Test: Aromaticity And Acid Base Properties


15 Questions MCQ Test Chemistry for JEE | Test: Aromaticity And Acid Base Properties


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QUESTION: 1

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.

Solution:

Aniline (pKvalue = 4.87)

 Pyridine (pKvalue = 5.2)

Piperidine (pKvalue = 11.22)

Pyrrole ( pKvalue =16.5)

Higher the pKa value (lower acidity), higher will be the basicity. 

So the order of basic strength will be 

I < II < III < IV 

 

QUESTION: 2

In the compound below, which nitrogen is protonated first when treated with HCI?

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

Which of the following compounds is more easily oxidised to a carbonyl when treated with MnO2?

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

Which among these is the simplest example for polycyclic arenes?

Solution:

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

QUESTION: 5

The compound shown below evolve hydrogen gas when refluxed with potassium metal, why?

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

Which compound below has maximum tendency to form a salt when treated with HBr?

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

What is the correct order of increasing acidic strength of the following?

Solution:

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.

QUESTION: 8

How many Kekule structures exist for benzene?

Solution:


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

QUESTION: 9

How many monobromo derivatives exists for anthracene?

Solution:

The correct answer is Option B. 

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

QUESTION: 10

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

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).
 

*Multiple options can be correct
QUESTION: 11

Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. Consider the following compounds.

The correct statement regarding properties of above mentioned compounds is/are

Solution:
  • 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.

*Multiple options can be correct
QUESTION: 12

Which of the following systems are aromatic?

Solution:

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
 

*Multiple options can be correct
QUESTION: 13

What is true about the 1,3,5,7-cyclooctatetraene?

Solution:

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.

*Multiple options can be correct
QUESTION: 14

What is true regarding the following compound?

Solution:

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.

QUESTION: 15

Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group

Solution:

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

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