MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 7

$S_{N}1$ , the first step involves the formation of:

0%
free radical
0%
carbanion
0%
carbocation
0%
final product

Explanation

This pathway for $S_N1$ mechanism is a multi-step process with the following characteristics:

		step 1: slow loss of the leaving group, LG, to generate a carbocation intermediate

		step 2 : rapid attack of a nucleophile on the electrophilic carbocation to form a new s bond

The product

$(D)$ of the reaction will be:

$C{ H }_{ 3 }Cl\xrightarrow { \ KCN\ } \left( A \right) \xrightarrow { \ { H }_{ 3 }{ O }^{ + }\ } \left( B \right) \xrightarrow { \ N{ H }_{ 3 }\ } \left( C \right) \xrightarrow { \Delta \ } \left( D \right)$

0%
$C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 }$
0%
$C{ H }_{ 3 }CN$
0%
$HCON{ H }_{ 2 }$
0%
$C{ H }_{ 3 }CON{ H }_{ 2 }$

Explanation

Reaction:

$C{ H }_{ 3 }-Cl\xrightarrow { KCN\ } C{ H }_{ 3 }-CN\xrightarrow { { H }_{ 3 }{ O }^{ \oplus } } C{ H }_{ 3 }-COOH \xrightarrow { N{ H }_{ 3 } } C{ H }_{ 3 }-CO\overset { \ominus }{ O } \overset { \oplus }{ N } { H }_{ 4 }\xrightarrow { \Delta } C{ H }_{ 3 }-CON{ H }_{ 2 }$

Option D is correct.

The most reactive towards

$S_{ N }{ 1 }$ is:

0%
$PhC{ H }_{ 2 }Cl$
0%
$PhCl$
0%
$PhCH(Cl)C{ H }_{ 3 }$
0%
$p-N{ O }_{ 2 }PhC{ H }_{ 2 }Cl$

Explanation

The rate of

$S_{N}{1 }$ reaction depends upon the stability of the carbocation formed.

The compound given in option C forms the most stable carbocation after removal of Cl. This carbocation is most stable due to resonance, hyperconjugation and

$+I$ effect of methyl group so it will react easily.

$Ph-\overset { \oplus }{ C } H-C{ H }_{ 3 }$

Option C is correct.

Most stable carbocation formed from

${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }CHBr$ and

${ C }_{ 6 }{ H }_{ 5 }C{ H }_{ 2 }Br$ would be:

0%
${ C }_{ 6 }{ H }_{ 5 }\overset { \oplus }{ C } { H }_{ 2 }$
0%
${ \left( C{ H }_{ 3 } \right) }_{ 3 }\overset { \oplus }{ C }$
0%
${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }\overset { \oplus }{ C }$
0%
${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }\overset { \oplus }{ C } H$

Explanation

Most stable carbocation formed will be

${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }\overset { \oplus }{ C }$ obtained from the hydrolysis of

${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }CHBr$ .

This is because triphenyl methyl carbocation

${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }C^+$ is highly stable due to delocalization of positive charge on three phenyl groups.

${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr \rightarrow { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }C^+$

Bond pair of electrons in

$C-X$ bond is nearer to:

0%
$X$
0%
$C$
0%
equidistant
0%
can't be predicted

Explanation

Halogens are more electronegative than carbon. Halogens are electron withdrawing group which pulls a bond pair of electrons towards it.

$C- X$

Thus, Bond pair of electrons in

$C-X$ is near to X

Which of the following compounds will given racemic mixture on nucleophilic substitution by

$O{H}^{\ominus}$ ion?
(a)

$C{ H }_{ 3 }-\underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ CH } -Br$
(b)

$C{ H }_{ 3 }-\overset { \underset { | }{ Br } }{ \underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ C } } -C{ H }_{ 3 }$
(c)

$C{ H }_{ 3 }-\underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ CH } -C{ H }_{ 2 }Br$

0%
$(a)$
0%
$(a), (b), (c)$
0%
$(b), (c)$
0%
$(a), (c)$

Explanation

$C :$

$1^{\circ}$ Alkyl halide can directly forms substitution product without inversion.

$B :$

$3^{\circ}$ Alkyl halide allows

$OH^{-}$ to attack only from one side d after leaving of halide

$OH^{-}$ attacks due steric hinderance.

$A :$

$2^{\circ}$ Alkyl halide allows

$OH^{-}$ to attack while leaving of

$Br^{-}$ hence inversion of molecule occurs forms racemic mixture.

Alkyl halide can be converted into alkene by:

0%
nucleophilic substitution reaction
0%
elimination reaction
0%
both nucleophilic substitution and elimination reaction
0%
rearrangement

Explanation

$R- CH_2 - CH_2 - X + KOH_{(alc)} \xrightarrow{\Delta} \underset{alkene}{RCH = CH_2 + }KX + H_2O$

The

$H_2O$ and KX molecules are eliminated during reaction, so we can say alkene is formed from alkyl halide by elimination reaction.

Benzene reacts with chlorine in presence of light to yield:

0%
chlorobenzene
0%
trichlorobenzene
0%
tetrachlorobenzene
0%
benzenehexachloride

Explanation

In the presence of ultraviolet light (but without a catalyst present), hot benzene will also undergo an addition reaction with chlorine or bromine. The ring delocalisation is permanently broken and a chlorine or bromine atom adds on to each carbon atom.

For example, if you bubble chlorine gas through hot benzene exposed to UV light for an hour, you get $1,2,3,4,5,6$ -hexachlorocyclohexane or benzenehexachloride.

Which of the following has strongest carbon - halogen bond ?

0%
$C{ H }_{ 3 }Cl$
0%
$C{ H }_{ 3 }Br$
0%
$C{ H }_{ 3 }F$
0%
$C{ H }_{ 3 }I$

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

$\xrightarrow{Br_2/hu}$

For the given reaction how many products are optically active (all isomers) :

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

Explanation

Optimally active isomers are-

$1-Chloro-2,2,4-trimethyl \ pentane$

$2-Chloro-2,2,4-trimethyl \ pentane$

$3-Chloro-2,2,4-trimethyl \ pentane$

$4-Chloro-2,2,4-trimethyl \ pentane$

Under identical conditions, the

$S_{N}1$ reaction will occur most efficiently with:

0%
tert-butyl chloride
0%
1-chlorobutane
0%
2-methyl-1-chloropropane
0%
2-chlorobutane

Explanation

$S_{N}1$ reaction is most effective in

$3^{\circ}$ carbon.

The order goes as

$Tert>secondary>primary>methyl$

$\underset {tert-butyl\ chloride (3^{\circ})}{(CH_{3})_{3} - C - Cl}$

$\underset {1 - chlorobutane (1^{\circ})}{CH_{3}CH_{2}CH_{2}CH_{2}Cl}$

$\underset {2 - methyl-1-chloropropane (1^{\circ})}{Cl - CH_{2} - CH(CH_{3})-CH_{3}}$

$\underset {2-chlorobutane (2^{\circ})}{CH_{3}CH_{2}CH(Cl)CH_{3}}$

Following reaction,

${({CH}_{3})}_{3}CBr+{H}_{2}O\longrightarrow {({CH}_{3})}_{3}COH+HBr$ is an example of:

0%
elimination rection
0%
free radical substitution
0%
nucleophilic substitution
0%
electrophilic substitution

Explanation

${({CH}_{3})}_{3}Br+{H}_{2}O\longrightarrow {({CH}_{3})}_{3}C-OH+HBr$

$Br$ is substituted by

$-{OH}^{-}$ (nucleophile)

${S}_{N}1$ (unimolecular substitution reaction)

Which of the following compounds have highest melting point?

0%
Only I
0%
Only II
0%
I and II
0%
II and III

t-Butyl chloride preferably undergo hydrolysis by:

0%
$S_N 1$ mechanism
0%
$S_N 2$ mechanism
0%
any of (A) and (B)
0%
none of these

The false statement regarding alkane is :

0%
this does not perform polymerization reaction.
0%
this does not gives elimination reaction.
0%
it does not disappear the colour of dilute $KMnO_{4}$ solution..
0%
it does not decolourise bromine water.

Explanation

Alkanes show elimination reactions. Higher alkanes eliminate hydrogen and give lower alkanes and alkenes on heating in the absence of air at high temperatures.

$R-CH_2-CH_3\xrightarrow{catalyst} R-CH=CH_2$

Which of the following statements is incorrect regarding benzyl chloride ?

0%
It gives white precipitate with alcoholic $AgNO_3$ .
0%
It is an aromatic compound with substitution in the side chain.
0%
It undergoes nucleophilic substitution reaction.
0%
It is less reactive than vinyl chloride.

Alkanes having odd carbons cannot be prepared as a major product in:
(A) Wurtz reaction
(B) Frankland reaction
(C) Kolbe's electrolysis
(D) Sabatier - sendersen reaction

0%
B, C and D
0%
A, C and D
0%
A, B and D
0%
A, B and C

Explanation

Alkanes having odd carbons can be prepared in Sabatier - sendersen reaction.
In this reaction, alkenes add hydrogen in the presence of platinum or nickel catalyst, to form alkanes. The reaction termed as hydrogenation, is an exothermic reaction.

$\displaystyle CH_3-CH=CH_2 +H_2 \xrightarrow [473 K]{Ni} CH_3-CH_2-CH_3$
Alkanes having odd carbons cannot be prepared in
(A) Wurtz reaction
(B) Frankland reaction
(C) Kolbe's electrolysis
Wurtz reaction and Kolbe's electrolysis can be used for the preparation of alkanes having even carbon atoms.

When Zn is used in Wurtz reaction in place of Na, the reaction is named as Frankland method. In wourtz reaction a solution of alkyl halide in ether on heating with sodium gives alkane.

Which of the following statements about benzyl chloride is incorrect?

0%
It is less reactive than alkyl halides
0%
It can be oxidised to benzaldehyde by boiling with copper nitrate solution
0%
It is a lachrymatory liquid and answers Beilstein's test
0%
It gives a white precipitate with alcoholic silver nitrate

The monohalogen derivatives of alkanes are called:

0%
alkene halide
0%
haloalkanes
0%
halogen alkyl
0%
none of the above

Which of the following dihalides on hydrolysis with aqueous

$KOH$ gives ethylene glycol?

0%
Ethylene Chloride
0%
Ethylidene Chloride
0%
Isopropylidene Chloride
0%
None of the above

Explanation

Ethyl can undergo both nucleophilic substitution as well as elimination reactions with strong alkali like

$KOH$ .

When it undergoes nucleophilic, ethyl alcohol is formed as a major product.

$C_{2}H_{5}Cl+KOH_{(aq)}\rightarrow C_{2}H_{5}OH+KCl$

Hence the correct option is A.

The criteria for optical isomerism are:

0%
the presence of at least one asymmetric carbon is essential
0%
the optical rotation is due to asymmetric carbon
0%
non-superimposible mirror image
0%
all of the above

IUPAC name of the above compound is:

0%
$3$ -Chlorotoluene
0%
$4$ -Chlorotoluene
0%
$4$ -Methylchlorobenzene
0%
$3$ -Methylchlorobenzene

Haloarenes react with halogens in the presence of _________ as the catalyst to give ortho-para isomers.

0%
ferric chloride
0%
sodium chloride
0%
potassium chloride
0%
magnesium chloride

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

0%
$2$ -fluorobutane
0%
$2$ -chlorobutane
0%
$2$ -bromobutane
0%
$2$ -idobutane

Explanation

2-iodobutane is more susceptible to nucleophilic substitution because the

$C-I$ bond is weak as compared to

$C-Br,\ C-Cl$ and

$C-F$ bond.

Which of the following statement is correct?

0%
The $C-X$ bond in haloalkanes is more polar than the $C-X$ in haloarenes.
0%
The $C-X$ bond in haloalkanes is less polar than the $C-X$ in haloarenes.
0%
Polarity of $C-X$ bond in haloalkane is equal to the polarity of $C-X$ bond in haloarenes
0%
Polarity of $C-X$ bond in both depends on conditions.

Explanation

In haloarenes due to resonance, charge distribution or rotation of charge on benzene nucleus takes place hence the

$C-X$ bond in haloalkanes ( which are not aromatic) is more polar than the

$C-X$ bond in halo arenes (aromatic)

Identify the correct statement about nomenclature of alkyl halide.

0%
The monohalogen derivatives of alkanes are called as alkyl halides.
0%
The common name of an alkyl halide is always written as two separate words.
0%
In IUPAC system, the monohalogen derivatives of alkanes are named as haloalkanes.
0%
All of above.

Explanation

The monohalogen derivatives of alkanes are called as alkyl halides denoted as

$R-X$ where (

$R-$ ) is the alkyl group and

$X-$ is the halogen atom. The common name of an alkyl halide is always written as two separate words. In IUPAC system, the monohalogen derivatives of alkanes are named as haloalkanes. for e.g., methyl chloride has IUPAC name chloromethane.

Which of the following is/are optically active compound/s?

0%
n-Butyl bromide
0%
sec-Butyl bromide
0%
Isobutyl iodide
0%
tert-Butyl chloride

Which of the following carbon-halogen bond is extremely polar though nonionic among given alkyl halides?

0%
$C-F$
0%
$C-Br$
0%
$C-Cl$
0%
$C-I$

Explanation

$C - F$ bond is extremely polar because Flourine is most Electronegative atom (E.N = 4.0) so the E.N difference is more for carbon and Flourine.

Polarity order :

$C - F > C - Br > C - Cl > C- I$

Which among the given haloalkanes develop colour when exposed to light?

0%
$R-Br$ and $R-Cl$
0%
$R-I$ and $R-F$
0%
$R-Cl$ and $R-Br$
0%
$R-Br$ and $R-I$

The difference in electronegativity of carbon-halogen bond is minimum in:

0%
$C-F$
0%
$C-Cl$
0%
$C-I$
0%
$C-Br$

Explanation

Down the group electronegativity decreases.

$C-I$ It has the least electronegativity difference among carbon-halogen bond.

Option C is correct.

$C-X$ bond is strongest in:

0%
$CH_3Cl$
0%
$CH_3Br$
0%
$CH_3F$
0%
$CH_3I$

$R-X$ the bond length is maximum in:

0%
$C-F$
0%
$C-Cl$
0%
$C-Br$
0%
$C-I$

Physical properties of any compound depend largely on:

0%
its mass
0%
the type of intermolecular and intramolecular forces of attraction
0%
both mass and type of intermolecular and intramolecular forces of attraction
0%
none of the above

Which of the following is not true for

$S_N1$ reaction?

0%
Favoured by polar solvents.
0%
$3^o -$ alkyl halides generally react through $S_N1$ reaction.
0%
The rate of the reaction does not depend upon the molar concentration of the nucleophile.
0%
$1^o -$ alkyl halides generally react through $S_N1$ reaction.

Explanation

Primary alkyl halides generally react through

$S_{N}{2}$ reaction.

Tertiary alkyl halides generally react through

$SN_1$ reaction.

As in

$S_{N}{1}$ mechanism the carbocation forms and the stability order for carbocation is

$3^0 > 2^0 > 1^0$

The option (D) is not true for

$S_N1$ reaction.

Option D is correct.

Dipole moment of chlorobenzene is less then cyclohexylchloride

0%
True
0%
False

Explanation

In chlorobenzene the C-Cl bond is

$sp^2$ hybrised, while in cyclohexylchloride C-Cl bond is

$sp^3$ hybridised. The

$sp^2$ hybridised C atom in chlorobenzene is more electronegative and less polar. Hence, the dipole moment of chlorobenzene is less then cyclohexylchloride.

Which of the following is not correct for haloalkanes?

0%
Density is directly proportional to the mass of compound
0%
Haloalkanes are slightly soluble in water
0%
Boiling Point depends upon the intermolecular forces of attraction
0%
None of these

The IUPAC name for tertiary butyl iodide is:

0%
4-iodobutane
0%
2-iodobutane
0%
1-iodo-3-methylpropane
0%
2-iodo-2-methylpropane

Explanation

Explanation:
There are certain steps that we follow during IUPAC nomenclature such as;

Step 1 :- Find and name the longest continuous carbon chain.

Step 2 :- Identify and name groups attached to this chain.

Step 3 :- Do the numbering

Step 4 :- Designate the location of each substituent group by an appropriate number and name.

Step 5 :- Assemble the name, listing groups in alphabetical order using the full name.

So, here in this case the longest continuous chain is of

$3 \ C$ so the base name would be

$propane$
There is a methyl group and Iodine group present at

$C - 2$ position so the substituent naming would be according to alphabetical order as;

$2-Iodo-2-Methyl$

Therefore, The IUPAC name of a compound is

$2-iodo-2-methylpropane$ .

Correct Option: $D$ .

2176694_580297_ans_e1cef8daf90a450fb2bc86a551d64c2e.png

Arrange these haloalkanes in correct order of density:

0%
$CH_3I < C_2H_5I < C_3H_7I$
0%
$CH_3I > C_2H_5I > C_3H_7I$
0%
$CH_3I < C_2H_5I > C_3H_7I$
0%
$CH_3I > C_2H_5I < C_3H_7I$

Arrange the alkyl in decreasing order of their mass.

0%
$CH_3Cl > H_2O > CH_3Br$
0%
$CH_3Cl>CH_3Br>H_2O$
0%
$CH_3Br> CH_3Cl > H_2O$
0%
$H_2O > CH_3Cl > CH_3Br$

Explanation

$CH_{3}Br$ Atomic mass

$\Rightarrow$

$12+3+80=95$

$CH_{3}Cl$ Atomic mass

$\Rightarrow$

$12+3+35.5=50.5$

$H_{2}O$ Atomic mass

$\Rightarrow$

$16+2=18$

According to given Atomic masses of compounds the decreasing order is

$CH_{3}Br>CH_{3}Cl>H_{2}O$ .

Arrange in correct order of density:

0%
$CH_2Cl_2 < CHCl_3 < CCl_4$
0%
$CH_2Cl_2 > CCl_4 > CHCl_3$
0%
$CCl_4 < CH_2Cl_2 < CHCl_3$
0%
$CH_2Cl_2 > CCl_4 < CHCl_3$

Choose the correct order of bond strength:

0%
$CH_3F > CR_3Cl > CR_3Br > CH_3I$
0%
$CH_3F < CR_3Cl < CR_3Br < CH_3I$
0%
$CH_3F > CR_3Cl > CR_3Br < CH_3I$
0%
$CH_3F > CR_3Cl < CR_3Br < CH_3I$

Explanation

Bond strength :

$C-F > C-CI > C-Br > C-I$

bond strength

$(kJ \, mol^{-1})$

$C-F \,\,\,\,\,\, 484$

$C-CI \,\,\,\, 338$

$C-Br \,\,\,\, 276$

$C-I \,\,\,\,\,\,\,\, 238$

Haloalkane in the presence of alcoholic

$KOH$ undergoes:

0%
elimination
0%
polymerisation
0%
dimerisation
0%
substitution

Explanation

Haloalkanes in the presence of alcoholic KOH undergoes elimination reaction or dehydrohalogenation reaction.

$CH_3CH_2CH_2Br+alc. KOH\rightarrow CH_3CH=CH_2+KBr+H_2O$

Pure form of alkyl halide is:

0%
odourless
0%
colourless
0%
both colourless and odourless
0%
neither colourless nor odourless

Halide containing halogen attached to

$sp^3$ carbon is/are:

0%
allylic halide
0%
vinylic halide
0%
benzylic halide
0%
aryl halide

Explanation

In allylic and benzylic halides, halogen is attached to the $sp^3$ hybridised carbon.

As $CH_3-CH_2Cl$ and $C_6H_5-CH_2Cl$

Which of the following does not exist in solid state at room temperature?

0%
$1,4-$ dichlorobenzene
0%
$1,3-$ dichlorobenzene
0%
$1,2-$ dichlorobenzene
0%
All of the above

The only methyl halide which is a liquid is at room temperature is :

0%
iodomethane
0%
bromomethane
0%
chloromethane
0%
fluoromethane

Which is used as an anaesthetic during surgery?

0%
Chloroquine
0%
Thyroxine
0%
Halothane
0%
Chloramphenicol

Explanation

$A :$ Chloroquine - used for malaria.

$B :$ Thyroxine - Hormone of Thyroid gland.

$C :$ Haloethane - Anesthetic during surgery.

$D :$ Chloramphenicol - Antibiotic for bacterial infections.

Halide containing halogen attached to

$sp^2$ carbon is/are:

0%
allylic halide
0%
benzylic halide
0%
aryl halide
0%
vinyl halide

Explanation

$A :$

$H_{2}C=CH-CH_{2}-Cl$ Allylic - chloride halide attached to

$sp^{3}$ hybridised carbon.

$B :$ Benzylic chloride halide attached to

$sp^{3}$ hybridised carbon.

$C :$ Benzyl chloride (Allyl chloride) halide attached to

$sp^{2}$ hybridised carbon.

$D :$

$CH_{2}=CHCl$ Vinyl chloride halide attached to

$sp^{2}$ hybridised carbon.

Alkyl halides develop colour on:

0%
addition of water
0%
addition of non polar solvent
0%
exposure to light
0%
exposure to air

Which of the following statements is correct about halogens?

0%
they are all capable of exhibiting more than one oxidation states
0%
they are all diatomic and form diatomic ions
0%
they are all diatomic and form univalent ions
0%
they are all reducing agents

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 7 - MCQExams.com

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CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 7 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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