Aromaticity MCQ Quiz in తెలుగు - Objective Question with Answer for Aromaticity - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:-

• The compounds are said to be aromatic if they satisfy the following conditions-
  • The molecule should be cyclic.
  • The molecule must be conjugated, i.e every atom in the molecule should be sp2 hybridized.
  • The number of pi electrons in the system is (4n + 2), where values of n can be 0, 1, 2...
  • The molecule has to be planar so that all pi electrons in the system are delocalized.
  • Aromatic compounds are exceptionally stable.

Geometrical isomerism:

• Geometrical isomerism is shown in molecules that have a carbon-carbon double bond C = C.
• Geometrical isomerism is shown only when each carbon atom of the double bond is attached to two different atoms or groups.
• Compounds of the type abC = Cad, abC = Cab, and abC = Cde will show geometrical isomerism, where a, b,d, e are alkyl groups.
• The cause of geometrical isomerism is restricted rotation about a C = C bond.

Cis and trans isomerism:

• In a geometrical isomer, if the two identical groups are situated on the same side of the C = C then that is called the cis isomer.
• In a geometrical isomer, if the two identical groups are situated on the opposite side of the C = C then that is called the trans isomer.

2 - Butene

• The molecule has the formula CH3-CH=CH-CH3.
• It is of abC = Cab type and will thus show cis-trans isomerism.

Explanation:-

• For geometric isomerism: At this carbon for both groups must be different and C–C bond rotation must be restricted.

• The above compound can show geometric isomerism as it has a carbon-carbon double bond C = C and each carbon atom of the double bond is attached to two different atoms or groups.

• Both the rings of the charged separated species of compound (ii) will be aromatic, as one of the rings has 2π electrons and the other one has 6π electrons.
• it obeys Huckel's (4n+2)π rule for n=0 and 1.
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.
• Thus, the charged separated species will be very stable and thus the C=C bond will act as a partially single bond.
• Hence, compound (ii) does not give geometrical isomerism.

(iii)

• One of the rings of the charged separated species will be antiaromatic, as the ring has 8π electrons and it obeys Huckel's 4nπ rule for n=2.
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.
• Thus, the charged separated species will be unstable and thus the C=C bond will not act as a partially single bond.
• Hence, compound (iii) gives geometrical isomerism.

Conclusion:-

• Hence, (i) and (iii) will show geometrical isomerism.

Concept:-

• A hydride donor is a chemical species that can donate a hydride ion (H^-) to another compound.
• A hydride ion is composed of a hydrogen atom with a negative charge (H^-).
• Hydride donors are important in various chemical reactions and are often used as reducing agents.
• In organic chemistry, hydride donors are commonly used to reduce functional groups by adding a hydride ion to them.

Huckel's rule of aromaticity:

• The compounds are said to be aromatic if they satisfy the following conditions:-
  • The molecule should be cyclic.
  • The molecule must be conjugated, i.e. every atom in the molecule should be sp2 hybridized.
  • The number of π electrons in the system is (4n + 2), where values of n can be 0, 1, 2...
  • The molecule has to be planar so that all π electrons in the system are delocalized.
  • Aromatic compounds are exceptionally stable.

Explanation:-

• Aromatization after hydride donation refers to a chemical reaction where a cyclic compound, typically a non-aromatic ring, is transformed into an aromatic ring system through the removal of a hydride ion (H^-) followed by the formation of a cation.
• This type of reaction is commonly observed in organic chemistry and can lead to the formation of aromatic compounds from non-aromatic precursors.
• Among the following the best hydride donor is

• Upon hydride donation, it forms

• The formed compound is aromatic.
• In the above compound, the lone pair of electrons of the N atom is planar to the ring and is in conjugation with the system.
•  The N atom will contribute 2 electrons to the system.
• Thus, the total number of π electrons in the system is

= 6π

• Hence, the compound follows Huckel's (4n+2)π electron rule for n=1
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.
• Hence, the compound is aromatic.
• The other three compounds will not result in the formation of an aromatic compound.

Conclusion:-

•  Hence, the best hydride donor is

Concept:-

• The compounds are said to be aromatic if they satisfy the following conditions-
  • The molecule should be cyclic.
  • The molecule must be conjugated, i.e every atom in the molecule should be sp2 hybridized.
  • The number of pi electrons in the system is (4n + 2), where values of n can be 0, 1, 2...
  • The molecule has to be planar so that all pi electrons in the system are delocalized.
  • Aromatic compounds are exceptionally stable.
• Conditions for compounds to be anti-aromatic
  • The compounds should possess a 4n number of pi electrons, where n = 0, 1, 2..
  • The molecule must be cyclic.
  • The molecule must be conjugated, i.e every atom in the molecule should be sp2 hybridized.
  • The molecule has to be planar.
  • Anti-aromatic compounds are unstable.
• Non-aromatic compounds have 4n + 2 or 4n number of pi electrons but they are not planar or fail any of the above-mentioned criteria.
• Homoaromatic compounds are those which contain one or more sp3 hybridized carbon atoms in an otherwise conjugated cycle. In order for the orbitals to overlap effectively so as to close a loop, the sp3 atoms are forced to lie almost vertically above the plane of the aromatic atoms.

​Explanation:-

• Ionization of 3,4-dichloro-1,2,3,4-tetramethylcyclobutene in SbF₅/SO₂ at -75^oC leads to a dication whose formation and special stability is attributable to aromaticity.
• The reaction pathway is shown below:

• In the above reaction, it is shown that the intermediate formed in the following reaction is a carbocation. The two Cl atom interacts with the lewis acid SbF₅ and forms a carbocation.
• The compound has 2π electrons. It obeys Huckel's (4n+2)π rule for n=0.
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.

• Hence, the compound is aromatic.

​Conclusion:-

• Hence, option 2 is the correct answer.

Concept:

Huckel's rule of aromaticity:-

• Conditions for compounds to be Non-aromatic:-
  • Non-aromatic compounds have 4n + 2 or 4n number of π electrons
  • There is no delocalized electron system in that molecule.
  • The system must be non-planar and unconjugated.

​Explanation:-

A. 

• The compound has 6π electrons. It obeys Huckel's (4n+2)π rule for n=1.
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.

Hence, the compound A is aromatic.

B. 

• The compound has 10π electrons. It obeys Huckel's (4n+2)π rule for n=2
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.

Hence, the compound B is aromatic.

C. 

• The compound has 12π electrons.
• Although every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.
• This compound should be anti-aromatic as it contains 12π electrons (obeys 4nπ rule for n=3).

Hence, the compound is anti-aromatic.

Conclusion:-

Hence, only A and B species are aromatic.

Concept:

Huckel's rule of aromaticity:

• The compounds are said to be aromatic if they satisfy the following conditions:-
  • The molecule should be cyclic.
  • The molecule must be conjugated, i.e. every atom in the molecule should be sp2 hybridized.
  • The number of π electrons in the system is (4n + 2), where values of n can be 0, 1, 2...
  • The molecule has to be planar so that all π electrons in the system are delocalized.
  • Aromatic compounds are exceptionally stable.
• Conditions for compounds to be anti-aromatic:-
  • The compounds should possess a 4n number of π electrons, where n = 0, 1, 2....
  • The molecule must be cyclic.
  • The molecule must be conjugated, i.e. every atom in the molecule should be sp2 hybridized.
  • The molecule has to be planar.
  • Anti-aromatic compounds are unstable.
• Conditions for compounds to be Non-aromatic:-
  • Non-aromatic compounds have 4n + 2 or 4n number of π electrons
  • There is no delocalized electron system in that molecule.
  • The system must be non-planar and unconjugated.

Explanation:-

• The given compounds are shown below:

• In compound M, the N atom possesses a lone pair that can participate in resonance and contributes to the aromaticity of the compound.
• The B atom in M in also sp² hybridized.
• The compound has 10π electrons. It obeys Huckel's (4n+2)π rule for n=2.
• Every member in the ring is sp2 hybridized and thus the molecule is planar.
• The system is conjugated.

• Hence, the compound M is aromatic.​

• In compound N, the B atom is sp2 hybridized and thus very member in the ring is sp2 hybridized, and thus the molecule is planar.
• The compound has 6π electrons. It obeys Huckel's (4n+2)π rule for n=1.
• Hence, the compound N is also aromatic.

​Conclusion:-

Hence, According to Hückel's rule compound M and N are both aromatic.

Concept:-

Acid Strength and Aromaticity:

• A substance that gives H+ ion in the solution and forms salts by combining with some metals is called acid. According to Arrhenius, acid is a substance that releases an H+ ion in the solution and a base is a substance that accepts an H+ ion from the solution.
• The pKa value indicates the strength of an acid in the solution. The lower the value of pKa, the more the acid's strength will be.
• Aromatic compounds follow Huckel's rule, according to which a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Aromatic compounds are very stable.
• Whereas, the anti-aromatic compounds follow the 4npi electrons rule. According to this a cyclic, planar, and conjugated species having (4n+2)pi electrons (n = 0,1,3...) is aromatic.
• Anti-aromatic compounds are very unstable.

Explanation:-

I. 

• The above compound undergoes deprotonation to form a carbanion as follows:

• The carbanion or conjugate base formed in the reaction is aromatic and planar with 10pi electrons. Which follows the huckle's (4n+2)pi electrons for n=2.

II. 

• The above compound undergoes deprotonation to form a carbanion as follows:

• The carbanion or conjugate base formed in the above reaction is aromatic with 14pi electrons. Which follows huckle's (4n+2)pi electrons for n=3.
• The conjugate base of compound I is more stable due to less steric effect compare to the conjugate base of compound II. Thus, compound I is more acidic than compound II.

III. 

• The above compound undergoes deprotonation to form a carbanion as follows:

• The carbanion or conjugate base formed in the above reaction is antiaromatic with 8pi electrons. Which follows huckle's 4npi electrons for n=2.
• Thus, compound III is the least acidic among the four acids.

IV. 

• The above compound undergoes deprotonation to form a carbanion as follows:

• The carbanion or conjugate base formed in the reaction is aromatic and planar with 6pi electrons. Which follows the huckle's (4n+2)pi electrons for n=1.
• As compound IV consists of five strong electron-withdrawing -CF₃ groups, the carbanion or conjugate base of compound IV will be the most stable among the four acids.
• Thus, the acidity order will be ​​IV > I > II > III.

​Explanation:-

• Hence, the correct acid strength of the following compounds will be

​​IV > I > II > III

Concept:-

• Conditions for compounds to be anti-aromatic
  • The compounds should possess a 4n number of pi electrons, where n = 0, 1, 2.
  • The molecule must be cyclic.
  • The molecule must be conjugated, i.e every atom in the molecule should be sp2 hybridized.
  • The molecule has to be planar.
  • Anti-aromatic compounds are unstable.

• Anti-aromatic compounds are cyclic organic molecules that follow Hückel's rule but have an odd number of π electrons in their π system, which makes them unstable and highly reactive. Unlike aromatic compounds, which are characterized by increased stability and resonance, anti-aromatic compounds are destabilized by their electron configuration, which results in non-planar geometries, high reactivity, and low solubility.
• Anti-aromatic compounds are generally considered to be less stable and less useful than their aromatic counterparts, but they can be important intermediates in organic synthesis and in some biochemical processes.
• Anti-aromatic compounds are highly reactive and unstable due to their electron configuration, which results in the dimerization of anti-aromatic compounds.

Conclusion:-

I : 

• The above compound consists of a simple cyclic four-member ring that can undergo dimerization as follows:

• Thus, it is not a stable anti-aromatic compound.

II. 

• The above compound is a derivative of a simple cyclobutadiene compound, which consists of 4π electrons and an antiaromatic compound. Thus, it should be unstable and should undergo a dimerization reaction.
• But, it is a stable cyclobutadiene derivative due to the push-pull effect or caprodative effect.
• The mechanism of the push-pull effect is shown below:
• Due to this push-pull effect, the compound does not undergo any dimerization reaction.
• Thus, it is a stable anti-aromatic compound.
• Compounds III and IV also consists of 4π electrons and are anti-aromatic compounds.
• Thus, these two compounds should be unstable and should undergo dimerization.
• But, dimerization is not possible for the case of compounds III and IV as bulky substituents are attached to the four-membered ring,
• Hence, although all of the compounds consist of 4π electrons and are anti aromatic. But only compound II, III, and IV are stable anti-aromatic compounds and does not undergo any dimerization reaction.

​Conclusion:-

• Hence, the most stable Anti-aromatic compound from the following compounds will be

 II, III, and IV.