JEE MainChemistryAldehydes Ketones and Carboxylic AcidsMCQ+4 / −1
Match List - I and List - II.
Choose the correct answer from the options given below :
| List-I | List-II | ||
|---|---|---|---|
| (a) | ![]() | (i) | |
| (b) | ![]() | (ii) | |
| (c) | ![]() | (iii) | Conc. |
| (d) | ![]() | (iv) | Red |
Choose the correct answer from the options given below :
- A(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
- B(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
- C(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
- D(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
View written solutionFree
Correct answer: A
We need to match the reagents in List-II with the named reactions/transformations in List-I.
Although List-I names are not visible in the text extraction, the reagents in List-II are standard and correspond to well-known reactions:
-
This is the Haloform reaction (for methyl ketones / compounds oxidizable to methyl ketones). -
This is Rosenmund reduction, which reduces acid chlorides to aldehydes. -
This is Clemmensen reduction, reducing carbonyl group to methylene. -
This is Hell-Volhard-Zelinsky (HVZ) reaction, i.e. -halogenation of carboxylic acids.
Now compare with the options.
- Rosenmund reduction = (ii)
- HVZ reaction = (iv)
- Haloform reaction = (i)
- Clemmensen reduction = (iii)
Thus the correct matching is:
This corresponds to Option A.
Final Answer
Option A is correct.
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