JEE Main 2020 · 2 September · evening · Single correct · +4 / −1
Two compounds A and B with same molecular formula (C₃H₆O) undergo Grignard’s reaction with methylmagnesium bromide to give products C and D. Products C and D show following chemical tests.
C and D respectively are :
- (a) CH₃; |; C =, H₃C, -, C, -, OH;; |; CH₃; D =, H₃C, -, CH₂, -, CH, -, CH₃; |; OH
- (b) C = H₃C-CH₂-CH₂-CH₂-OH;; D = H₃C-CH₃; |; C; |; CH₃-OH
- (c) C = H₃C-CH₂-CH₂-CH₂-OH;; D = H₃C-CH₂-CH-CH₃
- (d) C = H₃C-CH₂-C(|)(OH)H-CH₃;; D = H₃C-C(|(CH₃))-OH
Answer and solution
Answer (d) C = H₃C-CH₂-C(|)(OH)H-CH₃;; D = H₃C-C(|(CH₃))-OH
Solution
Chemistry ma’am · 4 steps
Find the two C₃H₆O carbonyls first, then let Lucas speed and iodoform sort the alcohols.
- Identify A and B. The two carbonyl compounds with formula C₃H₆O are propanal, CH₃CH₂CHO, and propanone (acetone), CH₃COCH₃. CH₃MgBr adds a CH₃ group to the carbonyl carbon, and hydrolysis then gives an alcohol.
- Grignard products. Propanal gives butan-2-ol, CH₃CH₂CH(OH)CH₃, a 2° alcohol. Acetone gives 2-methylpropan-2-ol, (CH₃)₃COH, a 3° alcohol. Both are alcohols, so both give a positive ceric ammonium nitrate test.
Alcohol Type Lucas test Iodoform test Butan-1-ol 1° no turbidity at room temperature negative Butan-2-ol 2° turbidity in about 5 minutes positive: has a CH₃CH(OH)⁻ group 2-Methylpropan-2-ol 3° turbidity immediately negative: no H on the carbinol carbon - Match the tests. C goes cloudy after five minutes and gives iodoform, so C is the 2° alcohol, butan-2-ol. D goes cloudy at once and fails the iodoform test, so D is the 3° alcohol, 2-methylpropan-2-ol.
- So. C is butan-2-ol and D is 2-methylpropan-2-ol, which is option (d).
The idea to keep. The Lucas test sorts alcohols by speed: 3° clouds at once, 2° in about 5 minutes, 1° not at room temperature. Iodoform needs a CH₃CH(OH)⁻ group, so a 3° alcohol always fails it.
Answer from the NTA final answer key. Solution by PrepWiser.