JEE Advanced 2018 · Paper 1 · Chemistry Q17

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ChemistrySingle correct+3 / -1Question 35 of 54

Passage

An organic acid P (C $_{11}$ H $_{12}$ O $_2$ ) can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

[Scheme: S ← P: 1) H $_2$ /Pd-C; 2) NH $_3$ /Δ; 3) Br $_2$ /NaOH; 4) CHCl $_3$ , KOH, Δ; 5) H $_2$ /Pd-C. P → Q: 1) H $_2$ /Pd-C; 2) SOCl $_2$ ; 3) MeMgBr, CdCl $_2$ ; 4) NaBH $_4$ . Q → R: 1) HCl; 2) Mg/Et $_2$ O; 3) CO $_2$ (dry ice); 4) H $_3$ O $^+$ .]

Reaction scheme: P converts to S (left arrow) with 1) H₂/Pd-C, 2) NH₃/Δ, 3) Br₂/NaOH, 4) CHCl₃, KOH, Δ, 5) H₂/Pd-C; P converts to Q (right arrow) with 1) H₂/Pd-C, 2) SOCl₂, 3) MeMgBr, CdCl₂, 4) NaBH₄; Q converts to R with 1) HCl, 2) Mg/Et₂O, 3) CO₂ (dry ice), 4) H₃O⁺.

The compound R is

2-(4-Isobutylphenyl)propanoic acid (ibuprofen)

Answer and worked solution

Official answer: A

Identify P.

Oxidation gives a dibasic acid that forms dacron with ethylene glycol. That is terephthalic acid, so P is para-disubstituted and carries a $\ce{COOH}$ .
Ozonolysis gives an aliphatic ketone (acetone), so the other substituent is $\ce{-CH=C(CH3)2}$ .
$\ce{(CH3)2C=CH-C6H4-COOH}$ is $\ce{C11H12O2}$ . ✓

P → Q.

$\ce{H2/Pd-C}$ saturates the side chain to isobutyl.
$\ce{SOCl2}$ forms the acid chloride.
$\ce{Me2Cd}$ (from MeMgBr/ $\ce{CdCl2}$ ) converts it to the methyl ketone, 4-isobutylacetophenone.
$\ce{NaBH4}$ reduces it to Q, 1-(4-isobutylphenyl)ethanol.

Q → R. HCl gives the benzylic chloride. Mg makes the Grignard. $\ce{CO2}$ then $\ce{H3O+}$ gives the acid. R is 2-(4-isobutylphenyl)propanoic acid, ibuprofen.

(A)

Topics:Carboxylic acids Acid derivatives Alcohols

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JEE Advanced 2018 · Paper 1, Chemistry question 17