JEE Advanced 2019 · Paper 2 · Chemistry Q18

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ChemistryList match+3 / -1Question 36 of 54

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.

(R) Isochromane (3,4-dihydro-1H-2-benzopyran).

(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).

(III), (S), (R)

(IV), (Q), (U)

(III), (T), (U)

(IV), (Q), (R)

Answer and worked solution

Official answer: D

Reaction (III).

KCN gives the nitrile.
$\ce{H3O+}$ /heat hydrolyses the nitrile and the ester, giving homophthalic acid (T).
$\ce{LiAlH4}$ reduces it to the diol (Q).
Acid gives isochromane (R).

So (III) leads to T, Q and R.

Reaction (IV). $\ce{LiAlH4}$ reduces both esters to the diol (Q), and acid closes the ether ring (R). So (IV) leads to Q and R.

The options.

(A) has S, which (III) does not form.
(B) has U.
(C) has U.
(D) is correct.

(D)

Topics:Acid derivatives Oxidation and reduction Ethers

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JEE Advanced 2019 · Paper 2, Chemistry question 18