$$2C_{6}H_{5}Cl+2Na\rightarrow X$$ In the given reaction $$X$$ is:

$$1$$-chloro-$$2$$-phenyl ethane

Bond order of $$C-X$$ bond is $$1$$ because :

$$Cl$$ achieve noble gas configuration by sharing one electron

$$C$$ not able to form multiple bonds

$$Cl$$ does not have vacant orbitals

$$C$$ achieve noble gas configuration by sharing one electron

$$\displaystyle S_{N^1}$$ & $$\displaystyle S_{N^2}$$ product are same in (excluding stereoisomer):

$$Ph-\underset {CH_3} {\underset {|}CH}-\underset {Cl} {\underset {|}CH}-CH_3$$

MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 5

In which of the following can

C - C l

$C-Cl$ bond be easily broken?

0%
Chlorobenzene
0%
Vinyl chloride
0%
Allyl chloride
0%
p-Chlorotoluene

Explanation

C - C l

$C-Cl$ bond is easily broken in allyl chloride

Reason:- It is form stable carbocation as an intermediate.

1452641_769521_ans_b250ea0604ac4e9f980cdad648dd2787.png

The reagents required for the following two-step transformation are?

0%
(i) $HBr$ , benzoyl peroxide, (ii) $CH_3CN$
0%
(i) $HBr$ , (ii) $NaCN$
0%
(i) $Br_2$ ; (ii) $NaCN$
0%
(ii) $NaBr$ ; (ii) $NaCN$

Which of the following is most reactive towards the

S_{N} 1

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

0%
0%
0%
0%
A and B reacts at similar rate

Which of the following reaction is used to increase the length of the carbon chain ?

0%
Clemmensen reduction
0%
Wolff-Kishner Reduction
0%
Wurtz reaction
0%
Kolbe’s electrolysis

Which of the following is used in paint removing?

0%
$CHCl_{3}$
0%
$CH_{2}Cl_{2}$
0%
$CCl_{4}$
0%
$CH_{3}Cl$

Explanation

C H C l_{3}

$CHCl_3$ is used as solvent and as an anaesthetic.

C C l_{4}

$CCl_4$ is used as a fire extinguisher.

C H_{2} C l_{2}

$CH_2Cl_2$ is used as paint removing.

C H_{3} C l

$CH_3Cl$ is used as a refrigerant.

Hence, the correct option is B.

In chlorination of benzene in presence of

F e C l_{3}

$FeCl_{3}$ , function of

F e C l_{3}

$FeCl_{3}$ is :

0%
to provide halogen.
0%
to form electrophile.
0%
resonance stabilisation.
0%
act as a catalyst.

Explanation

Since

F e C l_{3}

$FeCl_3$ is generated again after the reaction hence it is acting as catalyst.

2 C_{6} H_{5} C l + 2 N a \to X

$2C_{6}H_{5}Cl+2Na\rightarrow X$

In the given reaction

X

$X$ is:

0%
toulene
0%
biphenyl
0%
phenyl ethane
0%
$1$ -chloro- $2$ -phenyl ethane

Explanation

Fitting reaction: Aryl chloride reacts with Na in presence of dry ether and diaryl forms as product.(similar to Wurtz reaction).
Reaction:

2 C_{6} H_{5} C l + 2 N a \to P h - P h + 2 N a X

$2C_{6}H_{5}Cl+2Na\rightarrow Ph-Ph+2NaX$

What is the hybridisation of carbon that is attached to chlorine atom in n-primary alkyl halide?

0%
$sp^3$
0%
$sp^2$
0%
$sp$
0%
$sp^3d$

Explanation

n-primary alkyl halide is

R - C H_{2} - C l

$R-CH_2-Cl$ .
In alkyl halides (saturated alkyl halides), halogen is always attached to

s p^{3}

$sp^3$ carbon.

C H_{3} - C H (C l) - C H_{2} - C H_{2} - C H_{3}

$CH_3-CH(Cl)-CH_2-CH_2-CH_3$ and

C H_{3} - C H_{2} - C (C l) (C H_{3}) - C H_{3}

$CH_3-CH_2-C(Cl)(CH_3)-CH_3$ are both positional and chain isomerism of each other.

0%
True
0%
False

The reagent used for dehydrohalogenation is ____________.

0%
$H_2O_2$
0%
$K_2SO_4$
0%
alcoholic $KOH$
0%
none of the above

Alkyl halide reacting with metallic sodium in dry ether solution is called:

0%
Friedel-crafts reaction
0%
Sandmeyer reaction
0%
Wurtz reaction
0%
Gabriels reaction

Explanation

The Wurtz reaction is a coupling reaction in which two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane:

2 R X + 2 N a \to R - R + 2 N a^{+} X^{-}

$2RX+2Na\rightarrow R-R+2Na^+X^-$

Which of following will give syn-elimination?

0%
$\displaystyle CH_{3}-CH_{2}CH_{2}-OCOCH_{3}$
0%
$CH_3CH_2\underset {Cl}{\underset{|}C}H-CH_3$
0%
$CH_3CH_2CH_2-O-\underset {S} {\underset {||} C}-S-CH_3$
0%

Explanation

Due to formation of

1^{0}

$1^0$ carbocation, all these

(A, C, D)

$(A,C,D)$ will give syn-elimination.

The conversion (Image) Can be effected using

0%
$Br_2/CCl_4$
0%
$Br_2/H_2O$
0%
$Br_2/Fe$
0%
$Br_2/benzoyl peroxide$

IUPAC name of

(C H_{3})_{2} C H - C H_{2} - C H_{2} B r

$(CH_{3})_{2}CH - CH_{2} - CH_{2}Br$ is:

0%
1-Bromopentane
0%
2-Methyl-4-bromobutane
0%
1-Bromo-3-methylbutane
0%
2-Methyl-3-bromobutane

Explanation

I U P A C

$IUPAC$ name of

(C H_{3})_{2} C H - C H_{2} - C H_{2} B r

$(CH_{3})_{2}CH-CH_{2}-CH_{2}Br$ is 1-Bromo-3-methylbutane. The parent chain has 4 carbon atoms with a methyl substituent (

C H_{3}

$CH_{3}$ ) at

C - 3

$C-3$ and a bromine atom at

C - 1

$C-1$ .

The IUPAC name of the compound

C H_{3} - C H (C H_{3}) - C H_{2} - C H_{2} C l

$CH_3-CH(CH_3)-CH_2-CH_2Cl$ is:

0%
1-chloro-3-methylbutane
0%
2-methyl-4-chlorobutane
0%
2-methyl-1-chlorobutane
0%
1-chloropentane

Which of the following undergoes nucleophilic substitution exclusively by

S_{N} 1

$S_N{1}$ mechanism?

0%
ethyl chloride
0%
isopropyl chloride
0%
benzyl chloride
0%
chlorobenzene

Nature of

C - X

$C-X$ bond is:

0%
ionic
0%
covalent
0%
co-ordinate
0%
none of these

Explanation

Non-metals form covalent bonds. Electronegativity of halogens is greater than carbon hence nature of

C - X

$C-X$ bond is polar covalent, formed by sharing of one electron each.

Bond order of

C - X

$C-X$ bond is

1

$1$ because :

0%
$Cl$ achieve noble gas configuration by sharing one electron
0%
$C$ not able to form multiple bonds
0%
$Cl$ does not have vacant orbitals
0%
$C$ achieve noble gas configuration by sharing one electron

Explanation

Bond order of

C - X

$C-X$ bond is 1 because halogen atoms are one atom deficient to a noble gas configuration which is attached by sharing of one electron and formation of one sigma bond.

The main difference in C - X bond of a haloalkane and a haloarene is

0%
C - X bond in haloalkanes is shorter than haloarenes.
0%
In haloalkanes the C attached to halogen in C - X bond is $sp^{3}$ hybridised while in haloarenes it is $sp^{2}$ hybridised.
0%
C - X bond in haloarenes acquires a double bond character due to higher electronegativity of X than haloalkanes.
0%
haloalkanes are less reactive than haloarenes due to difficulty in C - X cleavage in haloalkanes.

Explanation

The main difference in C - X bond of a haloalkane and a haloarene is that in haloalkanes carbon of C - X is

s p^{2}

$sp^{2}$ hybridised while in haloarenes it is

s p^{3}

$sp^{3}$ hybridised.

Hence, Option "B" is the correct answer.

What is the correct IUPAC name for

0%
1-Bromo-2-ethylpropane
0%
1-Bromo-2-ethyl-2-methylethane
0%
1-Bromo-2-methylbutane
0%
2-Methyl-1-bromobutane

Match the isomers given in column I with their names given in column II and mark the appropriate choice.

0%
$(A) \rightarrow (iii), (B) \rightarrow (i), (C) \rightarrow (iv), (D) \rightarrow (ii)$
0%
$(A) \rightarrow (iv), (B) \rightarrow (iii), (C) \rightarrow (ii), (D) \rightarrow (i)$
0%
$(A) \rightarrow (i), (B) \rightarrow (ii), (C) \rightarrow (iii), (D) \rightarrow (iv)$
0%
$(A) \rightarrow (ii), (B) \rightarrow (iii), (C) \rightarrow (iv), (D) \rightarrow (i)$

Identify the products X and Y formed in the following reaction:

C H_{3} - C H_{2} - C H = C H - C H_{3} + H C l \to X + Y

$CH_3-CH_2-CH = CH-CH_3 + HCl \rightarrow X + Y$

In which of the following cases

C - C

$C-C$ bond length will be the longest?

0%
$CH_3-CF_3$
0%
$FCH_2-CH_2F$
0%
$F_2CH-CHF_2$
0%
$CF_3-CF_3$

S_{N^{1}}

$\displaystyle S_{N^1}$ &

S_{N^{2}}

$\displaystyle S_{N^2}$ product are same in (excluding stereoisomer):

0%
0%
0%
0%
$Ph-\underset {CH_3} {\underset {|}CH}-\underset {Cl} {\underset {|}CH}-CH_3$

Explanation

As in these reactants, there is no possibility for intermediate carbocation to rearrange so

B

$B$ and

C

$C$ will give same

S_{N} 1

$S_N1$ and

S_{N} 2

$S_N2$ product.

The compound

A

$A$ is :

0%
0%
0%
0%
none of the above

Explanation

E_{2}

$E_2$ reaction is stereospecific reaction and involves trans -elimination. Thus, the product obtained is cis-1,2-diphenyl-1-propene. Thus, the option A is the correct answer.

Choose all alkane that give only one monochloro derivative upon reaction with chlorine in sun light.

Explanation

In presence of sunlight, it will follow free radical mechanism.
Single product will be form if all carbon have same configuration so

A, B

$A, B$ and

D

$D$ will give one product only as all replaceable

H

$H$ are same here.

Which of the following sequences would yield m-nitro chlorobenzene

(Z)

$(Z)$ from benzene?

0%
Benzene $\xrightarrow [ ]{ { Cl }_{ 2 }/Fe{ Cl }_{ 3 } } (X)\xrightarrow [ { H }_{ 2 }{ SO }_{ 4 } ]{ H{ NO }_{ 3 } } (Z)$
0%
Benzene $\xrightarrow [ ]{ { H }_{ 2 }{ SO }_{ 4 }/H{ NO }_{ 3 } } (Z)$
0%
Benzene $\xrightarrow [ ]{ { H }_{ 2 }{ SO }_{ 4 }/H{ NO }_{ 3 } } (X)\xrightarrow [ ]{ { Cl }_{ 2 }/Fe{ Cl }_{ 3 } } (Z)$
0%
All of these

Iso-propyl bromide on Wurtz reaction gives:

0%
Hexane
0%
Propane
0%
2, 3-dimethylbutane
0%
Neo-hexane

Explanation

Hence, the correct option is

C

$\text{C}$

2-phenylethylbromide when heated with

N a O E t

$NaOEt$ , elimination takes place. No deuterium exchange takes place when the reaction is carried out in

C_{2} H_{5} O D

${C}_{2}{H}_{5}OD$ solvent. The mechanism will be:

0%
$E1$ elimination
0%
$E2$ elimination
0%
$E1cB$ elimination
0%
$E2$ or ${E1}cB$

Explanation

It is a primary bromide. So it will undergo elimination either by

E 2

$E2$ or

E 1 c B

$E1cB$ . Since there is no deuterium exchange in

C_{2} H_{5} O D

${C}_{2}{H}_{5}OD$ solvent,

C - H

$C-H$ bond is not broken to form carbanion.

Hence, the actual mechanism is

E 2

$E2$ only.

Which one of the following undergoes nucleophilic substitution exclusively by

S_{N} 1

$S_N1$ mechanism?

0%
Benzyl chloride
0%
Ethyl chloride
0%
Chlorobenzene
0%
Isopropyl chloride

Explanation

Aliphatic

S_{N} 1

$S_N1$ reaction is carried out in two steps. In first step carbocation is formed and its formation is based on the stability of carbocation.

C_{6} H_{5} + C H_{2} > C H_{3} + C H C H_{3} > C H_{3} + C H_{2}

$C_6H_5\overset{+}{C}H_2 > CH_3\overset{+}{C}H CH_3> CH_3 \overset{+}{C}H_2$

In the second step, the nucleophile is attracted to carbocation to give final products. Hence, an order of

S_{N} 1

$S_N1$ reaction is

C_{6} H_{5} + C H_{2} > C H_{3} C H - C H_{3} > C H_{3} C H_{2}

$C_6H_5\overset{+}{C}H_2 >CH_3 CH-CH_3 > CH_3 CH_2$

The aryl halides e.g., chlorobenzene are less reactive as compared to alkyl halides towards nucleophilic reagents in either

S_{N} 2

$S_N2$ or

S_{N} 1

$S_N1$ reaction because carbon-halogen bond in the aryl halide is strong (due to its double bond character).

So, the correct option is

A

$A$

On commercial scale, phenol is obtained from chlorobenzene. The chlorobenzene needed for the purpose is prepared by:

Explanation

In Raschig's process a mixture of benzene vapours, air and hydrogen chloride is passed over heated

C u {C l}_{2}

$Cu{Cl}_{2}$ at

500 K

$500K$ .

C_{6} H_{6} + C H C l_{3} + O_{2} (a i r) C u {C l}_{2} - -- \to 500 K 2 C_{6} H_{6} C l + 2 H_{2} O

${ C }_{ 6 }{ H }_{ 6 }+CHCl_3+{ O }_{ 2 }(air)\xrightarrow [ 500K ]{ Cu{ Cl }_{ 2 } } 2{ C }_{ 6 }{ H }_{ 6 }Cl+2{ H }_{ 2 }O$

Phenol is obtained from chlorobenzene as shown in the above reaction

Hence, the correct option is D

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%
Both Assertion and Reason are incorrect.

Explanation

C H_{2} = C H - C l

$CH_2=CH-Cl$ has some partial double bond character between carbon and chlorine atom. So, nucleophilic substitution is difficult to carry as it is difficult to break the partial double bond in vinyl chloride than ethyl chloride

C H_{3} C H_{2} - C l

$CH_3CH_2-Cl$ where there is no double bond character.

Vinyl group is not electron donating.

Which of the following halides undergoes nucleophilic substitution most readily?

0%
$p- H_3CC_6H_4Cl$
0%
$O- MeOC_6H_4Cl$
0%
$p-ClC_6H_4Cl$
0%
$C_6H_5CH(Cl)CH_3$

Explanation

C_{6} H_{5} C H (C l) C H_{3}

$\displaystyle C_6H_5CH(Cl)CH_3$ undergoes nucleophilic substitution most readily as Cl is attached to aliphatic C atom.

p - H_{3} C C_{6} H_{4} C l, O - M e O C_{6} H_{4} C l, p - C l C_{6} H_{4} C l

$\displaystyle p- H_3CC_6H_4Cl, O- MeOC_6H_4Cl, p-ClC_6H_4Cl$ , Cl is attached to aromatic carbon atom. The nucleophilic substitution of aryl chlorides requires drastic conditions.

Isopropyl bromide on Wurtz reaction gives:

0%
hexane
0%
propane
0%
$2, 3$ -dimethyl butane
0%
neo-hexane

Which is the correct increasing order of electronegativity of halides?

0%
$F,Cl,Br,I$
0%
$I,Br,Cl,F$
0%
$Br,Cl,F,I$
0%
$Cl,Br,I,F$

Explanation

The bond between carbon and the halogen atom (C-X) is polar. Since halogen atoms are more electronegative than carbon, there is partial negative charge on halogen atom and positive charge on carbon. This makes the molecule dipolar and polar nature of C-X bond is increase with increase in the electronegativity difference between carbon and halogenatom. Hence the order of polarity of $C-X$ bond is $C-F > C-Cl > C-Br > > C-I$

Which of the following is used for metal cleaning and finishing?

0%
$CHCl_{3}$
0%
$CHI_{3}$
0%
$CH_{2}Cl_{2}$
0%
$C_{6}H_{6}$

Explanation

Chloroform is used as solvent and as anaesthetic.

$CHI_3$ is used as antiseptic.

$CH_2Cl_2$ is used for metal cleaning and finishing.

$C_6H_6$ is used for printing and lithography, paint, rubber, dry cleaning, adhesives and coatings, detergents.

$C_{6}H_{13}Br + OH^{-} \longrightarrow C_{6}H_{13}OH + Br^{-}$ is an example of:

0%
Nucleophilic addition
0%
Nucleophilic substitution
0%
Electrophilic addition
0%
Electrophilic substitution
0%
Free radical substitution

Explanation

$C_6H_{13}Br+^{\ominus}OH\rightarrow C_6H_{13}OH+Br^{\ominus}$

It is hydrolysis of alkyl halides.

It is an example of

$S_N2$ reaction i.e. nucleophilic substitution reaction

The IUPAC name of haloalkane is always written as a ______ word.

0%
single
0%
double
0%
triple
0%
none of these

Heating many alkyl chlorides or bromides in water effects their conversion into alcohol through a

$S_{N}$ 1 reaction. Order each of the following sets of compounds with respect to solvolytic reactivity.

0%
$i>ii>iii$
0%
$ii>i>iii$
0%
$iii>ii>i$
0%
$iii>i>ii$

The ability of rotation of plane polarized light either towards right or left is called as:

0%
optical activity
0%
chirality
0%
superimposibility
0%
non superimposibility

Which of the following haloalkanes is most reactive?

0%
1-chloropropane
0%
1-bromopropane
0%
2-chloropropane
0%
2-bromopropane

In the dehydrohalogenation of ethyl chloride, which of the following change occurs?

0%
$sp^{2}$ carbon converts to $sp^{3}$ carbon
0%
$sp^{2}$ carbon converts to $sp$ carbon
0%
$sp^{3}$ carbon converts to $sp$ carbon
0%
$sp^{3}$ carbon converts to $sp^{2}$ carbon

Explanation

In dehydrohalogenation, alkyl halide converts into an alkene. Alkyl halide contains

$sp^3$ carbon and alkene contains

$sp^2$ carbon.

Hence, the correct option is 'D'.

Consider the following anions.

When attached to

$sp^{3}-$ hybridized carbon, their leaving group ability in nucleophilic substitution reaction decreases in the order :

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

Explanation

The rate of

$S_N1$ reaction is largely influenced by the nature of the leaving group. The more stable the leaving group, the more easily it will be lost to favour the carbocation formation. Their decreasing order is given by

in accordance with their pKa values.

1056493_989659_ans_c5f5fd0fc1e34c63828aca112435728f.PNG

Product

$(B)$ is :

0%
0%
0%
0%
None of these

Explanation

(Refer to Image 1) on reaction with

$NBS (N-Bromosuccinimide)$ , followed by reaction with sodium methanethiolate gives (Refer to Image 2)

$NBS$ is a chemical agent which is used in radical substitution, electrophilic substitution. It is a source of bromine radical. Also,

$CH_3SNa$ is the conjugate base of methanethiol. It is commercially available as white solid. It is a powerful nucleophile. Its hydrides in moist air produces methanethiol.

1119548_989614_ans_06a66090ac854ca2b3c51bd3e06eef27.png

Arrange the following alkyl halides in decreasing order of the rate of

$\beta$ -elimination reaction with alcoholic KOH.

${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br\\ \quad \quad \quad \quad \quad (i)$

${ CH }_{ 3 }-CH{ -CH }_{ 2 }Br\quad \\ \quad \quad \quad \ |\\ \quad \quad \ { CH }_{ 3 }\\ \quad \quad (ii)$

${ CH }_{ 3 }-{ CH }_{ 2 }{ -CH }-{ CH }_{ 3 }\quad \\ \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad Br\\ \quad \quad \quad \quad \quad (iii)$

0%
$(ii)\quad >\quad (iii)\quad >\quad (i)$
0%
$(iii)\quad >\quad (i)\quad >\quad (ii)$
0%
$(i)\quad >\quad (ii)\quad >\quad (iii)$
0%
$(ii)\quad >\quad (i)\quad >\quad (i)$

Explanation

$CH_3-CH_2-CH_2-Br$

$CH_3-\underset {|}{CH}-CH_2-Br$

$CH_3-CH_2-\underset{|}{CH}-CH_3$

(i)

$CH_3$

$Br$

(ii) (iii)

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

Decreasing order of rate of

$\beta$ -elimination reaction with alcoholic

$KOH$ .

Ortho and para hydrogen differ in

0%
atomic number
0%
mass number
0%
electron spin in two atoms
0%
nuclear spin of two atoms

Follows

0%
$E_2$ mechanism
0%
$\epsilon _1$ mechanism
0%
$S_N^1$ mechanism
0%
$S_N^2$ mechanism

Select whether the following combinations of reactants will react by substitution

$(S_{N^1} or S_{N^2}$ mechanism), elimination (

$E_1 or E_2$ mechanism).

$(CH_3)_3C - OH \xrightarrow[25^0C]{HBr 48\% in H_2O}$

0%
$S_{N^1}$
0%
$S_{N^2}$
0%
$E_1$
0%
$E_2$

Explanation

t-butanol undergoes

$S_N1$ (Nucleophilic substitution unimolar) when treated with acid.

$(CH_3)_3C-OH\xrightarrow [ HBr48%inH_2O ]{ 25^{\circ}C } H_3C-C-Br$

The reaction involves the protonation of -OH group followed by formation of carbocation. Generation of

$3^{\circ}$ cation is the rate determining step and is said to be unimolar i.e.

$S_N1$ .

1056456_991451_ans_892e0eaeadaf40ae908c00b90d99635d.PNG

$S_N1$ characteristic would be exhibited in the hydrolysis of;

0%
Chlorobenzene
0%
$1$ -chloronapthalene
0%
propenyl chloride
0%
$2$ -chloro- $1$ -propene

Explanation

${S_N}$ characteristic would exhibited in the hydrolysis of propanyl chloride because of greater stability of the carbonation intermediate

$C{H_2} = CHC{H_2}^ +$ through different resonating structure.

Which of the following is an example of alkyl halide?

0%
$p-$ chlorotoluene
0%
Chloro methane
0%
Allyl chloride
0%
Benzyl chloride

Explanation

solution:

alkyl halide means alkyl means only single bonded carbon having halide group in it halides like

$F, Cl ,Br , I$

option B

$CH_3-Cl$

this is alkyl chloride

hence the correct option: B

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 5 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 5 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.