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Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2

18 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2

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Attempt Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 | 18 questions in 36 minutes | Mock test for JEE preparation | Free important questions MCQ to study Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper for JEE Exam | Download free PDF with solutions

QUESTION: 1

Chlorobenzene can be prepared by reacting aniline with : (1984 - 1 Mark)

A.
hydrochloric acid
B.
cuprous chloride
C.
chlorine in presence of anhydrous aluminium chloride
D.
nitrous acid followed by heating with cuprous chloride

Solution:

QUESTION: 2

The reaction of toluene with chlorine in presence of ferricchloride gives predominantly : (1986 - 1 Mark)

A.
benzoyl chloride
B.
m-chlorotoluene
C.
benzyl chloride
D.
o- and p-chlorotoluene

Solution:

TIPS/Formulae :
The given reaction is an example of electrophilic substitution. Further, CH₃ group in toluene is o, p-directing

QUESTION: 3

The reaction conditions leading to the best yields of C₂H₅Cl are : (1986 - 1 Mark)

Solution:

Chlorination beyond monochlorination during the preparation of alkyl halides in presence of UV light can be suppressed by taking alkane in excess.
NOTE : For isomeric alkanes the one having largest straight chain has highest b.p. because of large surface area.

QUESTION: 4

n-Propyl bromide on treatment with ethanolic potassiumhydroxide produces (1987 - 1 Mark)

A.
Propane
B.
Propene
C.
Propyne
D.
Propanol

Solution:

Further dehydrohalogenation of CH₃CH = CH₂ can be done only by strong base like NaNH₂

QUESTION: 5

The number of structural and configurational isomers of abromo compound, C₅H₉Br, formed by the addition of HBr to 2-pentyne respectively are (1988 - 1 Mark)

A.
1 and 2
B.
2 and 4
C.
4 and 2
D.
2 and 1

Solution:

Addition of HBr of 2-pentyne gives two structural isomers (I) and (II)

Each one of these will exist as a pair of geometrical isomers. Thus, there are two structural and four configurational isomers

QUESTION: 6

1-Chlorobutane on reaction with alcoholic potash gives (1991 - 1 Mark)

A.
1-butene
B.
1-butanol
C.
2-butene
D.
2-butanol

Solution:

NOTE : Alkyl halides give alcohols with aq. KOH, which is a substitution reaction.

QUESTION: 7

The chief reaction product of reaction between n-butane and bromine at 130ºC is : (1995 S)

A.
CH₃CH₂CH₂CH₂Br
B.
C.
D.

Solution:

TIPS/Formulae :
The reaction proceeds via free radical mechanism.
As 2º free radical is more stable than 1º, so
CH₃CH₂CH(Br)CH₃ would be formed.

QUESTION: 8

Isobutyl magnesium bromide with dry ether and ethyl alcohol gives : (1995S)

Solution:

QUESTION: 9

(CH₃)₃ CMgCl on reaction with D²O produces :

A.
(CH₃)₃ CD
B.
(CH₃)₃OD
C.
(CD₃)₃CD
D.
(CD₃)₃OD.

Solution:

(CH₃ )₃C -MgCl + D₂O → (CH₃)₃C -D +Mg(OD)Cl

QUESTION: 10

A solution of (+) –2–chloro–2–phenylethane in tolueneracemises slowly in the presence of small amount of SbCl₅,due to the formation of (1999 - 2 Marks)

A.
carbanion
B.
carbene
C.
free-radical
D.
carbocation

Solution:

Occurrence of racemization points towards the
formation of carbocation as intermediate, which being planar can be attacked from either side.

QUESTION: 11

Identify the set of reagent / reaction conditions 'X' and 'Y' in the following set of transformations (2002S)

A.
X = dilute aqueous NaOH, 20°C; Y = HBr/acetic acid,20°C
B.
X = concentrated alcoholic NaOH, 80°C; Y = HBr/aceticacid, 20°C
C.
X = dilute aqueous NaOH, 20°C ; Y = Br₂/CHCl₃, 0°C
D.
X = concentrated alcoholic NaOH, 80°C; Y = Br₂/CHCl₃,0°C

Solution:

Dehydrobromination by strong base (alc. NaOH) followed by Markownikoff addition of HBr.

QUESTION: 12

How many structures for F are possible? (2003S)

A.
2
B.
5
C.
6
D.
3

Solution:

So F can have three possible structures

QUESTION: 13

What would be the product formed when 1-bromo-3- chlorocyclobutane reacts with two equivalents of metallic
sodium in ether? (2005 S)

Solution:

TIPS/Formulae :
It is an example of intramolecular Wurtz reaction.

NOTE : Br^– is a better leaving group than chloride. In this reaction alkali metal (Na) is electron donor.

QUESTION: 14

When phenyl magnesium bromide reacts with tert -butanol, the product would be (2005S)

A.
Benzene
B.
Phenol
C.
ter-butylbenzene
D.
ter-butyl phenyl ether

Solution:

Grignard reagents react with compounds containing active hydrogen to form hydrocarbons corresponding to alkyl (or aryl) part of the Grignard reagent.
C₆H₅MgBr + Me₃COH →C₆H₆ + Me₃COMgBr

QUESTION: 15

The reagent(s) for the following conversion,

A.
alcoholic KOH
B.
alcoholic KOH followed by NaNH₂
C.
aqueous KOH followed by NaNH₂
D.
Zn/CH₃OH

Solution:

NOTE : Elimination of HBr from CH₂= CHBr requires a stronger base because here, Br acquires partial double bond character due to resonance.

QUESTION: 16

The major product of the following reaction is – (2008)

Solution:

The product (a) will be formed.
Nucleophilic substitution of an alkyl halide is easier as compared to that of an aryl halide.
PhS^– is a strong nucleophile and dimethyl formamide is a highly polar aprotic solvent.
These reagent favour S_N2 reactions at 2° benzylic carbon.
NOTE : In a S_N2 reaction, the major product formed is inversion product.

*Multiple options can be correct

QUESTION: 17

Aryl halides are less reactive towards nucleophilicsubstitution reaction as compared to alkyl halides due to :(1990 - 1 Mark)

A.
The formation of less stable carbonium ion
B.
Resonance stabilization
C.
Longer carbon-halogen bond
D.
The inductive effect
E.
sp² hybridized carbon attached to the halogen.

Solution:

TIPS/Formulae :
Aryl halides are stable due to resonance stabilization.
The resonating structures

stabilise the aryl halide. These structures include a double bond between C and Cl which is shorter and thus stronger than the usual C – Cl single bond. The sp² hybridised carbon, being electronegative, makes the C – Cl bond shorter and stronger.

*Multiple options can be correct

QUESTION: 18

Benzyl chloride (C₆H₅CH₂Cl) can be prepared from toluene by chlorination with (1998 - 2 Marks)