Exams › JEE Main › Chemistry

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. Test | C | D Ceric ammonium nitrate Test | Positive | Positive Lucas Test | Turbidity obtained after five minutes | Turbidity obtained immediately Iodoform Test | Positive | Negative C and D respectively are :

1. C = H3C–C(CH3)2–OH ; D = H3C–CH2–CH(OH)–CH3
2. C = H3C–CH2–CH2–CH2–OH ; D = H3C–C(OH)(CH3)2
3. C = H3C–CH2–CH(OH)–CH3 ; D = H3C–C(CH3)2–OH
4. C = H3C–CH2–CH2–CH2–OH ; D = H3C–CH2–CH(OH)–CH3

Correct answer: C = H3C–C(CH3)2–OH ; D = H3C–CH2–CH(OH)–CH3

Solution

The correct option identifies C as a tertiary alcohol, which reacts quickly in the Lucas test, resulting in immediate turbidity, while D is a secondary alcohol that reacts more slowly, explaining the five-minute delay. Additionally, the Iodoform test is positive for C due to the presence of a methyl ketone structure, while D lacks this structure, resulting in a negative test.

Related JEE Main Chemistry questions

• Why is ortho-nitrophenol less soluble in water than the meta- and para-nitrophenol isomers?
• Which of the following structures represents vinylcarbinol?
• Which separation technique is most appropriate for a 1:1 mixture of ortho-nitrophenol and para-nitrophenol?
• Which of the following is NOT a starting material used for the preparation of iodoform?
• Iodoform is formed from all of the following except:
• Which reagent can convert CH$_3$CH$_2$OH into CH$_3$CHO?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →