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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Aldehydes and Ketones

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Aldehydes and Ketones

An organic compound " P " of molecular formula C_ 6 H_ 12 O_ 3...

An organic compound " P " of molecular formula C6H12O3\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 :An organic compound " P " of molecular formula C_ 6 H_ 12 O_ 3 gives positive Iodoform test but negative Tollen's test.An organic compound " P " of molecular formula C_ 6 H_ 12 O_ 3 gives positive Iodoform test but negative Tollen's test.An organic compound " P " of molecular formula C_ 6 H_ 12 O_ 3 gives positive Iodoform test but negative Tollen's test.An organic compound " P " of molecular formula C_ 6 H_ 12 O_ 3 gives positive Iodoform test but negative Tollen's test.

Options

  1. AThe number of species from the following which are paramagnetic and with bond order equal to one is H _ 2 , He _ 2 ^ +
  2. BThe number of species from the following which are paramagnetic and with bond order equal to one is H _ 2 , He _ 2 ^ +
  3. CThe number of species from the following which are paramagnetic and with bond order equal to one is H _ 2 , He _ 2 ^ +
  4. DThe number of species from the following which are paramagnetic and with bond order equal to one is H _ 2 , He _ 2 ^ +

Correct answer

A. The number of species from the following which are paramagnetic and with bond order equal to one is H _ 2 , He _ 2 ^ +

Step-by-step solution

P gives positive iodoform (has CH3CO−CH_3CO-) but negative Tollen's (no −CHO-CHO).
Q (after acid hydrolysis) gives both tests, meaning Q has both CH3CO−CH_3CO- and −CHO-CHO.
This indicates P has an acetal protecting an aldehyde group.
Option (1): CH3−CO−CH2−CH(OCH3)2CH_3-CO-CH_2-CH(OCH_3)_2
Has CH3CO−CH_3CO- → positive iodoform
Has acetal −CH(OCH3)2-CH(OCH_3)_2, no free aldehyde → negative Tollen's
Hydrolysis: −CH(OCH3)2→H+/H2O−CHO-CH(OCH_3)_2 \xrightarrow{H^+/H_2O} -CHO
Q = CH3−CO−CH2−CHOCH_3-CO-CH_2-CHO gives both iodoform and Tollen's test
Molecular formula: C6H12O3C_6H_{12}O_3
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