JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

Given below are two statements for the following reaction sequence.

Statement I: Compound '

Z

' will give yellow precipitate with NaOI.

Statement II: Compound 'Q' has two different types of ' H ' atoms (aromatic : aliphatic) in the ratio 1 : 3 .

In the light of the above statements, choose the correct answer from the option given below:

(A)Statement I is true but Statement II is false
(B)Both Statement I and Statement II are true✓ Correct
(C)Statement I is false but Statement II is true
(D)Both Statement I and Statement II are false

Answer: (B)

Solution

The dichloride

C X_{3} H X_{6} Cl X_{2}

(X) with excess

NaNH X_{2}

loses two molecules of HCl to give propyne,

CH X_{3} - C \equiv CH

(Y). Markovnikov hydration of propyne with dilute

H X_{2} SO X_{4}

and

Hg X^{2 +}

places the oxygen on the central carbon, so Z is acetone,

CH X_{3} COCH X_{3}

; being a methyl ketone it gives the yellow iodoform precipitate with

NaOI

, so the first statement is true. Passing propyne through a red-hot iron tube trimerises it to 1,3,5-trimethylbenzene (mesitylene),

C X_{9} H X_{12}

(Q), which has three aromatic hydrogens and nine methyl hydrogens, a ratio of

1 : 3

, so the second statement is true. Both statements are true.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

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