Courses
Courses for Kids
Free study material
Offline Centres
More

Assertion:
The \[p{K_a}\]value of (I) is lower than the \[p{K_a}\]of (II).
Reason:Non-aromatic compounds are more stable than anti-aromatic compounds.
                                             
A.Both assertion and reason are correct and the reason is the correct explanation for assertion
B.Both assertion and reason are correct and the reason is not correct explanation for assertion
C.Assertion is correct but reason is incorrect.
D.Assertion is incorrect but the reason is correct.

seo-qna
Last updated date: 24th Feb 2024
Total views: 18.6k
Views today: 0.18k
IVSAT 2024
Answer
VerifiedVerified
18.6k+ views
Hint: Aromatic and Anti-aromatic compounds can be differentiated on the basis of Huckel's Rule.

Complete step by step answer:
We can identify the and classify compounds as either aromatic or anti-aromatic based on the following rules:
1.For a compound to be Aromatic, the following conditions must be satisfied
-The compound should be cyclic
-The molecular geometry of the compound should be planar
-The compound should exhibit Resonance
-The compound should obey Huckle’s rule, and hence should have (4n+2) \[\pi \]electrons, where ‘n’ denotes the number of subsequent energy levels and can be equal to any whole number, i.e. 0,1,2,3,…
2.And for the compound to be Anti-Aromatic, the following conditions must be satisfied
-The compound should be cyclic
-The molecular geometry of the compound should be planar
-The compound should exhibit Resonance
-The compound should obey Huckle’s rule, and hence should have (4n) \[\pi \]electrons
3.If a compound does not have a continuous ring of conjugated p orbitals in a planar conformation, then it is nonaromatic

One important question you must be having is how we identify \[\pi \]electrons. For that, remember the following points. \[\pi \]electrons lie in p orbitals and \[s{p^2}\]hybridized orbitals, each have 1 p orbital. In a completely conjugated molecule, each carbon atom is \[s{p^2}\]hybridized, and hence the electrons present in these p orbitals are \[\pi \]electrons.
As for the compounds given to us,

Compound (I) satisfies the following conditions:
1.It is a cyclic compound
2.Its molecular geometry is planar
3.But it does not exhibit resonance

Shifting of electrons would bring us back to the same structure.
On the other hand, Compound (II) satisfies the following conditions:
1.It is a cyclic compound
2.Its molecular geometry is planar
3.It does exhibit resonance

This compound exhibits resonance in both directions.
And it has 4\[\pi \]electrons, which satisfies Huckle’s rule with the value of n=1.
Hence, we can conclude that Compound (I) is Non-Aromatic and Compound (II) is Anti-Aromatic.
-We know that, the trend in the stability of Aromatic, Non-Aromatic and Anti Aromatic compounds is:
Aromatic > Non-Aromatic > Anti-Aromatic
Hence, Anti aromatic compounds are less stable than Non-aromatic Compounds. This means that the conjugate bases of Anti aromatic compounds are more stable than Non-aromatic Compounds. And we know that the increasing stability of the conjugate base increases the acidity of the compound.
Hence Anti aromatic compounds are less acidic than Non-aromatic Compounds.
\[p{K_a}\] is a method used to indicate the strength of an acid. The lower the value of \[p{K_a}\], higher is the strength of the acid.
Now, relating the above conclusions to the given compounds, we get,
\[p{K_a}\] of Compound(I) < \[p{K_a}\]of Compound (II)
 Hence, both the reason and assertion are true and are interdependent.


Option A is the correct

Note:
You should not fret over finding the value of ‘n’ in Huckle’s rule because ‘n’ is not a characteristic property of any molecule. It is merely a constant which can be substituted with any natural number to satisfy the Huckel's Rule.