Match List-I with List-II
|
List-I (Reaction) |
List-II (Name) |
|
(A) |
(I) Aldol Reaction |
|
(B) |
(II) Cannizzaro Reaction |
|
(C) |
(III) Haloform Reaction |
|
(D) |
(IV) Hell Volhard Zelinski Reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
These four named reactions are told apart by what the reactant must contain and what the reagent does. The aldol reaction needs an α-hydrogen next to C=O and dilute alkali. The Cannizzaro reaction is the reaction of an aldehyde without α-hydrogen with concentrated alkali. The haloform reaction needs a \( \ce{CH3CO-} \) group and a hypohalite. The Hell–Volhard–Zelinsky reaction halogenates the α-carbon of a carboxylic acid with halogen and red phosphorus.
(A) \( \ce{CH3CHO} \) with dil. NaOH — This matches (I) Aldol Reaction. Ethanal has α-hydrogens; two molecules join to give a β-hydroxy aldehyde, the aldol (3-hydroxybutanal):
\( \displaystyle \ce{2CH3CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3CH(OH)CH2CHO} \)
(B) \( \ce{CH3CH2CH2COOH} \) with \( \ce{Br2} \)/red P, then water — This matches (IV) Hell Volhard Zelinski Reaction. The α-hydrogen of the carboxylic acid is replaced by bromine, giving 2-bromobutanoic acid:
\( \displaystyle \ce{CH3CH2CH2COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{Br2}\text{/red P}} \ce{CH3CH2CHBrCOOH} \)
(C) \( \ce{C6H5CHO} \) with conc. KOH — This matches (II) Cannizzaro Reaction. Benzaldehyde has no α-hydrogen; one molecule is reduced to the alcohol and another oxidised to the salt of the acid (self oxidation-reduction):
\( \displaystyle \ce{2C6H5CHO} \xrightarrow{\text{conc. KOH}} \ce{C6H5CH2OH + C6H5COOK} \)
(D) \( \ce{CH3COCH3} \) with NaOCl — This matches (III) Haloform Reaction. The methyl ketone is chlorinated at the methyl group and cleaved to chloroform and the salt of an acid with one carbon atom less:
\( \displaystyle \ce{CH3COCH3 + 3NaOCl -> CHCl3 + CH3COONa + 2NaOH} \)
Shortcut: (A)-(I) appears in Options (1), (2) and (3); (B)-(IV) then leaves only Option (1).
Note: It seems that there is a mistake in the original NTA paper, which prints item (B) as CH3CH3CH2COOH → CH2CH2CHBrCOOH when it should read CH3CH2CH2COOH → CH3CH2CHBrCOOH.