \ce{4-ClC6H4COCH3} \xrightarrow{\ce{Zn(Hg)/HCl}} \ce{4-ClC6H4CH2CH3} \)

If the acylation were done before the chlorination (B, D, A, C), the meta directing acetyl group would send Cl to the meta position, and the para product would not form.

Shortcut: The –OH must be removed first, so the sequence starts with B (Options 1 and 2). Cl has to be on the ring before acylation to get the para product, which leaves Option (2).