Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: carboxylic acids and their derivatives (202-SNU-RE)

Acid derivatives are the only family in organic chemistry that is arranged entirely on a scale: from the acyl chloride to the carboxylate, each one can give those that follow it and none can move back up the line without going through the acid.

This sheet leaves the course in your notes and gathers what decides an exam: the permitted direction of interconversions, the two-stage mechanism, and the difference between an equilibrium and a complete reaction.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A single hierarchy explains the whole chapter, the reactivity of the derivatives, and it follows exactly the order of the $pK_{a}$ values of the conjugate acids of their leaving groups.

This chapter is part of 202-SNU-RE, Organic Chemistry

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Carboxylate ion
Electron-withdrawing inductive effect
Acyl chloride
Acid anhydride
Ester
Amide
Nucleophilic acyl substitution
Tetrahedral intermediate
Fischer esterification
Saponification
Resonance of the amide
Reduction by hydrides

The essentials

The carboxyl group and its acidity

• The carboxylate ion is stabilized by TWO equivalent resonance forms: its two carbon-oxygen bonds are identical, $127$ pm long.
• In the non-ionized acid, they measure $123$ and $136$ pm: the resonance only really appears after the proton has left.
• Hence $pK_{a}\approx 4.8$ for a carboxylic acid, against $16$ for an alcohol: eleven orders of magnitude.
• $pK_{a}$ values to know: methanoic $3.75$ ; ethanoic $4.76$ ; benzoic $4.20$ ; chloroethanoic $2.87$ ; dichloroethanoic $1.35$ ; trichloroethanoic $0.66$ .

The two resonance forms of the carboxylate are superimposable: the charge is shared equally, and the two bonds become identical.

Each added chlorine gains more than one unit of $pK_{a}$ through the electron-withdrawing inductive effect, and the effect fades quickly with distance: a chlorine at the end of the chain changes almost nothing.

The reactivity hierarchy

• From most to least reactive: acyl chloride > anhydride > ester > amide > carboxylate.
• It follows exactly the order of the $pK_{a}$ values of the conjugate acids of the leaving groups: $HCl$ $-7$ , $RCOOH$ $4.8$ , $ROH$ $16$ , amine $36$ .
• A derivative can give all those that FOLLOW it in the list, never those that precede it.
• To move back up the hierarchy, you must go back through the acid, then use $SOCl_{2}$ .

The five derivatives, from most to least reactive: under each, the

pK_{a}

of the conjugate acid of its leaving group, which fixes the order.

This scale is not to be learned by heart: it is rebuilt by ranking the leaving groups from the most stable to the least stable, that is, from the weakest base to the strongest.

The mechanism: addition, then elimination

• Nucleophilic ACYL substitution, in two stages: ADDITION of the nucleophile, which gives a TETRAHEDRAL intermediate, then ELIMINATION of the leaving group, which restores the carbonyl.
• Never an $S_{N}2$ : the carbon is $sp^{2}$ and backside attack is impossible.
• It is the tetrahedral intermediate that decides: it expels the better leaving group of the two available.

The carbon becomes tetrahedral for one step, then flat again: it is this round trip that replaces the one-step substitution, which is impossible here.

This rule alone predicts the direction of interconversions: if the incoming nucleophile is a poorer leaving group than the one in place, the reaction goes forward; otherwise it goes back.

Fischer, saponification, amide and reductions

• FISCHER ESTERIFICATION: acid plus alcohol, acid catalysis, an EQUILIBRIUM with a constant close to $4$ . Labelling with oxygen-18 proves that the bond broken is that of the ACID.
• SAPONIFICATION: ester plus $HO^{-}$ , COMPLETE and irreversible, because the final step traps the acid as a carboxylate.
• AMIDE: the nitrogen lone pair makes the $C-N$ bond partly double, $133$ pm instead of $147$ , the group is PLANAR, and the nitrogen is NOT basic.
• $LiAlH_{4}$ : ester and acid give a primary ALCOHOL; amide and nitrile give an AMINE. $NaBH_{4}$ reduces none of these derivatives.

The nitrogen lone pair moves toward the carbonyl: the

C-N

bond becomes partly double, shorter, and the whole amide is flat.

An equilibrium can be shifted, a complete reaction cannot: that is why the water is driven off in Fischer esterification, and why nothing of the kind is done in saponification.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Writing a one-step substitution at the acyl carbon

2 points, and the whole mechanism is marked wrong

What not to write

“The nucleophile attacks from the back and the chloride leaves at the same time, as in an $S_{N}2$ .”

What to write

“The carbon is $sp^{2}$ : the nucleophile first ADDS, then the leaving group is ELIMINATED. Two steps, one tetrahedral intermediate.”

Why: A backside attack requires a tetrahedral carbon and a clear site behind the leaving group. Here, the $\pi$ electrons of the carbonyl offer a much easier path.

2. Believing that saponification is reversible

2 points, and the announced yield is wrongly limited

What not to write

“Saponification is the reverse of esterification, so it is an equilibrium.”

What to write

“It is COMPLETE: the final step is an acid-base reaction that turns the acid into a carboxylate, out of reach of the reverse reaction.”

The last step of saponification is acid-base: the carboxylate formed, a very weak base, does not go back.

Why: The carboxylate is the last in the hierarchy: it no longer reacts with the alcohol. It is this irreversible capture, and not the nature of the attack, that makes the reaction complete.

3. Getting the bond broken in esterification wrong

2 points, it is the experimental question of the chapter

What not to write

“It is the $O-H$ bond of the alcohol that breaks, as in an acid-base reaction.”

What to write

“Labelling with oxygen-18 shows that it is the $C-OH$ bond of the ACID that breaks: the oxygen of the ester comes from the alcohol.”

The labelled oxygen of the alcohol ends up in the ester: it is the

C-OH

bond of the ACID that breaks, and its

-OH

leaves in the water.

Why: The isotopic labelling experiment settles it unambiguously: the labelled alcohol gives a labelled ester. It is the historical proof of the addition-elimination mechanism.

4. Treating the nitrogen of an amide as basic

2 points, and an invented acid-base reaction

What not to write

“The amide carries a nitrogen with a lone pair, so it is a base like an amine.”

What to write

“This lone pair is delocalized toward the carbonyl: the $C-N$ bond measures $133$ pm instead of $147$ , and the nitrogen is NOT basic.”

Protonated on the oxygen, the amide keeps its resonance; protonated on the nitrogen, it would lose it: the nitrogen of an amide is not basic.

Why: Protonating the nitrogen would destroy the resonance that stabilizes the amide, which costs about $80$ kJ/mol. It is the same logic as for aniline, only more pronounced.

5. Moving back up the reactivity hierarchy

2 points, and an impossible synthesis proposed

What not to write

“The ester is turned into an acyl chloride by adding sodium chloride.”

What to write

“You never move back up the line: the ester must be hydrolysed to the acid, then $SOCl_{2}$ is used.”

You go down the scale in one step; to go back up, you go back through the acid, then

SOCl_{2}

Why: The tetrahedral intermediate always expels the BETTER leaving group. Getting a chloride into an ester would amount to asking it to stay when the alkoxide is harder to expel.

6. Reducing an ester with borohydride

2 points, and nothing happens in reality

What not to write

“ $NaBH_{4}$ reduces the ester to a primary alcohol.”

What to write

“ $NaBH_{4}$ touches neither esters nor acids: $LiAlH_{4}$ in anhydrous medium is needed.”

Why: The neighbouring oxygen stabilizes the carbonyl of the ester by resonance and makes it much less electrophilic than a ketone. Borohydride, being mild, is not strong enough.

7. Expecting an alcohol when reducing an amide

2 points, and a functional group lost in the synthesis

What not to write

“ $LiAlH_{4}$ reduces every carbonyl to an alcohol, so the amide gives an alcohol.”

What to write

“The amide and the nitrile give an AMINE: the nitrogen stays, only the oxygen leaves.”

LiAlH_{4}

gives an alcohol from an ester, but an AMINE from an amide: the nitrogen stays, only the oxygen leaves.

Why: The intermediate loses the oxygen rather than the nitrogen, because the amide anion is a much worse leaving group than the aluminated oxygen. It is the hierarchy that decides, once again.

8. Confusing the acidity of the acid and that of the alcohol

1.5 points, and the ranking of a series of acidities

What not to write

“The carboxylic acid and the alcohol both carry an $O-H$ , so their acidities are close.”

What to write

“ $4.8$ against $16$ : eleven orders of magnitude, because only the carboxylate is stabilized by resonance.”

Why: Acidity is read from the conjugate base. The carboxylate shares its charge over two equivalent oxygens, the alkoxide keeps it whole on a single one.

9. Making a carboxylate react like an acid

2 points on the synthesis question

What not to write

“The sodium carboxylate is heated with an alcohol to obtain the ester.”

What to write

“The carboxylate is the LAST in the hierarchy: it must be reprotonated to the acid, or you go through the acyl chloride.”

Why: A carbon that is already negatively charged can no longer attract a nucleophile. It is precisely this blockage that makes saponification irreversible.

Which method to choose

Which reagent to go from one derivative to another

Locate the two functional groups on the hierarchy, and the direction of the change.

If you go down the hierarchy, from the chloride to the ester or the amide → just add the corresponding nucleophile, alcohol or amine

Example: acyl chloride plus alcohol gives the ester, quickly
If you go from the acid to the ester → Fischer with acid catalysis, or go through the acyl chloride if the yield matters

Example: $K$ close to $4$ limits Fischer to about two thirds
If you want to go back up, from the ester to the acyl chloride → hydrolyse first to the acid, then use $SOCl_{2}$

Example: there is no shortcut
If you want to hydrolyse an ester → hot sodium hydroxide for a complete reaction, dilute acid for an equilibrium

Example: saponification gives the carboxylate, not the acid
If you want an alcohol or an amine → $LiAlH_{4}$ : alcohol from an ester or an acid, amine from an amide or a nitrile

Example: $NaBH_{4}$ is suitable in none of these cases

The permitted direction is read from the leaving groups: the reaction goes forward if the group that leaves is a BETTER leaving group than the one that enters.

Equilibrium or complete reaction

Check whether a product is removed from the medium by an irreversible step.

If Fischer esterification → equilibrium: shift it by driving off the water or forcing a reactant

Example: $K\approx 4$ gives a yield of $67\,\%$ from equal amounts
If saponification → complete: the carboxylate formed can no longer react in the reverse direction

Example: the final acid-base step locks everything in
If reaction of an acyl chloride → complete: the hydrogen chloride formed leaves the medium

Example: a base is added to neutralize it
If hydrolysis of an ester in acidic medium → equilibrium, exactly the reverse of Fischer

Example: the water is put in large excess to shift it

A yield above that of the equilibrium is impossible without shifting it: announcing $95\,\%$ for a Fischer esterification without saying how the water is removed costs the mark.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up an addition-elimination

When to use it: The question asks for the mechanism of an esterification, a saponification or a reaction of an acyl chloride.

1 Say whether the medium is acidic or basic, and protonate the carbonyl if needed.
2 Have the nucleophile attack the CARBON, and show the tetrahedral intermediate with its charge.
3 Eliminate the BETTER leaving group, justifying this choice.
4 End with the acid-base step that makes the product neutral, or that locks the reaction.

Concluding sentence

“The tetrahedral intermediate expels the alkoxide rather than the hydroxide ion, because it is the better of the two leaving groups.”

The trap: Making the leaving group disappear at the same time as the attack: that is writing an $S_{N}2$ , which does not exist here.

Marking: 1 mark for the tetrahedral intermediate, 1 mark for the justified choice of leaving group, 1 mark for the final step.

Calculating an equilibrium yield

When to use it: The question gives the equilibrium constant of an esterification and asks for the yield.

1 Set up a reaction table in amounts of substance, with $x$ the extent of reaction.
2 Write the constant with the equilibrium amounts, the volumes cancelling.
3 Solve, taking the square root when the coefficients allow it.
4 Conclude with a yield, that is, a percentage of the limiting reactant.

Concluding sentence

“ $\dfrac{x^{2}}{(1-x)^{2}}=4$ gives $\dfrac{x}{1-x}=2$ , so $x=\dfrac{2}{3}$ : the yield is $67\,\%$ .”

The trap: Forgetting that the square root simplifies the equation: many papers launch into a needless quadratic.

Marking: 1 mark for the table, 1 mark for the expression of $K$ , 1 mark for the solution, 0.5 mark for the yield as a percentage.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The direction of the interconversion

Check that the derivative formed lies AFTER the starting derivative in the hierarchy.

An acyl chloride obtained directly from an ester is impossible.
Is the intermediate tetrahedral?

Reread the mechanism: the carbon must pass through four single bonds before becoming flat again.

A mechanism with no intermediate drawn is a one-step substitution, hence wrong.
Does the yield exceed the equilibrium?

Compare the announced yield with what the constant allows. Beyond it, you must explain how the equilibrium was shifted.

$90\,\%$ for a simple esterification at $K=4$ is impossible without removing water.
The unit and the nature of the final product

After a saponification, the product is a CARBOXYLATE; an acidification is needed to obtain the acid.

Announcing the acid directly after the sodium hydroxide forgets a step.

The typical problem, taken apart

The yield of a Fischer esterification

$1$ mol of ethanoic acid is heated with $1$ mol of ethanol in the presence of a few drops of sulfuric acid. The equilibrium constant is $4$ .

Calculate the yield at equilibrium, then propose two ways of improving it.

The situation of the example: equal amounts of acid and alcohol, a few drops of sulfuric acid, and an equilibrium constant of 4.

Step 1

Reaction table: $1-x$ mol of acid and of alcohol remain, and $x$ mol of ester and $x$ mol of water have formed.

Why

The coefficients are all $1$ , which makes the table immediate. It is also what will allow the simplification by square root.

Step 2

$K=\dfrac{x\times x}{(1-x)(1-x)}=4$ .

Why

The volumes cancel since there are as many moles on each side. We therefore work directly in amounts of substance, which lightens the whole calculation.

Step 3

Taking the root: $\dfrac{x}{1-x}=2$ , so $x=2-2x$ and $x=\dfrac{2}{3}$ .

Why

Both sides being squares of positive quantities, the root is legitimate. It avoids a quadratic and any discussion of signs.

Step 4

Yield: $\dfrac{2/3}{1}\approx 67\,\%$ .

Why

The yield is relative to the limiting reactant, here both at once. It is the classic value for the esterification of an acid and a primary alcohol.

Step 5

Improvement: remove the water as it forms, or use the alcohol in large excess.

Why

An equilibrium is shifted by removing a product or adding a reactant. Acid catalysis, for its part, changes NOTHING about the yield: it only speeds things up.

Step 6

Another route: go through the acyl chloride, a complete reaction.

Why

The hydrogen chloride formed leaves the medium or is neutralized, which locks the reaction. The yield is then limited by no equilibrium.

Step 7

Check: $x=\dfrac{2}{3}$ does give back $\dfrac{(2/3)^{2}}{(1/3)^{2}}=4$ .

Why

The solution is put back into the starting expression. It is the quickest check, and it catches sign errors.

The conclusion, written out

“The yield at equilibrium is $67\,\%$ . To exceed it, the equilibrium must be shifted by driving off the water or forcing the alcohol, or you must change route by going through the acyl chloride, whose reaction is complete.”

The classic mistake on this problem: Announcing that the acid catalyst improves the yield: it only speeds up the approach to equilibrium, without shifting its position.

Learn by heart

• The carboxylate has TWO equivalent resonance forms: hence $pK_{a}$ close to $4.8$ against $16$ for an alcohol.
• Hierarchy: acyl chloride > anhydride > ester > amide > carboxylate. You never move back up it directly.
• It follows the $pK_{a}$ values of the conjugate acids of the leaving groups: $HCl$ $-7$ , acid $4.8$ , alcohol $16$ , amine $36$ .
• Mechanism: ADDITION then ELIMINATION, with a tetrahedral intermediate. Never an $S_{N}2$ .
• Fischer: EQUILIBRIUM, $K$ close to $4$ , yield $67\,\%$ . Saponification: COMPLETE.
• Labelling with oxygen-18 proves that the bond broken is that of the ACID.
• The amide is PLANAR and its nitrogen is not basic. $LiAlH_{4}$ : ester and acid to alcohol, amide and nitrile to AMINE.

Frequently asked questions

Why is a carboxylic acid so much more acidic than an alcohol?

Because its conjugate base, the carboxylate ion, has two equivalent resonance forms that share the negative charge between two oxygens. The alkoxide keeps it whole on a single atom. The gap reaches eleven pKa units, which explains why a carboxylic acid reacts with sodium hydrogencarbonate and an alcohol does not.

Why is the reaction never a one-step substitution?

Because the carbonyl carbon is flat and sp2 hybridized: backside attack, which a concerted substitution requires, is impossible. The nucleophile therefore adds to the double bond, which gives a tetrahedral intermediate, then the leaving group is eliminated and the carbonyl re-forms.

Why is saponification complete?

Because its last step is an acid-base reaction: the acid formed is immediately turned into the carboxylate ion by the excess base. The carboxylate is the last in the reactivity hierarchy and can no longer react with the alcohol. The product is therefore trapped, and the reverse reaction impossible.

Why is the nitrogen of an amide not basic?

Because its lone pair no longer really belongs to it: it is delocalized toward the neighbouring carbonyl, which shortens the carbon-nitrogen bond to one hundred and thirty-three picometres and makes the group flat. Picking up a proton would destroy this stabilization, which is worth about eighty kilojoules per mole.

How can the yield of an esterification be improved?

By shifting the equilibrium, never by adding catalyst, which only speeds things up. The water is removed as it forms, or one of the two reactants is used in large excess. The other solution is to change route and go through an acyl chloride, whose reaction with the alcohol is complete.

Practise it

Corrected exercises: Carboxylic acids and their derivatives, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Aldehydes and ketones Next sheet Amines

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-carboxylic-acids-derivatives. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 202-SNU-RE tutor in Montreal?

Get in touch for a first session. All of the reactivity of acid derivatives comes down to a single question, “is this leaving group a weak base?”, and that is the method set up here.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: carboxylic acids and their derivatives (202-SNU-RE)

The essentials

The carboxyl group and its acidity

The reactivity hierarchy

The mechanism: addition, then elimination

Fischer, saponification, amide and reductions

The mistakes that cost marks

1. Writing a one-step substitution at the acyl carbon

2. Believing that saponification is reversible

3. Getting the bond broken in esterification wrong

4. Treating the nitrogen of an amide as basic

5. Moving back up the reactivity hierarchy

6. Reducing an ester with borohydride

7. Expecting an alcohol when reducing an amide

8. Confusing the acidity of the acid and that of the alcohol

9. Making a carboxylate react like an acid

Which method to choose

Which reagent to go from one derivative to another

Equilibrium or complete reaction

How the answer is expected to be written

Writing up an addition-elimination

Calculating an equilibrium yield

Check before you hand in

The typical problem, taken apart

The yield of a Fischer esterification

Learn by heart

Frequently asked questions

Why is a carboxylic acid so much more acidic than an alcohol?

Why is the reaction never a one-step substitution?

Why is saponification complete?

Why is the nitrogen of an amide not basic?

How can the yield of an esterification be improved?

Corrected exercises: Carboxylic acids and their derivatives, 202-SNU-RE

See also

Looking for a 202-SNU-RE tutor in Montreal?