Q.Which acid of each pair shown here would you expect to be stronger?

Concept: Inductive Effect – electron-withdrawing groups (EWGs) like halogens stabilise the conjugate base by pulling electron density away from the carboxylate anion, making the acid stronger. The effect is stronger with more electronegative atoms and when the EWG is closer to the –COOH group.

Step 1 – Compare $CH_3CO_2H$ and $CH_2FCO_2H$

Fluorine is highly electronegative and exerts a strong –I effect. Replacing a hydrogen on the α-carbon with F stabilises the conjugate base, so $CH_2FCO_2H$ is the stronger acid.

Step 2 – Compare $CH_2FCO_2H$ and $CH_2ClCO_2H$

Fluorine is more electronegative than chlorine, so the –I effect of F is stronger than that of Cl. Hence $CH_2FCO_2H$ is stronger than $CH_2ClCO_2H$.

Step 3 – Compare $CH_2FCH_2CH_2CO_2H$ and $CH_3CHFCH_2CO_2H$

In the first acid, F is on the γ-carbon (three bonds away from –COOH); in the second, F is on the β-carbon (two bonds away). The inductive effect diminishes rapidly with distance, so the closer F makes the second acid stronger. …

The Concept: Why Inductive Effect Decides Acidity

Acidity is about how easily a molecule gives up a proton ($H^+$). When an acid $HA$ loses $H^+$, it leaves behind its conjugate base $A^-$. The more stable that conjugate base is, the stronger the acid.

For carboxylic acids ($RCO_2H$), the conjugate base is a carboxylate ion ($RCO_2^-$), which already has the negative charge spread over two oxygen atoms. But if the group $R$ pulls electron density toward itself (an electron-withdrawing inductive effect), it further stabilises the negative charge on the carboxylate — making the acid stronger. If $R$ pushes electrons away (electron-donating), it destabilises the negative charge and makes the acid weaker.

Halogens (F, Cl, Br, I) are electron-withdrawing by induction because they are more electronegative than carbon. The strength of this effect depends on two things:

• Electronegativity of the halogen (F > Cl > Br > I)
• Distance from the carboxyl group — the closer the halogen, the stronger its pull.

Now let’s apply this to each pair.

(i) $CH_3CO_2H$ vs $CH_2FCO_2H$

1. In $CH_3CO_2H$, the group attached to the carboxyl is a methyl ($CH_3$). Methyl is weakly electron-donating (it pushes electron density toward the carboxylate), which destabilises the conjugate base slightly. So acetic acid is a weak acid ($pK_a \approx 4.76$).

2. In $CH_2FCO_2H$, one hydrogen on the methyl has been replaced by fluorine. Fluorine is highly electronegative — it pulls electron density away from the carboxyl group through the sigma bonds. This withdrawal stabilises the negative charge on the conjugate base, making the acid stronger.

3. The inductive effect of fluorine is strong and operates through only one carbon-carbon bond. So $CH_2FCO_2H$ is significantly more acidic than $CH_3CO_2H$.

Result for (i): $CH_2FCO_2H$ is stronger.

(ii) $CH_2FCO_2H$ vs $CH_2ClCO_2H$

1. Both acids have a halogen on the alpha carbon (the carbon directly attached to the carboxyl). The difference is the halogen: fluorine vs chlorine.

2. Fluorine is more electronegative than chlorine (4.0 vs 3.2 on the Pauling scale). So fluorine pulls electron density more strongly than chlorine does.

3. Even though chlorine is larger and has more electrons, its inductive effect is weaker because electronegativity is the dominant factor here. The distance from the carboxyl is the same (one bond away), so the comparison is purely about how strongly each halogen withdraws.

Result for (ii): $CH_2FCO_2H$ is stronger.

(iii) $CH_2FCH_2CH_2CO_2H$ vs $CH_3CHFCH_2CO_2H$

1. Here the acids are longer-chain carboxylic acids. Both have a fluorine atom, but at different positions relative to the carboxyl group. …

Method: Inductive Effect Analysis

Concept: The strength of a carboxylic acid depends on how well it can stabilise the conjugate base (the carboxylate anion, $RCOO^-$). Electron-withdrawing groups (EWGs) pull electron density away from the carboxylate ion, stabilising the negative charge and making the acid stronger. Electron-donating groups (EDGs) do the opposite — they destabilise the anion and make the acid weaker.

Steps for each pair

1. Identify the substituent attached to the carbon chain near the $-COOH$ group.
2. Determine its inductive effect:
   • More electronegative atom → stronger $-I$ effect (withdraws electrons).
   • Closer the EWG to $-COOH$ → stronger the effect.
3. Compare acid strength: stronger $-I$ effect → stronger acid.

(i) $CH_3CO_2H$ vs $CH_2FCO_2H$

• $CH_3CO_2H$: The $-CH_3$ group is weakly electron-donating ($+I$ effect).
• $CH_2FCO_2H$: Fluorine is highly electronegative — strong $-I$ effect, pulling electrons away from the carboxylate anion.
• Result: $CH_2FCO_2H$ is stronger than $CH_3CO_2H$.

Answer: $CH_2FCO_2H$ is stronger.

(ii) $CH_2FCO_2H$ vs $CH_2ClCO_2H$

• Both have a halogen at the same position (α-carbon).
• Electronegativity order: $F > Cl$.
• Inductive effect order: $-I$ effect of $F$ > $-I$ effect of $Cl$.
• Result: $CH_2FCO_2H$ is stronger than $CH_2ClCO_2H$.

Answer: $CH_2FCO_2H$ is stronger.

(iii) $CH_2FCH_2CH_2CO_2H$ vs $CH_3CHFCH_2CO_2H$

• First acid: Fluorine is on the γ-carbon (3 bonds away from $-COOH$).
• Second acid: Fluorine is on the β-carbon (2 bonds away from $-COOH$).
• Key principle: Inductive effect decreases rapidly with distance. The closer the EWG, the stronger its stabilising effect.
• Result: $CH_3CHFCH_2CO_2H$ (F at β-position) is stronger than $CH_2FCH_2CH_2CO_2H$ (F at γ-position).

Answer: $CH_3CHFCH_2CO_2H$ is stronger.

Pair (iv): $F_3C{-}C_6H_4{-}COOH$ vs $H_3C{-}C_6H_4{-}COOH$ …

Here are the common mistakes students make when comparing acid strength using the Inductive Effect, along with how to avoid each.

Mistake 1: Forgetting that the Inductive Effect is distance-dependent

The error:

In pair (iii), students often think $CH_2FCH_2CH_2CO_2H$ is stronger because it has a fluorine atom — ignoring where the fluorine is placed.

Why it’s wrong:

The inductive effect weakens rapidly as the number of bonds between the electronegative atom and the acidic proton increases. Fluorine on the γ-carbon (3 bonds away) has almost no effect. In $CH_3CHFCH_2CO_2H$, fluorine is on the β-carbon (2 bonds away), so it still stabilises the conjugate base noticeably.

How to avoid:

Always count the bonds from the electronegative atom to the carboxyl carbon.

• 1 bond away → strong effect
• 2 bonds away → moderate effect
• 3 bonds away → very weak effect

Correct answer for (iii): $CH_3CHFCH_2CO_2H$ is stronger.

Mistake 2: Confusing electronegativity with acid strength order directly

The error:

In pair (ii), some students recall the haloALKANE reactivity order or the polarisability of larger halogens and conclude $CH_2ClCO_2H$ must be the stronger acid.

Why it’s wrong:

Acid strength here is controlled by the inductive (−I) effect on the carboxylate, and fluorine — the most electronegative element — withdraws electron density most strongly, stabilising the conjugate base best. Real pK$_a$ values agree: fluoroacetic acid ≈ 2.59 vs chloroacetic acid ≈ 2.87 — $CH_2FCO_2H$ is the stronger acid.

How to avoid:

Remember the trend for α-halo acids:

Order of acid strength: $F > Cl > Br > I$ (follows electronegativity — the −I inductive effect dominates)

This is exactly the order the electronegativity series predicts; there is no inversion for α-halo carboxylic acids.

Correct answer for (ii): $CH_2FCO_2H$ is stronger.

Mistake 3: Ignoring the number of electronegative atoms or substituent position

The error:

In pair (i), some students think both acids are similar because both have a —COOH group.

Why it’s wrong:

$CH_2FCO_2H$ has a fluorine atom directly on the α-carbon. This strongly withdraws electron density via the inductive effect, stabilising the conjugate base much more than the methyl group in $CH_3CO_2H$, which is electron-donating.

How to avoid:

• Compare the substituent attached to the α-carbon.
• Electron-withdrawing groups (EWG) like F, Cl, Br, I, NO₂, CN increase acidity.
• Electron-donating groups (EDG) like CH₃, C₂H₅ decrease acidity.

Correct answer for (i): $CH_2FCO_2H$ is stronger.

Mistake 4: Applying the Inductive Effect to meta/para positions incorrectly

The error:

Students treat every pair identically — or skip the aromatic pair entirely.

Why it’s wrong: …