Test: Aromaticity - NEET MCQ

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

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. 
 

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

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

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.

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.

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.

Cyclooctatetraene readily reacts with potassium metal to form the salt, which contains the dianion C₈H₈^2-. The dianion is both planar and octagonal in shape and aromatic with a Hückel electron count of 10.

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).

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).

• 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.

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.
 

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

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. .

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 .
 

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. .

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

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.

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.

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