CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

Question : 36

Total: 37

(a) An organic compound (A) having molecular formula C4H8O gives orange red precipitate with 2, 4-DNP reagent. It does not reduce Tollens' reagent but gives yellow precipitate of iodoform on heating with NaOH and I2. Compound (A) on reduction with NaBH4 gives compound (B) which undergoes dehydration reaction on heating with conc. H2SO4 to form compound (C). Compound (C) on ozonolysis gives two molecules of ethanal.
Identify (A), (B) and (C) and write their structures. Write the reactions of compound (A) with (i) NaOH∕I2 and (ii) NaBH4.
(b) Give reasons:
(i) Oxidation of propanal is easier than propanone.
(ii) α-hydrogen of aldehydes and ketones is acidic in nature.
OR
(a) Draw structures of the following derivatives:
(i) Cyanohydrin of cyclobutanone
(ii) Hemiacetal of ethanal
(b) Write the major product(s) in the following :
(i) CH3−CH=CH−CH2−CN ‌

‌ (i) DIBAL-H ‌

──────▸

‌ (ii) ‌H3O+

(ii) CH3−CH2−OH‌

CrO3

────▸

Solution:

(a) Compound A(C4H8O) gives positive, 2, 4-DNP test, it must be carbonyl compound. It gives iodoform test.

(i)

CH3−

∥

−CH2−CH3

(A)

NaOH∕I2

────▸

C2H5‌COOH+

CHI3

Iodoform

(iii)

CH3−

∥

−CH2−CH3

(A)

NaBH4

────▸

CH3−

∥

−CH2−CH3

Butan−2−ol

(b) (i) Oxidation of propanal is easier than propanone because aldehydes have one hydrogen atom attached to the carbonyl group while ketones have two alkyl or aryl groups attached to the carbonyl group. Propanal easily oxidised to form acid with same number of carbon atoms whereas propanone is difficult to be oxidise and form acids with less number of carbon atoms.

(ii) α-hydrogen of aldehydes and ketones is acidic in nature. They can be easily abstracted by suitable bases. Two molecules condense to form a β-hydroxyaldehyde or β-hydroxyketone which gets dehydrated in presence of acid upon heating to form α,β-unsaturated compound.

OR
(a) (i) Cyanohydrin of cyclobutanone

(ii) Hemiacetal of ethanol

(ii) CH3−CH2−OH‌

CrO3

────▸

[CH3−

∥

−H] →[CH3−

∥

−OH]
(c) By iodoform test : Propanone on treatment with I2∕NaOH undergoes iodoform test to give a yellow ppt. of iodoform.
CH3COCH3+3‌NaOI‌────▸

CHI3

yellow‌ppt.

+CH3‌COONa+2‌NaOH
Propanal does not give this test.

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