Q2 / 75 · JEE Main 2026 (21 January Shift 1) · Chemistry
An organic compound " P " of molecular formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}$ gives positive Iodoform test but negative Tollen's test. When " P " is treated with dilute acid, it produces " Q ". " Q " gives positive Tollen's test and also iodoform test. The structure of " P " is :
- Correct answerA
- B
- C
- D
Solution
P gives positive iodoform (has $CH_3CO-$) but negative Tollen's (no $-CHO$).
Q (after acid hydrolysis) gives both tests, meaning Q has both $CH_3CO-$ and $-CHO$.
This indicates P has an acetal protecting an aldehyde group.
Option (1): $CH_3-CO-CH_2-CH(OCH_3)_2$
Has $CH_3CO-$ → positive iodoform
Has acetal $-CH(OCH_3)_2$, no free aldehyde → negative Tollen's
Hydrolysis: $-CH(OCH_3)_2 \xrightarrow{H^+/H_2O} -CHO$
Q = $CH_3-CO-CH_2-CHO$ gives both iodoform and Tollen's test
Molecular formula: $C_6H_{12}O_3$
Q (after acid hydrolysis) gives both tests, meaning Q has both $CH_3CO-$ and $-CHO$.
This indicates P has an acetal protecting an aldehyde group.
Option (1): $CH_3-CO-CH_2-CH(OCH_3)_2$
Has $CH_3CO-$ → positive iodoform
Has acetal $-CH(OCH_3)_2$, no free aldehyde → negative Tollen's
Hydrolysis: $-CH(OCH_3)_2 \xrightarrow{H^+/H_2O} -CHO$
Q = $CH_3-CO-CH_2-CHO$ gives both iodoform and Tollen's test
Molecular formula: $C_6H_{12}O_3$
Hence, the correct answer is option: 'A'.