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← IIT-JEE 2011 · Paper 2
ChemistryMatrix match+8Question 20 of 60

IIT-JEE 2011 · Paper 2, Chemistry question 20

Match the reactions in column I with appropriate types of steps/reactive intermediate involved in these reactions as given in column II

Column I

  • A
    1-Phenylpentane-1,4-dione with aq NaOH gives 3-phenylcyclopent-2-en-1-one.
  • B
    4-Chloro-1-phenylbutan-1-one with CH₃MgI gives 2-methyl-2-phenyltetrahydrofuran.
  • C
    4-Hydroxy-1-phenylbutan-1-one (¹⁸O in the OH) with H₂SO₄ gives 5-phenyl-2,3-dihydrofuran with ¹⁸O in the ring.
  • D
    2-Methyl-5-phenylpentan-2-ol with H₂SO₄ gives 1,1-dimethyl-1,2,3,4-tetrahydronaphthalene.

Column II

  • P
    Nucleophilic substitution
  • Q
    Electrophilic substitution
  • R
    Dehydration
  • S
    Nucleophilic addition
  • T
    Carbanion
Mark every match for each entry
PQRST
A
B
C
D
Answer and worked solution

Official answer: A → R, S, T; B → P, S; C → R, S; D → Q, R

  • A Intramolecular aldol: OHX−\ce{OH-} forms the enolate (carbanion) of the methyl ketone, which adds to the aryl ketone C=O (nucleophilic addition). The β-hydroxy ketone then dehydrates to the enone. → R, S, T.
  • B CHX3MgI\ce{CH3MgI} adds to the ketone (nucleophilic addition). The alkoxide then displaces ClX−\ce{Cl-} intramolecularly (SN2\mathrm S_\mathrm N2) to close the tetrahydrofuran. → P, S.
  • C The ¹⁸OH adds to the protonated ketone to form a cyclic hemiketal (nucleophilic addition), which loses water to give the dihydrofuran; the ring keeps the ¹⁸O. → R, S.
  • D The tertiary alcohol loses water to give a tertiary carbocation, which attacks the ring (intramolecular Friedel-Crafts, electrophilic substitution). → Q, R.

Answer: A → R, S, T; B → P, S; C → R, S; D → Q, R

Topics:Reaction types and mechanismsAldol and related condensationsAromatic hydrocarbons and electrophilic substitution
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