Part 5. The chemistry of ALDEHYDES and KETONES - REDUCTION

 All revision notes on the chemistry of aldehydes and ketones

Part 5.5 Reduction of aldehydes, ketones and nitriles

Reduction of aldehydes to primary alcohols and ketones to secondary alcohols

Examples of nucleophilic addition of hydride to aldehydes and ketones to reduce them to alcohols

These reactions are essentially the reduction of the carbonyl group C=O to H-C-OH by the nucleophilic addition of the equivalent of a hydride ion.

R₂C=O  +  2[H]  ===>  R₂CHOH   (R = H or alkyl)

 

(1) On reduction, aldehydes give primary alcohols

RCHO + 2[H] ==> RCH₂OH (R = H, alkyl or aryl)

e.g. (i) ethanal to ethanol: CH₃CHO + 2[H] ==> CH₃CH₂OH

or (ii)  + 2[H] ===>

butanal + [hydrogen] ==> butan-1-ol

The product will not exhibit R/S isomerism, unless the original aldehyde exhibited R/S isomerism.

 

The hydrogen, as a hydride ion (H-) attacks the δ+ carbon and is derived from the reducing agent e.g.

 

NaBH₄ sodium tetrahydridoborate(III),  (sodium borohydride)

The reduction of aldehydes and ketones with NaBH₄ can be carried out in water as the solvent.

 

LiAlH₄ lithium tetrahydridoaluminate(III),  (lithium aluminium hydride)

LiAlH₄ is a more powerful reducing agent than NaBH₄ and reacts violently with water (and reacts with ethanol too), so the reaction must be carried out in an inert solvent like ethoxyethane ('ether') - which must be dry!

The simplified equations above apply.  (mechanism lower down the page)

LiAlH₄ is a more powerful reducing agent than NaBH₄ because the Al–H bond is weaker than the B–H bond. This fits in with the aluminium atom having a larger radius than the boron atom giving a longer weaker bond with hydrogen. Covalent radii: B = 0.080 nm (80 pm), Al = 0.125 nm (125 pm).

 

The initial product is the salt is the alkoxide salt of the alcohol, but this is hydrolysed by adding dilute sulphuric acid which frees the alcohol.

R₂CHO^-  +  H⁺  ===>  R₂CHOH   (R = H or alkyl)

 

(2) On reduction, ketones give secondary alcohols

ketone: R₂C=O + 2[H] ==> R₂CHOH (R = alkyl or aryl)

e.g. (i)  propanone to propan–2–ol: CH₃COCH₃ + 2[H] ==> CH₃CH(OH)CH₃

propan-2-ol does not exhibit R/S isomerism, no chiral carbon in the molecule.

or (ii)  + 2[H] ===>

butan-2-one + [hydrogen] ==> butan-2-ol

The butan-2-ol has a chiral carbon (asymmetric carbon atom) and can exhibit R/S isomerism ('optical isomers', mirror image forms). It is likely to be a racemate (50:50) mixture of the isomers because of the equal probability of nucleophilic attack from 'above' or 'below' the plane of the >C=O bonding system (see diagram below).

Note that butan-2 -one is an unsymmetrical ketone, which will always give product that can exhibit R/S isomerism, but symmetrical ketones will not give a product with a chiral carbon atom and cannot exhibit R/S isomerism e.g.

pentan-3-one is reduced to pentan-3-ol

CH₃CH₂COCH₂CH₃  + 2[H]  ===>  CH₃CH₂CH(OH)CH₂CH₃

 

The notes on the reducing agent methods are the same as for aldehydes above.

 

The reduction mechanism is very complicated, but can be considered in a simplistic way as involving the donation of a hydride ion (H^-) to the aldehyde/ketone.

An outline of the nucleophilic addition mechanism is given below

 

(3) Further comments on methods of reduction and comparison with the reduction of alkenes

(i) The use of the reducing agent NaBH₄ is particularly important because it is a milder reducing agent and selectively reduces an aldehyde or ketone group and NOT a carboxylic acid group (unlike LiAlH₄).

e.g. the reduction of a carboxylic acid with an aldehyde of ketone functional group.

CH₃-CO-CH₂-COOH  +  2[H]  == NaBH₄  ==>  CH₃-CH(OH)-CH₂-COOH

CH₃-CO-CH₂-COOH  +  6[H]  == LiAlH₄  ==>  CH₃-CH(OH)-CH₂-CH₂OH  +  H₂O

O=CH-CO-CH₂-COOH  +  2[H]  == NaBH₄  ==>  HOCH₂-CH(OH)-CH₂-COOH

O=CH-CO-CH₂-COOH  +  6[H]  == LiAlH₄  ==>  HOCH₂-CH(OH)-CH₂-CH₂OH  +  H₂O

(ii) In industry, aldehydes and ketones can be reduced with hydrogen gas and platinum catalyst

R₂C=O  +  H₂  == Pt ==>  R₂CHOH   (R = H or alkyl)

(iii) Alkenes can be reduced in the same way, but they are not reduced by NaBH₄ or LiAlH₄

R₂C=CR₂  +  H₂  == Pt ==>  R₂CH-CHR₂   (R = H or alkyl)

(iv) LiAlH₄ is a reducing agent that specifically reacts with polar pi bonds like those in carbonyl compounds.

It will reduce the C=O not only in aldehydes and ketones, but also in carboxylic acids (RCOOH) and acid derivatives (e.g. RCOCl, RCOOR) and also the C≡N group in nitriles.

(v) However, alkenes are not reduced by NaBH₄ or LiAlH₄ because unlike the C=O bond, the >C=C< double bond is not polar and will not be attacked by a 'hydride ion' - in fact, the pi bonding electrons of alkenes will tend to repel nucleophiles.

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What is the mechanism for the reduction of aldehydes and ketones with sodium tetrahydridoborate(III) or lithium tetrahydridoaluminate(III)?

Lithium tetrahydridoaluminate(III) (lithium aluminium hydride) or sodium tetrahydridoborate(III) (sodium tetraborohydride) reduce aldehydes to primary alcohols and ketones to secondary alcohols.

 very simply the reaction is: RR'C=O + 2[H] ==> RR'CHOH

Aldehyde: R = H, R' = H or alkyl)

or ketone: R and R' are either alkyl or aryl, but NOT H.

mechanism 40 – nucleophilic addition of a hydride ion (via NaBH₄ or LiAlH₄) to an aldehyde or ketone

[mechanism 40 above] is a considerable simplification of the full mechanism. R, R' = H, alkyl or aryl

 

In step (1) the AlH₄^– or BH₄^– ions act as nucleophiles and donate the 'equivalent' of an electron pair donating nucleophilic hydride ion :H^– to the positive carbon of the polarised carbonyl bond.

The hydrogen, as a hydride ion (H-) attacks the δ+ carbon.

A hydride ion is effectively the nucleophile - donating an electron pair to a partially positive carbon atom (though the full mechanism is quite complicated).

In step (2) the intermediate ion is a strong conjugate base and reacts with any proton donor e.g. water (but can be an alcohol ROH or acid H₃O⁺) to form the alcohol product.

 

Mechanism 83a shows the reduction of ethanal to ethanol (an aldehyde reduced to a primary alcohol) by a 'hydride ion' derived from the reducing agents NaBH₄ or LiAlH₄.

 

Mechanism 83b shows the reduction of butan-2-one to butan-1-ol (a ketone reduced to a secondary alcohol) by a 'hydride ion' derived from the reducing agents  NaBH₄ or LiAlH₄.

The real mechanism involves a step–wise replacement of the hydrogen atoms on the reducing reagent with alkoxide groups (RR'CH–O–, e.g. ethoxy CH₃CH₂–O– from reduced ethanal CH₃CHO).

This happens because all the intermediates are themselves nucleophilic agents. In the sequence X = B or Al and R₂ = RR' for simplicity)

XH₄^– + R₂C=O => [H₃XO–CHR₂]^– = R₂C=O => [H₂X(O–CHR₂)₂]^–

= R₂C=O => [HX(–OCHR₂)₃]^– = R₂C=O => [X(–OCHR₂)₄]^– 

Then the alkoxide complex reacts with any proton donor (depending on reagent/reaction conditions e.g. the water/ethanol/ acid) to free the alcohol.

[X(OCR₂)₄]^– + water/acid/alcohol ===> 4R₂CHOH + compound of X