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← IIT-JEE 2009 · Paper 1
ChemistrySingle correct+4 / -1Question 17 of 60

IIT-JEE 2009 · Paper 1, Chemistry question 17

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Paragraph for Question Nos. 16 to 18

A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.

P →2. HX+,HX2O  3. HX2SOX4, Δ1. MeMgBr Q →2. Zn, HX2O1. OX3 R →2. Δ1. OHX− S\mathbf P\ \xrightarrow[\text{2. }\ce{H+, H2O}\ \ \text{3. }\ce{H2SO4},\ \Delta]{\text{1. MeMgBr}}\ \mathbf Q\ \xrightarrow[\text{2. Zn, }\ce{H2O}]{\text{1. }\ce{O3}}\ \mathbf R\ \xrightarrow[\text{2. }\Delta]{\text{1. }\ce{OH-}}\ \mathbf S

The structures of the products Q and R, respectively, are

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Answer and worked solution

Official answer: A

Q. Cyclisation of the cation Ph−CH=C(Me)−CX+MeX2\ce{Ph-CH=C(Me)-C+Me2} onto the ortho carbon gives 1,1,2-trimethyl-1H-indene (C1 bears two Me, the C2=C3 double bond carries a Me at C2).

R. Ozonolysis cleaves C2=C3: C3 (attached to the ring) becomes CHO and C2 becomes a methyl ketone, still attached to CMeX2\ce{CMe2}:
R=2-(1,1-dimethyl-2-oxopropyl)benzaldehyde, OHC−CX6HX4−CMeX2−COCHX3.\text{R}=\text{2-(1,1-dimethyl-2-oxopropyl)benzaldehyde},\ \ce{OHC-C6H4-CMe2-COCH3}.

(A)