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Friday, 9 October

JEE Main 2026 · ChemistryMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2026, Shift 2, Chemistry Q66

Question 66 of 75 in this shift, Chemistry question 16 of 25, Section A.

An organic compound 'x' where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced 'y'. The most likely structure of 'y' is: [Molar mass of 'x' is 58 g mol−1^{-1}]
  1. (1)CH2=CH−C(=O)−OHCH_2 = CH - C(=O) - OH
  2. (2)CH3−CH=CH−CH=OCH_3 - CH = CH - CH = O
  3. (3)\begin{array}[t]{c} O \\ || \\ O - C - OH \\ | \\ CH_2 - OH \end{array}Official answer
  4. (4)\begin{array}[t]{c} O \\ || \\ CH_3 - C - OH \end{array}

Official answer

Option 3

NTA final key.

Chapter
Aldehydes, Ketones and Carboxylic Acids
Topic
Aldol and Cannizzaro
Idea tested
Intramolecular Cannizzaro Reaction

Same idea in other shifts

Asked 2× in all
  1. 3 Apr 2025, Shift 2 · Q68The major product (P) in the following reaction is : Ph−CO−CHO\mathrm{Ph{-}CO{-}CHO} →ΔKOH\xrightarrow[\Delta]{\mathrm{KOH}} P (Major Product)MediumDiagram basedHas a figure

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. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.