Match List-I with List-II
|
List-I (Name reaction.) |
List-II (Reaction) |
|
(A) Williamson ether synthesis |
(I) |
|
(B) Hydroboration oxidation |
(II) |
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(C) Reimer-Tiemann reaction |
(III) |
|
(D) Friedel Craft Acylation reaction |
(IV) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
These four reactions make or change alcohols, phenols and ethers, and each is recognised by its reagent: an alkoxide with an alkyl halide (Williamson synthesis), diborane followed by alkaline hydrogen peroxide (hydroboration-oxidation), chloroform with aqueous NaOH on a phenol (Reimer-Tiemann reaction), and an acyl chloride with anhydrous \( \ce{AlCl3} \) on an aromatic ring (Friedel-Crafts acylation).
(A) Williamson ether synthesis — This matches (IV), sodium isopropoxide with methyl iodide. The alkoxide ion displaces iodide from the primary halide by \( S_N2 \) attack, giving 2-methoxypropane:
\( \displaystyle \ce{(CH3)2CHONa + CH3I -> (CH3)2CH-O-CH3 + NaI} \)
(B) Hydroboration oxidation — This matches (III), but-1-ene with \( \ce{B2H6} \) and then \( \ce{H2O2} \)/\( \ce{OH-} \). Boron adds to the terminal carbon, and oxidation puts –OH in its place, so the product is butan-1-ol (anti-Markovnikov addition of water):
\( \displaystyle \ce{3CH3CH2CH=CH2 + BH3 -> (CH3CH2CH2CH2)3B} \)
\( \displaystyle \ce{(CH3CH2CH2CH2)3B} \xrightarrow{\ce{3H2O2}/\ce{OH-}} \ce{3CH3CH2CH2CH2OH + B(OH)3} \)
(C) Reimer-Tiemann reaction — This matches (II), phenol with chloroform and aqueous NaOH. A –CHO group enters the ortho position, giving salicylaldehyde after acidification:
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{, aq. NaOH}} o\text{-}\ce{HOC6H4CHO} \)
(D) Friedel Craft Acylation reaction — This matches (I), anisole with acetyl chloride and anhydrous \( \ce{AlCl3} \). The –\( \ce{OCH3} \) group directs the acyl group to the ortho and para positions, giving methoxyacetophenones:
\( \displaystyle \ce{C6H5OCH3 + CH3COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{CH3OC6H4COCH3 + HCl} \)
Shortcut: (A)-(IV) appears only in Option (3).