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Question: Choose the option in which given pairs of compunds having $I^{st}$ stronger base than $II^{nd}$...

Choose the option in which given pairs of compunds having IstI^{st} stronger base than IIndII^{nd}

A

Aniline vs. p-Nitroaniline

B

CH2=CH−CH=CH−CH2−CH_2=CH-CH=CH-CH_2^-, CH2=CH−CH2−CH_2=CH-CH_2^-

C

HOOC−CH2−COO−HOOC-CH_2-COO^-, HOOC−CH2−COOHHOOC-CH_2-COOH

D

o-Cresol vs. o-(Trifluoromethyl)phenol

Answer

Aniline vs. p-Nitroaniline

Explanation

Solution

The basicity of a compound is determined by its ability to accept a proton, which is related to the availability of lone pair electrons and the stability of the conjugate acid formed. A stronger base has a more available lone pair and/or a less stable conjugate acid.

  • Option (A): Aniline vs. p-Nitroaniline. The nitro group (NO2NO_2) is a strong electron-withdrawing group. It withdraws electron density from the benzene ring and the amino group, making the lone pair on nitrogen less available for protonation. Thus, aniline is a stronger base than p-nitroaniline.

  • Option (B): CH2=CH−CH=CH−CH2−CH_2=CH-CH=CH-CH_2^- vs. CH2=CH−CH2−CH_2=CH-CH_2^-. These are carbanions. Basicity is inversely related to the stability of the carbanion. The first carbanion (CH2=CH−CH=CH−CH2−CH_2=CH-CH=CH-CH_2^-) is resonance stabilized over 3 carbon atoms (C1,C3,C5C_1, C_3, C_5). The second carbanion (CH2=CH−CH2−CH_2=CH-CH_2^-) is resonance stabilized over 2 carbon atoms (C1,C3C_1, C_3). Greater resonance stabilization leads to greater stability of the carbanion. Thus, CH2=CH−CH=CH−CH2−CH_2=CH-CH=CH-CH_2^- is more stable and hence a weaker base than CH2=CH−CH2−CH_2=CH-CH_2^-. So, the first compound is not stronger than the second.

  • Option (C): HOOC−CH2−COO−HOOC-CH_2-COO^- vs. HOOC−CH2−COOHHOOC-CH_2-COOH. The first compound is a carboxylate anion, which is a relatively strong base. The second compound is a carboxylic acid, which is a very weak base. Therefore, HOOC−CH2−COO−HOOC-CH_2-COO^- is a stronger base than HOOC−CH2−COOHHOOC-CH_2-COOH.

  • Option (D): o-Cresol vs. o-(Trifluoromethyl)phenol. These are phenols. Basicity of phenols is very low. The methyl group (CH3CH_3) in o-cresol is electron-donating, which increases electron density on the oxygen atom, making it a stronger base. The trifluoromethyl group (CF3CF_3) in o-(trifluoromethyl)phenol is a strong electron-withdrawing group, which decreases electron density on the oxygen atom, making it a weaker base. Thus, o-cresol is a stronger base than o-(trifluoromethyl)phenol.

Options A, C, and D all satisfy the condition that the first compound is a stronger base than the second. If this is a single-choice question, option A is a common example used to illustrate the effect of substituents on amine basicity.