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CBSE 12th Standard Chemistry Subject Aldehydes , Ketones and Carboxylic Acids Ncert Exemplar 2 Mark Questions 2021

Published on: 25/05/2021

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Std: 12th Standard CBSE - Chemistry

chapter: - NCERT Exemplers

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Questions:

Generic 2

4 x 2 = 8

When liquid 'A' is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline the solid on treatment with sodium hydrogensulphite. Liquid 'B' also forms a white crystalline solid with sodium hydrogensulphite but it does not give test with ammoniacal silver nature. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.

Alkenes, and carbonyl compounds , both contain a \(\pi\) - bond but alkenes show electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition reactions. Explain.

Compound 'A' was prepared by oxidation 'B' with alkaline KMnO₄. Compound 'A' on reduction with lithium aluminium hydride gets converted back to compound 'B'. When compound 'A' is heated with compound B in the presence of H₂SO₄ it produces fruitly smell of compound C to which family the compounds 'A', 'B' and 'C' belong to?

Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Pentan-2-one and Pentan-3-one
(ii) Ethanal and Propanal

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Answers Key:

Generic 2

4 x 2 = 8

Since, the liquid A reduces ammoniacal silver nitrate (Tollen's reagent), hence A is aldehyde and B does not give test with ammnoniacal silver nitrate, thus, B is a ketone.
Further, B forms a white crystalline solid on treatment with sodium hydrogen sulphite. This suggests that B is a methyl ketone.

Carbon atom in carbonyl compounds acquires slight positive charge and is attacked by nucleophile.
Alkenes undergo electrophilic addition whereas aldehydes and ketones undergo nucleophilic addition because in alkenes the double bond joins two carbon atoms and there is no resultant polarity.

'A' is carboxylic acid, 'B' is alcohol, 'C' is easter.

(i) Add I₂ and NaOH. P-2- one will give yellow ppt of iodoform, whereas pentan-3- one will not react.
(ii) Add I2 and NaOH. Ethanal will yellow precicipate of iodoform, whereas proponal does not react.