Hydrocarbons Test

Result

Hydrocarbons Test - 18

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1 - 10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c), and (d), out of which ONLY ONE option is correct.
Q.
Arrange the following compounds in increasing order of their basic strength.

A
I < II < III < IV

B
IV < III < II < I

C
III < II < I < IV

D
IV < II < III < I

Solution

Aniline (pK_avalue = 4.87)
Pyridine (pK_avalue = 5.2)
Piperidine (pK_avalue = 11.22)
Pyrrole ( pK_avalue =16.5)
Higher the pK_a value (lower acidity), higher will be the basicity.
So the order of basic strength will be
I < II < III < IV
Question 2
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In the compound below, which nitrogen is protonated first when treated with HCI?

A
1 is protonated because it is sp² hybridised

B
2 is protonated because it is sp³ hybridised

C
1 is protonated because its lone pair is not involved in aromaticity

D
2 is protonated because its lone pair is not involved in aromaticity

Solution

1 is protonated because its lone pair is not involved in aromaticity and also lone pair on Nitrogen 2 will get involved in conjugation and will provide extra electron density on Nitrogen 1. Hence, Option C is correct.
Question 3
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Which of the following compounds is more easily oxidised to a carbonyl when treated with MnO₂?

A
Only I

B
Only II

C
Both are oxidised with comparable ease

D
Both are very resistant to oxidation

Solution

MnO₂ is used for the selective oxidation of allylic and benzylic carbon and in these compounds these carbons are not very easily oxidisable.
Question 4
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Which among these is the simplest example for polycyclic arenes?

A
Benzacephenanthrylene

B
Naphthalene

C
Pyrene

D
Dibenz-anthracene

Solution

Naphthalene has fused ring of aromaticity and has the simplest structure when compared with other polycyclic aromatic hydrocarbons.
Question 5
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The compound shown below evolve hydrogen gas when refluxed with potassium metal, why?

A
Loss of H₂ introduces a new pi-bond that makes system aromatic

B
Loss of H₂ converts the given compound into an aromatic cation

C
Loss of H₂ converts the given compound into an aromatic anion

D
Deprotonation of the above compound converts it into an aromatic anion

Solution

Metals like Potassium tend to loose their own electrons and the excess electrons will complete the aromaticity of the system.Thus, Deprotonation of the above compound converts it into an aromatic anion witn 6 pi electrons.
Question 6
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Which compound below has maximum tendency to form a salt when treated with HBr?

A

B

C

D

Solution

Option C has maximum tendency to form a salt when treated with HBr. Due to the polarity difference and the aromatic nature of ring after generation of carbocation on Carbon of Carbonyl group. It will give an aromatic system of 6 pi electrons.
Question 7
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What is the correct order of increasing acidic strength of the following?

A
Ill < I < IV < II

B
Ill < IV < II < I

C
Ill < II < IV < I

D
III

Solution

Compound I is having the highest acidic strength due to the -I effect of five CF₃substituents.
Compound II is having less acidic strength than I but more than the rest due to the extremely stable conjugate anion formed after deprrotonation.
So, Option B is correct.
Question 8
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How many Kekule structures exist for benzene?

A
1

B
2

C
3

D
4

Solution

Benzene has Five resonance structures. They are Two Kekule’s and Three Dewar’s structure.
Question 9
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How many monobromo derivatives exists for anthracene?

A
2

B
3

C
4

D
5

Solution

The correct answer is Option B.
There are 3 monobromo derivatives exists for anthracene:
1-Chloroanthracene
2-Chloroanthracene
and 9-Chloroanthracene
Question 10
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7-bromo-1,3,5-cycloheptatriene exists as ionic species in aqueous solution while 5-bromo-1,3-cyclopentadiene does not ionise even in presence of AgNO₃(aq) because

A
C—Br bond has partial double bond character in 5-bromo-1,3-cyclopentadiene

B
7-bromo-1,3,5-cyclopheptatriene ionises to form an aromatic anion while 5-bromo-1,3-cyclopentadiene ionises to form an anti-aromatic cation

C
7-bromo-1,3,5-cycloheptatriene ionises into an aromatic cation while 5-bromo-1,3 cyclopentadiene ionises into a highly unstable anti-aromatic cation

D
Both ionises to form anions but one form aromatic while other form anti-aromatic anion

Solution

The correct answer is Option C.
The C-Br bond in the case of 7-bromo-1, 3, 5-cycloheptatriene is broken easily because the intermediate carbocation formed is very stable (aromatic as it contains (4n + 2)π e- ie, follows Huckle rule) while it does not break easily in the case of 5-bromo-1, 3-cyclopentadiene because carbocation formed here is highly unstable as it is antiaromatic i.e., does not follow Huckel rule. (It contains 4π electrons).