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← IIT-JEE 2009 · Paper 1
ChemistrySingle correct+4 / -1Question 16 of 60

IIT-JEE 2009 · Paper 1, Chemistry question 16

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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 structure of the carbonyl compound P is

Choose one option
Answer and worked solution

Official answer: B

Work backwards from the answer to Q17. Q must be 1,1,2-trimethylindene, so P has the carbon skeleton Ph−CH=C(CHX3)−CO−CHX3\ce{Ph-CH=C(CH3)-CO-CH3}.

Check. P is a methyl ketone (positive iodoform test). MeMgBr gives the tertiary alcohol Ph−CH=C(Me)−C(OH)MeX2\ce{Ph-CH=C(Me)-C(OH)Me2}; with HX2SOX4\ce{H2SO4} the tertiary (allylic) carbocation attacks the benzene ring (the alkene is cis to Ph), closing the five-membered ring of the indene Q.

(B) 3-methyl-4-phenylbut-3-en-2-one