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Chemistry MCQs for NEET — Practice Questions with Answers

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Arrange in their increasing order of equilibrium constants for hydraion ?

${CH}_{3} - C_{(A)}^{\overset{O}{∥}} - {CH}_{3}$ , ${CH}_{3} - C_{(B)}^{\overset{O}{∥}} - H$ , ${CH}_{3} - C_{(C)}^{\overset{O}{∥}} - {CH}_{2} - Cl, H - C_{(D)}^{\overset{O}{∥}} - H, Cl - {CH}_{2 \underset{(E)}{}} - \overset{\overset{O}{∥}}{C} - H$

A < C < B < E < D

A < C < E < B < D

C < A < B < E < D

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Explanation

(b) Aldehyde is more reactive then ketone, -I effect increases reactivity towards nucleophilic attack.

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$Ph - {CH}_{2} - CN \to_{\underset{(3) H_{3} O^{\oplus} / ∆}{(2) {CH}_{3} - \overset{\overset{O}{∥}}{C} - Cl}}^{(1) EtONA}$ (P); Product (p) of the reaction will be :

$Ph - {CH}_{2} - \overset{\overset{O}{∥}}{C} - H$

$Ph - {CH}_{2} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$

$Ph - C \underset{\overset{l}{{CH}_{3}}}{H} - \overset{\overset{O}{∥}}{C} - H$

$Ph - C \underset{\overset{l}{{CH}_{3}}}{H} - \overset{\overset{O}{∥}}{C} - {CH}_{3}$

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$\underset{Symmetrical ketone}{(A)} \overset{{LiAlH}_{4}}{\to} (B) \to_{∆}^{H^{\oplus}}$ Diastereomers Reactent (A) is :

${CH}_{3} - \overset{\underset{∥}{O}}{C} - {CH}_{3}$

${CH}_{3} - \overset{\underset{∥}{O}}{C} - {CH}_{2} - {CH}_{3}$

${CH}_{3} - {CH}_{2} - \overset{\underset{∥}{O}}{C} - {CH}_{2} - {CH}_{3}$

${CH}_{3} - {CH}_{2} - {CH}_{2} - CH = O$

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$Ph - CH = CHCHO + {CH}_{3} CH = CHCHO \to_{ErOH, ∆}^{base}$ (A) 87%; product of this reaction is :

$Ph - {(CH = CH)}_{2} - CHO$

$Ph - {(CH = CH)}_{3} CHO$

$Ph - {(CH = CH)}_{4} CHO$

$Ph - CH = CH - CH = CH - {CH}_{3}$

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Explanation

The given reaction is a Wittig reaction, where an ylide (Ph-CH=CHCHO + CH3CH=CHCHO) reacts with an aldehyde or ketone to form an alkene. The product formed is Ph-CH=CH-CHO, which is a styryl aldehyde or cinnamaldehyde.

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CH₃CHO $\to_{5 ° C}^{10 % NaOH} \underset{∆}{\to} \to_{Ni}^{H_{2}}$ (A) ; Product (A) of the reaction is :

Propanol

ethanol

butanol

pentanol

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Explanation

(c) ${CH}_{3} CHO$ $\overset{NaOH}{\to}$ CH₃ $- {CH}_{3} - CH = CH - CHO \overset{H_{2} / Ni}{\to} {CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{2} - OH$

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Which of the following acid gives positive Tollen's reagent test.

$({CH}_{2})_{2} <_{{CO}_{2} H}^{{CO}_{2} H}$

${CH}_{2} <_{{CO}_{2} H}^{{CO}_{2} H}$

$\underset{|}{C} O_{2} H {CO}_{2} H$

HCO₂H

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Explanation

The Tollen's reagent test is a chemical test used to distinguish aldehydes from other carbonyl compounds like ketones. Only aldehydes give a positive test by forming a silver mirror on the inner walls of the test tube. Among the given options, only formic acid (HCO2H) is an aldehyde and will give a positive Tollen's test.

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Acetaldehyde and benzaldehyde can be differentiated by:

Fehling test

Iodoform test

Tollen's reagent

both (a) and (b)

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Explanation

Acetaldehyde and benzaldehyde, both being aldehydes, can be differentiated by Fehling's test and iodoform test. Fehling's test distinguishes aldehydes from ketones, while the iodoform test differentiates between methyl ketones and other carbonyl compounds.

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When vapours of acetic acid are passed over.....at $300 ° C$ we get acetone.

${Al}_{2} O_{3}$

$CuO$

$MoO$

$Cu$

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Explanation

(a) $2 {CH}_{3} COOH \to_{or {ThO}_{2}}^{{Al}_{2} O_{3} or MnO} {CH}_{3} {COCH}_{3} + H_{2} O + {CO}_{2}$

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Formaldehyde is not used in:

adhesives

bakelite

tooth powders

explosives

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Explanation

(d) In rest all HCHO is used.

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Ketones can be prepared by:

Rosenmund's reduction

Stephen's reduction

both (a) and (b)

none of the above

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Yes. You can attempt every Chemistry question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.