University Chemistry — Year 1 · Bachelor Year 1

24Oxidation and Reduction in Organic Chemistry

A bottle of wine left open turns, within days, into vinegar: bacteria use the oxygen of the air to oxidise ethanol into ethanoic acid. A ketone and the alcohol it comes from differ by two hydrogen atoms — and, counted properly, by two electrons. Organic oxidations and reductions do not look like the exchanges of electrons between metal ions of Chapter 13, because electrons travel with protons or with hydride ions, but the bookkeeping is the same. This chapter assigns oxidation levels to carbon, uses them to predict what each alcohol gives when oxidised, and introduces the hydride reagents that reduce carbonyl compounds, with an eye on which groups each reagent leaves untouched.

You already know

Oxidation numbers, half-equations and the balancing of redox equations are in Chapter 13; the classes of alcohols and of carbons in Chapter 18; acetals and the protection of carbonyl groups in Chapter 23; Grignard additions in Chapter 21.

Wine aging in a cellar. Kept from air, wine keeps its ethanol; exposed to air, acetic acid bacteria oxidise it to ethanoic acid.

24.1 Oxidation levels of carbon

Definition 24.1 (Oxidation level)

The oxidation level of a carbon atom is its oxidation number, obtained by the rules of Proposition 13.3: each bond to a more electronegative atom (O, N, halogen) counts $+1$ , each bond to hydrogen $-1$ , each bond to another carbon 0 (a double bond counts twice).

Proposition 24.2 (Oxidation numbers of carbon)

The oxidation number of a carbon equals (number of bonds to O, N or halogens) minus (number of bonds to H). Within one family of compounds of the same carbon skeleton, alcohol $\to$ aldehyde or ketone $\to$ carboxylic acid $\to$ carbon dioxide is a series of oxidations by two electrons each.

Proof. Breaking every bond and giving both electrons to the more electronegative partner: from H (less electronegative than C) carbon gains an electron, charge $-1$ per C–H; to O, N, halogens (more electronegative) it loses one, $+1$ per bond; C–C bonds are split equally. Going from $\ce{-CH2OH}$ ( $-1$ for a carbon bound to one other carbon) to $\ce{-CHO}$ ( $+1$ ) to $\ce{-COOH}$ ( $+3$ ), the number rises by 2 at each step: two electrons. ∎

The oxidation levels of the one-carbon family: methane, methanol, methanal, methanoic acid, carbon dioxide. Each arrow is a two-electron oxidation; read backwards, a reduction.

Method 24.3 (Balancing an organic half-equation)

Write the organic oxidant and reductant with the same skeleton.
Count the electrons from the change of oxidation numbers of the carbons that change (or simply: two per pair of H lost, or per O gained).
Balance oxygen with water, hydrogen with $\ce{H+}$ , charge with the electrons.

Ethanol to ethanoic acid: $\ce{CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-}$ . With acidified dichromate ( $\ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O}$ ), three ethanol molecules for two dichromate ions: $\ce{3CH3CH2OH + 2Cr2O7^2- + 16H+ -> 3CH3COOH + 4Cr^3+ + 11H2O}$ .

24.2 Oxidising alcohols

Proposition 24.4 (What alcohols give when oxidised)

A primary alcohol $\ce{RCH2OH}$ is oxidised to the aldehyde $\ce{RCHO}$ , then (in water, with a strong oxidant) to the carboxylic acid $\ce{RCOOH}$ ; a secondary alcohol $\ce{R2CHOH}$ to the ketone $\ce{R2CO}$ , which resists further oxidation; a tertiary alcohol $\ce{R3COH}$ is not oxidised without breaking a C–C bond.

Proof. Oxidation of the alcohol removes the hydrogen of the OH and a hydrogen from the carbinol carbon, forming C=O. A tertiary carbinol carbon has no such hydrogen. The aldehyde still carries an H on the carbonyl carbon; in water it is partly hydrated, $\ce{RCH(OH)2}$ , which is oxidised again in the same way to $\ce{RCOOH}$ . A ketone has no H on that carbon and stops. ∎

Strong oxidants in water — acidified potassium dichromate or permanganate — oxidise primary alcohols to acids. To stop at the aldehyde, it is distilled out as it forms (aldehydes boil lower than the alcohols they come from, having no O–H hydrogen bonds), or milder, anhydrous reagents are used, described in the Year 2 volume.

Safety

Potassium dichromate, the classic orange oxidant of alcohols, is classified among the substances that may cause cancer: it is handled only in small amounts, with gloves and goggles, and its chromium waste is collected separately; where possible a less hazardous oxidant is preferred.

24.3 Reducing carbonyl compounds with hydrides

Definition 24.5 (Hydride donor)

A hydride donor transfers a hydrogen atom with its bonding pair, $\ce{H-}$ , to an electrophilic carbon. Sodium borohydride $\ce{NaBH4}$ and lithium aluminium hydride $\ce{LiAlH4}$ are complex metal hydrides: the hydride is delivered from the $\ce{BH4-}$ or $\ce{AlH4-}$ ion, not as a free ion.

Reduction of a ketone by borohydride: a B–H pair forms the new C–H bond while the C=O π pair goes to oxygen; the alkoxide is protonated by the solvent (water or an alcohol) or at the work-up. Each BH4- can deliver its four hydrogens in turn. — Reduction of a ketone by borohydride: a B–H pair forms the new C–H bond while the C=O $\pi$ pair goes to oxygen; the alkoxide is protonated by the solvent (water or an alcohol) or at the work-up. Each $\ce{BH4-}$ can deliver its four hydrogens in turn.

Proposition 24.6 (Scope of the two hydrides)

Sodium borohydride, used in water or alcohols, reduces aldehydes to primary alcohols and ketones to secondary alcohols, and leaves esters, carboxylic acids, amides and C=C double bonds untouched. Lithium aluminium hydride, used in dry ether, also reduces esters and carboxylic acids to primary alcohols, and reacts violently with water.

Proof. The Al–H bond is more polarised than the B–H bond (aluminium is less electronegative than boron), so $\ce{AlH4-}$ is a far stronger hydride donor, able to attack the less electrophilic carbonyl of esters and of carboxylates; the same reactivity makes it react with any acidic hydrogen, including water. $\ce{BH4-}$ reacts only slowly with water and alcohols, and attacks only the most electrophilic carbonyls. An isolated C=C bond is not electrophilic and is reduced by neither. ∎

24.4 Selectivity

Definition 24.7 (Chemoselective reaction)

A chemoselective reaction acts on one functional group of a molecule in the presence of others that could, with another reagent, react too.

group	$\ce{NaBH4}$ (alcohol solvent)	$\ce{LiAlH4}$ (dry ether)
aldehyde $\ce{RCHO}$	primary alcohol	primary alcohol
ketone $\ce{R2CO}$	secondary alcohol	secondary alcohol
ester $\ce{RCOOR}$	no reaction	two alcohols
carboxylic acid $\ce{RCOOH}$	no reaction (salt)	primary alcohol
alkene $\ce{C=C}$	no reaction	no reaction
acetal	no reaction	no reaction

What the two hydrides reduce. Sodium borohydride is the chemoselective reagent; lithium aluminium hydride, the powerful one.

Example 24.8 (A keto ester)

Ethyl 4-oxopentanoate, $\ce{CH3CO(CH2)2COOEt}$ , with sodium borohydride gives ethyl 4-hydroxypentanoate: only the ketone is reduced. With lithium aluminium hydride, both groups are reduced: pentane-1,4-diol (and ethanol). To reduce only the ester, the ketone must first be protected as an acetal (Chapter 23).

In the lab — Reducing a ketone

Sodium borohydride is added in small portions to a cold solution of the ketone in ethanol: the reaction is exothermic, and hydrogen bubbles from the slow reaction of the reagent with the solvent. After stirring, dilute acid destroys the excess reagent (more hydrogen: no flame nearby) and the alcohol is extracted. Lithium aluminium hydride, in contrast, is used only in dry ether under an inert gas, and its excess is destroyed with great care.

24.5 Exercises

Exercise 24.1 ★

Give the oxidation number of each carbon in ethanol, ethanal, ethanoic acid, propanone and methanoic acid.

Solution

Solution of Exercise 24.1.

Ethanol: $\ce{CH3}$ $-3$ , $\ce{CH2OH}$ $-1$ . Ethanal: $\ce{CH3}$ $-3$ , CHO $+1$ . Ethanoic acid: $\ce{CH3}$ $-3$ , COOH $+3$ . Propanone: $\ce{CH3}$ $-3$ (twice), C=O $+2$ . Methanoic acid: $+2$ .

Exercise 24.2 ★

What does acidified dichromate give with propan-1-ol, propan-2-ol and 2-methylpropan-2-ol?

Solution

Solution of Exercise 24.2.

Propan-1-ol: propanal, then propanoic acid. Propan-2-ol: propanone. 2-Methylpropan-2-ol: no reaction (the orange colour stays).

Exercise 24.3 ★

Write the half-equations of the couples propanone/propan-2-ol and ethanal/ethanol in acid.

Solution

Solution of Exercise 24.3.

$\ce{CH3COCH3 + 2H+ + 2e- -> CH3CH(OH)CH3}$ ; $\ce{CH3CHO + 2H+ + 2e- -> CH3CH2OH}$ .

Exercise 24.4 ★

What do sodium borohydride and lithium aluminium hydride give with butanal, butanone, ethyl butanoate and butanoic acid?

Solution

Solution of Exercise 24.4.

$\ce{NaBH4}$ : butan-1-ol, butan-2-ol, no reaction, no reaction (the acid is only deprotonated). $\ce{LiAlH4}$ : butan-1-ol, butan-2-ol, butan-1-ol and ethanol, butan-1-ol.

Exercise 24.5 ★★

Balance the oxidation of propan-2-ol by permanganate in acid ( $\ce{MnO4-}$ / $\ce{Mn^2+}$ ) and compute the volume of permanganate at $0.020\,\mathrm{mol}/\mathrm{L}$ needed for $1.20\,\mathrm{g}$ of alcohol ( $M = 60.0\,\mathrm{g}/\mathrm{mol}$ ).

Solution

Solution of Exercise 24.5.

$\ce{5CH3CH(OH)CH3 + 2MnO4- + 6H+ -> 5CH3COCH3 + 2Mn^2+ + 8H2O}$ . $n = 1.20/60.0 = 0.0200\,\mathrm{mol}$ ; permanganate $\frac25 \times 0.0200 = 8.0 \times 10^{-3}\,\mathrm{mol}$ , $400\,\mathrm{mL}$ of the $0.020\,\mathrm{mol}/\mathrm{L}$ solution.

Exercise 24.6 ★★

How can propanal be obtained from propan-1-ol without being oxidised further? Use the boiling points: propan-1-ol $97\,{}^{\circ}\mathrm{C}$ , propanal $48\,{}^{\circ}\mathrm{C}$ .

Solution

Solution of Exercise 24.6.

Heat the alcohol with the oxidant in a flask fitted for distillation, at a temperature between the two boiling points: propanal ( $48\,{}^{\circ}\mathrm{C}$ ) distils as it forms and escapes further oxidation; propan-1-ol ( $97\,{}^{\circ}\mathrm{C}$ ) stays in the flask.

Exercise 24.7 ★★

Write the mechanism of the reduction of ethanal by $\ce{BH4-}$ in ethanol. Which atom of the product comes from the reagent?

Solution

Solution of Exercise 24.7.

A B–H pair of $\ce{BH4-}$ attacks the carbonyl carbon, the C=O $\pi$ pair going to oxygen; the alkoxide takes a proton from ethanol. The hydrogen on the carbon of the product ( $\ce{CH3CH2OH}$ , one of the two on C1) comes from the reagent; the one on oxygen from the solvent.

Exercise 24.8 ★★

Reducing butan-2-one with $\ce{NaBH4}$ gives a chiral alcohol. Is it optically active? Why?

Solution

Solution of Exercise 24.8.

Butan-2-ol is chiral, but the planar ketone is attacked equally on both faces: a racemic mixture, optically inactive.

Exercise 24.9 ★★

In the breath test for alcohol once used by police, orange dichromate turned green. Write the reaction with ethanol and explain the colour change.

Solution

Solution of Exercise 24.9.

$\ce{3CH3CH2OH + 2Cr2O7^2- + 16H+ -> 3CH3COOH + 4Cr^3+ + 11H2O}$ : orange dichromate (chromium $+$ VI) becomes green chromium(III).

Exercise 24.10 ★★★

Propose syntheses of: 2-methylbutan-2-ol from ethanal and a Grignard reagent followed by an oxidation and a second Grignard step; pentan-3-ol from propanal. Count the oxidations and reductions.

Solution

Solution of Exercise 24.10.

2-Methylbutan-2-ol: ethanal + $\ce{CH3CH2MgBr}$ gives butan-2-ol (an addition, no redox); oxidation to butanone; $\ce{CH3MgBr}$ on butanone, then work-up. One oxidation. Pentan-3-ol: propanal + $\ce{CH3CH2MgBr}$ in one step, no redox.

Exercise 24.11 ★★★

Reduce only the ester group of methyl 4-oxopentanoate, $\ce{CH3CO(CH2)2COOCH3}$ . Plan the three steps with a protection and give the product.

Solution

Solution of Exercise 24.11.

1. Protect the ketone as a cyclic acetal (ethane-1,2-diol, acid, Dean–Stark). 2. $\ce{LiAlH4}$ in dry ether reduces the ester to the primary alcohol (and methanol). 3. Aqueous acid removes the acetal: 5-hydroxypentan-2-one, $\ce{CH3CO(CH2)3OH}$ .

Exercise 24.12 ★★★

Show that in the oxidation of a primary alcohol to the acid, four electrons are exchanged, by the oxidation numbers and by the half-equation. Generalise to the oxidation of methane to carbon dioxide.

Solution

Solution of Exercise 24.12.

$\ce{RCH2OH}$ : carbinol carbon at $-1$ ; $\ce{RCOOH}$ : $+3$ ; four electrons, as in $\ce{RCH2OH + H2O -> RCOOH + 4H+ + 4e-}$ . Methane ( $-4$ ) to carbon dioxide ( $+4$ ): eight electrons, $\ce{CH4 + 2H2O -> CO2 + 8H+ + 8e-}$ .

24.6 Problem: From Cyclohexanol to Adipic Acid

Problem 24.1

Weekend problem — oxidation levels in cyclohexanol, cyclohexanone and adipic acid, the nitric acid oxidation and its electrons, a hydrogen peroxide route, and the mass of dinitrogen oxide released per tonne of adipic acid

Adipic acid, hexanedioic acid $\ce{HOOC(CH2)4COOH}$ , is made on a large scale for the nylons, mostly by oxidation of cyclohexanol (with cyclohexanone) by nitric acid, which releases dinitrogen oxide $\ce{N2O}$ , a powerful greenhouse gas. Molar masses ( $\mathrm{g}/\mathrm{mol}$ ): $\ce{C6H12O}$ 100.0, $\ce{C6H10O}$ 98.0, $\ce{C6H10O4}$ 146.0, $\ce{N2O}$ 44.0, $\ce{HNO3}$ 63.0, $\ce{H2O2}$ 34.0.

Part I — Oxidation levels.

Give the oxidation number of each carbon of cyclohexanol.
Same for cyclohexanone.
Same for adipic acid.
Which carbons are oxidised in going from cyclohexanol to adipic acid, and which bond is broken?
How many electrons does one molecule of cyclohexanol lose?
Why is cyclohexanone, a ketone, nevertheless oxidised further here?

Part II — The nitric acid route.

Write the half-equation of cyclohexanol to adipic acid in acid.
Give the oxidation number of N in $\ce{HNO3}$ and in $\ce{N2O}$ .
Write the half-equation $\ce{HNO3}$ / $\ce{N2O}$ .
Combine them; check that you obtain $\ce{C6H12O + 2HNO3 -> C6H10O4 + N2O + 2H2O}$ .
How many moles of electrons are exchanged per mole of adipic acid?
Compute the mass of nitric acid consumed per tonne of adipic acid.

Part III — A route with hydrogen peroxide.

Write the half-equation $\ce{H2O2}$ / $\ce{H2O}$ .
Write and balance the oxidation of cyclohexanol by hydrogen peroxide to adipic acid.
Compute the mass of $\ce{H2O2}$ per tonne of adipic acid.
What is the only by-product? Why is this route called greener?
Hydrogen peroxide must be activated by a catalyst. Using Exercise 13.12, explain why a thermodynamically favoured oxidation may still need one.
What would sodium borohydride do to cyclohexanone? Write the reaction.

Part IV — Masses and the greenhouse gas.

Compute the amount of adipic acid in one tonne.
Compute the amount of $\ce{N2O}$ formed with it by the nitric route.
Compute the mass of cyclohexanol needed per tonne, at $95\,\%$ yield.
State the mass of $\ce{N2O}$ released per tonne of adipic acid by the nitric route, in tonnes.

Solution

Solution of Problem 24.1.

1. C1 (bearing OH): 0; the five $\ce{CH2}$ : $-2$ . 2. C1 (C=O): $+2$ ; the five $\ce{CH2}$ : $-2$ . 3. The two COOH carbons: $+3$ ; the four $\ce{CH2}$ : $-2$ . 4. C1 ( $0 \to +3$ ) and its neighbour C6 ( $-2 \to +3$ ), which becomes the second carboxylic carbon; the C1–C6 bond is broken. 5. $3 + 5 = 8$ electrons. 6. Nitric acid is strong enough to break the C–C bond next to the carbonyl, which milder oxidants do not do. 7. $\ce{C6H12O + 3H2O -> C6H10O4 + 8H+ + 8e-}$ . 8. $+$ V and $+$ I. 9. $\ce{2HNO3 + 8H+ + 8e- -> N2O + 5H2O}$ . 10. Adding, eight electrons, $\ce{8H+}$ and three water molecules cancel: $\ce{C6H12O + 2HNO3 -> C6H10O4 + N2O + 2H2O}$ . 11. 8. 12. $10^6/146.0 = 6.85 \times 10^{3}\,\mathrm{mol}$ of acid; $2 \times 6.85 \times 10^3 \times 63.0 = 8.6 \times 10^{5}\,\mathrm{g}$ , $0.86\,\mathrm{t}$ of nitric acid. 13. $\ce{H2O2 + 2H+ + 2e- -> 2H2O}$ . 14. $\ce{C6H12O + 4H2O2 -> C6H10O4 + 5H2O}$ . 15. $4 \times 6.85 \times 10^3 \times 34.0 = 9.3 \times 10^{5}\,\mathrm{g}$ , $0.93\,\mathrm{t}$ . 16. Water: no nitrogen oxide. 17. A large equilibrium constant says nothing about the rate; $\ce{H2O2}$ reacts slowly without a catalyst, as its own decomposition does. 18. It reduces cyclohexanone back to cyclohexanol: $\ce{4C6H10O + BH4- -> B(OC6H11)4-}$ , then $\ce{B(OC6H11)4- + 4H2O -> 4C6H11OH + B(OH)4-}$ . 19. $6.85 \times 10^{3}\,\mathrm{mol}$ . 20. $6.85 \times 10^{3}\,\mathrm{mol}$ of $\ce{N2O}$ (one per acid). 21. $6.85 \times 10^3/0.95 \times 100.0 = 7.2 \times 10^{5}\,\mathrm{g}$ , $0.72\,\mathrm{t}$ . 22. $6.85 \times 10^3 \times 44.0 = 3.0 \times 10^{5}\,\mathrm{g}$ : $0.30\,\mathrm{t}$ of $\ce{N2O}$ per tonne of adipic acid.

Terms defined in this chapter

See all 852 terms in the glossary