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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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This mock test of Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 for JEE helps you for every JEE entrance exam. This contains 18 Multiple Choice Questions for JEE Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 exercise for a better result in the exam. You can find other Test: 35 Year JEE Previous Year Questions: Haloalkanes & Haloarene- 2 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

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

Solution:

QUESTION: 2

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

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)
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)
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)
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)
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 :
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)
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)
Solution:

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

QUESTION: 12

How many structures for F are possible?                 (2003S)
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)
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,
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 SN2 reactions at 2° benzylic carbon.
NOTE : In a SN2 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)
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)
Solution:

SOCl2​,SO2​Cl2​ both give chlorination by nucleophilic substitution on an electrophilic carbon.

In toulene ,there is no electrophilic centre

oxidation occur by controlled free radical reaction using Cl2​ in presence of high.

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