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AS.030.205 AROMATIC & NON-AROMATIC COMPOUNDS QUIZ 1 - 4 Q & A 2024
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AS.030.205 AROMATIC & NON-AROMATIC COMPOUNDS QUIZ 1 - 4 Q & A 2024
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AS.030.205 AROMATIC & NON-AROMATIC COMPOUNDS
Quiz 1 -4
Q&A
2024
,1. Which of the following compounds is aromatic
according to Huckel's rule?
a) Cyclopropenyl cation
b) Cyclobutadiene
c) Cyclopentadienyl anion
d) Cycloheptatrienyl cation
* e) All of the above
Rationale: Huckel's rule states that a compound is aromatic
if it is cyclic, planar, fully conjugated and has (4n+2) pi
electrons, where n is any integer. All of the options satisfy
these criteria, with n = 0 for a), b) and c), and n = 1 for d)
and e).
2. Which of the following statements is true about anti-
aromatic compounds?
a) They are more stable than aromatic compounds
b) They are less stable than non-aromatic compounds
* c) They are less stable than both aromatic and non-
aromatic compounds
d) They are more stable than non-aromatic compounds
e) They have the same stability as non-aromatic
compounds
Rationale: Anti-aromatic compounds are cyclic, planar,
fully conjugated and have (4n) pi electrons, where n is any
integer. These compounds are highly unstable because they
have an even number of electron pairs in the pi system,
which leads to a high energy state. They are less stable than
, both aromatic and non-aromatic compounds, which have
lower or zero resonance energy.
3. Which of the following compounds is non-aromatic?
a) Benzene
b) Pyridine
c) Furan
d) Cyclooctatetraene
* e) All of the above
Rationale: Non-aromatic compounds are either acyclic,
non-planar or not fully conjugated. Benzene, pyridine and
furan are all aromatic compounds because they satisfy
Huckel's rule. Cyclooctatetraene is non-aromatic because it
adopts a tub-shaped geometry that is not planar, thus
breaking the conjugation of the pi system.
4. What is the hybridization state of each carbon atom in an
aromatic compound?
a) sp
b) sp2
* c) sp3
d) sp3d
e) sp3d2
Rationale: Each carbon atom in an aromatic compound has
three sigma bonds and one p orbital that participates in the
pi system. Therefore, the hybridization state of each carbon
atom is sp2, which allows for planarity and conjugation.
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