Hydrocarbons Test

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Hydrocarbons Test - 15

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1 - 8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. How many different alkenes are formed when 2 -chlorobutane is treated with ethanolic solution of KOH?

A
1

B
2

C
3

D
4

Solution

With ethanolic KOH, we have E₂ elimination.

So, we have 3 products in all
Question 2
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What is the major dehydration product in the following reaction ?

A

B

C

D

Solution

Note that H₂SO₄, H3PO₄ and KHSO₄ are dehydrating agents with rearrangement possible. While P₂O₅, P₄O₁₀, ThO₂, POCl₃ and Al₂O₃ are dehydrating agents without any possibility of rearrangement.
Question 3
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Which of the following compounds will lose optical activity after the reaction ?

A

B

C

D

Solution

The correct answer is option B
Question 4
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In which of the following reactions , only single is omer of alkene is formed ?

A

B

C

D

Solution

As there is no hydrogen at adjacent position for E₂ elimination.
Question 5
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Consider the following reaction .
The correct statement concerning product of the above reaction is

A
only a single alkene is formed

B
a pair of geometrical isomers are formed

C
a pair of enantiomers in equal amount is formed

D
a pair of diastereomers in equal amount is formed

Solution

Both are the same alkene. So only one alkene is formed.
Question 6
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An optically active hydrocarbon X has molecular formula C₆H₁₂. X on catalytic hydrogenation gives optically inactive C₆H₁₄. X could be

A
3-methyl-1-pentene

B
3-methyl-2-pentene

C
4-methyl-2-pentene

D
2-ethyl-1 -butene

Solution

The optically active C6H1 2 hydrocarbon is 3-Methylpent-1-ene, having one chiral carbon shown in fig A.
On catalytic hydrogenation , the compound obtained is shown in fig B, and the molecule does not have chiral carbon.
The reaction is C6H1 2 → C6H1 4 and this reaction takes place inthe presence of H2 and Pd.
Question 7
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Which of the following reactions produces an alkene?

A

B

C

D

Solution

Option a) Diazene(N₂H₂) is a hydrogenating agent. So, there will be no reaction.
Option b) Al₂O₃+CrO₃ acts as a dehydrogenation catalyst and so an alkene is formed. (Here 1-propene is formed)
Option c) This reaction is Wolff Kishner reaction. Here, acetone would be converted to alkane.
Option d) Zn/CH₃COOH substitutes Cl with H and an alkane is formed.
Question 8
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What is the major product of the reaction given below ?
meso -2 ,3 -dichiorobutane + Nal (aq) →

A
c/s-2-butene

B
1,3-butadiene

C
trans-2-butene

D
1-butene

Solution
Question 9
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Direction (Q, Nos. 9 - 12) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Q.
Statement I : 2-butyne on reduction with Pd/CaCO₃ gives c/s-2-butene.
Statement II : Hydrogenation proceed through adsorption mechanism.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

A Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate or barium sulphate which is then poisoned with various forms of lead or sulphur. It is used for the hydrogenation of alkynes to cis alkenes (i.e. without further reduction into alkanes) When the triple bond is (2-butyne) hydrogenated over the Lindlar’s catalyst i.e.Pd/CaCO₃
it gives predominantly cis alkene(2-butene).
Question 10
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Statement I : 2, 3-dibromo butane with Zn-dust gives frans-2-butene as major product.
Statement II : frans-2-butene is more stable than c/s-2-butene.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

Zn reacts with 2,3 - dibromobutane to give alkene. As trans 2-butene is more stable than cis 2-butene(owing tosterric hinderancce), former is major product.
Question 11
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Statement I : 2-bromobutane with (CH₃)₃COK in tertiary butanol gives 1 -butene as major product.
Statement II : Very strong base (CH₃)₃COK in tertiary butanol brings about E-2 elimination reaction.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect
Question 12
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Statement I : Boiling point of an alkene is slightly greater than that of its hydrogenated alkane.
Statement II : Alkenes have stronger van der Waals’ force of attraction than alkanes.

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is correct but Statement I is incorrect

Solution

The correct answer is Option D.
The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes get stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane.