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The strongest acid among the following is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}\) (c) \(\mathrm{m}-\mathrm{CH}_{3} \mathrm{O} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{COOH}\) (d) \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH}\)

Short Answer

Expert verified
Answer: The strongest acid among the given options is (d) para-methoxybenzoic acid.

Step by step solution

01

Identify the given acids

The given acids are: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\): Benzoic acid (b) \(\mathrm{CH}_{3} \mathrm{COOH}\): Acetic acid (c) \(\mathrm{m}-\mathrm{CH}_{3} \mathrm{O} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{COOH}\): meta-methoxybenzoic acid (d) \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH}\): para-methoxybenzoic acid
02

Analyze the effect of attached groups on the acidity of carboxylic acids

The acidity of carboxylic acids is majorly affected by the electron-donating or electron-withdrawing nature of the groups attached to the phenyl ring. (a) In benzoic acid, no additional groups are attached to the phenyl ring. (b) In acetic acid, an electron-donating \(\mathrm{CH}_3\) group redirects electron density towards the carboxyl group. (c) In meta-methoxybenzoic acid, an electron-donating \(\mathrm{OCH}_3\) group is attached at the meta position of the phenyl ring. The resonance effect of this group is not significant, but it has some inductive effect. (d) In para-methoxybenzoic acid, an electron-donating \(\mathrm{OCH}_3\) group is attached at the para position of the phenyl ring. This group has a resonance effect in the molecule and redirects electron density towards the carboxyl group.
03

Compare the acidity based on the effect of the attached groups

As mentioned earlier, the more stable the resulting conjugate base after loss of \(\mathrm{H}^+\), the stronger the acid. The electron-donating groups like \(\mathrm{CH}_3\) and \(\mathrm{OCH}_3\) stabilize the conjugate base through a positive inductive effect (+I effect) causing an increase in the acidity of the molecules. Since the resonance effect of \(\mathrm{OCH}_3\) at the para position is more significant, the acidity will be more for the para-methoxybenzoic acid (d). Therefore, the strongest acid among the given options is: (d) \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{OC}_{6} \mathrm{H}_{4} \mathrm{COOH}\) (para-methoxybenzoic acid).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Benzoic Acid
Benzoic acid, with the chemical formula \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} \), is a simple carboxylic acid where a single carboxyl group \( -COOH \) is attached to a benzene ring. This aromatic carboxylic acid serves as a benchmark when comparing the acidity of different carboxylic acids.

The acidity of benzoic acid can be attributed to the ability of the carboxylate ion, formed upon the loss of a proton, to delocalize the negative charge over the aromatic ring through resonance stabilization. Since benzoic acid doesn't have any electron-donating or withdrawing groups directly attached to the benzene ring, its acidity serves as a standard when other substituent effects are taken into account.

Electron-Donating and Withdrawing Groups
In the context of carboxylic acids, electron-donating and withdrawing groups have a significant impact on acidity. Electron-donating groups, like the methyl \(\mathrm{CH}_3\) and methoxy \(\mathrm{OCH}_3\) groups, donate electron density through their atoms' p-orbitals or lone pairs. These groups typically reduce the acidity of carboxylic acids by destabilizing the conjugate base.

Conversely, electron-withdrawing groups, such as nitro \(\mathrm{NO}_2\) groups or halogens, attract electron density away from the carboxylic acid, stabilizing the conjugate base and increasing acidity. The location of these groups on the aromatic ring also plays a role, with the meta position offering less resonance interaction than the ortho or para positions, which are directly connected to the ring's conjugated system.
Resonance Effect in Acidity
The resonance effect significantly influences the acidity of aromatic carboxylic acids. It stabilizes the carboxylate anion formed when the acid donates a proton. For example, when a methoxy group \(\mathrm{OCH}_3\) is at the para position of the benzene ring, as seen with para-methoxybenzoic acid, it can engage in resonance with the ring, distributing the negative charge after deprotonation.

Because of the resonance stabilization, the negative charge is less localized and the conjugate base is more stable, making the carboxylic acid more acidic. This is a vital concept for understanding why substituents in different positions on the ring can lead to varying levels of acidity — with para positions often having a stronger effect due to extended resonance possibilities.
Inductive Effect on Acidity
The inductive effect is another crucial factor affecting carboxylic acid acidity. It involves the shifting of electron density through sigma bonds due to the electronegativity differences between atoms. Electron-withdrawing groups exert a negative inductive effect \( -I \), pulling electron density away from the carboxylic group and stabilizing the conjugate base, thus increasing acidity.

On the other hand, electron-donating groups exert a positive inductive effect \( +I \), which can destabilize the conjugate base by pushing electron density towards it, decreasing the acidity of the carboxylic acid.

While this effect is felt through bonds and diminishes with increasing distance from the acidic group, it's important to remember that the inductive effect is always present alongside resonance effects. This dual influence helps explain why different substituents on the benzene ring can alter the acidity of compounds like benzoic acid. In fact, the interplay between resonance and inductive effects is key to predicting the relative acid strengths of aromatic carboxylic acids.

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Most popular questions from this chapter

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Cd} \rightarrow\) product \((\mathrm{s})\). The major organic product obtained is (a) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Which of the following halides will not give good yield of the corresponding acid by the conversion to cyanide, followed by hydrolysis? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) (d) CC1CCCCC1

Eventhough a carboxylic acid contains \(-\mathrm{C}\) - group, it does not form oxime with hydroxyl amine because (a) hydroxyl amine is less reactive to carboxylic acids. (b) in this case hydroxyl amine acts as a reducing agent, oxidized to \(\mathrm{HNO}_{2}\). (c) the \(\sum \mathrm{C}=\mathrm{O}\) group enters into resonance with lone pairs of \(-\mathrm{OH}\) group. (d) the -OH group of carboxylic acid reacts with \(\mathrm{NH}_{2} \mathrm{OH}\).

Directions: Each question contains Statement- 1 and Statement- 2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement- 1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement- 1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Formic acid produces a white precipitate with \(\mathrm{HgCl}_{2}\). and Statement 2 Mercuric salt of formic acid is insoluble and therefore gets precipitated.

Statement 1 All aldehydes and ketones react with HCN to give a racemic mixture of the corresponding cyanohydrin. and Statement 2 Cyanide ion is an ambident nucleophile.

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