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Test: Aromaticity


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20 Questions MCQ Test Chemistry Class 11 | Test: Aromaticity

Test: Aromaticity for NEET 2023 is part of Chemistry Class 11 preparation. The Test: Aromaticity questions and answers have been prepared according to the NEET exam syllabus.The Test: Aromaticity MCQs are made for NEET 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Aromaticity below.
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Test: Aromaticity - Question 1

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. Arrange the following compounds in increasing order of polarity

Detailed Solution for Test: Aromaticity - Question 1

In case 1, the bond is broken in oxygen’s favor and it will attain its octet. Also, carbon becomes sp2 hybridized, so there is a chance of polarity.
In case 2, if the bond is broken in favor of oxygen, then the ring will become anti-aromatic which is highly unstable and the bond won’t be broken in that way. If the bond is broken in favor of carbon in the ring, then although the ring becomes aromatic but oxygen will bear +ve charge which is very unstable. So, there is no chance to break the bond. 
In case 3, if the double bond is broken in favor of oxygen, then oxygen will acquire a negative charge and the ring will become aromatic. So, it is a highly favorable case of double bond breaking.
Therefore, the order of polarity: - III>I>II

Test: Aromaticity - Question 2

In principle, what is true regarding benzene and 1, 3, 5-cyclohexatriene?

Detailed Solution for Test: Aromaticity - Question 2

The correct answer is Option D.
In 1,3,5-cyclohexatriene, there are three C = C having a bond length 134 pm and three C - C having a bond length 154 pm. In benzene, all the six C - C bonds have the same bond length 139 pm. 
 

Test: Aromaticity - Question 3

Enthalpy of hydrogenation of cyclohexene is -119 kJ/mol and that of benzene is -208 kJ/mol. Based on these information, resonance energy of benzene can be calculated to be

Detailed Solution for Test: Aromaticity - Question 3

The correct answer is Option C.
ΔHhyd (cyclohexene) = - 119 kJ/mol (1C = C)
ΔHexp (benzene) = -3 × 119 = -357 (3C = C)
ΔHcal (benzene) = -208 KJ/mol
∴ resonance energy = ΔHexp - ΔHcal
                                        = -357 + 208
                                 = -149 kJ/mol
 

Test: Aromaticity - Question 4

Select the species which is not aromatic.

Detailed Solution for Test: Aromaticity - Question 4


Here, we can see that Nitrogen gave its lone pair to make the system aromatic. The same case happens with option b and d. But with option c, Boron is not having any lone pair to donate. So option c is correct answer

 

Test: Aromaticity - Question 5

Consider the following bromides :

 

Detailed Solution for Test: Aromaticity - Question 5

In the compound I, Br will dispatch as Br+ so that a -ve charge appears on carbon which will give us 6π electrons. So ring will become aromatic.(4n+2 π electron is needed for aromaticity)
In compound II, Br will dispatch as Br-. So that carbon has +ve charge and all the double bond will circulate in the ring. This will maintain 4n+2 π electron and the molecule will remain as aromatic.

Test: Aromaticity - Question 6

Which is not true regarding 1, 3, 5, 7-cyclooctatetraene?

Detailed Solution for Test: Aromaticity - Question 6

The correct answer is Option D.

 

1, 3, 5, 7-cyclooctatetraene is a system with 8 electrons but it is a nonaromatic compound as it adopts a tub like shape to escape anti-aromaticity. Hence, d option is wrong.

Test: Aromaticity - Question 7

The following compound , when treated with excess of AgBF4 gives a red precipitate leaving a highly conducting filtrate, due to

Detailed Solution for Test: Aromaticity - Question 7

The correct answer is Option B.
An aromatic dication is formed, the no of  electrons become 2 thus it obeys 4n+2 rule hence it is aromatic and highly stable.

Test: Aromaticity - Question 8

When potassium metal is added to 1, 3, 5, 7-cyclooctatetraene, a highly conducting salt is formed without evolution of H2 gas because

Detailed Solution for Test: Aromaticity - Question 8

Cyclooctatetraene readily reacts with potassium metal to form the salt, which contains the dianion C8H82-. The dianion is both planar and octagonal in shape and aromatic with a Hückel electron count of 10.

 

Test: Aromaticity - Question 9

Direction (Q. Nos. 9 - 12) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

When considering electrophilic aromatic substitution reactions the halides are described as

Detailed Solution for Test: Aromaticity - Question 9

Halides are ortho, para directing groups but unlike most ortho, para directors, halides mildly deactivate the arene. This unusual behavior can be explained by two properties: Since the halogens are very electronegative they cause inductive withdrawal (withdrawal of electrons from the carbon atom of benzene).

Test: Aromaticity - Question 10

What makes the following compound aromatic?

Detailed Solution for Test: Aromaticity - Question 10

To make the given compound aromatic, we will remove an H+ from sp3 hybridized carbon atom to get 10 cyclic electrons (i.e. electrons move around the periphery of the ring).

*Multiple options can be correct
Test: Aromaticity - Question 11

What is true about the compound calicene?

Detailed Solution for Test: Aromaticity - Question 11
  • In calicene, the electrons move towared the five-membered ring because both rings are aromatic in the resonance contributor that has a negative charge on a carbon of the five-membered ring and a positive charge on a carbon of the three-membered ring
  • It is highly soluble in water and in solution it shows very high electrical conductivity

So, Option C is not correct and other Options are Correct.

*Multiple options can be correct
Test: Aromaticity - Question 12

Which of the following species is/are anti-aromatic?

Detailed Solution for Test: Aromaticity - Question 12

The correct options are A & D
Anti Aromatic compounds are those compounds that satisfy the rules of planarity and fully conjugation of the pi electrons inside the ring but fail to satisfy Huckel's rule of 4n+2 pi electrons.
The anti-aromatic compound contains 4nπ electrons.Option C, is a Non-Aromatic Compound as it does not satisfy the condition of Planarity.
Here the options A and D are anti-aromatic.
 

Test: Aromaticity - Question 13

Direction (Q. Nos. 13 - 17) This section is based on Statement i and Statement II. Select the correct answer from the codes given below.

Q. 

Statement I : Benzene has very high stability than a general triene.

Statement II : Benzene is a completely conjugated system.

Detailed Solution for Test: Aromaticity - Question 13

Benzene has very high stability than a general triene due to Aromaticity and not just completely conjugated system. Hence, Both Statement I and Statement II are correct but, Statement II is not the correct explanation of Statement  I

Test: Aromaticity - Question 14

Statement I: Pyrene, although aromatic, decolorises brown colour of bromine water.

Statement II: It has one pi-bond extra which is not part of the aromatic system.

Detailed Solution for Test: Aromaticity - Question 14

Pyrenes are strong electron donor materials and has one pi-bond extra which is not the part of aromatic system Thus, decolourises brown colour of bromine water. Hence, Option A is correct.

Bromine Test: Bromine solution is brown. In this test when bromine solution is added to the unsaturated hydrocarbon the brown colour disappears if the hydrocarbon is unsaturated. Bromine forms an addition product with the unsaturated hydrocarbon. .

Test: Aromaticity - Question 15

Statement I : The follow compound is optically active.

Statement II : It has a chiral carbon.

Detailed Solution for Test: Aromaticity - Question 15

The correct answer is option C

Statement 1 is correct because Plane Of Symmetry and Centre Of Symmetry are absent in the compound .
Statement 2 is clearly incorrect as there is No 'Sp3 ' Hybridized Carbon atom .
 

Test: Aromaticity - Question 16

Statement I : Furan is an aromatic system, has resonance energy comparable to that of benzene.

Statement II : Furan decolourises the brown colour of bromine water solution.

Detailed Solution for Test: Aromaticity - Question 16

Furan has resonance energy comparable to benzene. But the double bonds are involved in aromaticity, so it won’t decolourise Bromine water

Bromine Test: Bromine solution is brown. In this test when bromine solution is added to the unsaturated hydrocarbon the brown colour disappears if the hydrocarbon is unsaturated. Bromine forms an addition product with the unsaturated hydrocarbon. .

Test: Aromaticity - Question 17

Arrange the halogens F2, Cl2, Br2, I2 in order of their increasing reactivity with alkanes.

Detailed Solution for Test: Aromaticity - Question 17

The correct answer is Option A.
Since reactivity decreases down the group as the electronegativity of the halogen decreases down the group. Thus, rate of reaction of alkanes with halogens is 
I2 < Br2 < Cl2 <F2

*Answer can only contain numeric values
Test: Aromaticity - Question 18

Direction (Q. Nos. 18 - 20) This section contains 3 questions. When worked out will result in an integer from 0 to 9 (both inclusive)
Cyclobutene when refluxed in presence of potassium metal, evolve hydrogen gas and an aromatic system is formed. How many pi-electrons are involved in the above formed aromatic system?


Detailed Solution for Test: Aromaticity - Question 18

Metals have a tendency of loosing electrons. So, Potassium Metal will give extra electrons to cyclobutene thus, making it aromatic in nature with 6 pi- electrons.

Test: Aromaticity - Question 19

When two or more different substituents are attached with a benzene ring the number 1 position in the ring is given to a high priority group. Which one of the following groups has the highest - priority?

Detailed Solution for Test: Aromaticity - Question 19

Option C is correct, because in Nomenclature of Organic Compounds. Priority is given to that functional group that is on the top of the Priority Order of Functional Groups. Among the given Functional Groups COOH is having the highest priority so, the number 1 position in the ring is given to it.

See the source image

*Answer can only contain numeric values
Test: Aromaticity - Question 20

The compound below has four phenyl rings, but very less stable due to an opposing factor of stability. Therefore, this compound absorb bromine in dark.
How many bromine molecules, when added to this molecule, would make it stable and prevent further bromine addition?

[IIT JEE 2005]


Detailed Solution for Test: Aromaticity - Question 20

Three bromine molecules would make it stable and would prevent further bromine addition.

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