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Chemistry MCQs for NEET — Practice Questions with Answers

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${CH}_{3} - c \equiv CH \to_{dil . H_{2} {SO}_{4}}^{{HgSO}_{4}} (A)$

${CH}_{3} - c \equiv CH \to_{(2) H_{2} {SO}_{4} / {HO}^{-}}^{(1) {BH}_{3} THF} (B)$

Product (A) and (B) is differentiated by:

2-4 DNP

NaOI

Na-metal

NaHSO₃

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Explanation

${CH}_{3} - C \equiv CH \to_{H_{2} O}^{{Hg}^{2 +} / H^{+}} {CH}_{3} - \overset{\overset{OH}{|}}{C} = {CH}_{2} ⇌ {CH}_{3} - \overset{\overset{O}{| |}}{C} - {CH}_{3} A ketone (gives + ve iodoform test with NaoI)$

${CH}_{3} - C \equiv CH \to_{(2) H_{2} O_{2} / {OH}^{-}}^{(1) {BH}_{3} THF} {CH}_{3} - CH = {CH}_{2} - OH ⇌ {CH}_{3} - {CH}_{2_{}} = O (an aldehyde) (- ve Iodoform test beacuse - \underset{\overset{| |}{O}}{C} - {CH}_{3} group is absent)$

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$Ph - CH = CH - \overset{\overset{O}{| |}}{C} - {CH}_{3} \to Ph - CH = CH - {CO}_{2} H$
Above conversion can be achieved by :

${KMnO}_{4}, ∆ followed by H^{+}$

$I_{2} / NaOH followed by H^{+}$

H₂/Pt

LiAlH₄

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Explanation

(b) Iodoform test.

(b.1) Oxidative ozonolysis by not KMnO₄

(b.2) Iodoform test. Only methyl ketones. No reaction with double bond.

(b.3) H₂/Pt can reduce only double bond.

(b.4) LAH oxidizes ketone to $2 °$ alcohol not to carboxylic acid.

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$R - \overset{\overset{O}{ll}}{C} - H \overset{R - {NH}_{2}}{\to} R - CH = N - R$ . This reaction gives best yield at:

pH 1 - 2

pH 4 - 5

pH 10 - 11

pH 13 - 14

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Carbonyl compounds can generally be converted to hydrocarbons by :

H₂/Pt

LiAlH₄

$N_{2} H_{4} - KOH / ∆$

$K_{2} {Cr}_{2} O_{7} - H_{2} {SO}_{4}$

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Explanation

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Which statement about the aldol condensation is correct ?

A Lewis acid is commonly used as a catalyst

The initial step is probably the formation of a carbanion

A Lewis base is employed to induce carbocation formation

The carbon chain is lengthened through the elimination of 1 mole of water

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A compound gives a positive test with I₂/NaOH and is extracted from benzene by saturated NaHSO₃, It may be :

${CH}_{3} {({CH}_{2})}_{4} CHO$

${CH}_{3} {({CH}_{2})}_{3} {COCH}_{3}$

${CH}_{3} {CH}_{2} {COCH}_{2} {CH}_{3}$

${CH}_{3} {({CH}_{2})}_{4} {CH}_{2} OH$

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Explanation

(b) Methyl Ketone give positive Tollen's test.

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In which of the following compounds the methylenic hydrogens are the most acidic ?

${CH}_{3} {COCH}_{2} {CH}_{3}$

${CH}_{3} {CH}_{2} {COOC}_{2} H_{5}$

${CH}_{3} {CH}_{2} CH {({COOC}_{2} H_{5})}_{2}$

${CH}_{3} {COCH}_{2} CN$

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Explanation

${CH}_{3} - \overset{\overset{O}{ll}}{C} - \underset{(acidic)}{{CH}_{2}} - C \equiv N$

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The presence of unsaturation in organic compounds can be tested with :

Schiff's reagent

Tollen's reagent

Fehling's reagent

Baeyer's reagent

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Explanation

(d) Cold dil. alk KMnO₄ is Baeyer's reagent.

(d.1) Schiff's reagent is used to test presence of Aldehydes.
(d.2) Tollen's reagent is used to test presence of Aldehydes.
(d.3) Fehling's reagent is used to test presence of Aldehydes.
(d.4) Baeyers's reagent is Cold dil. alk KMnO₄ is used to test presence of unsaturation.

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$Ph - {CH}_{3} \overset{{CrO}_{2} {Cl}_{2}}{\to} (A) \overset{conc . KOH}{\to} Ph - {CH}_{2} OH + (B)$

Product (B) of above the raction is :

$Ph - {CO}_{2} H$

$Ph - {CO}_{2}^{-}$

$Ph - CHO$

$Ph - {CH}_{3}$

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Explanation

(b) ${PhCH}_{3} \to_{Etard}^{{CrO}_{2} {Cl}_{2}} PhCHO \to_{Cannizaro}^{Conc . KOH} {PhCOO}^{\oplus} + {PhCH}_{2} OH$

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${CH}_{3} - \overset{\overset{O}{∥}}{C} - H \overset{HCN}{\to} (A) \overset{H_{3} O^{\oplus}}{\to} (B) \underset{△}{\to} (C) \overset{{LiAlH}_{4}}{\to} (D) \overset{{HIO}_{4}}{\to} HCHO + (E)$

Compound (C) can show geometrical isomerism, Product (E) of the reaction will be :

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

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

${CH}_{3} - CHO$

HCHO

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