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Aldehydes, Ketones and Carboxylic Acids Set-6 MCQ Online Practice Test | Class 12 | Quiz Tweek

Aldehydes, Ketones and Carboxylic Acids Set-6

Practice Important MCQs for Aldehydes, Ketones and Carboxylic Acids — Chemistry, Class 12.

This chapter is crucial for both CBSE and competitive exams because it forms the core concepts behind carbonyl reactivity and common reaction pathways of aldehydes and ketones—especially nucleophilic addition, aldol/Cannizzaro reactions, enolate chemistry, and hydration. Mastery of these topics also directly strengthens problem-solving for derivative formations (like cyanohydrins), acidity trends of $\alpha$ -hydrogens, and equilibrium-based questions that commonly appear in JEE/NEET.

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Q1. (A mixture of benzaldehyde ( $\mathrm{C_6H_5CHO}$ ) and excess acetone ( $\mathrm{(CH_3)_2CO}$ ) is treated with catalytic NaOH and heated. The major isolated organic product is:)

(A)

( $\mathrm{C_6H_5CH_2OH\;+\;C_6H_5COOH}$ )

(B)

( $\mathrm{C_6H_5CH(OH)CH_2COCH_3}$ )

(C)

( $\mathrm{C_6H_5CH=CHCOCH_3}$ )

(D)

( $\mathrm{CH_3COCH=C(CH_3)_2}$ )

Q2. (Arrange the following carbonyl compounds in decreasing order of reactivity towards nucleophilic addition (e.g., $^{-}\!CN$ or $H^{-}$ ): formaldehyde ( $\mathrm{HCHO}$ ), acetaldehyde ( $\mathrm{CH_3CHO}$ ), benzaldehyde ( $\mathrm{C_6H_5CHO}$ ), acetone ( $\mathrm{(CH_3)_2CO}$ ).)

(A)

( $\mathrm{HCHO\;>\;CH_3CHO\;>\;C_6H_5CHO\;>\;(CH_3)_2CO}$ )

(B)

( $\mathrm{CH_3CHO\;>\;HCHO\;>\;C_6H_5CHO\;>\;(CH_3)_2CO}$ )

(C)

( $\mathrm{HCHO\;>\;C_6H_5CHO\;>\;CH_3CHO\;>\;(CH_3)_2CO}$ )

(D)

( $(CH_3)_2CO\;>\;CH_3CHO\;>\;HCHO\;>\;C_6H_5CHO$ )

Q3. (Which of the following has the most acidic $\alpha$ ‑hydrogen (lowest pKa for the hydrogen on carbon adjacent to the carbonyl)?)

(A)

( $\mathrm{CH_3CHO}$ )

(B)

( $\mathrm{(CH_3)_2CO}$ )

(C)

( $\mathrm{CH_3COOCH_2CH_3}$ )

(D)

( $\mathrm{CH_3COCH_2COCH_3}$ )

Q4. (Statement I: Benzaldehyde gives Cannizzaro reaction on treatment with concentrated NaOH, while acetaldehyde under similar conditions gives mainly aldol products.
Statement II: Cannizzaro reaction requires the absence of $\alpha$ ‑hydrogen and proceeds by hydride transfer between two aldehyde molecules; aldol condensation requires enolate formation (hence presence of $\alpha$ ‑hydrogen). Choose the correct option.)

(A)

(Both Statement I and Statement II are true, and Statement II correctly explains Statement I.)

(B)

(Both Statement I and Statement II are true, but Statement II does not correctly explain Statement I.)

(C)

(Statement I is true but Statement II is false.)

(D)

(Statement I is false but Statement II is true.)

Q5. (Consider 2‑methylcyclohexanone. When subjected to alkylation with methyl iodide following enolate formation under two different conditions, which pair of products will be obtained as the major products? (i) LDA, THF, −78 °C, then MeI (ii) NaOEt, EtOH, room temp, then MeI)

(A)

((i) $\mathrm{2,6\text{-}dimethylcyclohexanone}$ ; (ii) $\mathrm{2,6\text{-}dimethylcyclohexanone}$ )

(B)

((i) $\mathrm{2,6\text{-}dimethylcyclohexanone}$ ; (ii) $\mathrm{2,2\text{-}dimethylcyclohexanone}$ )

(C)

((i) $\mathrm{2,2\text{-}dimethylcyclohexanone}$ ; (ii) $\mathrm{2,6\text{-}dimethylcyclohexanone}$ )

(D)

((i) mixture of both regioisomers in ~1:1 ratio ; (ii) mixture of both regioisomers in ~1:1 ratio)

Q6. Which of the following correctly compares the acidity of $4$ -nitrobenzoic acid ( $4$ -NO $_2$ C $_6$ H $_4$ COOH) and $3$ -nitrobenzoic acid ( $3$ -NO $_2$ C $_6$ H $_4$ COOH)?

(A)

$3$ -NO $_2$ C $_6$ H $_4$ COOH is stronger because the meta $-$ NO $_2$ exerts a larger $-I$ effect stabilizing the conjugate base.

(B)

Both acids have nearly the same acidity because the inductive effect of $-$ NO $_2$ is position independent.

(C)

$4$ -NO $_2$ C $_6$ H $_4$ COOH is stronger because the para $-$ NO $_2$ stabilises the conjugate base by resonance in addition to the inductive effect.

(D)

$3$ -NO $_2$ C $_6$ H $_4$ COOH is stronger due to intramolecular hydrogen bonding between $-$ NO $_2$ and $-$ COOH in the meta position.

Q7. Arrange the following compounds in decreasing acidity of their $\alpha$ -hydrogens: $CH_3CHO,\ CH_3COCH_3,\ CH_3COOC_2H_5$ .

(A)

$CH_3CHO > CH_3COCH_3 > CH_3COOC_2H_5$

(B)

$CH_3COCH_3 > CH_3CHO > CH_3COOC_2H_5$

(C)

$CH_3COOC_2H_5 > CH_3CHO > CH_3COCH_3$

(D)

$CH_3COCH_3 > CH_3COOC_2H_5 > CH_3CHO$

Q8. When benzaldehyde ( $C_6H_5CHO$ ) and acetone ( $CH_3COCH_3$ ) are mixed in equimolar amounts and treated with dilute NaOH, the major organic product expected is:

(A)

Benzyl alcohol ( $C_6H_5CH_2OH$ ) and benzoate ( $C_6H_5COO^-$ ) from a Cannizzaro reaction of benzaldehyde.

(B)

Self-condensation product of acetone (diacetone alcohol/mesityl oxide) as the major product.

(C)

Benzoin (product of benzaldehyde self-condensation under CN^- catalysis).

(D)

An $\alpha,\beta$ -unsaturated ketone (benzylideneacetone) $C_6H_5CH=CHCOCH_3$ (cross aldol product).

Q9. Arrange the following carbonyl compounds in decreasing extent of hydration (formation of gem-diol) in aqueous solution: formaldehyde ( $HCHO$ ), benzaldehyde ( $C_6H_5CHO$ ), $p$ -nitrobenzaldehyde ( $p$ -NO $_2$ C $_6$ H $_4$ CHO), acetone ( $CH_3COCH_3$ ).

(A)

$HCHO > CH_3COCH_3 > p\text{-}NO_2C_6H_4CHO > C_6H_5CHO$

(B)

$HCHO > p\text{-}NO_2C_6H_4CHO > C_6H_5CHO > CH_3COCH_3$

(C)

$p\text{-}NO_2C_6H_4CHO > C_6H_5CHO > HCHO > CH_3COCH_3$

(D)

$C_6H_5CHO > p\text{-}NO_2C_6H_4CHO > HCHO > CH_3COCH_3$

Q10. Propanal ( $CH_3CH_2CHO$ ) is first converted into its cyanohydrin by HCN. The cyanohydrin is then treated with excess $CH_3MgBr$ followed by acidic work-up. The major product formed is:

(A)

$CH_3CH_2C(OH)C(=O)CH_3$ (3-hydroxy-2-pentanone)

(B)

$CH_3CH_2CH(OH)CH_3$ (2-butanol)

(C)

$CH_3CH_2CH(OH)CN$ (the original cyanohydrin, unchanged)

(D)

$CH_3CH_2C(=O)CH_2CH_3$ (2-pentanone)

Q11. Calculate the pH of a $0.10\ \mathrm{M}$ solution of acetic acid ( $CH_3COOH$ ) at 25°C given $K_a = 1.8\times10^{-5}$ . (Use the approximation $[H^+]\approx\sqrt{K_aC}$ .)

(A)

$3.00$

(B)

$2.50$

(C)

$2.87$

(D)

$2.30$

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