MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 3

In Friedel craft alkylation of benzene with methyl chloride, the catalyst used is:

0%
Anhydrous $AlCl_{3}$
0%
Conc. sulphuric acid
0%
Dilute sulphuric acid
0%
Anhydrous $ZnCl_{2}$

Explanation

Anhydrous

$AlCl_3$ which acts as Lewis acid and catalyst in this reaction.

Option A is correct.

Which of the following reagents, when heated with ethyl chloride, forms ethylene?

0%
$Aq.\ KOH$
0%
$Zn/HCl$
0%
$Alc.\ KOH$
0%
$HI$

Explanation

Ethyl chloride reacts with aq.

$KOH$ and alcohol forms.
Ethyl chloride reacts with

$Zn/HCl$ and alkane forms.
Ethyl chloride reacts with alc.

$KOH$ and ethene forms.
The reaction of the process is:

${ C }_{ 2 }{ H }_{ 5 }Cl \xrightarrow { alc.KOH } { CH }_{ 2 }=C{ H }_{ 2 }+HCl$

Ethyl chloride is one of the reactants in:

0%
Wurtz reaction
0%
Williamson's synthesis
0%
Friedel-craft's alkylation
0%
All the above

Explanation

Ethyl chloride used as a reactant in all the reactions.

Williamson's synthesis: Sodium ethoxide reacts with alkyl halide and ether forms as product.

$RONa+RX\rightarrow ROR +NaX$

Wurtz reaction:

$RX+2Na+RX\rightarrow R-R+2NaX$

Friedel - Craft alkylation: (as shown in figure)

Option D is correct.

Which of the following statements are true about

$CHCl_{3}$ ?

i) It is used as an anesthetic agent.

ii) It is used as a solvent.

iii) It has

$sp^{2}$ hybridized carbon.

iv) It has a distorted tetrahedral shape.

The correct statements are:

0%
i, ii, iv
0%
iii, iv
0%
i, ii, iii
0%
i, ii

$S_{N}1$ reaction is favoured by:

0%
non-polar solvents
0%
bulky group on the carbon atom attached to the halogen atom
0%
small groups on the carbon atom attached to halogen atom
0%
all the above

Explanation

$SN^1$ mechanism proceeds through carbocation formation and as stability of carbocation increases, it favours

$SN^1$ mechanism.

And as we add bulky group to carbon attached with halogen, carbocation formed will become more stable due to +I effect of bulky alkyl group so it favours

$SN^1$ mechanism.

Identify the set of reagent/reaction conditions 'X' and 'Y' in the following set product:

$CH_3-CH_2-CH_2Br\xrightarrow{\text{X}}CH_3-CH=CH_2 \xrightarrow{\text{Y}}CH_3-\underset{|}CH-CH_3$

$Br$

0%
$X=$ dilute aqueous NaOH $20 ^ { \circ } \mathrm { C }; \mathrm { Y }=$ HBr/acetic acid $20 ^ { \circ } C$
0%
$X=$ concentrated alcoholic $\mathrm { NaOH } , 80 ^ { \circ } \mathrm { C } ; \mathrm { Y }=$ HBr/acetic acid $20 ^ { \circ } C$
0%
$X=$ dilute aqueous NaOH $20 ^ { \circ } \mathrm { C };$ $\mathrm Y=\mathrm { Br } _ { 2 } / \mathrm { CHCl } _ { 3, } \ 0 ^ { \circ } \mathrm { C }$
0%
$X=$ concentrated alcoholic $\mathrm { NaOH } , 80 ^ { \circ } \mathrm { C } ; \mathrm { Y }=$ $\mathrm { Br } _ { 2 } / \mathrm { CHCl } _ { 3 } , 0 ^ { \circ } \mathrm { C }$

On reaction with benzene, which halide is least reactive for Friedel crafts reaction ?

0%
Isopropylchloride
0%
Vinyl chloride
0%
Benzylchloride
0%
methylchloride

$CH\equiv CH\overset{H_2O/Hg^{2+}}{\underset{H_2SO_4}{\rightarrow}} X \overset{LiAlH_4}{\rightarrow} Y \overset{P_4/Br_2}{\rightarrow} Z$ where Z is:

0%
Ethylene bromide
0%
Ethanol
0%
Ethyl bromide
0%
Ethylidene bromide

The product of the following reaction

$C_6H_5Cl \xrightarrow [(Room\ temp.)]{KOH(aq)} product$

0%
$C_6H_5OH$
0%
$C_6H_4(OH)Cl$
0%
$C_6H_4(OH)_2$
0%
No reaction

Interhalogen wwhich is very unstable among the following >

0%
BrCi
0%
ICI
0%
IF
0%
IBr

The order of decreasing reactivities of these alcohols towards nucleophilic substitution with

$HBr$ is:

0%
$III> I> IV> II$
0%
$III> I> II> IV$
0%
$I> III> IV> II$
0%
$I> III> II> IV$

Which of the following cannot shows

$S_{N}1$ reaction?

2-Bromopentane is heated wtih potassium ethoxide in ethanol. The major product is:

0%
trans-2-pentene
0%
2-ethoxypentane
0%
1-pentane
0%
cis-2-pentene

In which of the following reaction

$1^o$ alkyl halide is a major product ?

0%
${ CH }_{ 3 }-\overset { \underset { | }{ H } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -Ph\xrightarrow [ ]{ { Cl }_{ 2 }/hv } Product$
0%
${ CH }_{ 3 }-\underset { \underset { { CH }_{ 3 } }{ | } }{ CH } -CH_{ 3 }\xrightarrow [ ]{ { Br }_{ 2 }/hv+\Delta } Product$
0%
${ CH }_{ 3 }-\underset { \underset { { CH }_{ 3 } }{ | } }{ CH } -CH_{ 3 }\xrightarrow [ ]{ { Cl }_{ 2 }/hv } Product$
0%
${ CH }_{ 3 }-\underset { \underset { { CH }_{ 3 } }{ | } }{ CH } -CH=CH_{ 2 }\xrightarrow [ ]{ { Cl }_{ 2 }/hv+\Delta } Product$

The major product of the following reaction is:

Which one of the following is likely to give a precipitate with

$AgNO_3$ solution?

0%
$(CH_3)_3CCl$
0%
$CHCl_3$
0%
$CH_2=CH-Cl$
0%
$CCl_4$

Which of the following factors does not favour

$S_{N}1$ mechanism?

0%
Strong nucleophile
0%
Polar solvent
0%
Low conc. of nucleophile
0%
$3^{0}$ halide

Explanation

$S_N1$ mechanism proceeds through carbocation formation, as stability of carbocation increases, reaction favours

$SN^1$ mechanism.

$3^0$ carbocation are more stable than others and also polar solvent increases stability of carbocation so they both favours

$S_N1$ mechanism.

$S_N1$ reactions, rate of the reaction depends upon the concentration of the substrate only not on the strength of nucleophile.

$C - Cl$ bond of chlorobenzene in comparison to

$C - Cl$ bond of methyl chloride is:

0%
longer and weaker
0%
shorter and weaker
0%
shorter and stronger
0%
longer and stronger

Explanation

The

$C-Cl$ bond in

$C_6H_5Cl$ is shorter than in

$CH_3Cl$ and therefore, stronger.

The lone pair on chlorine are dispersed throughout the benzene ring by resonance. This gives the

$C-Cl$ bond a double bond character.

Option C is correct.

The reaction conditions leading to the best yield of

$C_{2}H_{5}Cl$ are:

0%
$C_{2}H_{6}(excess)+Cl_{2}\rightarrow$
0%
$C_{2}H_{6}+Cl_{2}\rightarrow$
0%
$C_{2}H_{6}+Cl_{2}(excess)\rightarrow$
0%
none of these

Explanation

Halogenation of alkanes will lead to the formation of alkyl halide.

In this reaction, if we use

$Cl_2$ in excess, then dichloro and trichloro forms as product and if ethane is used in excess, ethyl chloride forms as major product.

So, the reaction takes place as follows:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

(excess)

Option A is correct.

The IUPAC name of chloretone is :

0%
Trichloroacetone.
0%
Trichloronitromethane.
0%
1,1,1-Trichloro-2-methyl-2-propanone.
0%
1,1,1-Trichloro-2-methyl-2-propanol.

Explanation

IUPAC name of chloretone is 1,1,1-Trichloro-

$2$ -methyl-

$2$ -propanol.

Hence, the correct option is D.

Product is :

Select the correct statement:

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1,4-dibromobutane react with excess of magnesium in ether to generate di-Grignard reagent.
0%
1,2-dichlorocyclohexane treated with excess of Mg in ether produces cyclohexene.
0%
Vicinal dihalides undergo dehalogenation to give alkene when heated with Zn dust or Mg.
0%
1,3-dichloropropane by treatment with Zn dust or Mg forms cyclopropane.

Find out the incorrect statement.

0%
It gives total 9 allylic brominated products
0%
All allylic brominated products are optically active
0%
Substrate has 7 allylic hydrogens
0%
NBS gives Br constantly to reaction mixture.

Explanation

In this reaction NBS gives

$Br$ constantly to reaction mixture which attacks at allylic position and forms bromo products.
here, substrate has

$7$ allylic hydrogens and forms

$9$ allylic brominated products out of which this product is optically inactive.

Identify the product

$X$ .

Predict the product formed in the reaction.

$X\xrightarrow{Na + dry\hspace{1mm}ether}Z$

What can be used as

$X$ if

$Z$ is

$\displaystyle \:CH_{3}-CH_{3}$ ?

0%
$CH_{3}-Cl$
0%
$HCHO$
0%
$CH_3CHO$
0%

Explanation

Chloromethane

${ CH }_{ 3 }Cl$

$(X)$ undergoes Wurtz reaction in the presence of metal

$Na$ and solvent dry ether which couples to give ethane

${ CH }_{ 3 }-{ CH }_{ 3 }$

$(Z)$ .

Select the incorrect statement about the product mixture in the following reaction.

0%
It is optically active.
0%
It is racemic mixture.
0%
It is a resolvable mixture.
0%
It is a mixture of erythro compounds.

Which will give white ppt. with

$AgNO_{3}$ ?

0%
0%
0%
0%
Both A and C

Explanation

As both A and C provide chloride ion, due to formation of stable carbocation (allylic and benzylic), they both give white ppt. of

$AgCl$ with silver nitrate.

Assertion : Rate of ethanolysis of

$1^o$ halide by

$S_N1$ mechanism is fast.
Reason : Carbocation is stabilised by resonance.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

The ethanolysis of primary alkyl halide generally proceeds
via $S_N2$ mechanism. In case it proceeds via $S_N1$ mechanism, the intermediate
carbocation is unstable, since it has no adjacent carbon atom for resonance.

Hence, both assertion and reason are incorrect.

Neopentyl chloride on reaction with ethanolic

$KOH$ is likely to:

0%
Neopentyl alcohol
0%
Pentylene
0%
$2$ -Methyl- $2$ -butene
0%
Undergo no reaction

Explanation

Neopentyl chloride on reaction with ethanolic

$KOH$ is likely to undergo no reaction as it lacks alpha

$H$ atom. Due to this elimination is not possible.

Option D is correct.

Which of the following reactions is not stereospecific?

0%
${S_N}{2}$
0%
Addition of ${Br}_{2}$ to ethylene in $C{Cl}_{4}$
0%
Electrophilic substitution
0%
Glycol formation from alkenes with alkaline $KMn{O}_{4}$

Explanation

${S_N}{2}$ is stereospecific and stereoselective
b. Addition of

${Br}_{2}$ is stereospecific i.e anti-addition of two

$Br$ atoms.
c. Electrophilic substitution is not stereospecific.
d. Hydroxylation is stereopsecific, syn-addition of two

$(-OH)$ groups.

In the reaction of p-chlorotoluene with

$K{NH}_{2}$ in liquid

${NH}_{3}$ , the major product is:

0%
o-toluidine
0%
m-toluidine
0%
p-toluidine
0%
p-chloroaniline

$1$ -Chlorobutane on reaction with alcoholic potash gives:

0%
$1$ -Butene
0%
$1$ -Butanol
0%
$2$ -Butene
0%
$2$ -Butanol

Explanation

${1}^{o}RX$ undergoes

$E_2$ elimination with alc.

$KOH$ in which no rearrangement takes place.

The reaction of

$t$ -butyl bromide with sodium methoxide mainly produces:

0%
isobutane
0%
isobutylene
0%
$t$ -butyl methyl ether
0%
sodium tert-butoxide

$n$ -Propyl bromide on treatment with ethanolic potassium hydroxide produces:

0%
propane
0%
propene
0%
propyne
0%
propanol

Explanation

$\underset { n-Propyl\:bromide }{ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br } \xrightarrow [ ]{ Ethanolic\ KOH } \underset { Propene }{ { CH }_{ 3 }CH={ CH }_{ 2 } } +KBr+{ H }_{ 2 }O$

Structural isomerism and stereoisomerism should be considered when answering this question.

$2$ -bromopentane is heated with an excess of ethanolic sodium hydroxide. How many different hydrocarbons are produced?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$CH_3-\overset{\overset{Br}{|}}CH-CH_2-CH_2-CH_3\xrightarrow{alc.NaOH}\underset{\text{Cis and trans}}{CH_3-CH=CH-CH_2-CH_3}+CH_2=CH-CH_2-CH_2-CH_3$

It is

$\beta$ -elimination reaction. In this reaction one of the product is present in cis and trans form.

Thus we get 3 product.

When hydrochloric acid gas is treated with propene in the presence of benzoyl peroxide, it gives:

0%
2-chloropropane
0%
allyl chloride
0%
no reaction
0%
n-propyl chloride

Which of the following is potential energy diagram for

$S_N1$ reaction ?

Explanation

Solution:- (B)

$S_N1$ is a two step reaction in which first step is rate determining step. Hence the peak of first step is higher than second step.

What are the suitable reactant for the following ether synthesis?

$CH_3 - \underset{CH_3\!\!\!\!}{\underset{|}{\overset{CH_3\!\!\!\!\!\!}{\overset{|}{C}}}} - O-CH_2-Ph$

0%
$CH_3 - \underset{CH_3\!\!\!\!}{\underset{|}{\overset{CH_3\!\!\!\!\!\!}{\overset{|}{C}}}} - \bar{O} \overset{\oplus}{N}a +Ph - CH_2 - Cl$
0%
$Ph - \underset{\!\!\!\!\!Cl}{\underset{\!\!\!\!\!|}{CH}} - CH_2 - CH_3 + CH_3 - \overset{\bar{O}\, \overset{\oplus}{N}a\!\!}{\overset{\!\!\!\!\!\!|}{CH}} - CH_3$
0%
$CH_3 - \underset{CH_3\!\!\!\!}{\underset{|}{\overset{CH_3\!\!\!\!\!\!}{\overset{|}{C}}}} - Cl+PhCH_2\overset{\ominus}{O} \overset{\oplus}{N}a$
0%
$CH_3 - \underset{CH_3\!\!\!\!}{\underset{|}{\overset{CH_3\!\!\!\!\!\!}{\overset{|}{C}}}} - \bar{O} \overset{\oplus}{N} a + CH_3 - Cl$

Which of the following are correct for

$S N^1$ reaction?

0%
The reaction intermediate is carbocation
0%
In this reaction the complete inversion takes place
0%
It is favoured by stability of carbocation
0%
None of these

Explanation

Characteristcics of

$S_N1$ reactions:

1) The reaction intermediate is a carbocation.

2) It is favored by the stability of carbocation.

3) Product formed is an enantiomeric mixture.

Hence, Options "A" & "C" are the correct answers.

An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be:

0%
Optically active mixture
0%
Pure enantiomer
0%
Meso-compound
0%
Racemic mixture

Explanation

$(d)or(+)RCOOH+(dl)or (\pm)R'OH)\rightarrow (dd)or(+)(+)RCOOR'(I)$ and

$(dl)or(+)(-)RCOOR'(II)$

When optically active acid reacts with racemic mixture of alcohol, it forms two types (I and II) of isomeric esters. In each, the configuration of chiral centre of acid would remain the same. So, the mixture would be optically active

Option A is correct.

An alkyl chloride produces a single alkene on reaction with sodium ethoxide and ethanol. The alkene further undergoes hydrogenation to yield 2-methylbutane. Identify the alkyl chloride from amongst the follwoing

0%
$ClCH_2CH(CH_3)CH_2CH_3$
0%
$ClCH_2CH_2CH_2CH_3$
0%
$ClCH_2CH(CH_3)CH_2CH_3$
0%
$CH_3C(Cl)(CH_3)CH_2CH_3$

Explanation

Complete set of reaction involving that alkyl halide is

So, the alkyl chloride is

$ClCH_2-CH(CH_3)-CH_2-CH_3$

Hence, Option "C" is the correct answer.

The IUPAC name of the compound is

0%
trans $-3-$ iodo $-4-$ chloro $-3-$ pentene
0%
cis $-2-$ chloro $-3-$ iodo $-2-$ pentene
0%
trans $-2-$ chloro $-3-$ iodo $-2-$ pentene
0%
cis $-3-$ iodo $-4-$ chloro $-3-$ pentene

Cholorobenzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is

0%
Fitting reaction
0%
Wurtz - Fitting reaction
0%
Sandmeyer reaction
0%
Gattermann reaction

Which of the following will be the major and minor product?

0%
(A) is major product and (B) is minor products.
0%
(A) is minor product and (B) is major product
0%
Both (A) and (B) are major products.
0%
Only (B) is formed and (A) is not formed.

Which of the following reactions will yield

$2,2-$ dibromopropane?

0%
$CH_2=CHBr+HBr$
0%
$CH_3C \equiv CCH_3+2 HBr$
0%
$CH_3C \equiv CH+2 HBr$
0%
$CH_3CH = CHBr+HBr$

State True or False:
The formation of

$CH_{3}Cl$ by the gas phase ultraviolet irradiation of a mixture of

$CH_{4}$ and

$Cl_{2}$ involves free radical intermediate.

0%
True
0%
False

Explanation

The mechanism of formation of

$CH_3Cl$ by the gas phase ultraviolet irradiation of a mixture of

$CH_4$ and

$Cl_2$ is given above and involves free-radical intermediate.

Thus, option A is correct.

1935265_1878136_ans_4de83ccc33b54997bdfcc4038d39df18.JPG

$S{_N}{1}$ reaction on optically active substrates mainly gives:

0%
retention in configuration
0%
inversion in configuration
0%
racemic product
0%
no product

Explanation

since the nucleophile attacks the carbocation only after the leaving group has departed, in

${S_N}^1$ , there is no need for back-side attack. The carbocation and its substituents are all in the same plane (Figure 1), meaning that the nucleophile can attack from either side. As a result, both enantiomers are formed in a

${S_N}^1$ reaction, leading to a racemic mixture of both enantiomers.

Hence, the correct option is

$\text{C}$

Aryl halide is less reactive than alkyl halide towards nucleophilic substitution because:

0%
of less stable carbonium ion
0%
due to large $C-Cl$ bond energy
0%
of inductive effect
0%
of resonance stabilization and $s{p}^{2}$ hybridization of $C$ attached to halide

Explanation

Aryl halide is less reactive than alkyl halide towards nucleophilic substitution because of the resonance stabilization and

$sp^2$ hybridization of

$C$ attached to halide. Due to resonance, carbon-chlorine bond acquires partial double bond character, hence it becomes shorter and stronger, and thus, cannot be easily replaced by nucleophiles. This effect is further enhanced by the

$sp^2$ nature of the

$C$ atom to which chlorine is attached.

Which of the following alkyl halides will undergo

${S}_{N}{1}$ reaction most readily?

0%
${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-F$
0%
${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Cl$
0%
${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br$
0%
${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-I$

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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