Aromaticity & Acid Base Properties - Free MCQ Practice Test with solutions,

So, correct option is A. 

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.

MnO₂ is used for the selective oxidation of allylic and benzylic carbon and in these compounds these carbons are not very easily oxidisable.

Naphthalene has fused ring of aromaticity and has the simplest structure when compared with other polycyclic aromatic hydrocarbons.

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.

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.

Compound I is having the highest acidic strength due to the -I effect of five CF₃ substituents.

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.

Benzene has Five resonance structures. They are Two Kekule’s and Three Dewar’s structure.

The correct answer is Option B. 

There are 3 monobromo derivatives exists for anthracene:
1-Chloroanthracene
2-Chloroanthracene
and 9-Chloroanthracene

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).
 

• Both have all their C—C bonds of equal length due to conjugation.
• I does not decolorises brown colour of bromine water solution but II does as The π bonds in Cyclooctatetraene (Compound II) react as usual for olefins, rather than as aromatic ring systems.
• I is planar but II is not as it adopts a tub conformation.
• Cyclooctatetraene shows various other addition reactions including Sulfonation.

Hence, Option A, B and D are correct.

The correct answers are Options C and D.
Aromatic compounds are those which follow Huckel's rule i.e, they have (4n+2) π electrons, n must be an integer.
In option C, there are 5 π bonds which means 10 π electrons; so 4n+2 = 10 i.e, n= 2 which is an integer.
In option D, Nitrogen has a lone pair which contains 2 electrons therefore this compound also have 10 π electrons; so n= 2.

1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.
Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:
2 K + C8H8 → K2C8H8
The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.

The given compound will turn itself to
To gain aromaticity, it will transfer the electrons as follow:-

It is clear that the compound is heterocyclic(the ring constitutes other than C and H). Due to -ve charge on outer O atom, it has high affinity for BF₃. However, NaBH₄  has no reaction with this. As the compound will turn itself to latter, there is no aldehyde or ketone group present in the compound.

Isocyanide is a compound and it is not a functional group.