Reactions of Carboxylic Acids: Making Esters, Acid Chlorides, Anhydrides and Amides
Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
A carboxylic acid (R-COOH) can swap its -OH part to make four "derivatives": ester (with alcohol + conc. H2SO4), acid chloride (with SOCl2/PCl5/PCl3), anhydride (heat with P2O5, loses water), and amide (heat the ammonium salt, loses water). In each case the -OH of -COOH is replaced by -OR, -Cl, -OCOR, or -NH2. Memory hook: think "OH leaves, a new group comes in" — the order of how easily they react back is acid chloride > anhydride > ester > amide (Cl leaves easiest, N leaves hardest).
One carboxylic acid (R-COOH) makes four derivatives. In each, the -OH is replaced: -OR (ester), -Cl (acid chloride), -OCOR (anhydride), or -NH2 (amide). Reactivity order: acid chloride > anhydride > ester > amide.
Your doubts, answered
How does a carboxylic acid turn into an ester?
You heat the carboxylic acid with an alcohol and a small amount of concentrated H2SO4 as catalyst. The -OH of the acid and the -H of the alcohol leave together as water, and R-COO-R' (the ester) forms. This is called Fischer esterification. It is a reversible reaction, so we remove water (or use excess alcohol) to push it forward. NCERT: alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters.
What reagent converts a carboxylic acid to an acid chloride?
Use SOCl2 (thionyl chloride), PCl5, or PCl3. These replace the -OH of -COOH with -Cl, giving R-COCl (acyl chloride). SOCl2 is the best in NEET answers because the by-products are gases (SO2 and HCl) that escape, leaving a pure product. Remember: acid chloride is the MOST reactive derivative, so it is often made first and then used to make esters, amides, or anhydrides.
How do you make an amide from a carboxylic acid?
First the acid reacts with ammonia (NH3) to give the ammonium salt R-COO- NH4+. On strong heating, this salt loses water and forms the amide R-CO-NH2. So the two-step idea is: acid + NH3 makes the ammonium salt, then heat removes water to give the amide. Amide is the LEAST reactive derivative because nitrogen holds onto the carbon strongly.
How does an anhydride form from a carboxylic acid?
Heat two carboxylic acid molecules with a dehydrating agent like P2O5 (phosphorus pentoxide). One water molecule is removed between the two -COOH groups, joining them as R-CO-O-CO-R (acid anhydride). Example: acetic acid gives acetic anhydride. When two -COOH groups are on the same molecule and are close (like phthalic acid), they can form a cyclic anhydride on heating.
Why is esterification reversible but acid-chloride formation is not?
In esterification, water is a product and water can attack the ester to reform the acid (hydrolysis) — so it reaches equilibrium. That is why we use conc. H2SO4 to remove water and shift it right. When making an acid chloride with SOCl2, the by-products (SO2, HCl) are gases that leave the flask, so the reaction cannot go backward — it goes to completion.
What is the order of reactivity of the derivatives?
Acid chloride > anhydride > ester > amide. This tells you which way you can convert: a more reactive derivative can be turned into a less reactive one, but not easily the reverse. So an acid chloride can make an anhydride, ester, or amide, but an amide will not easily make an acid chloride. This order is a favourite NEET reasoning point.
⚠️ The NEET trap ✗ Thinking NaBH4 or the acidic esterification conditions also reduce or attack the ester/COOH group, so students change every functional group in the molecule. ✓ NaBH4 is a mild reducing agent — it reduces aldehydes and ketones to alcohols but does NOT touch esters or carboxylic acids. So in a molecule with both a ketone and a -COOCH3 ester side chain, only the ketone becomes -OH; the ester stays as it is. 🧠 Weak reducer (NaBH4) = weak reach: it grabs C=O of aldehydes/ketones only, leaving ester and COOH alone.
Real NEET questions
NEET 2021
Match List-I with List-II and choose the correct answer.
List-I: (a) C6H6 + CO, HCl / anhy. AlCl3, CuCl -> C6H5CHO (b) R-CO-CH3 + NaOX -> R-COO- + CHX3 (c) R-CH2-OH + R'COOH / conc. H2SO4 -> (d) R-CH2-COOH + (i) X2/Red P (ii) H2O ->
List-II: (i) Hell-Volhard-Zelinsky (ii) Gattermann-Koch (iii) Haloform (iv) Esterification
A · (a)-(i),(b)-(iv),(c)-(iii),(d)-(ii)
B · (a)-(ii),(b)-(iii),(c)-(iv),(d)-(i) ✓
C · (a)-(iv),(b)-(i),(c)-(ii),(d)-(iii)
D · (a)-(iii),(b)-(ii),(c)-(i),(d)-(iv)
Solution: Focus on (c): an alcohol (R-CH2-OH) plus a carboxylic acid (R'COOH) with conc. H2SO4 gives an ester — this is Esterification (iv). The others: (a) benzene + CO/HCl over AlCl3/CuCl is Gattermann-Koch (ii), (b) methyl ketone + NaOX is Haloform (iii), (d) acid + X2/Red P then water is Hell-Volhard-Zelinsky (i). So the match is (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i), option B.
NEET 2019
The major product of the following reaction is: benzene-1,2-dicarboxylic acid (phthalic acid) + NH3 / strong heating -> ?
A · Ring with one -COOH and one -CONH2 (mono-amide)
B · Phthalimide (benzene fused to a five-membered cyclic imide) ✓
C · Ring with one -COOH and one -NH2 (anthranilic-acid type)
D · Ring with two -NH2 groups (ortho-diamine)
Solution: Phthalic acid has two -COOH groups next to each other. With NH3 it first forms the ammonium salt/diamide. On strong heating, both carboxyl groups close onto a single nitrogen and lose water and ammonia (cyclodehydration), forming a five-membered cyclic imide fused to the benzene ring — phthalimide. This is the same acid-to-amide/imide idea: heat the ammonium salt, water leaves. Answer: B.
NEET 2021
A cyclohexanone ring bearing a -CH2COOCH3 (methyl-ester) side chain and a CH3 group is treated with NaBH4 in C2H5OH. The product is:
A · Both the ring C=O and the ester reduced to alcohols
B · Only the ring C=O reduced to -OH; the methyl-ester side chain unchanged ✓
C · Only the ester reduced; ring ketone unchanged
D · No reaction occurs
Solution: NaBH4 is a mild reducing agent. It reduces aldehydes and ketones to alcohols but does NOT reduce esters or carboxylic acids. So only the ring ketone becomes a secondary alcohol (cyclohexanol ring); the -CH2COOCH3 ester side chain stays intact. This proves esters are much less reactive than ketones toward reducing agents. Answer: B.
Solved Aldehydes, Ketones And Carboxylic Acid NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What are the four derivatives of a carboxylic acid?
Ester (R-COOR'), acid chloride (R-COCl), acid anhydride (R-CO-O-CO-R), and amide (R-CONH2). In every one, the -OH of the -COOH group is replaced by a new group: -OR, -Cl, -OCOR, or -NH2.
Why do we add concentrated H2SO4 in esterification?
It acts as an acid catalyst and as a dehydrating agent. Because esterification is reversible and makes water, removing water with H2SO4 pushes the equilibrium toward the ester side, giving more product.
Which is the most reactive carboxylic acid derivative?
The acid chloride. The reactivity order is acid chloride > anhydride > ester > amide. That is why acid chlorides are used to make the other derivatives, but amides are hard to convert back.
How is acetylation related to these reactions?
Acetylation means adding an acetyl group (CH3CO-). When an alcohol or phenol reacts with acetic anhydride or acetyl chloride, it forms an acetate ester. NCERT example: acetylation of salicylic acid gives aspirin.
Does esterification need an alcohol every time?
To make an ester, yes — you need an alcohol (or phenol) plus the carboxylic acid, acid chloride, or anhydride. But to make an acid chloride, anhydride, or amide from the acid, you use SOCl2/PCl5, P2O5/heat, or NH3/heat instead.
