A is meso 2, 3-butan-di-ol formed by syn addition

A is meso 2, 3-butan-di-ol formed by anti addition

A is a racemic mixture of d and l, 2, 3-butan-di-ol formed by anti addition

A is a racemic mixture of d and l, 2-3-butan-di-ol formed by syn addition

Both (A) and (B) in equal amount

None, as no reaction takes place

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JEE Exam > JEE Tests > Chemistry for JEE Main & Advanced > Hydrocarbon - 2 - JEE MCQ

Hydrocarbon - 2 - JEE MCQ

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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Hydrocarbon - 2

Hydrocarbon - 2 for JEE 2024 is part of Chemistry for JEE Main & Advanced preparation. The Hydrocarbon - 2 questions and answers have been prepared according to the JEE exam syllabus.The Hydrocarbon - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Hydrocarbon - 2 below.

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Hydrocarbon - 2 - Question 1

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The product(s) obtained via oxymercuation (HgSO₄ + H₂SO₄) of 1-butyne would be -

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Hydrocarbon - 2 - Question 2

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A.
A is meso 2, 3-butan-di-ol formed by syn addition
B.
A is meso 2, 3-butan-di-ol formed by anti addition
C.
A is a racemic mixture of d and l, 2, 3-butan-di-ol formed by anti addition
D.
A is a racemic mixture of d and l, 2-3-butan-di-ol formed by syn addition

Detailed Solution for Hydrocarbon - 2 - Question 2

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Hydrocarbon - 2 - Question 3

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Hydrocarbon - 2 - Question 4

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The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are -

A.
BrCH₂CH₂CH₂CH₂CH₃ and CH₃CH₂C≡CH
B.
BrCH₂CH₂CH₃ and CH₃CH₂ CH₂C≡CH
C.
BrCH₂CH₂CH₂ CH₂CH₃ and CH₃Cz ≡ CH
D.
BrCH₂CH₂CH₂CH₃ and CH₃C₂ C≡CH

Detailed Solution for Hydrocarbon - 2 - Question 4

Hydrocarbon - 2 - Question 5

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Identify a reagent from the following list which can easily distinguish between 1-butyne and 2-butyne

A.
bromine, CCl₄
B.
H₂, Lindlar catalyst
C.
HgSO₄/dil. H₂SO₄
D.
ammoniacal Cu₂Cl₂ solution

Detailed Solution for Hydrocarbon - 2 - Question 5

Hydrocarbon - 2 - Question 6

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Of the following, unsaturated hydrocarbons are-

A.
ethyne
B.
cyclohexane
C.
n-propane
D.
ethane

Detailed Solution for Hydrocarbon - 2 - Question 6

CH≡CH and CH₂=CH₂ both are unsaturated hydrocarbon.

Hydrocarbon - 2 - Question 7

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A.
Zn(Hg)/conc. HCl
B.
NH₂NH₂,OH^–
C.
H₂/Ni
D.
NaBH₄

Detailed Solution for Hydrocarbon - 2 - Question 7

Hydrocarbon - 2 - Question 8

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CH₃CH=CHCH₃+CHCl₃ + t–BuOK → A A is -

Detailed Solution for Hydrocarbon - 2 - Question 8

Hydrocarbon - 2 - Question 9

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The reaction of propene with HOCl proceeds via the addition of -

A.
H⁺ in first step
B.
Cl⁺ in first step
C.
OH¯ in first step
D.
Cl⁺ and OH^– in single step

Detailed Solution for Hydrocarbon - 2 - Question 9

Hydrocarbon - 2 - Question 10

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A.
Cu–C≡C–Cu
B.
CH₂=CH–C≡CH
C.
CH=C–Cu
D.
Cu–C≡C–NH₄

Detailed Solution for Hydrocarbon - 2 - Question 10

Hydrocarbon - 2 - Question 11

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Baeyer’s reagent is -

A.
alkaline permanganate solution
B.
acidified permanganate solution
C.
neutral permanganate solution
D.
aqueous bromine solution

Detailed Solution for Hydrocarbon - 2 - Question 11

Baeyer's reagent is alkaline permanganate solution

Hydrocarbon - 2 - Question 12

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What would be the product when 2-pentene reacts with HBr -

A.
2-bromo pentane
B.
3-bromo pentane
C.
Both A and B
D.
1-bromo pentane

Detailed Solution for Hydrocarbon - 2 - Question 12

Hydrocarbon - 2 - Question 13

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Acetylene may be prepared using Kolbe's electrolytic method employing -

A.
Pot. acetate
B.
Pot. succinate
C.
Pot. fumarate
D.
Pot propanoate

Detailed Solution for Hydrocarbon - 2 - Question 13

Factual

Hydrocarbon - 2 - Question 14

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Assertion : Addition of HBr to 1-butene gives two optical isomers.

Reason : The product contains one asymmetric carbon.

A.
If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B.
If both Assertion and Reason are true but Reason is not correct explanation of the Assertion.
C.
If Assertion is true but the Reason is false.
D.
If Assertion is false but Reason is true.

Detailed Solution for Hydrocarbon - 2 - Question 14

Hydrocarbon - 2 - Question 15

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Which of the following statement is correct about alkanes -

A.
They are coloured and soluble in water
B.
Their boiling point decreases with the increment in molecular weight
C.
They are colourless, odourless & tasteless.
D.
None of these

Detailed Solution for Hydrocarbon - 2 - Question 15

They are colourless, odourless & tasteless

Hydrocarbon - 2 - Question 16

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The product of reaction between one mole of acetylene and two mole of HCHO in the presence of Cu₂Cl₂ -

A.
HOCH2–C≡C–CH₂OH
B.
H₂C=CH–C≡C–CH₂OH
C.
HC=C–CH₂OH
D.
None of these

Detailed Solution for Hydrocarbon - 2 - Question 16

Hydrocarbon - 2 - Question 17

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Detailed Solution for Hydrocarbon - 2 - Question 17

Hydrocarbon - 2 - Question 18

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Consider the following reactions -

Identify the structure of the major product 'X'

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Hydrocarbon - 2 - Question 19

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The highest boiling point is expected for-

A.
iso-butane
B.
n-octane
C.
2,2,3,3-tetramethyl butane
D.
n-butane

Detailed Solution for Hydrocarbon - 2 - Question 19

n-octane has highest boiling point

Hydrocarbon - 2 - Question 20

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Which reagent is used for clemmensen reduction -

A.
Na–Hg/HCl
B.
Zn–Hg/HCl
C.
SnCl₂–HCl
D.
N₂H₄/KOH

Detailed Solution for Hydrocarbon - 2 - Question 20

Hydrocarbon - 2 - Question 21

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A.
CH₂ = CH₂
B.
CH₃CH = CH₂
C.
Both (A) and (B) in equal amount
D.
None, as no reaction takes place

Detailed Solution for Hydrocarbon - 2 - Question 21

Hydrocarbon - 2 - Question 22

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On acid catalysed hydration, 2-phenyl propene gives -

A.
3-phenyl-2-propanol
B.
2-phenyl-1-propanol
C.
1-phenyl-3-propanol
D.
2-phenyl-2-propanol

Detailed Solution for Hydrocarbon - 2 - Question 22

Hydrocarbon - 2 - Question 23

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Alcoholic solution of KOH is a specific reagent for -

A.
Dehydration
B.
Dehydrogenation
C.
Dehydro halogenation
D.
Dehalogenation

Detailed Solution for Hydrocarbon - 2 - Question 23

Alcoholic solution of KOH is a specific reagent for dehydrohalogenation.

So option (C) is correct.

Hydrocarbon - 2 - Question 24

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In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti- Markovnikov addition to alkene because -

A.
both are highly ionic
B.
one is oxidising and the other is reducing
C.
one of the step is endothermic ion both the cases
D.
All the steps are exothermic in both cases

Detailed Solution for Hydrocarbon - 2 - Question 24

Hydrocarbon - 2 - Question 25

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Which is expected to react most readily with bromine -

A.
CH₃CH₂CH₃
B.
CH₂ = CH₂
C.
CH ≡ CH
D.
CH₃–CH = CH₂

Detailed Solution for Hydrocarbon - 2 - Question 25

Hydrocarbon - 2 - Question 26

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Hydrogenation of the above compound in the presence of poisoned pallodium catalyst gives

A.
An optically active compound
B.
An optically inactive compound
C.
A racemic mixture
D.
A diastereomeric mixture

Detailed Solution for Hydrocarbon - 2 - Question 26

Hydrocarbon - 2 - Question 27

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Acetylene may be prepared using Kolbe's electrolytic method employing -

A.
Pot. acetate
B.
Pot. succinate
C.
Pot. fumarate
D.
Pot propanoate

Detailed Solution for Hydrocarbon - 2 - Question 27

Factual

Hydrocarbon - 2 - Question 28

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The number of structural and configurational isomers of a bromo compound, C₅H₉Br formed by the addition of HBr to 2-pentyne respectively -

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

Detailed Solution for Hydrocarbon - 2 - Question 28

Hydrocarbon - 2 - Question 29

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A.
sp - sp²
B.
sp³ - sp³
C.
sp - sp³
D.
sp² - sp³

Detailed Solution for Hydrocarbon - 2 - Question 29

Hydrocarbon - 2 - Question 30

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A.
2
B.
3
C.
4
D.
5

Detailed Solution for Hydrocarbon - 2 - Question 30

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