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NEET Practice Questions (MCQs) with Answers & Solutions

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Benzoyl Chloride is prepared from benzoic acid by:

Cl₂, hv

SO₂Cl₂

SOCl₂

Cl₂, H₂O

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Explanation

-OH group is replaced by PCl₅,PCl₃ or SOCl₂.

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In the following sequences of reactions;

${CH}_{3} {CH}_{2} {CH}_{2} Br \overset{KOH (alc .)}{\to}$ (A) $\overset{HBr}{\to}$ (B) $\overset{KOH (aq)}{\to}$ (C)

the product (C) is:

propene

propyne

propan-1-ol

propan-2-ol

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Explanation

(d) ${CH}_{3} {CH}_{2} {CH}_{2} Br \to_{- HBr}^{KOH (alc .)}$ ${CH}_{3} CH = {CH}_{2}$ $\overset{HBr}{\to}$ ${CH}_{3} {CHBrCH}_{3}$ $\overset{KOH (aq)}{\to}$ ${CH}_{3} {CHOHCH}_{3}$

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Iodoform can be prepared from all except :

isopropyl alcohol

3-methyl-2-butanone

isobutyl alcohol

ethyl methyl ketone

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The alkyl halide which does not give white precipitate with alcoholic ${AgNO}_{3}$ solution is:

ethyl chloride

allyl chloride

isopropyl chloride

vinyl chloride

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Explanation

In vinyl chloride, C-Cl bond has partial double bond character due to resonance

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The reaction of toluene with ${Cl}_{2}$ in presence of ${FeCl}_{3}$ gives predominantly :

m- chlorobenzene

benzoylchloride

benzyl chloride

O^- and p-chlorotoluene

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Explanation

(d) $- {CH}_{3}$ gp. is o- and p-directing. ${FeCl}_{3}$ is halogen carrier.

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The products of reaction of alcoholic silver nitrite with ethyl bromide are :

Ethane

Ethene

Ethyl alcohol

Nitroethane

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Explanation

$C_{2} H_{5} Br$ + ${AgNO}_{2} \to C_{2} H_{5} - {NO}_{2}$ + AgBr

Nitro ethane

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Order of nucleophilicity in polar protic solvent is

$I^{-} > {Br}^{-} > F^{-} > {Cl}^{-}$

$F^{-} > {Cl}^{-} > {Br}^{-} > I^{-}$

$I^{-} > {Cl}^{-} > F^{-} > {Br}^{-}$

$I^{-} > {Br}^{-} > {Cl}^{-} > F^{-}$

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Explanation

$I^{-} > {Br}^{-} > {Cl}^{-} > F^{-}$

HI is stronger acid in POLAR MEDIUM, not in non polar medium,so stronger acid have stronger nucleophile anion

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Phenylmagnesium bromide reacts with methanol to give

A mixture of toluene and Mg(OH)Br

A mixture of phenol and Mg(Me)Br

A mixture of anisole and Mg(OH)Br

A mixture of benzene and Mg(OMe)Br

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Explanation

$C_{6} H_{5} - Mg - Br + {CH}_{3} OH \to C_{6} H_{6} + Mg ({OCH}_{3}) Br$

Benzene

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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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