JEE Advanced 2017 · Paper 2 · Chemistry Q14

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ChemistryMultiple correct+4 / -2Question 32 of 54

Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is $\ce{C8H8O}$ . Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction.

(i) P $\xrightarrow{\text{i) } \ce{O3/CH2Cl2};\ \text{ii) } \ce{Zn/H2O}}$ Q ( $\ce{C8H8O}$ )

(ii) R $\xrightarrow{\text{i) } \ce{O3/CH2Cl2};\ \text{ii) } \ce{Zn/H2O}}$ S ( $\ce{C8H8O}$ )

The option(s) with suitable combination of P and R, respectively, is(are)

P = 4-methylstyrene; R = α-methylstyrene (2-phenylpropene)

P = 1-(3-methylphenyl)prop-1-ene; R = 3-methyl-2-phenylbut-2-ene

P = 1-(2-methylphenyl)-2-methylprop-1-ene; R = 2-(2-methylphenyl)but-2-ene

P = 4-methylstyrene; R = 4-methyl-α-methylstyrene

Answer and worked solution

Official answer: A, B

Identify Q and S.

Q ( $\ce{C8H8O}$ ) undergoes Cannizzaro but not haloform, so it is an aldehyde with no α-H: a methylbenzaldehyde, $\ce{CH3C6H4CHO}$ .
S ( $\ce{C8H8O}$ ) undergoes haloform but not Cannizzaro, so it is acetophenone, $\ce{C6H5COCH3}$ .

Check the options.

(A) 4-Methylstyrene gives 4-methylbenzaldehyde (+ HCHO). α-Methylstyrene gives acetophenone (+ HCHO). ✓
(B) $\ce{3-CH3C6H4CH=CHCH3}$ gives 3-methylbenzaldehyde (+ $\ce{CH3CHO}$ ). $\ce{C6H5C(CH3)=C(CH3)2}$ gives acetophenone (+ acetone). ✓
(C) The second alkene gives 2-methylacetophenone, which is $\ce{C9H10O}$ . ✗
(D) The second alkene gives 4-methylacetophenone, also $\ce{C9H10O}$ . ✗

(A), (B)

Topics:Alkenes Named reactions

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JEE Advanced 2017 · Paper 2, Chemistry question 14