Aniline has a p Ka=4.63 ; 3 -bromoaniline has a pKa= 3.58; p Ka=-log(Ka). Which is the stronger

step 1 In this question given PQA of aniline is 4 .63 and PK of 3 bromalanine is 3 .58. We have to expl

Aniline has a p Ka=4.63 ; 3 -bromoaniline has a pKa= 3.58; p...

Key Concepts

Substituent Effects on Basicity

Substituents attached to the aromatic ring can significantly influence the basicity of aromatic amines by either donating or withdrawing electron density. Electron-withdrawing groups, such as the bromo group in 3-bromoaniline, decrease the electron density on the nitrogen atom, thereby reducing the ability of the amine to accept protons. This results in a lower pKa for its conjugate acid and a correspondingly weaker base compared to aniline without such substituents.

Conjugate Acid-Base Relationship

Every base has a conjugate acid, and there is an inverse relationship between the strength of a base and the acidity of its conjugate acid. In this case, the aniline and 3-bromoaniline are compared by examining the pKa values of their respective conjugate acids. A compound whose conjugate acid has a higher pKa is the stronger base because it holds onto its proton less tightly, making it less acidic.

pKa as a Measure of Acidity

The pKa value is the negative logarithm of the acid dissociation constant (Ka) and indicates the strength of an acid. A lower pKa value means that the compound is a stronger acid because it more readily donates protons. In the context of bases, the pKa of the conjugate acid is used: a higher pKa value for the conjugate acid of a base implies that the base is stronger because its conjugate acid is weaker and less likely to donate a proton back to the base.

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