JEE Main 2026 · 28 January · morning · Single correct · +4 / −1
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 options given below :
- (a) Statement I is true but Statement II is false
- (b) Both Statement I and Statement II are true
- (c) Statement I is false but Statement II is true
- (d) Both Statement I and Statement II are false
Answer and solution
Answer (b) Both Statement I and Statement II are true
Solution
Chemistry ma’am · 4 steps
Excess NaNH₂ on a dichloride means an alkyne. Find propyne first and both statements follow.
- Find Y. C₃H₆Cl₂ is a dihalide of propane, either vicinal or geminal. Excess NaNH₂ removes two HCl in a double dehydrohalogenation, so Y is propyne, CH₃-C≡CH.
- Statement I. Dilute H₂SO₄ with Hg²⁺ hydrates propyne by Markovnikov's rule. The enol tautomerises to acetone, so Z is CH₃COCH₃. A methyl ketone gives the yellow iodoform precipitate CHI₃ with NaOI, so Statement I is true.
3 CH₃-C≡CH →(red hot Fe tube) C₆H₃(CH₃)₃ (mesitylene, C₉H₁₂)
- Statement II. Three propyne molecules trimerise in the red hot iron tube to give 1,3,5-trimethylbenzene, which matches C₉H₁₂. It has 3 equivalent ring H and 9 equivalent methyl H, so the aromatic to aliphatic ratio is 3:9 = 1:3. Statement II is true.
- So. Both statements are true, which is option (b).
The idea to keep. Alkyne chains to remember: hydration with Hg²⁺ gives a ketone (aldehyde only from ethyne), and a red hot iron tube trimerises the alkyne into a benzene ring.
Answer from the NTA final answer key. Solution by PrepWiser.