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Chapter 17: Problem 1

Other groups in addition to carbonyl groups enhance the acidities of adjacent \(\mathrm{C}-\mathrm{H}\) bonds. For instance, nitromethane, \(\mathrm{CH}_{3} \mathrm{NO}_{2}\), has \(\mathrm{p} K_{a}=10\); ethanenitrile, \(\mathrm{CH}_{3} \mathrm{CN}\), has a \(\mathrm{p} K_{a} \cong 25 .\) Explain why these compounds behave as weak acids. Why is \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) a stronger acid than \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3} ?\)

Short Answer

Expert verified
Nitromethane and acetone are stronger acids due to electron-withdrawing groups stabilizing their conjugate bases.

Step by step solution

01

Understanding Acid Strength

The strength of an acid is determined by its ability to donate a proton (H鈦�) and is generally presented as its pK鈧� value. The lower the pK鈧�, the stronger the acid. In the case of organic molecules, electron-withdrawing groups increase acidity by stabilizing the negative charge on the conjugate base after the proton donation.
02

Analyzing Nitromethane and Ethanenitrile

Nitromethane ( CH鈧僋O鈧� ) has a pK鈧� of 10, indicating it is a relatively stronger acid compared to ethanenitrile ( CH鈧僀N ) with a pK鈧� of about 25. The nitro group ( NO鈧� ) in nitromethane is a strong electron-withdrawing group, which stabilizes the negative charge on the conjugate base more effectively than the cyano group ( CN ) in ethanenitrile, leading to increased acidity.
03

Comparing Acetone and Methyl Acetate

Acetone ( CH鈧僀OCH鈧� ) has a carbonyl group ( C=O ), which is more effective at stabilizing a negative charge compared to methyl acetate ( CH鈧僀O鈧侰H鈧� ). In methyl acetate, the ester group donates electron density through resonance, making the compound less acidic than acetone. Therefore, the conjugate base of acetone is more stable, leading to stronger acidity.
04

Conclusion

The presence of electron-withdrawing groups in molecules like nitromethane and acetone stabilizes the conjugate base, enhancing acidity. Thus, acetone is a stronger acid than methyl acetate due to the resonance stabilization by the carbonyl group.

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Key Concepts

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

Electron-Withdrawing Groups
In organic chemistry, the electron-withdrawing groups play a pivotal role in determining the acidity of molecules. These groups tend to pull electron density away from other parts of the molecule. This ability to attract electrons enhances the overall acidity of a compound.
How They Affect Acidity:
- Electron-withdrawing groups stabilize the negative charge that forms on the conjugate base after a proton is lost.- This stabilization makes it easier for the molecule to give up a proton, thereby increasing acidity.
Examples:
- The nitro group (\(\mathrm{NO}_{2}\)) is a potent electron-withdrawing group that significantly increases acidity.- The cyano group (\(\mathrm{CN}\)) is another such group, although it is not as strong as the nitro group, which reflects in the acidity difference between nitromethane and ethanenitrile.
pKa Values
The \(\mathrm{pK}_{a}\) value of a substance is a numerical representation of its acidity. A low \(\mathrm{pK}_{a}\) indicates a strong acid, while a higher value suggests a weaker acid.
Why \(\mathrm{pK}_{a}\) Matters:
- It allows chemists to compare the relative acidity of different molecules.- Understanding \(\mathrm{pK}_{a}\) values helps predict chemical behavior, such as how a molecule might react in a biological system or industrial process.
In the context of the problem, nitromethane with a \(\mathrm{pK}_{a}\) of 10 is a stronger acid than ethanenitrile with \(\mathrm{pK}_{a}\approx 25\). This stark difference is mainly due to how well the conjugate base is stabilized by electron-withdrawing groups.
Conjugate Base Stabilization
After an acid donates a proton, it forms a conjugate base. The stability of this conjugate base is crucial in determining the strength of the original acid.
Key Points of Stabilization:
- A more stable conjugate base means a stronger parent acid. - Stability can be achieved through electron-withdrawing groups which delocalize the negative charge.
In the example of acetone and methyl acetate, acetone has a more stable conjugate base due to the presence of its carbonyl group. This makes acetone a stronger acid compared to methyl acetate.
Resonance Effects
Resonance effects influence acid strength through the delocalization of electrons. When electrons are delocalized, the molecule achieves greater stability by distributing charge over multiple atoms.
Impact on Acidity:
- Acids with conjugate bases that benefit from resonance are generally stronger. - Resonance can either increase or decrease acid strength based on whether it involves electron donation or withdrawal.
For instance, while the carbonyl group in acetone supports resonance that stabilizes the conjugate base, the ester group in methyl acetate donates electron density, reducing resonance's stabilizing effect and therefore, decreasing acidity.
Carbonyl Group
The carbonyl group (\(\mathrm{C}=\mathrm{O}\)) is fundamental in organic acids, significantly affecting their acidity. Known for its strong electron-withdrawing nature, it is often central to the stabilization of a conjugate base.Advantages of the Carbonyl Group:
- It stabilizes adjacent negative charges, which is crucial after the loss of a proton.- By polarizing and withdrawing electrons, it increases the likelihood of proton donation.This explains the stronger acidity of acetone compared to methyl acetate. In acetone, the carbonyl group's influence is unopposed, leading to a more stable conjugate base. Conversely, in methyl acetate, electron density is slightly pushed back into the molecule, thus reducing acidity.

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Most popular questions from this chapter

What features of the base-catalyzed dehydration of 3 -hydroxybutanal make it a more favorable and faster reaction than would be expected for a base-catalyzed dehydration of 2 -butanol? Give your reasoning.

On what basis can you account for the fact that \(\mathrm{HCN}\) adds to the carbonyl group of 3 -butenal and to the double bond of 3 -buten-2-one? Would you expect the carbonyl or the double-bond addition product of \(\mathrm{HCN}\) to 3 -buten- 2 -one to be more thermodynamically favorable? Give your reasoning.

A detailed study of the rate of bromination of 2 -propanone in water, in the presence of ethanoic acid and ethanoate \(\quad\) ions, \(\quad\) has \(\quad\) shown \(\quad\) that \(v=\left\\{6 \times 10^{-9}+5.6 \times 10^{-4}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+1.3 \times 10^{-7}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]+7\left[\mathrm{OH}^{-}\right]+3.3 \times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]+3.5 \quad\right.\) in which \(\left.\times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\right\\}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\) the rate \(v\) is expressed in \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{sec}^{-1}\) when the concentrations are in \(\mathrm{mol} \mathrm{L}^{-1}\). a. Calculate the rate of the reaction for a 1 M solution of 2 -propanone in water at \(\mathrm{pH} 7\) in the absence of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) b. Calculate the rate of the reaction for \(1 \mathrm{M} 2\) -propanone in a solution made by neutralizing \(1 \mathrm{M}\) ethanoic acid with sufficient sodium hydroxide to give \(\mathrm{pH} 5.0\) ( \(K_{a}\) of ethanoic acid \(=1.75 \times 10^{-5}\) ). c. Explain how the numerical values of the coefficients for the rate equation may be obtained from observations of the reaction at various \(\mathrm{pH}\) values and ethanoate ion concentrations. d. The equilibrium concentration of enol in 2 -propanone is estimated to be \(\sim 1.5 \times 10^{-4} \% .\) If the rate of conversion of \(1 \mathrm{M}\) 2-propanone to enol at \(\mathrm{pH} 7\) (no \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) present) is as calculated in Part a, calculate the rate of the reverse reaction from enol to ketone at \(\mathrm{pH} 7\) if the enol were present in \(1 \mathrm{M}\) concentration. e. Suggest a mechanistic explanation for the term \(3.5 \times 10^{-6}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\) in the rate expression.

Predict the principal products to be expected in each of the following reactions; give your reasoning: a. \(\mathrm{CH}_{3} \mathrm{CHO}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) b. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CHCOCH}_{3} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) c. \(\mathrm{CH}_{2} \mathrm{O}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCHO} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) d. \(\mathrm{CH}_{2} \mathrm{O}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCHO} \stackrel{\mathrm{Ca}(\mathrm{OH})_{2}}{\longrightarrow}\)

a. Explain why 2 -butanone is halogenated preferentially on the ethyl side with an acidic catalyst. (Review of Section \(11-\) 3 should be helpful.) b. What product would predominate in the acid-catalyzed bromination of 1 -phenyl-2-propanone? Give your reasoning.
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