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Advanced Level Organic Chemistry: Reduction of aldehydes & ketones

Part 5. The chemistry of ALDEHYDES and KETONES

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry reduction of aldehydes and ketones to alcohols and reduction of nitriles to amines

Part 5.5 Reduction of aldehydes, ketones and nitriles


INDEX of ALDEHYDES and KETONES revision notes

All Advanced A Level Organic Chemistry Notes

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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)  aldehydes and ketones nomenclature (c) doc b + 2[H] ===> alcohols and ether structure and naming (c) doc b

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)  aldehydes and ketones nomenclature (c) doc b + 2[H] ===> alcohols and ether structure and naming (c) doc b

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).

trigonal planar bonding around carbonyl carbon in aldehydes ketones nucleophilic attack on postive carbon doc brown's advanced A level organic chemistry notes

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 CN 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 of reduction of aldehydes and ketones by the hydride ion LiAlH4 NaBH4 advanced organic chemistry revision notes doc brown

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 of 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 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 of 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

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


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Reduction of hydroxynitriles derived from hydrogen cyanide addition to aldehydes and ketones

R2C=O  +  HCN  ===>  R2C(OH)CN   (R = H, alkyl or aryl)

RCH(OH)CN  +  4[H]  ===>  RCH(OH)CH2NH2

Hydroxynitriles are reduced to hydroxylamines (which have two functional groups - alcohol and amine).

The reaction can be done with the reducing agents:

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

Refluxing the nitrile with sodium metal in ethanol.

Hydrogen gas using a platinum catalyst (industrial method)


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INDEX of ALDEHYDE and KETONE revision notes

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