Choosing the reagent: selectivity in carbonyl chemistry

One molecule, two groups

Think of ethyl 4-oxopentanoate. It has a ketone and an ester. Add one reagent to it, and ask: which group reacts? A reagent that is right for a simple ketone can wreck a molecule that also has an ester. The skill is to choose the reagent that picks the group you want. We call this chemoselectivity.

You need two things for this: the order of reactivity of the groups, and the list of what each reagent can do.

The ladder of reactivity

A nucleophile attacks the carbonyl carbon. The more positive the carbon, and the better the group that can leave, the faster the reaction. So the order of reactivity is:

• In an acyl chloride, the chlorine pulls electrons and does not donate them back. The carbon is very positive.
• An ester has an oxygen with a lone pair that donates to the carbonyl by resonance. The carbon is less positive, so the ester is less reactive than a ketone.
• An amide has a nitrogen that donates more strongly. It is the least reactive.
• In base, a carboxylic acid loses its proton and becomes a carboxylate ion. A negative ion repels the nucleophile.

Hydride reagents

All of them give a hydride ion, $\text{H}^-$, to the carbonyl carbon. They differ in strength.

• Sodium borohydride, $\text{NaBH}_{4}$, is mild. It reduces aldehydes and ketones, and acyl chlorides. It leaves esters, acids, amides, nitriles and C=C alone. It works in ethanol or water.
• Lithium aluminium hydride, $\text{LiAlH}_{4}$, is strong. It reduces all the carbonyl groups: aldehyde, ketone, ester, acid, acyl chloride and amide (to an amine). It reduces a nitrile to an amine. It does not touch an isolated C=C. It must be used in dry ether, and then dilute acid is added.
• DIBAL-H (diisobutylaluminium hydride) is bulky. In the cold, at about $-78\ ^\circ\text{C}$, it takes an ester or a nitrile only as far as the aldehyde. A second, warmer treatment would give the alcohol.
• Hydrogen with palladium reduces a C=C bond easily, and a ketone or aldehyde more slowly.

An ester needs two hydrides. The first gives an aldehyde, which is reduced at once, so an ester gives two alcohols: $\text{RCOOR}' + _{4}[\text{H}] \longrightarrow \text{RCH}_{2}\text{OH} + \text{R}'\text{OH}$. A ketone needs one hydride. Each $\text{LiAlH}_{4}$ holds four.

Grignard reagents

A Grignard reagent, RMgX, is a carbon nucleophile and a strong base. It needs dry ether, because water destroys it. It also reacts with any acidic hydrogen in the molecule, such as $\text{O}\text{-}\text{H}$ or $\text{N}\text{-}\text{H}$, before it attacks a carbonyl group.

• Methanal: a primary alcohol. Another aldehyde: a secondary alcohol. A ketone: a tertiary alcohol. These use one equivalent.
• Carbon dioxide: a carboxylic acid with one carbon more.
• A nitrile: a ketone, after hydrolysis. One equivalent.
• An ester or an acyl chloride: a tertiary alcohol with two identical R groups, and the ester part leaves as an alcohol or chloride. It uses two equivalents, because the ketone formed in the middle is more reactive than the ester and reacts at once.

Set the substrate to 4, the ester, and 20 mmol. The solution is $14.7\ \text{mL}$ of a $3\ \text{mol/L}$ solution. Choose 1, the ketone, and the volume halves.

Practise with new numbers

Hiding a group: acetals as protecting groups

Suppose you want to reduce the ester of ethyl 4-oxopentanoate to an alcohol with $\text{LiAlH}_{4}$, but keep the ketone. The hydride would reduce both. The way out has three steps.

1. Protect. Ethylene glycol with a little acid turns the ketone into a cyclic acetal (a ketal). An acetal is stable to base and to hydrides.
2. React. $\text{LiAlH}_4$, then water, reduces the ester to a primary alcohol. The acetal is not touched.
3. Deprotect. Dilute aqueous acid turns the acetal back into the ketone.

The product is 5-hydroxypentan-2-one. The cost is two extra steps and some loss of yield at each, so a chemist uses a protecting group only when no selective reagent exists.

Try it in Python

The program stores what each reagent reduces. For a keto ester it prints what is reduced and what is left.

Common mistakes

• Using sodium borohydride on an ester. It does not reduce it.
• Giving an ester one equivalent of Grignard reagent. It uses two, and the product has two identical R groups.
• Using a Grignard reagent on a molecule with an O-H group. The O-H is attacked first.
• Using water with lithium aluminium hydride. It reacts violently, and the solvent must be dry.
• Skipping the deprotection. The acetal stays until acid removes it.

Remember these

• Reactivity: acyl chloride, aldehyde, ketone, ester, acid, amide.
• NaBH$_4$: aldehyde and ketone. LiAlH$_4$: every carbonyl group. DIBAL-H in the cold: ester or nitrile to aldehyde.
• RMgX: 1 equivalent for an aldehyde, a ketone or a nitrile, 2 for an ester or an acyl chloride.
• An acetal protects a carbonyl group from bases and hydrides, and acid removes it.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which reduces methyl 4-oxopentanoate selectively at the ketone: NaBH$_4$ or LiAlH$_4$?

2. How many equivalents of CH$_3$MgBr does ethyl ethanoate use, and what is the product?

3. Why is an acid not reduced by sodium borohydride?

4. A molecule has a ketone and an O-H group. Why can it not be treated directly with a Grignard reagent?

References

• Clayden, Greeves and Warren, Organic Chemistry, the chapters on nucleophilic addition to the carbonyl group, on nucleophilic substitution at the carbonyl group and on chemoselectivity and protecting groups.
• OpenStax, Chemistry 2e, Chapter 20, section 20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters.
• NCERT, Chemistry Part II for Class XII, Chapter 8.