Question

For each of the following pairs tell which acid is the stronger and why: (a) ${\mathrm{CH}}_3-\mathrm{COOH}$ or $\mathrm{H}-\mathrm{COOH}$ (b) ${\mathrm{CH}}_3-\mathrm{COOH}$ or $\mathrm{CL}-{\mathrm{CH}}_2-\mathrm{COOH}$ (c) $\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$ or $\mathrm{Br}-{\mathrm{CH}}_2-\mathrm{COOH}$ (d) $\mathrm{CL}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{COOH}$ or ${\mathrm{CH}}_3-\mathrm{CHCL}-\mathrm{COOH}$ (e) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$ or ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ (f) ${\mathrm{CH}}_3-\mathrm{OH}$ or ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}$

Short answer

(a) $\mathrm{H}-\mathrm{COOH}$ is stronger because carbon is less electronegative than hydrogen. (b) $\mathrm{Cl}-{\mathrm{CH}}_2-\mathrm{COOH}$ is stronger due to the inductive effect of chlorine. (c) $\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$ is stronger, due to higher electronegativity of fluorine. (d) ${\mathrm{CH}}_3-\mathrm{CHCl}-\mathrm{COOH}$ is stronger because chlorine is closer to the acidic proton. (e) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$ (phenol) is stronger than ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ (ethanol) due to resonance stabilization. (f) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}$ (propanoic acid) is stronger than ${\mathrm{CH}}_3-\mathrm{OH}$ (methanol) as organic acids are generally stronger than alcohols.

Step-by-step solution

a) Comparing ${\mathrm{CH}}_3-\mathrm{COOH}$ and $\mathrm{H}-\mathrm{COOH}$

Let's compare the electronegativity of carbon in ${\mathrm{CH}}_3$ and hydrogen in $\mathrm{H}$. Carbon has a much lower electronegativity compared to hydrogen which means it has a weaker stabilizing effect on the negative charge resulting from lost protons. Thus, ${\mathrm{CH}}_3-\mathrm{COOH}$ is a weaker acid compared to $\mathrm{H}-\mathrm{COOH}$.

b) Comparing ${\mathrm{CH}}_3-\mathrm{COOH}$ and $\mathrm{Cl}-{\mathrm{CH}}_2-\mathrm{COOH}$

In this case, the inductive effect of chlorine plays a significant role in acid strength. Chlorine is more electronegative than hydrogen, which helps in stabilizing the negative charge resulting from the loss of a proton. As a result, $\mathrm{Cl}-{\mathrm{CH}}_2-\mathrm{COOH}$ is a stronger acid than ${\mathrm{CH}}_3-\mathrm{COOH}$.

c) Comparing $\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$ and $\mathrm{Br}-{\mathrm{CH}}_2-\mathrm{COOH}$

Both fluorine and bromine have higher electronegativities compared to carbon. However, fluorine is the most electronegative element among them. This electronegativity stabilizes the negative charge better when the proton is lost, which leads to $\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$ being a stronger acid than $\mathrm{Br}-{\mathrm{CH}}_2-\mathrm{COOH}$.

d) Comparing $\mathrm{Cl}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{COOH}$ and ${\mathrm{CH}}_3-\mathrm{CHCl}-\mathrm{COOH}$

Here, we have two halogens attached to different positions of the acid. In the first case, $\mathrm{Cl}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{COOH}$, chlorine is further away from the acidic proton, leading to a weaker inductive effect. On the other hand, in ${\mathrm{CH}}_3-\mathrm{CHCl}-\mathrm{COOH}$, chlorine is closer the acidic proton, leading to a stronger inductive effect and stabilizing the negative charge. Thus, ${\mathrm{CH}}_3-\mathrm{CHCl}-\mathrm{COOH}$ is a stronger acid compared to $\mathrm{Cl}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{COOH}$.

e) Comparing ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$ and ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$

In the first case, ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$ has a phenoxide ion, which is stabilized through the resonance effect within the benzene ring (the electrons from the hydrogen bond can delocalize in the ring). In contrast, there is no such resonance stabilization in ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$. Therefore, ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$ (phenol) is a much stronger acid than ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$ (ethanol).

f) Comparing ${\mathrm{CH}}_3-\mathrm{OH}$ and ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}$

Methanol (${\mathrm{CH}}_3-\mathrm{OH}$) is an alcohol, while the other molecule is an organic acid. Generally, organic acids have a higher acidity than alcohols due to the stability of the resultant carboxylate anion. In this case, propanoic acid (${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}$) is a much stronger acid than methanol (${\mathrm{CH}}_3-\mathrm{OH}$).

Key concepts

Inductive Effect

Understanding the inductive effect is crucial when comparing the strengths of acids in organic chemistry. It involves the transmission of charge through a chain of atoms in a molecule, thanks to differences in electronegativity. For instance, when a substituent atom such as chlorine is bonded to a carbon atom near a carboxylic acid group, it pulls electron density towards itself, stabilizing the negative charge that results when the acid donates a proton.

This effect diminishes with distance, so atoms closer to the acidic hydrogen have a more pronounced effect on acid strength. The inductive effect is why \texttt{$\mathrm{Cl}-{\mathrm{CH}}_2-\mathrm{COOH}$} is a stronger acid than \texttt{${\mathrm{CH}}_3-\mathrm{COOH}$}, as chlorine's electronegativity helps stabilize the conjugate base, making proton loss easier.

Electronegativity

Electronegativity is the tendency of an atom to attract electron density towards itself in a chemical bond. It's a central concept when discussing acid strength in organic molecules. Typically, higher electronegativity correlates with stronger acids because the more electronegative atom can better stabilize negative charges.

For example, in comparing \texttt{$\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$} and \texttt{$\mathrm{Br}-{\mathrm{CH}}_2-\mathrm{COOH}$}, fluorine's higher electronegativity makes \texttt{$\mathrm{F}-{\mathrm{CH}}_2-\mathrm{COOH}$} a stronger acid, because it stabilizes the anion that forms when the molecule loses a proton.

Resonance Stabilization

Resonance stabilization is key to understanding why some organic acids are stronger than others. It's the delocalization of electrons in a molecule that can have multiple valid Lewis structures, or resonance forms. This delocalization distributes the negative charge over the entire molecule, increasing stability.

A prime example is phenol (\texttt{${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}$}), where the negative charge of the deprotonated phenoxide ion is spread over the aromatic ring through resonance. This effect is why phenol is a stronger acid than ethanol (\texttt{${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{OH}$}), which lacks such resonance stabilization.

Carboxylic Acids

Carboxylic acids are a class of organic acids characterized by their \texttt{$-COOH$} group. Their acidity is higher than alcohols and many other organic compounds because the carboxylate ion (the anion that forms when a carboxylic acid loses a proton) is resonance stabilized.

This resonance allows the negative charge to be distributed between the two oxygen atoms, lowering the energy of the system and stabilizing the anion. Due to this stability, carboxylic acids like propanoic acid (\texttt{${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}$}) readily lose their proton, making them stronger acids than alcohols, such as methanol (\texttt{${\mathrm{CH}}_3-\mathrm{OH}$}).

Organic Chemistry Acids and Bases

In organic chemistry, acid and base strength is interpreted through the lens of resonance, inductive effect, and electronegativity. Acids are substances that can donate a proton, while bases can accept a proton. Acid strength is often gauged by the stability of the conjugate base that forms after a proton is lost; more stable conjugate bases derive from stronger acids.

By analyzing factors such as the distance of electronegative atoms from the acidic proton (inductive effect), the ability of substituents to delocalize charges (resonance stabilization), and the intrinsic tendency of an atom to attract electrons (electronegativity), students can predict and explain the relative strengths of acids and bases in organic compounds.