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Chemistry MCQs for NEET — Practice Questions with Answers

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Debromination of meso-dibromobutane gives mainly

n-butane

1-butene

Cis-2-butene

Trans-2-butene

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Explanation

Debromination is a type of anti elimination. Meso compounds on anti elimination gives trans compound.

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What is the major product of the following reaction?

$H_{2} C = CH - {CH}_{2} - OH \to_{excess}^{HBr} Product$

${CH}_{3} - \overset{Br}{\overset{|}{CH}} - {CH}_{2} - Br$

$H_{2} C = CH - {CH}_{2} - Br$

${CH}_{3} - \overset{Br}{\overset{|}{CH}} - {CH}_{2} - OH$

${CH}_{3} - \overset{O H}{\overset{|}{CH}} - {CH}_{2} - OH$

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Explanation

${CH}_{2} = CH - {CH}_{2} - OH \overset{H - Br}{\to} {CH}_{2} = CCH - {CH}_{2} - H_{2} O^{\oplus} \overset{- H_{2} O}{\to} {CH}_{2} = CH - {CH}_{2}^{\oplus} \overset{{Br}^{-}}{\to} {CH}_{2} = CH - {CH}_{2} - Br \overset{H - Br}{\to} {CH}_{3} - \overset{\oplus}{C} H - {CH}_{2} - Br \overset{{Br}^{-}}{\to} {CH}_{3} - \underset{Br}{\underset{|}{CH}} - {CH}_{2} - Br$

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Addition of KI accelerates the hydrolysis of primary alkyl halides because:

KI is soluble in organic solvents

the iodide ion is a weak base and a poor leaving group

the iodide ion is a strong base

the iodide ion is a powerful nucleophile as well as a good leaving group

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Explanation

$I^{-}$ is a stranger nucleophile in $H_{2} O$ & good leaving group due to weaker base

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Which one of the following compound will be least susceptible to elimination of hydrogen bromide ?

Br-CH₂-CH₂-NO₂

Br-CH₂-CH₂-CH3

Br-CH₂-CH₂-CN

Br-CH₂-CH₂-Co₂Et

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Explanation

(b) At b-position -M gp. prefer elimination.

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Alkyl halides can be obtained by all methods except:

${CH}_{3} {CH}_{2} OH + Hcl / {ZnCl}_{2} \overset{}{\to}$

${CH}_{3} - {CH}_{2} - {CH}_{3} - {CH}_{2} \overset{{Cl}_{2} / UV light}{\to}$

$C_{2} H_{5} OH + NaCl \overset{}{\to}$

${CH}_{3} COOAg + {Br}_{2} / {CCl}_{4} \overset{}{\to}$

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Explanation

Alkyl halides cannot be obtained by the reaction of an alcohol with sodium chloride (NaCl). This reaction does not occur under normal conditions. The other options involve methods like addition of HX to alkenes, free radical halogenation, and conversion of alkyl carboxylates to alkyl halides.

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In order to prepare 1-chloropropane, which of the following reactants can be employed?

Propene and HCl in the presence of peroxide

Propene and Cl₂ followed by treatment with aq. KOH

Propanol-1 and SOCl₂/pyridine

Any of the above can be used

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Which of the following molecules would have a carbon-halogen bond most susceptible to nucleophilic substitution ?

2-fluorobutane

2-chlorobutane

2-bromobutane

2-iodobutane

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Explanation

The carbon-halogen bond strength decreases in the order C-F > C-Cl > C-Br > C-I. This trend makes the carbon-iodine bond the weakest and most susceptible to nucleophilic substitution among the given options. Therefore, 2-iodobutane would undergo nucleophilic substitution most readily.

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When benzyl chloride is treated with ethanolic KCN, the major product formed is:-

benzyl ethyl ether

benzyl alcohol

benzyl cyanide

benzyl isocyanide

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Explanation

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Ethyl bromide reacts with lead-sodium alloy to form:-

tetraethyl lead

tetraethyl bromide

both (a) and (b)

none of these

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Explanation

(a) 4C₂H₅Br+4Na+Pb→(C₂H₅)₄Pb+4NaBr

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Which is gem dihalide?

CH₃.CHBr₂

CH₂Br.CH₂Br

CH₃.CHBr.CH₂Br

None of these

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Explanation

(a) A gem dihalide possesses two halogens on same carbon atom.

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