Carboxylic acids — acidity, derivatives (acid chlorides, anhydrides, esters, amides), Hell-Volhard-Zelinsky, esterificat: Question bank

Before you start, a quick vocabulary anchor so no word below is unearned:

Conjugate base = what's left after an acid loses its $H^{+}$ . For $R C O O H$ it is the carboxylate $R C O O^{-}$ .
α-carbon = the carbon directly attached to the $- C O O H$ group. Its hydrogens are the α-hydrogens.
EWG / EDG = electron-withdrawing / electron-donating group — pulls electron density toward itself / pushes it away.

True or false — justify

Is a carboxylic acid more acidic than the alcohol with the same carbon skeleton, and is the reason "the O–H bond is weaker"?

False on the reason. Acidity is set by how stable the conjugate base is, not raw bond strength — carboxylate spreads its charge over two oxygens (resonance), so the anion is low-energy and the proton leaves easily. See Resonance and delocalisation.

The two C–O bonds in a carboxylate ion have different lengths because one is a double bond and one is single.

False. Both are the same length (~127 pm), intermediate between

C = O

and

C - O

— this measured equality is direct physical proof that resonance is real, not just paper bookkeeping.

Trichloroacetic acid (

C l_{3} C C O O H

) is more acidic than acetic acid.

True. The three chlorines are EWGs that pull negative charge off the carboxylate through bonds (the Inductive effect), spreading the charge further and stabilising the anion — lower-energy anion means stronger acid.

Phenol has resonance stabilisation of its anion, so it should be about as acidic as a carboxylic acid.

False. Phenoxide's negative charge lands mostly on carbon atoms (poor at holding charge), while carboxylate's sits on two oxygens (electronegative, happy to hold charge). So phenol (

p K_{a} \approx 10

) is far weaker than

R C O O H

(

p K_{a} \approx 4

–

5

You can convert an amide directly and easily into an acid chloride by adding a chloride source.

False. You can only slide down the reactivity ladder (

C l >

anhydride

>

ester

>

amide), never easily back up — the amide's carbonyl is "satisfied" by strong N lone-pair donation and its would-be leaving group (

^{-} N R_{2}

) is terrible.

The Fischer esterification oxygen in the ester's

- O - R^{'}

linkage comes from the alcohol, not the acid.

True. The classic ¹⁸O-labelled-alcohol experiment finds the label in the ester; the acid contributes the

- O H

that leaves as water.

HVZ works on any carboxylic acid you hand it.

False. It requires at least one α-hydrogen, because the mechanism needs an enolisable α-carbon; something like

(C H_{3})_{3} C - C O O H

(no α-H) simply cannot react.

Spot the error

"In Fischer esterification the alcohol loses its

- O H

as part of the water molecule."

The error: it's the acid that loses

- O H

. In the mechanism a proton is added to that oxygen turning it into

- O H_{2}^{+}

(a good leaving group), and that water departs from the carbon that used to be the acid — proven by ¹⁸O labelling.

"HVZ adds bromine to the carbon two positions from the

- C O O H

The error: only the α-carbon (directly next to

- C O O H

) is halogenated, because only it can become the nucleophilic enol carbon that attacks

B r_{2}

. β and further carbons are untouched.

"Acid chlorides and Fischer esterification make esters by the exact same conditions."

The error: Fischer is reversible, slow, and needs an acid catalyst;

R C O C l + R^{'} O H

is fast, irreversible, catalyst-free because

C l^{-}

is an excellent leaving group. Different mechanisms' costs, same product family.

"The red phosphorus in HVZ is a spectator — it doesn't change during the reaction."

The error: P (as

P B r_{3}

) actively converts the acid into its acid bromide, which is what enolises. It's a genuine catalyst, regenerated in the last step, not an inert bystander.

"Adding an electron-donating alkyl group to acetic acid makes it more acidic."

The error: EDGs like

- C H_{3}

- C H_{2} C H_{3}

push electron density toward the carboxylate, concentrating the negative charge and destabilising the anion — so propanoic acid (

p K_{a} 4.87

) is slightly weaker than acetic (

4.76

"Nucleophilic acyl substitution is just the nucleophile adding to the carbonyl and stopping there."

The error: it is addition–elimination. The nucleophile adds to give a tetrahedral intermediate, then the

C = O

re-forms by ejecting the leaving group

L^{-}

. Stopping at addition is what aldehydes/ketones do (they have no leaving group). See Nucleophilic acyl substitution and Aldehydes and ketones.

Why questions

Why does protonating the carbonyl oxygen (not the alcohol) start Fischer esterification?

Protonating

C = O

pulls electron density off the carbonyl carbon, making it much more

δ +

(more electrophilic), so the weak alcohol nucleophile can finally attack it. Protonating the alcohol would just make it less nucleophilic — the wrong direction.

Why is an amide the least reactive acid derivative even though it looks similar to an ester?

Nitrogen is less electronegative and overlaps better with the carbonyl, so it donates its lone pair strongly into

C = O

, satisfying the carbon; and

^{-} N R_{2}

is an awful leaving group. Both effects make the amide carbonyl calm and unreactive.

Why does an EWG lose its acid-strengthening power as you move it further from the

- C O O H

The Inductive effect is transmitted through σ-bonds and weakens sharply with each bond. A chlorine on the α-carbon helps a lot; the same chlorine three carbons away barely shifts the

p K_{a}

Why does removing water (or using excess alcohol) increase ester yield in Fischer esterification?

The reaction is a reversible equilibrium; removing a product (water) or flooding a reactant (alcohol) shifts it toward the ester side — this is Le Chatelier's principle applied to synthesis.

Why can't a plain carboxylic acid enolise easily enough to be halogenated directly, forcing us to use red P?

The acid's own

O - H

proton dominates its acid–base behaviour, so it forms very little enol at the α-carbon. Converting it to the acid bromide removes that competition and lets the α-carbon enolise readily — see Keto–enol tautomerism.

Edge cases

What happens if you attempt HVZ on formic acid (

H C O O H

Nothing useful — formic acid has no α-carbon at all (the H sits on the carboxyl carbon itself), so there is no enolisable position and no α-halogenation product.

What happens in HVZ if the α-carbon carries only one hydrogen, e.g.

(C H_{3})_{2} C H - C O O H

It still works but can be halogenated only once, giving

(C H_{3})_{2} C B r - C O O H

; after that the α-carbon has no remaining hydrogen, so further substitution stalls.

If both the acid and alcohol in a Fischer reaction are the same size and you use no excess and remove no water, what's the outcome?

You reach an equilibrium mixture of acid + alcohol + ester + water — yield is limited (often near ~65%), not complete, precisely because nothing pushes the reversible equilibrium forward.

What is the limiting acidity trend as you go from acetic → mono → di → trichloroacetic acid?

Acidity rises steadily (

p K_{a}

4.76 \to 2.86 \to 1.29 \to 0.65

) as each added chlorine withdraws more charge, but each extra

C l

adds less than the previous one — the inductive boost has diminishing returns.

What happens when you try to make an ester by base hydrolysis conditions instead of acid — do you get the ester back?

No — base drives the reverse, hydrolysing the ester into carboxylate + alcohol (saponification), and the carboxylate anion won't re-esterify. See Saponification.

Does a carboxylate ion (

R C O O^{-}

) still react as a nucleophilic-acyl-substitution electrophile?

Barely — the negative charge is spread onto the carbonyl carbon's neighbours, making it much less

δ +

and a poor electrophile. That's why we activate acids (protonation, or convert to

R C O C l

) before substituting.

Recall Quick self-test

One-line challenge: name the single structural feature that (a) makes $R C O O H$ acidic, (b) makes its derivatives substitutable, and (c) enables HVZ. Answer ::: The carbonyl ( $C = O$ ) — it stabilises the carboxylate by resonance (a), makes the acyl carbon electrophilic for substitution (b), and lets the α-carbon enolise (c).