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Saturday, 31 October

JEE Main 2022 · ChemistryMultiple choiceDiagram basedHardApplication

JEE Main 30 July 2022, Shift 2, Chemistry Q44

Question 44 of 90 in this shift, Chemistry question 14 of 30, Section A.

Major product of the following reaction is [figure: skeletal line drawing of a chain with a terminal C=C double bond at the left end, a carbon bearing two short line substituents, a CH2, then a carbon with =O and OH (printed O and OH); arrow with I2\mathrm{I_2} above and NaHCO3\mathrm{NaHCO_3} below]

The figure, in words

Stem: skeletal chain with terminal C=C, a carbon with two short line substituents, CH2, C(=O)OH as printed (O, OH); reagents over arrow: I2 / NaHCO3. Options A-D are skeletal drawings (two rings, two open chains) each with an I.
  1. (1)[figure: six-membered ring containing one ring O, the ring carbon next to O carries =O (printed O), a ring carbon bearing two short line substituents, and a ring carbon bearing I (printed I)]
  2. (2)[figure: five-membered ring containing one ring O, the ring carbon next to O carries =O (printed O), a ring carbon bearing two short line substituents, and a ring carbon bearing a CH2 line to I (printed I)]Official answer
  3. (3)[figure: open chain with I (printed I) on a branching carbon, a carbon bearing two short line substituents, a CH2, then a carbon with =O and OH (printed O and OH)]
  4. (4)[figure: open chain with I (printed I) at the left end, a carbon bearing two short line substituents, a CH2, then a carbon with =O and OH (printed O and OH)]

Official answer

Option 2

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.