Q.Arrange the following in decreasing order of their acidic strength and give reason for your answer.
$CH_3CH_2OH$, $CH_3COOH$, $ClCH_2COOH$, $FCH_2COOH$, $C_6H_5CH_2COOH$

Concept: Acidity of organic compounds depends on the stability of the conjugate base after losing $H^+$. More stable the conjugate base, stronger the acid.

Step 1 – Compare functional groups. Carboxylic acids ($-COOH$) are far more acidic than alcohols ($-OH$) because the carboxylate anion is resonance-stabilised. So $CH_3CH_2OH$ is the weakest.

Step 2 – Effect of substituents on carboxylic acids. Electron-withdrawing groups (like halogens) on the $\alpha$-carbon stabilise the conjugate base by inductive effect, increasing acidity. More electronegative the halogen, stronger the inductive effect: $F > Cl$. …

Why This Approach Works

Acidic strength is all about how easily a compound gives up a proton ($H^+$). The easier it is to lose $H^+$, the stronger the acid. But the real story is in what’s left behind — the conjugate base. A more stable conjugate base means a stronger acid.

For organic acids, two major forces govern this stability:

• Inductive effect: Electron-withdrawing groups (like halogens) pull electron density away from the negative charge on the conjugate base, stabilizing it. The stronger the electron-withdrawing effect, the stronger the acid.
• Resonance effect: Delocalization of the negative charge over multiple atoms (like in a carboxylate ion) dramatically stabilizes the conjugate base.

Alcohols ($ROH$) lack resonance stabilization for their conjugate base ($RO^-$), making them very weak acids. Carboxylic acids ($RCOOH$) have resonance-stabilized carboxylate ions ($RCOO^-$), making them much stronger. And when you add electron-withdrawing groups near the carboxyl group, you boost acidity further.

Let’s rank these five compounds step by step.

1. Identify the functional groups and their inherent acidity

   $CH_3CH_2OH$ is an alcohol. Its conjugate base ($CH_3CH_2O^-$) has the negative charge localized on oxygen — no resonance stabilization. Alcohols typically have $pK_a \approx 16$.

   The other four are all carboxylic acids ($RCOOH$). Their conjugate bases ($RCOO^-$) are resonance-stabilized: the negative charge is delocalized over two oxygen atoms. Carboxylic acids typically have $pK_a \approx 4–5$.

   So $CH_3CH_2OH$ is by far the weakest acid here. It goes last.

2. Compare the unsubstituted carboxylic acid: $CH_3COOH$

   Acetic acid has a methyl group ($CH_3$), which is weakly electron-donating (through hyperconjugation and inductive effect). This slightly destabilizes the conjugate base (pushes electron density toward the already negative carboxylate). Its $pK_a$ is 4.76.

   This will be the weakest among the carboxylic acids in this list — but still far stronger than the alcohol.

3. Introduce halogen substitution: $ClCH_2COOH$ and $FCH_2COOH$

   Halogens are electron-withdrawing by inductive effect. They pull electron density away from the carboxylate group, stabilizing the negative charge on the conjugate base. This makes the acid stronger.

   The strength of the inductive effect depends on electronegativity and distance. Here, both halogens are on the $\alpha$-carbon (one bond away), so distance is the same. The key difference is electronegativity:

   • Fluorine (electronegativity ≈ 4.0) is more electronegative than chlorine (≈ 3.0).
   • So $FCH_2COOH$ has a stronger electron-withdrawing effect than $ClCH_2COOH$. Therefore, $FCH_2COOH$ is a stronger acid than $ClCH_2COOH$.
   Tip

   The inductive effect of halogens decreases as: $F > Cl > Br > I$. This matches their electronegativity order. For $\alpha$-halo acids, the $pK_a$ values are: $FCH_2COOH$ (2.66), $ClCH_2COOH$ (2.87), $BrCH_2COOH$ (2.90), $ICH_2COOH$ (3.18). So F > Cl holds clearly.

4. Place $C_6H_5CH_2COOH$ (phenylacetic acid) …

Method: Inductive Effect & Resonance-Based Acidity Comparison

Step 1 – Identify the functional group

All compounds have a carboxyl group ($-COOH$) except ethanol ($CH_3CH_2OH$), which has a hydroxyl group ($-OH$).

Carboxylic acids are stronger acids than alcohols because the carboxylate anion ($RCOO^-$) is stabilised by resonance, while alkoxide ($RO^-$) is not.

Result: $CH_3CH_2OH$ is the weakest acid — place it last.

Step 2 – Compare the carboxylic acids using inductive effect

The acidity of $R-COOH$ increases when the R-group withdraws electrons (stabilises the conjugate base).

• Electron-withdrawing groups (EWG) like $F$, $Cl$ increase acidity.
• Electron-donating groups (EDG) like $CH_3$ decrease acidity.
• Phenyl group ($C_6H_5-$) is weakly electron-withdrawing due to its −I effect (inductive withdrawal), but less than halogens.

Step 3 – Arrange the EWGs by strength

Fluorine is more electronegative than chlorine → $FCH_2COOH$ is stronger than $ClCH_2COOH$.

Order of inductive withdrawal:

$F > Cl > C_6H_5 > CH_3$ (where $CH_3$ is actually electron-donating)

Step 4 – Final decreasing order of acidic strength …

Common Mistakes Students Make on Acidic Strength Ordering (and How to Avoid Them)

Mistake #1: Confusing Alcohols with Carboxylic Acids

The error: Students often place $CH_3CH_2OH$ (ethanol) somewhere in the middle, thinking it has comparable acidity to weak carboxylic acids.

Why it’s wrong: Alcohols are extremely weak acids compared to carboxylic acids. The $pK_a$ of ethanol is ~16, while even the weakest carboxylic acid here ($CH_3COOH$) has $pK_a \approx 4.76$.

How to avoid: Remember the resonance stabilization rule — the conjugate base of a carboxylic acid ($RCOO^-$) is stabilized by resonance between two equivalent oxygen atoms. The alkoxide ion ($RO^-$) has no resonance. This makes carboxylic acids ~10¹² times stronger than alcohols.

Key takeaway: Alcohols always come last in acidic strength order among these compounds.

Mistake #2: Ignoring Inductive Effect of Halogens

The error: Students rank $ClCH_2COOH$ and $FCH_2COOH$ incorrectly, often placing chloroacetic acid above fluoroacetic acid because chlorine is "more electronegative" in general chemistry memory.

Why it’s wrong: Fluorine is the most electronegative element ($\chi = 4.0$ vs $Cl = 3.0$). The stronger the $-I$ effect (electron-withdrawing inductive effect), the more the negative charge on the conjugate base is stabilized, making the acid stronger.

How to avoid: Always compare electronegativity of the substituent directly attached to the carbon chain. For halogens:

• $F > Cl > Br > I$ in terms of inductive effect strength
• So $FCH_2COOH$ is stronger than $ClCH_2COOH$

Correct order: $FCH_2COOH > ClCH_2COOH$

Mistake #3: Forgetting the Distance Effect of the Benzyl Group

The error: Students think $C_6H_5CH_2COOH$ (phenylacetic acid) is stronger than acetic acid because "benzene ring is electron-withdrawing."

Why it’s wrong: The benzene ring is attached via a $CH_2$ spacer. Inductive effects diminish rapidly with distance. The $-I$ effect of the phenyl group is weak through one carbon. In fact, phenylacetic acid ($pK_a \approx 4.31$) is only slightly stronger than acetic acid ($pK_a \approx 4.76$).

How to avoid: Draw the structure. Count the number of bonds between the substituent and the $-COOH$ group. The inductive effect is significant only when the substituent is directly attached or at most one carbon away.

Mistake #4: Mixing Up Inductive and Resonance Effects

The error: Students try to apply resonance stabilization to $C_6H_5CH_2COOH$ and conclude it's very strong.

Why it’s wrong: The benzene ring cannot directly conjugate with the carboxylate anion because of the $CH_2$ spacer. There is no resonance between the ring and the $-COO^-$ group.

How to avoid: Check if the substituent is directly attached to the carboxyl carbon or the alpha carbon. Only then can resonance operate. Here, the $CH_2$ group breaks conjugation.

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