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Friday, 9 October

JEE Main 2023 · ChemistryMultiple choiceDiagram basedHardMulti-step

JEE Main 11 April 2023, Shift 1, Chemistry Q75

Question 75 of 90 in this shift, Chemistry question 15 of 30, Section A.

[figure: 2-alkylcyclohexanone (R on the ring carbon adjacent to C=O, R = alkyl)] →KMnO4\xrightarrow{\mathrm{KMnO_4}} 'A' (Major Product) →(ii) H3O+(i) NH2NH2, KOH\xrightarrow[\text{(ii) } \mathrm{H_3O^+}]{\text{(i) } \mathrm{NH_2NH_2,\ KOH}} 'B' (Major Product) (R = alkyl) 'A' and 'B' in the above reactions are:

The figure, in words

Starting material is 2-R-cyclohexanone (R = alkyl); options are open-chain ring-cleavage products A and B.
  1. (1)[figure: A = R-CH(CO2H)-CH2CH2CH2CH2-CHO (acid on the carbon bearing R, terminal CHO); B = R-CH(CO2H)-CH2CH2CH2CH2-CH3]
  2. (2)[figure: A = R-CH(CO2H)-CH2CH2CH2CH2-CO2H; B = R-CH(CONHNH2)-CH2CH2CH2CH2-CONH-NH2]
  3. (3)[figure: A = R-C(=O)-CH2CH2CH2CH2-CO2H (keto acid); B = R-CH2CH2CH2CH2CH2CH2-CO2H]Official answer
  4. (4)[figure: A = R-C(=O)-CH2CH2CH2CH2-CHO (keto aldehyde); B = R-CH2CH2CH2CH2CH2CH2-CH3]

Official answer

Option 3

NTA final key.

Chapter
Aldehydes, Ketones and Carboxylic Acids
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.