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Chapter 1: Problem 4

Account for the fact that carboxylic acids are more acidic compared to an alcohol.

Short Answer

Expert verified
Answer: Carboxylic acids are more acidic compared to alcohols because their conjugate bases (carboxylate ions) are more stable due to resonance stabilization. This stabilization allows the carboxylic acid to donate a proton more readily than an alcohol, resulting in a higher acidity.

Step by step solution

01

Understanding Acids and Bases

According to the Brønsted-Lowry theory, an acid is a substance that donates a proton (H+) to a base, and a base is a substance that accepts a proton. The stronger the acid, the more readily it donates a proton. Similarly, the more stable the conjugate base (the species formed after the acid donates its proton), the stronger the acid will be.
02

Analyzing Alcohol's Structure

Alcohol is an organic compound having a hydroxyl (OH) group bonded to a carbon atom. The general formula of alcohol is R-OH, where R represents an alkyl or substituted alkyl group. In the case of alcohol, the acidic proton is attached to the oxygen atom of the hydroxyl group.
03

Analyzing Carboxylic Acid's Structure

Carboxylic acid is an organic compound having a carbonyl (C=O) group and a hydroxyl group attached to the same carbon atom, generally represented as R-COOH. The acidic proton in carboxylic acids is attached to the oxygen atom of the hydroxyl group, just like in alcohols.
04

Comparing the Stability of Conjugate Bases

When alcohols and carboxylic acids lose a proton, they form their respective conjugate bases. Alcohols form alkoxide ions (R-O^{-}), while carboxylic acids form carboxylate ions (R-COO^{-}). The stability of these conjugate bases plays a significant role in determining the acidity of the parent compound. Generally, the more stable the conjugate base, the more acidic the parent compound.
05

Resonance Stabilization of Carboxylate Ions

The carboxylate ion (R-COO^{-}) formed after a carboxylic acid loses a proton is resonance stabilized. Resonance stabilization means that the negative charge on the ion is delocalized between the two oxygen atoms through electron movement, which allows distribution and stabilization of the negative charge. In contrast, the negative charge on alkoxide ions (R-O^{-}) remains localized on the oxygen atom and does not have any resonance structures. This lack of charge delocalization makes alkoxide ions less stable compared to carboxylate ions.
06

Conclusion

Carboxylic acids are more acidic compared to alcohols because their conjugate bases (carboxylate ions) are more stable due to resonance stabilization. This stabilization allows the carboxylic acid to donate a proton more readily than an alcohol, resulting in a higher acidity.

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Most popular questions from this chapter

Between \(\mathrm{CH}_{3} \mathrm{O} \mathrm{CH}_{2} \mathrm{CH}_{2}(\mathrm{I})\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O} \mathrm{CH}_{2}(\mathrm{II})\) (a) I is more stable because it is \(1^{\circ}\) carbocation. (b) II is more stable because it is \(1^{\circ}\) carbocation. (c) II is more stable because the cation is stabilized by the lone pair on oxygen. (d) Both are equally stable since both the cations are \(1^{\circ}\) and stabilized by oxygen.

A description that does not fit in with the application of resonance among the following is (a) Resonance is used to explain the stability of phenoxide ion. (b) Resonance is used to explain the poor reactivity of halogen in chlorobenzene. (c) Resonance is used to explain the deactivating influence of bromine in bromo benzene. (d) Resonance is used explain the neutral behaviour of benzamide.

The IUPAC name of \(\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CHCl}-\mathrm{CH}_{2}-\mathrm{COOH}\) is (a) 3-Chloro-4-oxopentanoic acid (b) 4-Oxo-3-Chloropentanoic acid (c) 2-Oxo-3-Chloropentanoic acid (d) 1 -Carboxy-3-Chloro-4-ketobutane

Identify the correct statement among the following (a) The minimum number of Carbon atoms in a branched alkane is three. b.is an unsaturated compound. (c)is known as ethane-1,2-dioic acid. (d) is known as 2 -ethylbutane.

Match the elements of Column I to elements of Column II. There can be single or multiple matches. Column I (a) Free radicals (b) Photochemical reaction (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}\) (d) No bond resonanceColumn II (p) halogenation of alkanes (q) homolytic cleavage (r) hyperconjugation (s) chain reaction
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