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JEE Advanced 2019 · Paper 2, Chemistry question 17

Passage

Answer Q.17 and Q.18 by appropriately matching the lists based on the information given in the paragraph

List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I. (The entries of both lists are drawn structures; see the figures.)

(I) 2-(1,3-Dioxolan-2-yl)phenylacetonitrile treated with (i) DIBAL-H, (ii) dil. HCl, (iii) NaBH₄, (iv) conc. H₂SO₄.(II) 2-Allylbenzoic acid treated with (i) O₃, (ii) Zn, H₂O, (iii) NaBH₄, (iv) conc. H₂SO₄.(III) Methyl 2-(chloromethyl)benzoate treated with (i) KCN, (ii) H₃O⁺, Δ, (iii) LiAlH₄, (iv) conc. H₂SO₄.(IV) Methyl 2-(2-methoxy-2-oxoethyl)benzoate treated with (i) LiAlH₄, (ii) conc. H₂SO₄.(P) 2-(2-Oxoethyl)benzoic acid.(Q) 2-[2-(Hydroxymethyl)phenyl]ethanol.(R) Isochromane (3,4-dihydro-1H-2-benzopyran).(S) 2-(2-Hydroxyethyl)benzoic acid.(T) Homophthalic acid (2-carboxyphenylacetic acid).(U) Isochroman-1-one (3,4-dihydro-2-benzopyran-1-one).

Which of the following options has correct combination considering List-I and List-II?

List-I

  • I
    2-(1,3-Dioxolan-2-yl)phenylacetonitrile treated with (i) DIBAL-H, (ii) dil. HCl, (iii) NaBH₄, (iv) conc. H₂SO₄.
  • II
    2-Allylbenzoic acid treated with (i) O₃, (ii) Zn, H₂O, (iii) NaBH₄, (iv) conc. H₂SO₄.
  • III
    Methyl 2-(chloromethyl)benzoate treated with (i) KCN, (ii) H₃O⁺, Δ, (iii) LiAlH₄, (iv) conc. H₂SO₄.
  • IV
    Methyl 2-(2-methoxy-2-oxoethyl)benzoate treated with (i) LiAlH₄, (ii) conc. H₂SO₄.

List-II

  • P
    2-(2-Oxoethyl)benzoic acid.
  • Q
    2-[2-(Hydroxymethyl)phenyl]ethanol.
  • R
    Isochromane (3,4-dihydro-1H-2-benzopyran).
  • S
    2-(2-Hydroxyethyl)benzoic acid.
  • T
    Homophthalic acid (2-carboxyphenylacetic acid).
  • U
    Isochroman-1-one (3,4-dihydro-2-benzopyran-1-one).
Choose one option
Answer and worked solution

Official answer: C

Reaction (I).

  • DIBAL-H turns the nitrile into an imine-aluminium complex.
  • Dilute HCl hydrolyses it to an aldehyde and also opens the acetal, giving the dialdehyde.
  • NaBHX4\ce{NaBH4} reduces this to the diol (Q).
  • Acid closes the ring to isochromane (R).

So (I) leads to Q and R.

Reaction (II).

  • Ozonolysis of the allyl group gives 2-(2-oxoethyl)benzoic acid (P).
  • NaBHX4\ce{NaBH4} reduces only the aldehyde, giving 2-(2-hydroxyethyl)benzoic acid (S).
  • Acid lactonises it to isochroman-1-one (U).

So (II) leads to P, S and U.

(C)

Topics:Acid derivativesOxidation and reductionEthers
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