MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 1

An unknown alkyl halide

$[$ X

$]$ reacts with alcoholic KOH and produces a hydrocarbon

$(C_4H_8)$ as the major product. Ozonolysis of this hydrocarbon affords one mole of propanaldehyde and one mole of formaldehyde. Suggest which organic compound among the following has the correct structure of the above alkyl halide

$[$ X

$]$ ?

0%
$CH_3CH_2CH_2CH_2Br$
0%
$CH_3CH(Br)CH_2CH_3$
0%
$CH_3CH(Br)CH(Br)CH_3$
0%
$Br(CH_2)_4Br$

Explanation

As the ozonolysis gives propanaldehyde and formaldehyde thus the hydrocarbon is

$\underset{1-butene}{CH_3CH_2CH=CH_2(C_4H_8)}$

$1-$ butene can be obtained by

$CH_3-CH_2-CH_2-CH_2-Br\overset{Alc. KOH}{\underset{-HBr}{\rightarrow}}CH_3CH_2CH=CH_2$ .

In alkaline hydrolysis of tertiary halide by aqueous alkali if concentration of alkali doubled, then the reaction:

0%
will be doubled
0%
will be halved
0%
will remain constant
0%
None of the above

Explanation

In alkaline hydrolysis of tertiary halide by aqueous alkali, if concentration of alkali is doubled, then the reaction will remain constant because t-alkyl halides with aqueous alkali give

${S}_{N}1$ reaction and the rate of

${S}_{N}1$ reaction is not based upon the concentration of nucleophile (i.e., alkali).

An alkane

$C_6H_{14}$ gives only two monochloro derivatives on chlorination. Its possible structure is:

0%
$CH_3CH_2CH_2CH_2CH_2CH_3$
0%
$CH_3-{\underset{\displaystyle CH_3}{\underset{|}CH}}-CH_2CH_2CH_3$
0%
$CH_3CH_2-{\underset{\displaystyle CH_3}{\underset{|}CH}} -CH_2CH_3$
0%
$CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3$

Explanation

Monochloro derivation means substitution of single chlorine atom by replacing it with any hydrogen attached to the carbon atom.

(D) will give only two monochloroderivative:

$CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3 \xrightarrow{{Cl}_{2}} CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-\underset{\underset{\displaystyle Cl}|}{{\overset{\displaystyle \,\,\,\,CH_3}{\overset{|}C}}}-CH_3 + Cl-CH_2-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3$

Identify

$X$ and

$Y$ in the following reaction.

$H_ 2\underset { \underset {\displaystyle Br }{ | } }{ C } -\underset { \underset { \displaystyle Br }{ | } }{ CH_2}+KOH\xrightarrow [ ]{ alcohol } X\xrightarrow [ ]{ NaNH_ 2 } Y$

0%
$X: CH_3CHBr, \,\,Y: CH_2=CH_2$
0%
$X: CH_2OH-CH_2OH, \,\, Y: CH_2=CH_2$
0%
$X: CH_2=CHBr, \,\, Y: CH\equiv CH$
0%
$X: CH\equiv CBr, \,\, Y: CH\equiv CH$

Explanation

The reactant is symmetrical alkanes. On reacting with alcoholic KOH, dehydrohalogenation reaction will occur. It will give monobromoalkene.

$Br{H}_{2}C-C{H}_{2}Br + KOH \xrightarrow{alcohol} C{H}_{2}=CHBr$

$C{H}_{2}=CHBr$ on reaction with

$NaN{H}_{2}$ releases sodium bromide and ammonia and forms ethyne.

$C{H}_{2}=CHBr \xrightarrow{NaN{H}_{2}} CH \equiv CH + NaBr + N{H}_{3}$

$2-$

$Bromopentane is treated with alcoholic$

$KOH$ $ solution. What will be the major product formed in this reaction and what is this type of elimination called?

0%
Pent $-1-$ ene, $\beta$ - Elimination
0%
Pent $-2-$ ene, $\beta$ - Elimination
0%
Pent $-1-$ ene, Nucleophilic substitution
0%
Pent $-2-$ ene, Nucleophilic substitution

Explanation

Pent-2-ene is major product known as Saytzeff's Product and it is more stable alkene.

The process is known as $\beta-$ elimination as it involves elimination of $\beta-$ Hydrogen.

1061058_928797_ans_560601d1af3d4e31b7aca637709e25cb.PNG

In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the species :

0%
$X$
0%
$X^-$
0%
$X^+$
0%
$X^{\bullet}$

Explanation

Halogen carrier generates an electrophile with the help of halogen. e.g. during chlorination

${Cl}^{+}$ is generated.

$Al{X}_{3} + {X}_{2} \longrightarrow {Al{X}_{4}}^{-} + {X}^{+}$

The correct decreasing order of following alkyl halides according to the rate of

$\beta$ - elimination reaction with alcoholic KOH will be:
(i)

${ CH }_{ 3 }-\overset { \underset { | }{ H } }{ \underset { \underset { { CH }_{ 3 } }{ | } }{ C } } -{ CH }_{ 2 }Br$
(ii)

$CH_3-CH_2-Br$
(iii)

$CH_3-CH_2-CH_2-Br$

0%
(i) > (ii) > (iii)
0%
(iii) > (ii) > (i)
0%
(ii) > (iii) > (i)
0%
(i) > (iii) > (ii)

Explanation

Stability of alkene after

$\beta$ -elimination directly depends on the number of

$\beta$ -substituents.

Since number of

$\beta$ - methyl substituent in (i) is 2, (iii) is 1 and (ii) is 0, thus the correct order of stability is:

$(i)>(iii)>(ii)$ .

Which of the following compounds will react with Na to form 4,5- diethyloctane?

0%
$CH_3CH_2CH_2CH_2Br$
0%
$CH_3CH_2CH_2-{\underset{CH_3}{\underset{|}C}H}-CH_2CH_2Br$
0%
$CH_3CH_2CH_2CH_2-{\underset{Br}{\underset{|}C}H}-CH_3$
0%
$CH_3CH_2CH_2-{\underset{Br}{\underset{|}C}H}-CH_2CH_3$

Which of the following is a primary halide?

0%
iso-propyl iodide
0%
sec-butyl iodide
0%
tert-butyl bromide
0%
neo-hexyl chloride

$S_N1$ reaction is fastest in?

0%
$CH_3CH_2Br$
0%
$CH_3-{\underset{CH}{\underset{|}{CH}}}-Cl$
0%
0%
$CH_3-{\underset{Ch_3}{\underset{|}{\underset{CH_2}{\underset{|}{CH}}}}}-Cl$

Explanation

The

$S_N1$ , reaction proceeds in two steps. First step involves slow separation of halogen to form carbonium ion. Second step involves attack of nucleophile on carbonium ion from both sides to form alcohol.

So any species which is capable of forming stable carbonium ion would undergo nuleophilic substitution by

$S_N1$ . Benzyl is very stablilised. Hence benzyl chloride follow

$S_N1$ most readily. So C is the correct option.

Which of the following compounds can yield only one monochlorinated product upon free radical chlorination?

0%
2, 2-Dimethyipropane
0%
2-Methylpropane
0%
2-Methylbutane
0%
n-Butane

Which of the following statement regarding the

$S_N1$ reaction shown by alkyl halide is correct?

0%
The added nucleophile plays no kinetic role in $S_N1$ reaction.
0%
The $S_N1$ reaction involves the inversion of the configuration of the optically active substrate
0%
The $S_N1$ reaction on the chiral starting material ends up with racemisation o f the product.
0%
The more stable the carbocation intermediate the faster the $S_N1$ reaction.

Explanation

$S_N1$ reaction, rate of substitution depends on the concentration of substrate. In case of

$S_N1$ reaction there is racemisation.

$S_n1$ reaction will be faster, if carbocation formed is more stable .

Which of the following alkyl halides is hydrolysed by

$S_N1$ mechanism?

0%
$CH_3Cl$
0%
$CH_3CH_2Cl$
0%
$CH_3CH_2CH_2Cl$
0%
$(CH)_3CCl$

Explanation

More stable carbocation formed from alkyl halide than reactivity of that alkyl halide will increase to hydrolysis via

$SN^1$ mechanism.

$3^o$ carbocation

$>2^o$ carbocation

$>1^o$ carbocation(order of stability.).

1032862_937583_ans_1868648386e04595883117fa7cef13c7.PNG

2,4-hexadiyne

$(C_6 H_6)$ is allowed to react with

$Li$ in

$NH_3$ (liq). The product obtained is treated with 1 equivalent of

$Cl_2$

$CCl_4$ . Which of the following constitutional isomers are possible products ?

0%
I and II
0%
II and III
0%
I and IV
0%
I and III

Major product is:

Explanation

Here, the base takes

$-Br$ leaving a canbocation. The rearranges to form a stable compound, so, methyl group is shifted and hence, double bond is formed.
1132592_1154044_ans_f4767b2345c64ff6b69a29e604794410.png

1132592_1154044_ans_f4767b2345c64ff6b69a29e604794410.png

Which of the following isomer of C

$_{5}$ H

$_{11}$ Cl exhibits optical isomerism?

0%
1-chloropentane
0%
2-chloro-2-methylpentane
0%
2-chloro-3-methylbutane
0%
1-chloro-2,2-dimethylpropane

${CH}_{3}-I+2Na+I-{CH}_{3}\xrightarrow [ ]{dry \; ether }$ _______

0%
${C}_{2}{H}_{5}-{C}_{2}{H}_{5}$
0%
${CH}_{4}$
0%
${C}_{3}{H}_{8}$
0%
${CH}_{3}-{CH}_{3}$

Explanation

When methyl Iodide is treated with sodium in the presence of dry ether, Ethane is formed .It is

called Wurtz reaction.

Dry ether

$CH_3−I + 2Na + I−CH_3 \rightarrow CH_3−CH_3 + 2NaI$

So option

$D$ is correct

Which of the following alkyl bromides undergoes most rapid solvolysis in methanol solution to give corresponding methyl ether ?

What is the minimum quantity (in grams) of methyl bromide required for preparing one mole of ethane by wurtz reaction?

0%
188
0%
109
0%
326
0%
284

Pure methane can be produced by:

0%
Wurtz reaction
0%
Kolbe's electrolytic
0%
soda lime decarboxylation
0%
reduction with $H_2$

Explanation

Soda line decarboxylation reaction can produce pure methane.

$CH_3 COONa + NaOH \xrightarrow[Cao]{\Delta } CH_4 + Na_2 \,CO_3$

1192780_1186196_ans_c7a06b1842bf4f569af5f2a34cc96c5e.jpg

Which type of isomerism can not be shown by benzaldoxime ?

0%
Optical isomerism
0%
Functional group isomerism
0%
Geometrical isomerism
0%
Configuration isomerism

Explanation

Geometrical isomerism

In the syn-form,

H and OH are present on the same side.

In the anti-form,

H and OH are present on the opposite side.

The isomerism shown by Benzaldoxime is

$geometrical$ . Benzaldoxime exists in the synform and anti-form.

Given Reaction.
Y in reaction is:

0%
Hexane
0%
Cyclohexane
0%
Cyclohexylcyclohexane
0%
Cyclohexylether

Explanation

(Refer to Image)

When

$Mg$ in presence of ether reacts known as Grignard reagent and this form salt with halogen. Then

$H_2O$ and

$HCl$ provide

$H^+$ in place of

$MgBr$ . This is an example of substitution reaction.

1091021_1189053_ans_59457fc9ca2f42d88c5db635517ed4c9.png

What is the CORRECT order of reactivity of the following compounds in atomic nucleophilic substitution reaction when treated with aq. NaOH?

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

Consider the reaction -The major product

$A$ is?

The following stability of the following carbon cations:

0%
III > II >I
0%
II > III > I
0%
I > II > III
0%
III > I > II

Which of the following is a correct order of reactivity of chlorides for given compounds ?

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

Chlorobenzene is prepared commercially by

0%
Raschig process
0%
Wurtz Fitting reaction
0%
Friedel-Craft's reaction
0%
Grignard reaction

To prepare a pure sample of normal hexane using sodium metal as one reactant, the other reactant or reactants will be:

0%
Ethyl chloride and n-butyl chloride
0%
Methyl bromide and n-pentyl bromide
0%
n-propyl bromide
0%
ehtyl bromide and n-butyl bromide

Which is intermediate of above reaction :

Explanation

Other halogenating agents are

$lCl, HOCl$ etc,
order of effectiveness

$-Cl_2 < BrCl > Br_2 > lCl > l_2$
1147496_1348274_ans_5092b3de2e3c42a585d8963afa94de88.png

When octa-O-methyl d-cellobiose is hydrolysed by aqueous acid, two O-methylated glucose derivatives are formed: One is tetramethyl derivative and the other is trimethyl derivative. Why is a single methyl substituent lost in this process ?

0%
One methoxy group is lost by $\beta$ -elimination
0%
One methoxy group is an ester and the others are all ethers
0%
One methoxy group is a part of an acetal,the others are all ethers
0%
One glucose is an $\alpha$ -methyl glycoside, the other is a $\beta$ -methyl glycoside

The major product of the following reaction is :

2-choloro-2-methylpentane on reaction with sodium methoxide in methanol yields

0%
0%
0%
0%
None of these

$wC_{2}H_{6}+xI_{2}+HIO_{3}\rightarrow y C_{2}H_{5}I+z H_{2}O$ the value of

$w,x,y,z$ are:

0%
$5,2,5,3$
0%
$5,5,2,3$
0%
$5,3,5,2$
0%
$2,2,2,1$

Explanation

$5{C}_{2}{H}_{6} + 2{I}_{2} + HI{O}_{3} \rightarrow 5{C}_{2}{H}_{5}I + 3{H}_{2}O$

The given reaction is an iodonation of ethane to get ethanoic acid. Here

$HIO_3$ act as an oxidizing agent.

After balancing the reaction the value of

$w,x,y,z$ are

$w,x,y,z = 5,2,5,3$

Option A is correct

Preparation of alkyl halides in laboratory is least preferred by :

0%
Halide exchange
0%
Direct halogenation of alkanes
0%
Treatment of alcohols
0%
Addition of hydrogen halides to alkenes

In the question arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

0%
(I) < (II) < (III)
0%
(III) < (II) < (II)
0%
(I) < (III) < (II)
0%
(III) < (I) < (II)

Explanation

Presence of electron withdrawing group (-NO

$_2$ ) increases the reactivity towards nucleophilic substitution and the effect of o-position is greater than at m-position due to resonance.

So, the order becomes (I) < (III) < (II)

Hence, Option "C" is the correct answer.

Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces

0%
4-phenylcyclopentene
0%
2-phenylcyclopentene
0%
1-phenylcyclopentene
0%
3-phenylcyclopentene

State whether the statement is true or false. Rewrite the false statement correctly.
Wurtz reaction can be used to prepare alkanes with odd number of carbon atoms.

0%
True
0%
False

Explanation

Wurtz reaction can be used to prepare alkanes with both odd and even number of carbon atoms.

Examples:

$CH_3-CH_2Br+2Na+CH_3Br\xrightarrow {Dry\ ether} CH_3-CH_2-CH_3+2NaBr$

$CH_3Br+CH_3Br+2Na\xrightarrow {Dry\ ether} CH_3-CH_3+2NaBr$

The product of the reaction is :

Explanation

It is an addition reaction which goes through stable

$3^0$ carbocation intermediate. (Markonikov's rule)

The complete reaction is shown above.

Option B is correct.

An alkyl halide on reaction with sodium in the presence of ether gives 2,2,5,5-tetramethylhexane. The alkyl halide is:

0%
1-chloro pentane
0%
1-chloro-2,2-dimethyl propane
0%
3-chloro-2,2-dimethyl butane
0%
2-chloro-2-methyl-butane

Explanation

Alkyl halide on reaction with sodium in presence of ether to give alkane is called Wurtz reaction.
The reaction takes place as:

$2ClCH_2-C(CH_3)_2-CH_3+2Na\rightarrow CH_3-C(CH_3)_2-CH_2-CH_2-C(CH_3)_2-CH_3+ 2NaCl$

Addition of ethanol aqueous, to the benzyl chloride does not increase the rate of the
hydrolysis but changes only the composition of the final products. This indicates that the reaction is proceeding through:

0%
$S_{N}1$ mechanism
0%
$S_{N}2$ mechanism
0%
$S_{E}1$ mechanism
0%
$S_{E}2$ mechanism

Explanation

Rate of

$S_{N}1$ reactions depends only on one reactant and rate of

$S_{N}2$ reactions depends on both reactants.
here reaction's rate is depending only on one reactant so it follows

$S_{N}1$ mechanism.

Statement 1:

$S_{N}2$ reaction is carried out in the presence of polar aprotic solvents.

Statement 2: Polar aprotic solvents does not contain acidic hydrogen.

0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1.
0%
Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation for Statement 1.
0%
Statement 1 is correct and statement 2 is wrong.
0%
Statement 1 is wrong and statement 2 is correct.

The major product obtained in the photo catalysed bromination of

$2$ -methylbutane is:

0%
$1$ -bromo- $2$ -methylbutane
0%
$1$ -bromo- $3$ -methylbutane
0%
$2$ -bromo- $3$ -methylbutane
0%
$2$ -bromo- $2$ -methylbutane

Explanation

The major product obtained in the photo catalysed bromination of 2-methylbutane is 2-bromo-2-methylbutane.

Hence, the correct option is

$\text{D}$

Which one of the following occurs by an addition elimination mechanism?

0%
Toluene, Br $_2$ , FeBr $_3 \rightarrow$ $\rho$ -bromotoluene
0%
Benzene, HNO $_3$ , H $_2$ SO $_4 \rightarrow$ nitrobenzene
0%
Benzene, CO, HCl, AlCl $_3$ , CuCl $\rightarrow$ benzaldehyde
0%
2, 4-dinitrochlorobenzene, NaOH, heat, then H $^+ \rightarrow$ 2, 4-dinitrophenol

The following synthesis can not be carried out by:

0%
$\xrightarrow [(1)]{Cl_2}$ $\xrightarrow [(2)]{ICl/ CH_3COOH}$ $\xrightarrow [(3)]{ICl/CH_3COOH}$ $\xrightarrow [(4)]{Zn dust}$
0%
$\xrightarrow [(1)]{HOBr/H^+}$ $\xrightarrow [(2)]{Cl_2/Fe}$ $\xrightarrow [(3)]{ICl/ZnCl_2(excess)}$ $\xrightarrow [(4)]{NaOH, \Delta}$
0%
$\xrightarrow [(1)]{HOBr/H^+}$ $\xrightarrow [(2)]{Cl_2/Fe}$ $\xrightarrow [(3)]{ICl/ZnCl_2(excess)}$ $\xrightarrow [(4)]{Zn dust, CH_2COOH}$
0%
$\xrightarrow [(1)]{Br_2/CH_3COOH}$ $\xrightarrow [(2)]{Cl_2/Fe}$ $\xrightarrow [(3)]{ICl/CH_3COOH (excess)}$ $\xrightarrow [(4)]{NaNH_2}$

Explanation

The given synthesis cannot be carried out by options

$A, B$ and

$D$ . It can be carried out by option

$C$ .

The product of the reaction

$P$ is:

0%
0%
0%
0%
None of the above

Explanation

Here, due to possibility of a cyclic stable intermediate,

$KOH$ reacts with reactant and

$\alpha - C$ w.r.t. to

$O$ , forms a five-membered ring product.

In the given reaction, the products are:

Explanation

Due to the presence of a strong base, it will prefer elimination reaction and alkene will be formed as major product and SN2 product will be formed as a secondary product.

Hence, the correct option is

$B$

In the reaction

$C_{7}H_{8}\xrightarrow {3Cl_{2}/\Delta}X\xrightarrow {Br_{2}/Fe} Y \xrightarrow {Zn/HCl}Z$ .

The compound

$Z$ is :

0%
o-bromotoluene
0%
m-bromotoluene
0%
p-bromotoluene
0%
3-bromo-2, 2, 6-trichlorotoluene

Explanation

$Z$ is m-bromotoluene.

Toluene is heated with chlorine to form (trichloromethyl)benzene. The aromatic nucleus is then brominated with bromine in presence of iron to form 1-bromo-3-(trichloromethyl)benzene.

Treatment with

$Zn/HCl$ gives m-bromotoluene (compound

$Z$ ).

So, option B is correct.

Consider the following reaction, expected to be occuring by

$' E-i cb'$ mechanism

$PH-CH_2 - \underset { \overset { | }{ Br } }{ CH } -CH_2+C_2H_5OK\xrightarrow [ ]{ C_2H_2OD }$
Which is least expected compound in the reaction mixture:

0%
$PhCH = CHCH_3$
0%
$\underset { \overset { | }{ D } }{ PhCH } -\underset { \overset { | }{ Br } }{ CH } -CH_3$
0%
$Ph-CH2-\overset { \underset { | }{ Br } }{ \underset { \overset { | }{ D } }{ C } } -CH3$
0%
$Ph-\underset { \overset { | }{ D } }{ C } =CH-CH_3$

Which of the given molecules are optically active?

0%
(i) and (iii)
0%
(ii) and (iii)
0%
(i), (ii) and (iii)
0%
(i) and (ii)

Deduce structure

$(A)$ .

$(A)\ { C }_{ 8 }{ H }_{ 10 }\xrightarrow [ ]{ KMn{ O }_{ 4 } } (B)\quad { C }_{ 8 }{ H }_{ 6 }{ O }_{ 4 }\xrightarrow [ Fe ]{ { Br }_{ 2 } } (C)\quad { C }_{ 8 }{ H }_{ 5 }{ BrO }_{ 4 }$ (one product only)

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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