Aromaticity MCQ Quiz in தமிழ் - Objective Question with Answer for Aromaticity - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 19, 2025
Latest Aromaticity MCQ Objective Questions
Top Aromaticity MCQ Objective Questions
Aromaticity Question 1:
Which of the following is/are aromatic compound?
Answer (Detailed Solution Below)
Aromaticity Question 1 Detailed Solution
Concept:
Aromaticity is defined by the Huckel Rule, which states that a compound is aromatic if it has a cyclic, planar structure with conjugated pi-electrons and follows the (4n + 2) rule, where n is a non-negative integer. Based on this, compounds can be classified as:
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Aromatic Compounds: These have 4n + 2 pi-electrons and show high stability due to resonance. The compound must be cyclic, planar, and fully conjugated.
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Non-aromatic Compounds: These are either non-cyclic, non-planar, or do not have fully conjugated pi-electrons. As a result, they do not follow the Huckel Rule and do not exhibit aromatic stability.
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Anti-aromatic Compounds: These have 4n pi-electrons and are less stable than aromatic compounds due to electronic instability. These compounds are also cyclic and planar with conjugated pi-electrons but are destabilized by the 4n electron count.
Explanation:
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This compound is aromatic. It has 6 pi-electrons (4n + 2, where n = 1), is cyclic, planar, and conjugated, fulfilling the criteria of Huckel's rule.
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This compound is anti-aromatic as it has 4 pi-electrons (4n, where n = 1), making it less stable and following the 4n rule.
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This is non-aromatic due to its non-planar structure, which disrupts conjugation and prevents it from being aromatic or anti-aromatic.
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This compound is aromatic with 6 pi-electrons (4n + 2, where n = 1), and it is cyclic, planar, and conjugated, adhering to Huckel's rule.
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Conclusion:
The correct answer is I and IV, as these compounds fulfill the criteria for aromaticity under Huckel's rule.
Aromaticity Question 2:
Which molecule in each of the following pairs has a higher dipole moment?
Pair X:
Pair Y:
Answer (Detailed Solution Below)
Aromaticity Question 2 Detailed Solution
The correct answer is 1 and 4
Concept:-
Dipole moment:
- A dipole moment occurs in any molecule in which there is a separation of charges.
- A dipole moment generates in ionic bonds as well as in covalent bonds.
- Dipole moments arise due to the difference in electronegativity between two bonded atoms.
- A bond dipole moment is a measure of the polarity of a chemical bond between two atoms in a molecule.
- The bond dipole moment is a vector quantity since it has both magnitude and direction.
- For example,
- Mathematically dipole moment can be expressed
μ = 𝛿 × d
Where, μ = bond dipole moment,
𝛿 = magnitude of the partial charges 𝛿+ and 𝛿–
And d = distance between 𝛿+ and 𝛿– (bond length)
- It is measured in the Debye unit denoted by ‘D’.
- The larger the electronegativity difference between the two atoms, the larger the bond’s dipole moment and polarity.
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 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.
Explanation:-
For pair
For (1) aromaticity is generated from charge separation, so (1) has high dipole moment
For pair
(4) is aromatic, so have higher dipole than (3).
Conclusion:-
So, 1 and 4 pairs has a higher dipole moment.
Aromaticity Question 3:
The Hybridistaion of Nitrogen and Oxygen in the given compound
Answer (Detailed Solution Below)
Aromaticity Question 3 Detailed Solution
The correct answer is sp2 , sp3 , sp2 ,sp3
Concept:-
- Hybridization: This concept refers to the combination of atomic orbitals within an atom to form new hybrid orbitals. The hybrid orbitals can explain the geometry of molecules in terms of bond lengths and angles. The nature of the groups attached to an atom, along with their spatial orientation, can help determine the hybridization state of that atom.
- Resonance: Resonance is the phenomenon where a molecule or an ion can be represented by more than one Lewis structure. Resonance structures help convey the concept of delocalized electrons, which are electrons that are not associated with a single atom or a covalent bond. The actual structure of a molecule is an average of its resonance structures.
- Delocalization: Delocalization refers to the spreading of electron density over a larger volume. The process of delocalization often increases the stability of molecules and ions, as it results in the electron density being spread over more atoms. For example, in molecules like aniline, the lone pair of electrons on the nitrogen atom can delocalize into the aromatic ring, increasing the overall stability of the molecule.
Explanation:-
The change in hybridization of nitrogen in aniline and 1-phenylmethylamine is due to the conjugation and delocalization of the lone pair of electrons on the nitrogen.
In aniline, the nitrogen is directly bonded to the aromatic ring. Its lone pair of electrons can contribute to resonance with the π electrons of the benzene ring, delocalizing into the ring's π system. Thus, the nitrogen in aniline is commonly considered to be sp2 hybridized, with one of the sp2 orbitals containing the lone pair participating in resonance with the ring.
On the contrary, in 1-phenylmethylamine (or N-methylaniline), the nitrogen atom is bonded to a methyl group in addition to the phenyl ring. The nitrogen's lone pair is primarily localized on the nitrogen because it's geometrically unable to interact significantly with the benzene ring due to the presence of the methyl group. It's thus better described as sp3 hybridized and does not participate in the same kind of resonance as in aniline.
Similarly, phenol has sp2 and other have the sp3 hybridization
Aromaticity Question 4:
Increasing order of the reactivity of the following heterocycles towards electrophile is
Answer (Detailed Solution Below)
Aromaticity Question 4 Detailed Solution
The correct option is option 2 i.e. III
Concept:-
The reactivity of heterocyclic compounds toward electrophiles, such as pyrrole, furan, and thiophene, can be influenced by their electron density and aromaticity. Let's discuss each of them:
- Pyrrole: This five-membered heterocycle contains a nitrogen atom. Because of the nitrogen's lone pair of electrons, pyrrole is electron-rich and tends to be highly reactive toward electrophiles, particularly at the C2 and C5 positions - the carbon atoms adjacent to the nitrogen. Pyrrole's aromatic behavior makes it more reactive than benzene under electrophilic aromatic substitution reactions.
- Furan: Furan is a five-membered heterocycle with an oxygen atom. Like pyrrole, furan is electron-rich due to the oxygen's lone pair of electrons. It's very reactive toward electrophiles, particularly at the C2 and C5 positions - the carbon atoms adjacent to the oxygen. However, pyrrole is slightly more reactive compared to furan due to nitrogen's ability to donate electrons better than oxygen.
- Thiophene: Thiophene is a five-membered heterocycle with a sulfur atom. Because sulfur's pi electrons are not as readily delocalized into the ring as oxygen's or nitrogen's, thiophene is less reactive than either furan or pyrrole. However, it's more reactive toward electrophiles than benzene. Electrophilic attacks primarily occur at the C2 and C5 positions - the carbon atoms adjacent to the sulfur.
Explanation:-
- Oxygen, being more electronegative than nitrogen, distributes more negative charge density upon itself and less upon the ring, thus stabilizing the carbocation intermediate less, making furan less reactive towards EAS than pyrrole.
- Sulfur's 3p. orbital overlaps less effectively with carbon's 2p, orbitals, thereby sharing electron density more poorly than furan's oxygen does, stabilizing the carbocation intermediate less, making thiophene less reactive towards EAS than furan.
- Pyrrole, furan, and thiophene are all more reactive than benzene with EAS because the lone pair on the heteroatom can donate electron density into the ring by resonance, thus stabilizing the carbocation intermediate more effectively.
so, pyrrole > furan > thiophene
Conclusion:-
the relative reactivity of these heterocycles toward electrophiles could generally be ordered as follows:
Pyrrole > Furan > Thiophene
Aromaticity Question 5:
Which of the following compound gives geometrical isomerism?
Answer (Detailed Solution Below)
Aromaticity Question 5 Detailed Solution
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.
Aromaticity Question 6:
Which of the following compound is the best hydride donor?
Answer (Detailed Solution Below)
Aromaticity Question 6 Detailed Solution
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
Aromaticity Question 7:
The intermediate formed in the following reaction will be
Answer (Detailed Solution Below)
Aromaticity Question 7 Detailed Solution
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 -75oC 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.
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Hence, the compound is aromatic.
Conclusion:-
- Hence, option 2 is the correct answer.
Aromaticity Question 8:
Which of the following species is/are aromatic?
Answer (Detailed Solution Below)
Aromaticity Question 8 Detailed Solution
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.
Aromaticity Question 9:
According to Hückel's rule
Answer (Detailed Solution Below)
Aromaticity Question 9 Detailed Solution
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.
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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.
Aromaticity Question 10:
The most stable Anti-aromatic compound from the following compounds will be
I :
II :
III :
IV :
Answer (Detailed Solution Below)
Aromaticity Question 10 Detailed Solution
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.