JEE Advanced 2014 · Paper 2 · Chemistry Q37

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ChemistryList match+3 / -1Question 37 of 60

Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists.

Peroxyester R-CO-O-O-R′ with curved arrows on the O-O bond; four thermal pathways. P: loses CO₂ to give R• + R′O•. Q: loses CO₂ to give R• + R′O•, then R′O• → R• + X• + carbonyl compound (gas). R: gives RCO₂• + R′O•, then loses CO₂ to give R• + X′• + carbonyl compound (gas). S: gives RCO₂• + R′O•, then loses CO₂ to give R• + R′O•.

Methyl phenylperacetate, C₆H₅CH₂-CO-O-O-CH₃.

C₆H₅CH₂-CO-O-O-C(CH₃)₂CH₂C₆H₅ (1,1-dimethyl-2-phenylethyl phenylperacetate).

C₆H₅-CO-O-O-C(CH₃)₂C₆H₅ (cumyl perbenzoate).

List-I

P
Pathway P
Q
Pathway Q
R
Pathway R
S
Pathway S

List-II

P → 1; Q → 3; R → 4; S → 2

P → 2; Q → 4; R → 3; S → 1

P → 4; Q → 1; R → 2; S → 3

P → 3; Q → 2; R → 1; S → 4

Answer and worked solution

Official answer: A

What decides the pathway.

Concerted loss of $\ce{CO2}$ (P, Q) needs R to give a stable radical: benzyl ( $\ce{PhCH2-CO-O-O-}$ ). A benzoyloxy radical ( $\ce{PhCO2.}$ , R = Ph) loses $\ce{CO2}$ slowly, so phenyl peresters break stepwise (R, S).
Fragmentation of $\ce{R'O.}$ into a carbonyl compound and a radical (Q, R) needs a tertiary alkoxyl that can expel a stabilised radical: $\ce{PhCH2C(CH3)2O. -> PhCH2. + (CH3)2CO}$ , or cumyloxyl $\ce{PhC(CH3)2O. -> CH3. + PhCOCH3}$ . A methoxyl radical does not fragment.

Matching.

P: concerted, $\ce{R'O.}$ stays: benzyl perester of methanol, $\ce{PhCH2CO-O-OCH3}$ → 1.
Q: concerted, then $\ce{R'O.}$ fragments: $\ce{PhCH2CO-O-OC(CH3)2CH2Ph}$ → 3.
R: stepwise, then $\ce{R'O.}$ fragments: $\ce{PhCO-O-OC(CH3)2Ph}$ → 4.
S: stepwise, $\ce{R'O.}$ stays: $\ce{PhCO-O-OCH3}$ → 2.

(A) P → 1, Q → 3, R → 4, S → 2

Topics:Reactive intermediates

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JEE Advanced 2014 · Paper 2, Chemistry question 37