Carboxylic Acid Derivatives

ChemistryCarboxylic Acids And Their DerivativesFor JEE aspirants

Carboxylic acid derivatives are compounds $R - CO - Z$ in which the -OH of the acid is replaced by Z = -Cl (acid chlorides), -OCOR (anhydrides), -OR' (esters) or $- N H_{2}$ (amides). All of them react by nucleophilic acyl substitution, and their reactivity falls in the order acid chloride > anhydride > ester > amide, which also decides which derivative can be made from which. Carboxylic acid derivatives supply many named reactions tested in JEE Main, JEE Advanced and NEET, including Rosenmund reduction, Claisen condensation, saponification and the Hofmann bromamide reaction.

On this page1The four derivatives2Reactivity order3Acyl chlorides4Anhydrides5Esters6Amides7Exam practice

Key Formulas - Quick Reference

★ Must learn Reactivity: RCOCl > $(RCO)_{2} O$ > RCOOR' > $RCON H_{2}$
★ Must learn $RCOCl + H_{2} Pd/BaS O_{4} RCHO + HCl$ (Rosenmund)
$C_{6} H_{6} + RCOCl anhyd. AlC l_{3} C_{6} H_{5} COR + HCl$ (Friedel-Crafts acylation)
★ Must learn $RCOO R^{'} + NaOH \to RCOONa + R^{'} OH$ (saponification, irreversible)
$RCOO R^{'} + R^{''} OH ⇌ H^{+} RCOO R^{''} + R^{'} OH$ (transesterification)
★ Must learn $2C H_{3} COO C_{2} H_{5} NaO C_{2} H_{5} C H_{3} COC H_{2} COO C_{2} H_{5}$ (Claisen)
$RCOO R^{'} 2 R^{''} MgX, H_{3} O^{+} R R_{2}^{''} C - OH$ (3° alcohol)
$RCON H_{2} P_{2} O_{5}, Δ RCN + H_{2} O$
★ Must learn $RCON H_{2} + B r_{2} + 4NaOH \to RN H_{2} + N a_{2} C O_{3} + 2NaBr + 2 H_{2} O$ (Hofmann)
$RCON H_{2} LiAl H_{4} RC H_{2} N H_{2}$

1. What are Carboxylic Acid Derivatives?

There are four carboxylic acid derivatives, generally represented as $R - CO - Z$ , where Z is a halogen (usually Cl), -OCOR', -OR' or $- N H_{2}$ (or -NHR', $- N R_{2}^{'}$ ). All of them contain the acyl group and are readily converted back into the acid by hydrolysis.

Z = halogen (usually Cl): acid chlorides (acyl chlorides).
Z = -OR': esters.
Z = -OCOR': carboxylic acid anhydrides.
Z = $- N H_{2}$ : amides; when Z is -NHR' or $- N R_{2}^{'}$ , they are N-substituted amides.

Carboxylic acid derivatives are prepared from carboxylic acids; those preparation methods appear under the chemical reactions of carboxylic acids. Carboxylic acids themselves also react with strong nucleophiles and electrophiles, as the next example shows.

Solved Example 1

Identify A, B and C: (a)

PhLi (1) excess C O_{2} (2) H^{+} A

(b)

(C H_{3})_{3} CCOOH (1) 2 eq. C H_{3} Li (2) H^{+} B

(c)

(C H_{3})_{3} CCOOH A g_{2} O, then B r_{2}, Δ C

Solution:

(a) The carbanion-like carbon of phenyllithium attacks $C O_{2}$ ; acidification gives A = benzoic acid, $C_{6} H_{5} COOH$ .

(b) The first $C H_{3} Li$ removes the acidic proton; the second adds to the carboxylate, and hydrolysis of the dilithium salt gives a methyl ketone: B = $(C H_{3})_{3} C - CO - C H_{3}$ (3,3-dimethylbutan-2-one).

(c) The silver salt with bromine undergoes the Hunsdiecker reaction: C = $(C H_{3})_{3} C - Br$ (2-bromo-2-methylpropane) + $C O_{2}$ .

2. Relative Reactivity of Acid Derivatives

Reactivity towards nucleophilic acyl substitution follows the order RCOCl > $(RCO)_{2} O$ > RCOOR' > $RCON H_{2}$ . Substitution takes place in two steps: (a) the nucleophile adds to the electron-deficient carbonyl carbon, forming a tetrahedral intermediate, and (b) the intermediate eliminates the leaving group, regenerating the C=O. Step (a) is favoured by electron withdrawal and hindered by +I groups or bulky groups; step (b) depends on the leaving group. Two facts explain the order:

Basicity of the leaving group: the weaker the base, the better it leaves. Basicity follows $N H_{2}^{-} > R O^{-} > RCO O^{-} > C l^{-}$ , so reactivity runs the other way, and $C l^{-}$ , the weakest base, makes acid chlorides the most reactive.
Resonance: the lone pair on the atom joined to the carbonyl carbon is delocalised into C=O, giving that bond partial double-bond character and stabilising the derivative. The more stabilisation, the lower the reactivity. Stabilisation is least for acid chlorides, because the large chlorine 3p orbital overlaps poorly with carbon's 2p orbital and chlorine's strong -I effect withdraws electrons; it is greatest for amides, where nitrogen donates strongly.

Figure 1: Reactivity of acid derivatives depends on leaving-group basicity and lone-pair donation into the carbonyl group.

Nucleophilic acyl substitution is catalysed by acids, because protonating the carbonyl oxygen makes the carbon more electrophilic for step (a).

Solved Example 2

Acetyl chloride reacts with water more readily than methyl chloride does. Explain.

Solution:

Alkyl halides are much less reactive than acyl halides in nucleophilic substitution. Attack on the tetrahedral carbon of RX goes through a hindered transition state, and a bond must be partly broken to let the nucleophile attach. In $C H_{3} COCl$ , the nucleophile attacks the flat carbonyl carbon through a relatively unhindered transition state, and the substitution happens in two steps: the first is like addition to a carbonyl compound, and the second is loss of chloride.

Figure 2: Hydrolysis of acetyl chloride by nucleophilic addition-elimination.

Solved Example 3

Suggest likely mechanisms for: (a)

O H^{-} + R - CO - C X_{3} H_{2} O RCO O^{-} + CH X_{3}

(b) 2-fluorobenzophenone

N H_{2}^{-} /N H_{3}

benzamide + fluorobenzene

Solution:

(a) Hydroxide adds to the carbonyl carbon. The tetrahedral intermediate re-forms C=O by expelling $^{-} C X_{3}$ , which is a good leaving group for a carbanion because three halogens stabilise the negative charge. Proton exchange gives $RCO O^{-}$ and $CH X_{3}$ (the last step of the haloform reaction).

(b) Amide ion adds to the carbonyl carbon. The intermediate expels the aryl carbanion that is better stabilised: the 2-fluorophenyl anion, stabilised by the -I effect of fluorine next to the negative carbon. This leaves $C_{6} H_{5} CON H_{2}$ , and the aryl anion takes a proton from $N H_{3}$ to give fluorobenzene.

Solved Example 4

The ease of alkaline hydrolysis is greatest for:
(a) methyl 4-nitrobenzoate
(b) methyl 4-chlorobenzoate
(c) methyl benzoate
(d) methyl 4-methoxybenzoate

Solution:

Hydroxide attacks the carbonyl carbon, so the ester with the most electron-deficient carbonyl carbon reacts fastest. The strongly electron-withdrawing para $N O_{2}$ group gives the most electron deficiency; $- OC H_{3}$ gives the least. Answer: (a)

Exam Trick: the pKa ladder

Rank the derivatives by the $p K_{a}$ of the leaving group's conjugate acid: HCl (-7), RCOOH (4.8), R'OH (16), $N H_{3}$ (38). The lower the pKa, the faster the reaction, and a derivative can be made only from one above it. Order: CAEA, "Cats Always Eat Apples" (Chloride, Anhydride, Ester, Amide).

Because conversions only run downhill, going uphill needs a detour through the acid:

Figure 3: Derivatives convert only downhill (RCOCl → anhydride → ester → amide); to go uphill, return to RCOOH and make the acid chloride first.

Key idea

One mechanism (addition, then elimination) and one ladder, RCOCl > anhydride > ester > amide, explain every reaction on this page.

3. Acyl Chlorides

Preparation

Acid chlorides are prepared from carboxylic acids with $SOC l_{2}$ , $PC l_{3}$ or $PC l_{5}$ , often with pyridine to remove HCl.

RCOOH + SOC l_{2} pyridine RCOCl + S O_{2} + HCl

RCOOH + PC l_{5} ⟶ RCOCl + POC l_{3} + HCl

3RCOOH + PC l_{3} ⟶ 3RCOCl + H_{3} P O_{3}

Acylation reactions

Acyl chlorides are the most reactive acid derivatives, so they are often chosen as the starting material for preparing any other acid derivative. The acyl group is transferred to the nucleophile, and HCl (or a chloride salt) is released.

R - COCl + R^{'} - CO O^{-} N a^{+} ⟶ R - CO - O - CO - R^{'} + NaCl

R - COCl + R^{'} OH ⟶ R - COO R^{'} + HCl

R - COCl + 2N H_{3} ⟶ R - CON H_{2} + N H_{4} Cl

R - COCl + R^{'} N H_{2} ⟶ R - CONH R^{'} + HCl

R - COCl + R^{'} R^{''} NH ⟶ R - CON R^{'} R^{''} + HCl

C H_{3} COCl + H_{2} N C_{6} H_{5} ⟶ C_{6} H_{5} NHCOC H_{3} + HCl

R - COCl + H_{2} O ⟶ R - COOH + HCl

Acetylation of salicylic acid with acetyl chloride gives acetylsalicylic acid (aspirin): the phenolic -OH becomes $- OCOC H_{3}$ .

Figure 4: Reaction map of acyl chlorides, the most reactive acid derivatives.

Friedel-Crafts acylation

C_{6} H_{6} + RCOCl anhyd. AlC l_{3} C_{6} H_{5} COR + HCl

Acetyl chloride and aniline give acetanilide in the same way. With acetyl chloride the Friedel-Crafts product is acetophenone, $C_{6} H_{5} COC H_{3}$ ; with benzoyl chloride it is benzophenone, $C_{6} H_{5} CO C_{6} H_{5}$ .

Reaction with alkenes

Acetyl chloride adds to the double bond of an alkene in the presence of a catalyst ( $AlC l_{3}$ or $ZnC l_{2}$ ) to form a chloro ketone, which on heating eliminates HCl to give an unsaturated ketone.

C H_{3} CH = C H_{2} + C H_{3} COCl ZnC l_{2} C H_{3} CH (Cl) C H_{2} COC H_{3}

Δ C H_{3} CH = CHCOC H_{3} + HCl

Reduction

R - COCl + H_{2} Pd/BaS O_{4} R - CHO + HCl

R - COCl + 4 [H] LiAl H_{4} or NaB H_{4} R - C H_{2} OH + HCl

The first reaction is the Rosenmund reduction: $BaS O_{4}$ partly poisons the palladium so the reaction stops at the aldehyde. With $LiAl H_{4}$ , hydride first gives the aldehyde, which is reduced further to the primary alcohol.

Exam Trick: Rosenmund has brakes

$BaS O_{4}$ (sometimes with quinoline or sulfur) is the brake on palladium: the reduction stops at RCHO. No brake ( $LiAl H_{4}$ , $NaB H_{4}$ ) means the aldehyde is reduced on to $RC H_{2} OH$ . Formaldehyde cannot be made this way because HCOCl is unstable.

Reactions with KCN, silver nitrate and diazomethane

R - COCl + KCN ⟶ R - CO - CN

H_{3} O^{+} R - CO - COOH

The product is an $α$ -keto acid; when R = $C H_{3}$ , it is pyruvic acid. Aliphatic acid chlorides are readily decomposed by water, so an aqueous solution gives a white precipitate of AgCl with $AgN O_{3}$ .

R - COCl + H_{2} O + AgN O_{3} ⟶ R - COOH + AgCl ↓ + HN O_{3}

Arndt-Eistert synthesis. Reaction of an acyl chloride with diazomethane, then silver oxide and water, converts it into a carboxylic acid with one more carbon atom.

R - COCl + 2C H_{2} N_{2} ⟶ R - CO - CH N_{2} + C H_{3} Cl + N_{2}

R - CO - CH N_{2} A g_{2} O, - N_{2} R - CH = C = O

H_{2} O R - C H_{2} - COOH

Figure 5: Arndt-Eistert synthesis: diazoketone, Wolff rearrangement to a ketene, then water.

Solved Example 5

What would be the product if R-COCl reacts with

Na N_{3}

followed by hydrolysis?

Solution:

Azide displaces chloride to form an acyl azide, $R - CO - N_{3}$ . On warming it loses $N_{2}$ while R migrates from carbon to nitrogen (Curtius rearrangement), giving an isocyanate. Hydrolysis gives the primary amine and $C O_{2}$ .

R - COCl Na N_{3} R - CO N_{3}

Δ, - N_{2} R - N = C = O H_{2} O R - N H_{2} + C O_{2}

Solved Example 6

Hydrogenation of benzoyl chloride in the presence of Pd and

BaS O_{4}

gives:
(A) benzyl alcohol
(B) benzaldehyde
(C) benzoic acid
(D) phenol

Solution:

This is the Rosenmund reduction; the poisoned catalyst stops at the aldehyde. Answer: (B)

Solved Example 7

Compound (A) has a neutralisation equivalent of 116. It forms a semicarbazone and a phenylhydrazone and gives a positive iodoform test. (A) reacts with

SOC l_{2}

to give (B), which on Rosenmund reduction gives (C). Clemmensen reduction of (C) gives n-pentane. What are (A), (B) and (C)?

Solution:

Semicarbazone and phenylhydrazone formation show a >C=O group; the iodoform test shows a $C H_{3} - CO$ - group. A neutralisation equivalent of 116 shows one -COOH. Subtracting $C H_{3} CO$ (43) and COOH (45) from 116 leaves 28, which is two $C H_{2}$ groups.

A = $C H_{3} - CO - C H_{2}$ $- C H_{2} - COOH$ (4-oxopentanoic acid, levulinic acid); B = $C H_{3} - CO - C H_{2}$ $- C H_{2} - COCl$ ; C = $C H_{3} - CO - C H_{2}$ $- C H_{2} - CHO$ (4-oxopentanal). Clemmensen reduction converts both carbonyl groups into $C H_{2}$ , giving $C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}$ .

Quick Recall: tap to check

Which reagent turns benzoyl chloride into benzaldehyde?

H_{2}

over Pd/

BaS O_{4}

(Rosenmund reduction).

Why does acetyl chloride fume in moist air?

It is hydrolysed at once to acetic acid and HCl gas.

What does RCOCl give with

C H_{2} N_{2}

, then

A g_{2} O

H_{2} O

RC H_{2} COOH

: one extra carbon (Arndt-Eistert).

4. Acid Anhydrides

Preparation

Acid anhydrides are considered to be derived from two molecules of carboxylic acid by removal of one molecule of water.

R - COOH + R^{'} - COCl

pyridine R - CO - O - CO - R^{'} + C_{5} H_{5} N H^{+} C l^{-}

R - CO O^{-} N a^{+} + R^{'} - COCl ⟶ R - CO - O - CO - R^{'} + NaCl

In the second reaction the carboxylate ion acts as the nucleophile at the acyl carbon of the acid chloride. Cyclic anhydrides form simply by heating the dicarboxylic acid when a five- or six-membered ring results: succinic acid (300 °C) gives succinic anhydride and phthalic acid (230 °C) gives phthalic anhydride.

Reactions

Anhydrides are good acylating agents; their reactions are less vigorous than those of acyl chlorides. They are used to prepare esters and amides, and they are hydrolysed back to acids.

(R - CO)_{2} O + R^{'} OH ⟶ R - COO R^{'} + R - COOH

(R - CO)_{2} O + 2N H_{3} ⟶ R - CON H_{2} + R - CO O^{-} N H_{4}^{+}

(R - CO)_{2} O + H_{2} O ⟶ 2R - COOH

Figure 6: An anhydride transfers only one of its two acyl groups.

Phthalic anhydride with ammonia gives ammonium phthalamate, which acid converts into phthalamic acid, a molecule that is both an amide and an acid.

Solved Example 8

p-Anisaldehyde

(C H_{3} O - C_{6} H_{4} - CHO)

+ (X)

C H_{3} COONa, H_{3} O^{+}

p-methoxycinnamic acid. The compound (X) is:
(A)

C H_{3} COOH

(B)

BrC H_{2} - COOH

(C)

(C H_{3} CO)_{2} O

(D) CHO-COOH

Solution:

This is the Perkin reaction: an aromatic aldehyde condenses with an acid anhydride in the presence of the sodium salt of the same acid to give an $α, β$ -unsaturated (cinnamic) acid. Answer: (C)

Solved Example 9

An organic compound (A) on treatment with ethyl alcohol gives a carboxylic acid (B) and a compound (C). Hydrolysis of (C) under acidic conditions gives (B) and (D). Oxidation of (D) with

KMn O_{4}

also gives (B). (B) on heating with

Ca (OH)_{2}

gives (E),

C_{3} H_{6} O

. (E) does not give Tollens' test or reduce Fehling's solution but forms a 2,4-dinitrophenylhydrazone. Identify (A) to (E).

Solution:

(E) forms a 2,4-DNP derivative but is not an aldehyde, so it is a ketone: acetone. Heating the calcium salt of (B) gives acetone, so (B) is acetic acid. (D) oxidises to acetic acid, so (D) is ethanol, and (C), which hydrolyses to acetic acid and ethanol, is ethyl acetate. (A) gives an acid and an ester with ethanol, so it is an anhydride.

A = $(C H_{3} CO)_{2} O$ , B = $C H_{3} COOH$ , C = $C H_{3} COO C_{2} H_{5}$ , D = $C_{2} H_{5} OH$ , E = $C H_{3} COC H_{3}$ .

(C H_{3} CO)_{2} O + C_{2} H_{5} OH ⟶ C H_{3} COOH + C H_{3} COO C_{2} H_{5}

(C H_{3} COO)_{2} Ca Δ C H_{3} COC H_{3} + CaC O_{3}

5. Esters

Preparation

Esters are derivatives in which the -OH of the carboxyl group is replaced by -OR, where R may be an alkyl or aryl group.

RCOOH + C_{2} H_{5} OH ⇌ H^{+} RCOO C_{2} H_{5} + H_{2} O

R - COOH + C H_{2} N_{2} ether RCOOC H_{3} + N_{2}

R - COCl + C_{2} H_{5} OH C_{5} H_{5} N R - COO C_{2} H_{5} + HCl

(R - CO)_{2} O + C_{2} H_{5} OH ⟶ R - COO C_{2} H_{5} + R - COOH

Hydrolysis of esters

Acid-catalysed esterification is reversible. Following the esterification mechanism backwards gives the mechanism of acid-catalysed ester hydrolysis.

R - COO R^{'} + H_{2} O ⇌ H^{+} R - COOH + R^{'} OH

Base-promoted hydrolysis (saponification). Esters are also hydrolysed by base. The reaction is called saponification because most soaps are made this way. Refluxing an ester with aqueous NaOH gives an alcohol and the sodium salt of the acid.

R - COO R^{'} + NaOH H_{2} O R - CO O^{-} N a^{+} + R^{'} OH

Saponification is essentially irreversible, because the carboxylate ion is inert towards nucleophilic substitution.

Figure 7: Mechanism of saponification (

B_{AC} 2

). The final proton transfer makes it irreversible.

Saponification equivalent: if an ester is hydrolysed with a known excess of base, the amount of base used up can be measured. It gives the saponification equivalent (equivalent weight) of the ester, the ester counterpart of the neutralisation equivalent of an acid. For an ester with one -COOR' group it equals the molar mass.

Acid hydrolysis (

H_{3} O^{+}

)

Catalytic acid; every step is an equilibrium.

Reversible: the reverse is Fischer esterification. Products RCOOH + R'OH.

Saponification (NaOH)

One mole of base is used up.

Irreversible: the final proton transfer gives $RCO O^{-}$ , which alcohol cannot attack.

Mechanistic pathways of ester hydrolysis

Ester hydrolysis can follow the pathways $A_{AC} 1$ , $A_{AC} 2$ , $A_{AL} 1$ and $B_{AC} 2$ . A or B stands for acid- or base-catalysed, AC or AL for acyl-oxygen or alkyl-oxygen cleavage, and 1 or 2 for unimolecular or bimolecular.

Figure 8: Mechanistic codes for ester hydrolysis and the bond each one breaks.

$A_{AC} 2$ : the ester is protonated on carbonyl oxygen, water attacks in the slow step, and ethanol is lost from the tetrahedral intermediate (ethyl acetate + $H_{3} O^{+}$ ).
$A_{AL} 1$ : after protonation, the alkyl-oxygen bond breaks to give a stable carbocation, which water captures (tert-butyl acetate gives acetic acid + tert-butyl alcohol).
$A_{AC} 1$ : for very hindered esters such as methyl 2,4,6-trimethylbenzoate, the protonated ester loses methanol in the slow step to form an acylium ion, which water then attacks.
$B_{AC} 2$ : hydroxide adds to the carbonyl carbon, alkoxide leaves, and proton exchange gives carboxylate + alcohol (saponification).

Transesterification and ammonolysis

Esters can also be prepared by transesterification, where one alcohol displaces another from an ester. The mechanism is similar to esterification.

R - COO R^{'} + R^{''} - OH heat ⇌ HA R - COO R^{''} + R^{'} - OH

C H_{2} = CHCOOC H_{3} + C H_{3} C H_{2} C H_{2} C H_{2} OH

⇌ C H_{2} = CHCOOC H_{2} C H_{2} C H_{2} C H_{3} + C H_{3} OH

Transesterification is an equilibrium reaction. To shift it to the right, a large excess of the alcohol whose ester is wanted is used, or one product is removed from the reaction mixture. Esters also react with ammonia or amines (ammonolysis) to give amides.

R - COO R^{'} + HN R_{2}^{''} ⟶ R - CON R_{2}^{''} + R^{'} OH

Reduction of esters

C H_{3} C H_{2} C H_{2} COOC H_{3}

H_{2}, CuO \cdot CuC r_{2} O_{4} 150^{\circ} C, high pressure C H_{3} C H_{2} C H_{2} C H_{2} OH + C H_{3} OH

C H_{3} C H_{2} C H_{2} COOC H_{3}

LiAl H_{4} / H^{+} or Na/ C_{2} H_{5} OH C H_{3} C H_{2} C H_{2} C H_{2} OH + C H_{3} OH

Both catalytic hydrogenation and chemical reduction give two alcohols. Sodium amalgam reduces only the keto group of a keto ester, while $LiAl H_{4}$ reduces both groups:

C H_{3} COC H_{2} COO C_{2} H_{5} Na/Hg C H_{3} CH (OH) C H_{2} COO C_{2} H_{5}

C H_{3} COC H_{2} COO C_{2} H_{5}

LiAl H_{4} C H_{3} CH (OH) C H_{2} C H_{2} OH + C_{2} H_{5} OH

$LiAl H_{4}$ likewise reduces phthalic anhydride to benzene-1,2-dimethanol.

Reaction with Grignard reagents

The reaction of an ester with a Grignard reagent is a good method for preparing 3° alcohols. A ketone forms first, but ketones react readily with Grignard reagents, so the final product is a tertiary alcohol.

R - COO R^{'} R^{''} MgX R - CO - R^{''}

R^{''} MgX R - C (OMgX) R_{2}^{''} H^{+} R - C (OH) R_{2}^{''}

Figure 9: An ester reacts with two moles of Grignard reagent to give a tertiary alcohol.

Claisen condensation

When ethyl acetate reacts with sodium ethoxide, it undergoes a condensation reaction. After acidification, the product is a $β$ -keto ester, ethyl acetoacetate (acetoacetic ester).

2C H_{3} COO C_{2} H_{5} NaO C_{2} H_{5}

[C H_{3} CO \overline{C} HCOO C_{2} H_{5}] N a^{+} + C_{2} H_{5} OH

HCl C H_{3} COC H_{2} COO C_{2} H_{5}

This is the Claisen condensation. For esters it is the exact counterpart of the aldol condensation: both involve nucleophilic attack by a carbanion on an electron-deficient carbonyl carbon. In the aldol condensation this attack leads to addition (typical of aldehydes and ketones); in the Claisen condensation it leads to substitution (typical of acyl compounds).

Ethoxide removes an $α$ -hydrogen, forming a resonance-stabilised ester enolate.
The enolate attacks the carbonyl carbon of a second ester molecule.
The tetrahedral intermediate expels ethoxide, giving the $β$ -keto ester.
Ethoxide removes the acidic hydrogen between the two carbonyl groups. This step is highly favourable and draws the overall equilibrium towards the product.
Rapid acidification with $H_{3} O^{+}$ gives the neutral $β$ -keto ester, which exists in keto and enol forms.

Figure 10: Mechanism of the Claisen condensation of ethyl acetate.

When planning a Claisen condensation, use an alkoxide with the same alkyl group as the alkoxyl group of the ester, to avoid transesterification. An intramolecular Claisen condensation is called the Dieckmann condensation; it is useful mainly for making five- and six-membered rings. For example, diethyl hexanedioate gives ethyl 2-oxocyclopentane-1-carboxylate.

JEE Advanced

Why a Claisen condensation needs two $α$ -hydrogens. The addition-elimination steps are uphill; the reaction is pulled forward only because ethoxide removes the acidic CH between the two carbonyls of the product ( $p K_{a}$ about 11). Ethyl 2-methylpropanoate, $(C H_{3})_{2} CHCOO C_{2} H_{5}$ , has only one $α$ -H, so its product has no acidic hydrogen left and the equilibrium stays on the reactant side with NaOEt. A much stronger base (sodium triphenylmethide, $P h_{3} C^{-} N a^{+}$ ) is needed, which deprotonates the ester completely before it condenses.

Acyloin condensation and reaction of keto esters with ammonia

Heating an ester with sodium in xylene couples two molecules into an $α$ -hydroxy ketone (acyloin). Sodium transfers electrons to the ester carbonyl, the radical anions dimerise, two alkoxide ions are lost to give a 1,2-diketone, which is reduced further to an enediolate; acidification and tautomerisation give the acyloin.

2R - COO R^{'} (i) Na, xylene (ii) H_{3} O^{+} R - CO - CH (OH) - R

With ammonia, a $β$ -keto ester is attacked first at the more positive keto carbonyl carbon; loss of water gives a $β$ -amino $α, β$ -unsaturated ester.

R - CO - C H_{2} COO C_{2} H_{5} + N H_{3}

⟶ R - C (N H_{2}) = CHCOO C_{2} H_{5} + H_{2} O

Solved Example 10

Write the structures of E and F:

E (C_{11} H_{14} O_{2}) O H^{-} F + C H_{3} C H_{2} CO O^{-}

, and E

(i) KMn O_{4} /O H^{-} (ii) H^{+}

benzene-1,3-dicarboxylic acid (isophthalic acid).

Solution:

Saponification releasing propanoate shows E is a propanoate ester, so F is an alcohol with formula $C_{8} H_{10} O$ . Oxidation of E gives isophthalic acid, so the ring carries two carbon side chains meta to each other.

E = 3-methylbenzyl propanoate, $C H_{3} - C_{6} H_{4} - C H_{2}$ $- O - CO - C H_{2} C H_{3}$ (meta); F = 3-methylbenzyl alcohol, $C H_{3} - C_{6} H_{4} - C H_{2} OH$ (meta).

Solved Example 11

A neutral liquid of formula

C_{7} H_{14} O_{2}

is hydrolysed to an acid (A) and an alcohol (B). Acid (A) has a neutralisation equivalent of 74. Alcohol (B) is not oxidised by an acid solution of

N a_{2} C r_{2} O_{7}

. What is the formula and name of the original compound?

Solution:

A neutralisation equivalent of 74 fits propanoic acid, $C H_{3} C H_{2} COOH$ . An alcohol that resists dichromate oxidation is tertiary; with the remaining four carbons it is tert-butyl alcohol, $(C H_{3})_{3} COH$ . The ester is tert-butyl propanoate, $C H_{3} C H_{2} COOC (C H_{3})_{3}$ .

C H_{3} C H_{2} COOC (C H_{3})_{3} + H_{2} O

⟶ C H_{3} C H_{2} COOH + (C H_{3})_{3} COH

Solved Example 12

An ester of molecular formula

C_{8} H_{16} O_{2}

on hydrolysis gives an acid (A) and an alcohol (B). Oxidation of (B) with

N a_{2} C r_{2} O_{7}

gives an acid (C). The sodium salts of (A) and (C), fused with solid NaOH, both give propane. What is the ester?

Solution:

Soda-lime decarboxylation removes one carbon, so both (A) and (C) are butanoic acid. (B) oxidises to butanoic acid, so it is butan-1-ol. The ester is butyl butanoate.

C H_{3} C H_{2} C H_{2} COOC H_{2} C H_{2} C H_{2} C H_{3}

H_{2} O/ H^{+} C H_{3} C H_{2} C H_{2} COOH + C H_{3} C H_{2} C H_{2} C H_{2} OH

Solved Example 13

β

-Propiolactone (oxetan-2-one) reacts with

C H_{3} OH

to give A. A is:
(a)

HO - C H_{2} - C H_{2} - COOC H_{3}

(b)

C H_{3} O - C H_{2} - C H_{2} - COOH

Solution:

Methanol attacks the carbonyl carbon of the strained four-membered lactone, and the ring opens by acyl-oxygen cleavage. The ring oxygen becomes an -OH and the carbonyl becomes a methyl ester: methyl 3-hydroxypropanoate. Answer: (a)

Key idea

Esters: hydrolysis (reversible with acid, irreversible with base), two Grignard additions to a 3° alcohol, and enolate chemistry (Claisen) that builds

β

-keto esters.

Quick Recall: tap to check

What does ethyl acetate give with excess

C H_{3} MgBr

, then

H_{3} O^{+}

2-Methylpropan-2-ol (tert-butyl alcohol) and ethanol.

Which ester hydrolyses by alkyl-oxygen cleavage in acid?

tert-Butyl acetate (

A_{AL} 1

, through the tert-butyl cation).

6. Amides

Preparation

In the laboratory, amides are prepared by the reaction of ammonia with acid chlorides or acid anhydrides. Primary and secondary amines react in the same way; excess ammonia or amine neutralises the HCl formed.

R - COCl + 2N H_{3} ⟶ R - CON H_{2} + N H_{4} Cl

(R - CO)_{2} O + 2 R^{'} N H_{2} ⟶ R - CONH R^{'} + R - CO O^{-} R^{'} N H_{3}^{+}

R - COO R^{'} + N H_{3} ⟶ R - CON H_{2} + R^{'} OH

The third route is ammonolysis of an ester, and the last is partial hydrolysis of a nitrile under controlled conditions.

R - C \equiv N + H_{2} O H^{+} or O H^{-} R - CON H_{2}

Physical properties

All amides except formamide are crystalline solids at room temperature. They have relatively high melting and boiling points because amide molecules are associated by intermolecular hydrogen bonds. The lone pair on nitrogen is delocalised into the carbonyl group, giving nitrogen a partial positive charge; this makes the N-H hydrogens better hydrogen-bond donors, increasing hydrogen bonding and raising the melting and boiling points above those of other acid derivatives.

Basic and acidic character

Amides are very feebly basic, because the lone pair on nitrogen is involved in resonance with the carbonyl group. They form unstable salts with strong inorganic acids, such as $RCON H_{2} \cdot HCl$ , in which the proton is held mainly on oxygen. For the same resonance reason, nitrogen carries a partial positive charge, so amides are also feebly acidic: they dissolve mercuric oxide to form a covalent mercury compound in which mercury is probably linked to nitrogen, and they react with sodium.

2RCON H_{2} + HgO ⟶ (RCONH)_{2} Hg + H_{2} O

C H_{3} CON H_{2} + Na ⟶ C H_{3} CON H^{-} N a^{+} + \frac{1}{2} H_{2}

Figure 11: Amide resonance, intermolecular hydrogen bonding and amphoteric behaviour.

Hydrolysis, reduction and dehydration

Amides are hydrolysed when heated with aqueous acid or aqueous base.

R - CON H_{2} + H_{3} O^{+} Δ H_{2} O R - COOH + N H_{4}^{+}

R - CON H_{2} + O H^{-} Δ H_{2} O R - CO O^{-} + N H_{3}

Amides are reduced by Na/ $C_{2} H_{5} OH$ or by $LiAl H_{4}$ to primary amines.

R - CON H_{2} + 4 [H] Na/ C_{2} H_{5} OH or LiAl H_{4} R - C H_{2} - N H_{2} + H_{2} O

When heated with $P_{2} O_{5}$ , amides are dehydrated to nitriles (cyanides). $PC l_{5}$ or $SOC l_{2}$ also convert amides into nitriles.

R - CON H_{2} P_{2} O_{5}, - H_{2} O R - C \equiv N

R - CON H_{2} PC l_{5} R - CC l_{2} N H_{2} - 2HCl R - C \equiv N

With nitrous acid, amides lose nitrogen and give the carboxylic acid; the electrophile is $N O^{+}$ , formed from $HN O_{2}$ .

R - CON H_{2} + HONO ⟶ R - COOH + N_{2} + H_{2} O

Hofmann bromamide rearrangement

Amides with no substituents on nitrogen react with $B r_{2}$ or $C l_{2}$ in NaOH solution to give amines with one carbon atom less, through the Hofmann rearrangement.

R - CON H_{2} + B r_{2} + 4NaOH

⟶ RN H_{2} + 2NaBr + N a_{2} C O_{3} + 2 H_{2} O

C_{6} H_{5} CON H_{2} + B r_{2} + 4KOH

⟶ C_{6} H_{5} N H_{2} + K_{2} C O_{3} + 2KBr + 2 H_{2} O

Figure 12: Mechanism of the Hofmann bromamide rearrangement.

Exam Trick: Hofmann hacks, Beckmann swaps

Hofmann hacks one carbon: $RCON H_{2}$ ( $n$ C) → $RN H_{2}$ ( $n - 1$ C). Beckmann swaps: the group anti to -OH moves to N, so the carbon count stays the same. Both keep the configuration of the migrating group.

Hofmann bromamide (

B r_{2}

, NaOH)

$C H_{3} C H_{2} CON H_{2}$ → $C H_{3} C H_{2} N H_{2}$

One carbon fewer; the C=O carbon leaves as carbonate.

Reduction (

LiAl H_{4}

)

$C H_{3} C H_{2} CON H_{2}$ → $C H_{3} C H_{2} C H_{2} N H_{2}$

Same carbon count; C=O becomes $C H_{2}$ .

Solved Example 14

Acetamide reacts with NaOBr in alkaline medium to form:
(A)

N H_{3}

(B)

C H_{3} N H_{2}

(C)

C H_{3} CN

(D)

C_{2} H_{5} N H_{2}

Solution:

This is the Hofmann bromamide reaction, which removes one carbon: $C H_{3} CON H_{2}$ gives methylamine. Answer: (B)

Solved Example 15

A (C_{8} H_{8} O) N H_{2} OH/HCl

B and C. B

H^{+}

alc. KOH

(C_{7} H_{6} O_{2})

; C

H^{+}

alc. KOH

(C_{6} H_{7} N)

. E can also be made from G and

C H_{3} COCl

. B, C, D and E are all isomers of formula

C_{8} H_{9} NO

. Identify A to G.

Solution:

A forms an oxime, so it is a carbonyl compound: acetophenone, $C_{6} H_{5} COC H_{3}$ . It gives two geometrical isomers of the oxime, B and C. In acid, each undergoes the Beckmann rearrangement, in which the group anti to -OH migrates to nitrogen.

Figure 13: Beckmann rearrangement of the two acetophenone oximes (Solved Example 15).

A = acetophenone; B = oxime with $C H_{3}$ anti to OH; C = oxime with $C_{6} H_{5}$ anti to OH; D = $C_{6} H_{5} CONHC H_{3}$ (N-methylbenzamide); E = $C H_{3} CONH C_{6} H_{5}$ (acetanilide); F = $C_{6} H_{5} COOH$ ; G = $C_{6} H_{5} N H_{2}$ .

Solved Example 16

Formic acid and acetic acid may be distinguished by reaction with:
(A) sodium
(B) dilute acidic permanganate
(C) 2,4-dinitrophenylhydrazine
(D) sodium ethoxide

Solution:

Formic acid contains an aldehyde-like H-C=O unit and is oxidised to $C O_{2}$ , decolourising acidified $KMn O_{4}$ ; acetic acid is not oxidised. Both acids react with Na and sodium ethoxide, and neither forms a 2,4-DNP derivative. Answer: (B)

7. Chapter Summary

Figure 14: Mind map of carboxylic acid derivatives: one mechanism (addition-elimination), one reactivity ladder, many named reactions.

Quick Recall: tap to check

Which derivative is the usual starting point for making the others, and why?

The acid chloride: it is at the top of the reactivity ladder.

What does an anhydride give with an alcohol?

One ester molecule plus one carboxylic acid molecule.

Which group migrates in the Beckmann rearrangement?

The group anti to the -OH of the oxime.

8. Solved Examples: Exam Practice

Solved Example 17

Which compound reacts fastest with water at room temperature?
(A)

C H_{3} COCl

(B)

(C H_{3} CO)_{2} O

(C)

C H_{3} COO C_{2} H_{5}

(D)

C H_{3} CON H_{2}

Solution:

Answer: (A). Chloride is the weakest base of the four leaving groups and chlorine donates least into C=O, so acetyl chloride is hydrolysed instantly; the amide needs prolonged boiling.

Solved Example 18

Ethyl acetate is treated with excess

C H_{3} MgBr

and then with dilute acid. The main organic product is:
(A) acetone
(B) 2-methylpropan-2-ol
(C) propan-2-ol
(D) butan-2-ol

Solution:

Answer: (B). The first $C H_{3} MgBr$ replaces $- O C_{2} H_{5}$ to give acetone, which is more reactive than the ester and adds a second methyl group: $(C H_{3})_{3} COH$ . Ethanol is the by-product.

Solved Example 19

Write the balanced equation for the Hofmann reaction of benzamide and state the change in carbon count.

Solution:

C_{6} H_{5} CON H_{2} + B r_{2} + 4NaOH C_{6} H_{5} N H_{2} + N a_{2} C O_{3} + 2NaBr + 2 H_{2} O

Benzamide (7 C) gives aniline (6 C): the carbonyl carbon leaves as $N a_{2} C O_{3}$ .

Solved Example 20

0.88 g of a neutral ester needs exactly 10.0 mL of 1.00 M NaOH for complete saponification. Find its molar mass and suggest two structures.

Solution:

Moles of NaOH = $0.0100 L \times 1.00 mol L^{- 1} = 0.0100$ mol. For one -COOR group, moles of ester = 0.0100 mol, so the saponification equivalent (molar mass) is $\frac{0.88}{0.0100} = 88 g mo l^{- 1}$ .

An ester $Cn H_{2} n O_{2}$ with molar mass 88 has $n = 4$ : $C_{4} H_{8} O_{2}$ . Possible structures: ethyl acetate, $C H_{3} COO C_{2} H_{5}$ , or methyl propanoate, $C H_{3} C H_{2} COOC H_{3}$ (also propyl or isopropyl formate).

Practice Questions

Complete: $C H_{3} COCl$ + $C_{2} H_{5} OH$ →Answer: $C H_{3} COO C_{2} H_{5}$ + HCl
What does benzoyl chloride give with $H_{2}$ , Pd/ $BaS O_{4}$ ?Answer: benzaldehyde
Acetic anhydride + 2 $N H_{3}$ →Answer: acetamide + ammonium acetate
Ethyl benzoate is boiled with NaOH solution. Give the products.Answer: sodium benzoate and ethanol
Acetamide is heated with $P_{2} O_{5}$ . Name the product.Answer: ethanenitrile (acetonitrile), $C H_{3} CN$
Propanamide with $LiAl H_{4}$ , then water, gives?Answer: propan-1-amine, $C H_{3} C H_{2} C H_{2} N H_{2}$
Diethyl hexanedioate with NaOEt, then acid, gives?Answer: ethyl 2-oxocyclopentane-1-carboxylate (Dieckmann)

Common Mistakes to Avoid

Watch out

Reversing the reactivity order. It is acid chloride > anhydride > ester > amide; an amide cannot be converted directly into an acid chloride or ester by simple substitution.
Saying that $C l^{-}$ leaves easily "because chlorine is electronegative" only. The key reasons are that $C l^{-}$ is a very weak base and that Cl donates poorly into C=O.
Using $LiAl H_{4}$ to make an aldehyde from an acid chloride. It goes on to the alcohol; use Rosenmund reduction ( $H_{2}$ , Pd/ $BaS O_{4}$ ).
Calling saponification reversible. The carboxylate ion formed does not react with alcohol, so it is irreversible; acid hydrolysis is reversible.
Counting one mole of Grignard reagent for an ester. Two moles add, giving a 3° alcohol with two identical groups.
Using sodium methoxide with an ethyl ester in a Claisen condensation. Mismatched alkoxide causes transesterification.
Forgetting that only one acyl group of an anhydride is transferred; the other becomes the carboxylic acid or its salt.
Keeping the carbon count in the Hofmann reaction. The amine has one carbon fewer than the amide; $LiAl H_{4}$ reduction keeps the carbon count.
In the Beckmann rearrangement, moving the group syn to -OH. The group anti to -OH migrates.

Frequently Asked Questions

Why are acid chlorides more reactive than esters and amides?

Chloride is a very weak base and therefore an excellent leaving group, and chlorine's lone pairs overlap poorly with the carbonyl carbon, so acid chlorides get little resonance stabilisation. Esters and amides have better-donating oxygen and nitrogen atoms and much poorer leaving groups, so they react more slowly.

What is nucleophilic acyl substitution?

It is the reaction type of all acid derivatives. A nucleophile first adds to the carbonyl carbon to form a tetrahedral intermediate, and then the leaving group is expelled as the carbonyl bond re-forms. The net result is replacement of the leaving group by the nucleophile.

Why is saponification irreversible but acid hydrolysis of esters reversible?

In saponification the acid formed is immediately deprotonated to a carboxylate ion, which is negatively charged and not attacked by the alcohol. In acid hydrolysis every step is an equilibrium, and the reverse reaction is simply Fischer esterification.

What is the difference between Claisen and Dieckmann condensation?

Both join an ester enolate to an ester carbonyl to give a beta-keto ester. The Claisen condensation is intermolecular, such as two ethyl acetate molecules giving ethyl acetoacetate. The Dieckmann condensation is intramolecular and forms five- or six-membered cyclic beta-keto esters from diesters.

How does the Hofmann bromamide reaction reduce the carbon chain?

Bromine and base convert the amide into an N-bromoamide anion. The alkyl or aryl group then moves from the carbonyl carbon to nitrogen, forming an isocyanate. Hydrolysis removes that carbonyl carbon as carbon dioxide, leaving a primary amine with one carbon fewer.

Why are amides so weakly basic compared with amines?

In an amide the nitrogen lone pair is delocalised into the carbonyl group by resonance, so it is not free to accept a proton. Amines have a localised lone pair. Amides therefore form only unstable salts with strong acids and can even show weak acidity.

Which acid derivative reactions are important for JEE?

JEE Main and Advanced often test the reactivity order, mechanisms of nucleophilic acyl substitution and saponification, Claisen and Dieckmann condensations, esters with Grignard reagents, Hofmann and Curtius rearrangements, Arndt-Eistert homologation and Rosenmund reduction, frequently inside multi-step problems.

What should NEET students focus on in acid derivatives?

For NEET, learn the reactivity order, Rosenmund reduction, Friedel-Crafts acylation, preparation of esters and amides, saponification, reduction of amides and esters with LiAlH4, dehydration of amides with P2O5, and the Hofmann bromamide degradation giving an amine with one carbon less.

Previous year questions on Carboxylic Acid Derivatives

11 questions from past papers, each with a step-by-step solution.

Show all 11 questions

NEET 2023, Chemistry Q33

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