Carbonyl Chemistry Section 74 of 123

Hydrates & cyanohydrins

the why on every wrong step

12 min readfree

Put some acetaldehyde, CH3CHO, in water and wait. Just over half of it turns into CH3CH(OH)2, with two OH groups on the carbon that used to be the carbonyl carbon. Do the same with acetone and almost none of it changes.

Both flasks run the addition mechanism from the previous section, with water as the nucleophile. This section follows two small, simple nucleophiles, water and cyanide, and at the end a third, bisulfite. Water shows how far an addition goes, and why. Cyanide makes a new carbon–carbon bond. An alcohol adds in the same way, and then a second alcohol replaces the OH; that takes the next section.

Hydrates: addition of water

The product of adding water is the gem-diol, or hydrate, that the Aldehydes & ketones section at the start of this chapter used as evidence for the reactivity order: two OH groups on one carbon.

R2C=O + H2O  ⇌  R2C(OH)2

The mechanism is the one from the previous section, with water, or in base hydroxide, as the nucleophile. Pure water adds slowly; a trace of base or of acid speeds it up a great deal. In base, hydroxide attacks the carbon and the alkoxide takes a proton from water, which gives the hydroxide back.

HYDROXIDE ADDS TO THE CARBONthe π electrons move onto OOCH₃HCOH−THE ALKOXIDE TAKES H⁺ FROM WATERwater becomes HO⁻ againOHCH₃OHC−HOHTHE HYDRATE, AND HO⁻ IS BACKa gem-diol: two OH on one carbonOHHCH₃OHC+ HO⁻THE NET CHANGEHO⁻ is used, then handed backCH₃CHO + H₂OHO⁻ (catalyst)CH₃CH(OH)₂
Hydration in base, drawn on acetaldehyde. Follow the hydroxide: it enters in the first panel and leaves in the third.

In acid, the proton goes on first, neutral water attacks the activated carbon, and a second water takes the extra proton away. The acid comes back out.

ACID PUTS H⁺ ON THE OXYGENan O lone pair takes H⁺ from H₃O⁺OCH₃HCHOH₂+WATER ADDS TO THE ACTIVATED CARBONthe π electrons move onto O⁺OCH₃HCH+OHHA SECOND WATER TAKES OFF H⁺the O–H electrons stay on that OOHHCH₃OCHH+OH₂THE HYDRATE, AND H₃O⁺ IS BACKthe acid was a catalystOHHCH₃OHC+ H₃O⁺
The acid mechanism from Nucleophilic addition, run on acetaldehyde with water as the nucleophile. The proton goes on first and comes off last.

What is new here is the reverse direction. Every step runs backward as easily as forward: the hydrate can lose water and give the carbonyl back. So the reaction settles at an equilibrium. A catalyst changes how fast the equilibrium is reached. It does not change where it lies.

Where it lies depends on the carbonyl. Aldehydes & ketones gave the numbers; this table is the ruler the rest of this section measures against. For most carbonyls the equilibrium lies to the left:

CarbonylPercent hydrate at equilibrium
Acetone0.1%
Acetaldehydejust over half (K ≈ 1.2–1.4)
Formaldehyde~99.9%
Chloral, CCl3CHO~100%, and the hydrate is a crystalline solid you can weigh out
acetone, (CH₃)₂C=Otwo methyls0.1%acetaldehyde, CH₃CHOone methyljust over halfformaldehyde, H₂C=Ono alkyl group~99.9%chloral, CCl₃CHOthree Cl pulling~100%0%25%50%75%100%percent of the carbonyl present as its hydrate, in water at equilibrium
The table as bars, on a straight 0–100% scale. Acetone’s bar is drawn wider than 0.1% so that it shows.

From acetone to formaldehyde is three orders of magnitude in percent hydrate, and the only change is two methyl groups replaced by H. In equilibrium constants the gap is closer to six orders. Percentages squash the top of the scale: 99.9% and 99.99% look almost the same, yet the second equilibrium constant is ten times the first.

Two causes set the spread, the same two that put aldehyde > ketone toward every nucleophile.

  • Sterics. Addition turns a flat sp² carbon, with its groups 120° apart, into a crowded sp³ one, with its groups about 109.5° apart. The groups are pushed closer together. Two methyls resist that; two hydrogens do not.
  • Electronics. Alkyl groups donate electron density into the carbonyl carbon, which stabilizes the starting material. Electron-withdrawing groups do the opposite. Chloral carries three chlorines that pull electron density away, so it is fully hydrated.

The two causes are independent, and when they disagree the larger one wins. Hexafluoroacetone, (CF3)2C=O, is more crowded than acetone, because each CF3 group is bulkier than a CH3. It is nonetheless essentially completely hydrated. Its six fluorines, each one carbon away from the carbonyl carbon, pull electron density away, and that pull outweighs the crowding.

HEXAFLUOROACETONE: ELECTRONICS WINessentially all hydrateOCF₃CF₃CCF₃ bulkier than CH₃six F, one C away, pullOHOHCF₃CF₃Cthe pull outweighsthe crowding
Sterics and electronics disagree here, and electronics wins.

Aldehydes & ketones already ranked benzaldehyde below acetaldehyde because its C=O is conjugated with the ring. Hydration shows the size of that effect: benzaldehyde forms far less hydrate, because addition makes the carbon sp³ and cuts the link.

BENZALDEHYDE: THE RING HOLDS THE C=Ofar less hydrate than acetaldehydeOHCC=O conjugatedwith the ringOHOHHCsp³ carbon:conjugation broken
Conjugation stabilizes the aldehyde, so addition costs more.

In one case the angle argument runs backward. In cyclopropanone, the carbonyl carbon is a corner of a three-membered ring, so the ring angle at that carbon is about 60°. An sp² carbon wants 120°, so it is 60° off. After addition the carbon is sp³ and wants 109.5°. The same 60° angle is now only about 50° off.

So addition relieves angle strain here instead of creating it, and cyclopropanone is essentially fully hydrated even though it is a ketone. Ring strain, from Alkanes & Conformations, is doing the work the alkyl count usually does.

ACETONE: ADDITION CROWDS THE CARBONequilibrium lies far to the left (0.1%)OCH₃CH₃C120°sp², flatOHOHCH₃CH₃C109.5°sp³, methyls closerCYCLOPROPANONE: ADDITION EASES THE RINGequilibrium lies far to the rightOC60°sp² wants 120°60° offOHOHC60°sp³ wants 109.5°about 50° off
The same angle change, read two ways. Open-chain, it squeezes the groups together. In the three-membered ring, it brings the carbon closer to the angle it wants.

Every case above comes down to one comparison. Anything that makes the carbonyl more stable, or the hydrate more crowded, holds the equilibrium back. Anything that strains the carbonyl or pulls electrons from its carbon pushes it forward.

Why the hydrate matters even when it is the minor species

A compound present in small amounts can still decide the product, if it is the only form that can react further. The hydrate is the standard example.

Oxidizing a carbon means trading a C–H bond for a bond to oxygen. A chromium(VI) oxidant, the kind this section is about, needs a carbon that holds both an H and an OH, as Alcohols, Ethers & Related Chemistry showed when it counted the C–H bonds on an alcohol's carbon. A primary alcohol has that arrangement. The aldehyde it is oxidized to does not: its carbonyl carbon has an H but no OH.

Two chromium(VI) reagents show what water does to that. Jones reagent is Cr(VI) in aqueous sulfuric acid, usually with acetone as the solvent. PCC (pyridinium chlorochromate) is a Cr(VI) salt used in dry dichloromethane, CH2Cl2, with no water present. Both oxidize a primary alcohol to the aldehyde. Jones then carries on to the carboxylic acid. PCC stops at the aldehyde.

The difference is the hydrate. In the water of the Jones reagent, the aldehyde forms some hydrate, and the hydrate is an alcohol again: an H and an OH on one carbon. The oxidant attacks it a second time. In dry dichloromethane no hydrate can form, so the aldehyde gives the oxidant nothing to grip. Oxidation & Reduction treats both reagents in full, and the aldehyde oxidation section at the end of this chapter builds on this one.

1 · THE PRIMARY ALCOHOLCr(VI) removes this H and the O–H HRHHOHC2 · THE ALDEHYDEdry (PCC in CH₂Cl₂): it stops hereROHCno O–H on this carbon3 · IN WATER: THE HYDRATEan alcohol again: H and OH on one CROHHOHC4 · WATER (JONES): OXIDIZED AGAINthe carboxylic acidROOHC
Read the panels in order. The second oxidation needs the third panel, and only water can make it.

The hydrate does not have to be the main species for this to work. As the oxidant uses up hydrate, more aldehyde adds water to replace it, until all of it has been oxidized. An equilibrium does not have to favor an intermediate for that intermediate to decide the product.

The effect gets more striking the smaller the fraction is. Glucose sits almost entirely as a ring with no aldehyde group; well under a tenth of a percent is the open-chain aldehyde at any moment. Tollens’ silver(I), an oxidant that attacks aldehydes, still oxidizes all of it, because the ring keeps opening to replace the aldehyde used up. The aldehyde oxidation section returns to glucose.

Cyanohydrins: addition of cyanide

Mix acetone with HCN and a little sodium cyanide, and the product is acetone cyanohydrin, (CH3)2C(OH)CN. A cyanohydrin carries an OH and a C≡N (a nitrile group) on the same carbon. Unlike hydration, this addition makes a new carbon–carbon bond: the cyanide carbon ends up bonded to the old carbonyl carbon.

R2C=O + HCN  ⇌  R2C(OH)CN

The mechanism is the first example of the previous section: cyanide attacks, and HCN protonates the alkoxide. What is new here is why the reagent is written HCN with a catalytic amount of cyanide, HCN plus a little KCN or NaCN, and why neither component works alone.

HCN by itself is a feeble nucleophile. The carbon that would do the attacking holds a hydrogen, not a lone pair, so it has no free electrons to attack with.

Cyanide by itself does attack, but the product is an alkoxide. With no proton source to put an H on that oxygen, the alkoxide simply pushes the cyanide back out, and the addition runs backward.

Together they work. When HCN hands its proton to the alkoxide, it becomes a new CN−. The cyanide is made again on every turn, which is why a small amount is enough: it is genuinely catalytic.

CYANIDE ADDS TO THE CARBONthe π electrons move onto OOCH₃CH₃CCN−THE ALKOXIDE TAKES H⁺ FROM HCNHCN gives up H⁺ and becomes CN⁻OCH₃CH₃CCN−HCNTHE CYANOHYDRIN, AND CN⁻ IS BACKOH and C≡N on one carbonOHCH₃CH₃CCN+ CN⁻IN STRONG BASE IT RUNS BACKWARDSO⁻ re-forms C=O and pushes out CN⁻OCH₃CH₃CCN−after base hasremoved the O–H H⁺
Acetone and HCN with a little cyanide. The cyanide that attacks in the first panel is replaced in the second, which is why a small amount is enough.

The pH decides how much of each form is in the flask. The pKa of HCN is 9.2, and at a pH equal to the pKa an acid is half protonated and half not (Acids & Bases). So the reaction is run in a buffer near pH 8–10, which keeps both HCN and CN− present at once.

Strong base takes a cyanohydrin apart. Base removes the proton from the OH, and the alkoxide re-forms the C=O and pushes out cyanide, which is the last panel of the figure above. So cyanohydrins are made in a mildly basic buffer and destroyed by strong base. That is the reversibility of nucleophilic addition, showing up as a rule for handling the product.

What makes the reaction worth doing is what the nitrile can become. On acetone cyanohydrin:

  • Hydrolysis (aqueous acid, heat) turns C≡N into COOH, giving an α-hydroxy carboxylic acid.
  • Reduction with LiAlH4 turns C≡N into CH2NH2, giving a β-amino alcohol.
  • Dehydration removes the OH and an H from the neighboring carbon, putting a C=C there and giving an α,β-unsaturated nitrile.
THE CYANOHYDRINthe nitrile carbon is bonded to the C–OHOHCH₃CH₃CCNH₃O⁺, HEAT: C≡N BECOMES COOHan α-hydroxy acidOHCH₃CH₃COOHCC1αan acid countsfrom its COOHcarbon: C1LiAlH₄: C≡N BECOMES CH₂NH₂a β-amino alcoholOHCH₃CH₃CCH₂NH₂αβan alcohol countsfrom its C–OHcarbon: αLOSE H₂O: A C=C FORMSan α,β-unsaturated nitrileCH₂CH₃CNCαβC1an H from a CH₃and the OH leave
One skeleton, three products. The Greek letters are read from the group that names each product.

Greek letters name a carbon by its distance from a reference group: α is the carbon next to it, β the one after that. The α and β labels on these products are often marked wrong, because the products count from different reference groups.

In the acid, the reference is the carboxyl group. The COOH carbon is C1, and the letters start after it. Its neighbor, the carbon holding the OH, is α. So the product is an α-hydroxy acid.

In the amino alcohol, the reference is the alcohol, and the letters start on the carbon holding the OH. That carbon is itself α, so the CH2NH2 beside it is β. So the product is a β-amino alcohol. Nothing about the skeleton moved between the two products; only the carbon the counting starts from did.

In the unsaturated nitrile, the nitrile carbon is C1. The old carbonyl carbon is α, the CH2 is β, and the new C=C joins those two.

Working backward from a target to its starting materials is called retrosynthesis, as in the Grignard worked example of the previous section. In those terms, a cyanohydrin is a one-carbon extension that arrives with a functional group already on it. When a target has a carbon holding an OH right next to a COOH, a CH2NH2 or a C≡N, cut the bond between those two carbons. One piece is an aldehyde or ketone; the other is cyanide.

Worked example — making 2-hydroxy-2-methylpropanoic acid

Draw the target first: a carbon holding an OH, two methyls and a COOH, (CH3)2C(OH)COOH.

Step 1 — find the pattern. The carbon holding the OH is bonded straight to the COOH carbon. That is the cyanohydrin pattern.

Step 2 — cut the bond between them. The OH carbon, with its two methyls, becomes a C=O: acetone. The COOH carbon came from cyanide.

Step 3 — check it forward. Acetone with HCN and a little cyanide gives acetone cyanohydrin. Aqueous acid and heat turn its C≡N into COOH. That is the target.

Other ways of adding carbon lead to other groups, and the choice is about what you want at the end. All three use the Grignard reagents of Organometallic Bonding & Grignard Reagents:

  • Formaldehyde plus a Grignard reagent adds one carbon and gives a primary alcohol. The previous section previewed this one.
  • CO2 plus a Grignard reagent adds one carbon and gives a carboxylic acid.
  • A nitrile plus a Grignard reagent gives a ketone. This joins the Grignard’s R group to the nitrile carbon, so it adds the nitrile carbon and everything on it, not just one carbon. Nitriles themselves belong to Carboxylic Acids & Derivatives.

Bisulfite adducts, and a use for a bad reaction

Sodium bisulfite, NaHSO3, adds to aldehydes and to unhindered ketones. The product is called an adduct, meaning two molecules joined with nothing lost. Here the sulfur forms the new bond to the carbonyl carbon, and the adduct is a sodium salt.

As a salt, the adduct dissolves in water and not in organic solvents. That switch in solubility is the useful part. Shake a mixture in an organic solvent with aqueous bisulfite, and the carbonyl compound moves into the water layer while everything else stays behind.

BISULFITE ADDS THROUGH SULFURa salt: the new bond joins C and SCH₃CHO+ Na⁺ HSO₃⁻OHHCH₃SO₃⁻CNa⁺SHAKE WITH AQUEOUS NaHSO₃acid or base later releases the aldehydeorganic layereverything elsewater layerthe adduct salt
The adduct is a salt, so it goes where salts go: into the water.

The addition is reversible with acid or with base, so the carbonyl compound can be released again afterward. That makes the whole sequence a purification. Where the adduct crystallizes out instead, it can simply be filtered off.

It also tells carbonyls apart. Aldehydes and methyl ketones form the adduct readily; hindered ketones do not. So bisulfite can separate them, and this is the same steric effect that sets the hydration equilibrium.

What carries forward

WATERRROHOHCa gem-diol (hydrate)usually far to the leftHCN + a little CN⁻RROHC≡NCa cyanohydrinmakes a C–C bondNaHSO₃RROHSO₃⁻Ca bisulfite adducta salt, used to purify
Three nucleophiles, one addition. Only the group beside the OH changes.

Water, cyanide and bisulfite all do the same thing: they add reversibly to a carbonyl. How far each goes is decided by sterics and electronics, in the same direction every time. That is why the hydration percentages are worth knowing as a ruler rather than as facts. And one of the three makes a C–C bond, which is why cyanohydrins appear in syntheses and hydrates do not.