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Saturday, 31 October

JEE Main 2022 · ChemistryMultiple choiceDiagram basedMediumMulti-step

JEE Main 26 July 2022, Shift 1, Chemistry Q74

Question 74 of 90 in this shift, Chemistry question 14 of 30, Section A.

The products formed in the following reaction, A and B are [figure: six-membered ring with a C=O at the top vertex; the bottom ring carbon bears H and CHO] →[Ag(NH3)2]+ OH−\xrightarrow{[\mathrm{Ag(NH_3)_2}]^+\ \mathrm{OH^-}} A →NaBH4\xrightarrow{\mathrm{NaBH_4}} B

The figure, in words

Stem: six-membered ring with =O at the top vertex and, at the bottom vertex, H and CHO substituents; reagents [Ag(NH3)2]+ OH- then NaBH4. Options are pairs of drawn ring structures A and B.
  1. (1)A = [figure: six-membered ring, C=O (O) at top vertex, bottom ring carbon bearing H and CH2OH], B = [figure: six-membered ring, top ring carbon bearing H and OH, bottom ring carbon bearing H and CH2OH]
  2. (2)A = [figure: six-membered ring, top ring carbon bearing H and OH, bottom ring carbon bearing H and CHO], B = [figure: six-membered ring, top ring carbon bearing H and OH, bottom ring carbon bearing H and CH2OH]
  3. (3)A = [figure: six-membered ring, C=O (O) at top vertex, bottom ring carbon bearing H and COOH], B = [figure: six-membered ring, top ring carbon bearing H and OH, bottom ring carbon bearing H and COOH]Official answer
  4. (4)A = [figure: six-membered ring, C=O (O) at top vertex, bottom ring carbon bearing H and COOH], B = [figure: six-membered ring, top ring carbon bearing H and OH, bottom ring carbon bearing H and CH2OH]

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 108 of the 174 JEE Main shifts held from 2017 to 2026.