Question

Arrange the following compounds in increasing order of their acidic strengths: \[ CH_3CH_2COOH,\quad CH_3CH(CH_3)COOH,\quad CH_3CH_2CH_2COOH,\quad BrCH_2CH_2COOH \]

Hint:

−I group (like halogen) increases acidity. +I alkyl groups decrease acidity.

Answer 1

Step 1: Understand the Factors Affecting Acidity.
The acidic strength of carboxylic acids is influenced by the stability of the conjugate base (carboxylate anion, \( \text{RCOO}^- \)). The more stable the conjugate base, the stronger the acid. Substituents on the carbon chain can either stabilize or destabilize the conjugate base, thereby affecting the acidity.
- Electron-withdrawing groups (such as \( \text{Br} \)) stabilize the negative charge on the conjugate base, increasing the acid strength.
- Electron-donating groups (such as \( \text{CH}_3 \)) destabilize the conjugate base, decreasing the acid strength.

Step 2: Analyze the Given Compounds.
1. \( CH_3CH_2COOH \) (Propanoic acid): This is a simple carboxylic acid with no electron-withdrawing or electron-donating groups attached to the carbon chain.
2. \( CH_3CH(CH_3)COOH \) (Isopropanoic acid): This compound has a methyl group (\( \text{CH}_3 \)) attached to the carbon adjacent to the carboxyl group. The methyl group is electron-donating, making this acid less acidic than propanoic acid.
3. \( CH_3CH_2CH_2COOH \) (Butanoic acid): This compound has a longer alkyl chain than propanoic acid, and similar to propanoic acid, it does not have any significant electron-withdrawing groups. Longer alkyl chains usually make acids weaker by donating electrons, which destabilizes the conjugate base.
4. \( BrCH_2CH_2COOH \) (2-Bromopropanoic acid): The bromine atom (\( \text{Br} \)) is an electron-withdrawing group, which stabilizes the conjugate base, making this acid the strongest of the four.

Step 3: Arranging in Increasing Order of Acidity.
- The strongest acid is \( \text{BrCH}_2\text{CH}_2\text{COOH} \) due to the electron-withdrawing effect of the bromine atom.
- Next is \( \text{CH}_3\text{CH}_2\text{COOH} \), followed by \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \), which is slightly less acidic due to the longer alkyl chain donating electrons.
- The weakest acid is \( \text{CH}_3\text{CH(CH}_3)\text{COOH} \), as the methyl group is electron-donating.

Conclusion.
The increasing order of acidic strength is:
\[ CH_3CH_2CH_2COOH < CH_3CH_2COOH < CH_3CH(CH_3)COOH < BrCH_2CH_2COOH \]