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What will be the major product when m-cresol is reacted with propargyl bromide (HC ≡ C - CH$$_2$$Br) in presence of K$$_2$$CO$$_3$$ in acetone?
This reaction follows the mechanism of a classic Williamson Ether Synthesis, which proceeds via an $$\text{S}_\text{N}2$$ nucleophilic substitution pathway:
Option A: Electrophilic Ring Substitution Products
This statement is incorrect. Under these conditions, the base deprotonates the oxygen atom, making it the most nucleophilic site. Ring alkylation (Friedel-Crafts style) does not compete with the rapid $$\text{S}_\text{N}2$$ displacement on oxygen.
Option B (Correct Matrix): Propargyl m-Tolyl Ether
This statement is correct. The nucleophilic phenoxide ion directly bonds with the propargyl group ($$-\text{CH}_2\text{--C}\equiv\text{CH}$$), giving an ether product where the intact m-cresol ring skeleton is preserved with a newly appended propargyloxy side chain: $$\text{CH}_3\text{--C}_6\text{H}_4\text{--O--CH}_2\text{--C}\equiv\text{CH}$$.
Option C & D: Elimination or Alkyne Addition Products
These statements are incorrect. Mild bases like $$\text{K}_2\text{CO}_3$$ in acetone do not cause elimination reactions on terminal propargyl halides, nor is there any active reagent or catalyst present to add across the triple bond ($$\text{C}\equiv\text{C}$$).
The reaction cleanly transforms the phenolic group into an ether linkage through an $$\text{S}_\text{N}2$$ substitution without altering the aromatic ring or the terminal alkyne structure.
Answer: Option B — Propargyl m-tolyl ether
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