Chemical properties of aldehydes and ketones A) Nucleophilic addition reactions
This reaction is the most common reactions of aldehydes and ketones. The carbonyl carbon carries a small degree of positive charge. Nucleophile such as CN
−can attack the carbonyl carbon and uses its lone pair to form a new carbon – nucleophile ‘σ ’ bond, at the same time two electrons from the carbon – oxygen double bond move to the most electronegative oxygen atom. This results in the formation of an alkoxide ion. In this process, the hybridisation of carbon changes from sp to sp2 3 .
R C O
R
sp2hybridised carbon sp3 hybridised carbon tetrahedral alkoxide ion
Nu R
C
Nu
O
R
The tetrahedral intermediate can be protonated by water or an acid to form an alcohol.
R C
Nu
O
R
H
R C
Nu
OH
R
In general, aldehydes are more reactive than ketones towards nucleophilic addition reactions due to +I and steric effect of alkyl groups.
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Examples
1) Addition of HCN
Attack of CN−on carbonyl carbon followed by protonation gives cyanohydrins.
C O
H3C
H H3C
C
CN
O
H H3C
C
CN
OH
H
CN H
ethanal (acetaldehyde)
tetrahedral intermediate
2-hydroxy propanenitrile (acetaldehyde cyano hydrin)
The cyanohydrins can be converted into hydroxy acid by acid hydrolysis. Reduction of cyanohydrins gives hydroxy amines
2) Addition of NaHSO3
C O
H3C
H3C
OSO2 H Proton transfer
H3C H3C O
OSO2 Na
CH3
acetone (propanone)
bisulphite addition compound
H3C C
OH
OSO2 H
C
This reaction finds application in the separation and purification of carbonyl compound. The bisulphate addition compound is water soluble and the solution is treated with mineral acid to regenerate the carbonyl compounds .
3) Addition of alcohol
When aldehydes / ketones is treated with 2 equivalents of an alcohol in the presence of an acid catalyst to form acetals.
example
When acetaldehyde is treated with 2 equivalent of methanol in presence of HCl, 1,1, - dimethoxy ethane is obtained.
C O
H3C
H
2 CH3 OH HCl
H3C CH OCH3
OCH3
ethanal (acetaldehyde)
1,1 - dimethoxy (acetaldehyde dimethyl acetal)
ethane
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Mechanisum
H3C H C
O H Cl
H3C H C
O H
H O CH3
H3C C
O
H
O
H
H
H3C
H2O
H3C C
O
OCH3
H
H
H Cl -Cl
H3C C
O
OCH 3
H
HH
H3C H C
O CH3
H3C O H
H3C C
O
OCH3
H
CH3H
-H3O+
H3C C
O
OCH3
H
CH3
H2O
-H3O+
4) Addition of ammonia and its derivatives
When the nucleophiles, such as ammonia and its derivative 2H N-G ..
is treated with carbonyl compound, nuceophilic addition takes place, the carbonyl oxygen atom is protonated and then elimination takes place to form carbon – nitrogen double bond C = N G
When G – alkyl, aryl, OH, NH , C H NH, NHCONH etc…2 6 5 2
R R C
O
H2N G
-H2O
R C
O
R
N G
H H
R C
OH
R
N G
H
R R C
N G
G Ammonia derivatives Carbonyl derivatives Product name
OH Hydroxyl amine N OHC = Oxime
–NH2 Hydrazine N NH2=C Hydrazone
HN C6H5 Phenyl
hydrazine N NHC = C
6 H
5 Phenyl
hydrazone
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| G | Ammonia derivatives | Carbonyl derivatives | Product name |
|---|---|---|---|
| OH | Hydroxyl amine | C= NO H | Oxime |
| –NH2 | Hydrazine | C = NN H | Hydrazone |
| HN C H6 5 | Phenyl hydrazine | C= NN H C H6 | Phenyl hydrazone |
157
NH C NH 2
O Semi
carbabazide N NHC = C NH2
O
Semi carbazone
NH
NO2
NO2
2,4 – dinitrophenyl
hydrazine C = N NH
NO2
NO2
2,4 – dinitrophenyl
hydrazone
i) Reaction with hydroxyl amine
Aldehyde and ketones react with hydroxylamine to form oxime.
Example:
H
C = O + H2 N OH CH3 C = N OH + H2O
H H+
Acetaldehyde Hydroxylamine Acetaldoxime
CH3
(ethanal) (N - ethylidene hydroxylamine)
ii) Reaction with hydrazine
Aldehydes and ketones react with hydrazine to form hydrazone.
Example:
CH3
C = O + H2 N NH2 CH3 C = N NH2 + H2O
CH3 H+
Acetone Hydrazine Acetone hydrazone
CH3
iii) Reaction with phenyl hydrazine
Aldehydes and ketones react with phenyl hydrazine to form phenyl hydrazone.
Example:
CH3
C = O + H2 N NHC6H5 CH3 C = N NHC6H5 + H2O
CH3
H+
Acetone Phenyl hydrazine Acetone phenyl hydrazone
CH3
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|——|——|
| ONH CN H2 | Semi carbabazide | OC = NN H C NH2 | Semi carbazone | |
|---|---|---|---|---|
| NO2NH NO | 2,4 – dinitrophenyl hydrazine | NOC = N NH 2 NO2 | 2,4 – dinitrophenyl hydrazone |
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5) Reaction with NH3
i) Aliphatic aldehydes (except formaldehyde) react with an ethereal solution of ammonia to form aldimines.
O
CH3 C + H NH2
H
Acetaldehyde OH
H
Acetaldehyde ammonia
- H2O Aldimine
CH3 C NH2 CH3 CH = NH
ii) Formaldehyde reacts with ammonia to form hexa methylene tetramine, which is also known as Urotropine.
6HCHO + 4 NH3 (CH2)6N4 + 6 H2O
Formaldehyde Hexamethylene tetramine
Structure
N
H2C CH2
CH2
CH2
N
CH 2CH
2
N N
Uses (i) Urotropine is used as a medicine to treat urinary infection. (ii) Nitration of Urotropine under controlled condition gives an explosive RDX (Research
and development explosive). It is also called cyclonite or cyclotri methylene trinitramine. iii) Acetone reacts with ammonia to form diacetone amine.
CH3 C O +
CH3
CH2 C CH3
O
CH3
OAcetone Diacetone amineAcetone
H NH2
H CH2 C CH3
CH3 C NH2
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159
iv) Benzaldehyde form a complex condensation product with ammonia.
C6H5 C O + H2 N H
H
Benzaldehyde
C6H5 C O + H2 N H
H
+ O C C6H5
H
C
C6H5
H
Ammonia Hydrobenzamide
C6H5 CH N
C6H5 CH N
Benzaldehyde
B) Oxidation of aldehydes and ketones
a) Oxidation of aldehydes Aldehydes are easily oxidised to carboxylic acid containing the same number of carbon
atom, as in parent aldehyde. The common oxidising agents are acidified K2Cr2O7, acidic or alkaline KMnO4 or chromic oxide.
Example
CH3 C = O
O
Acetaldehyde Aceticacid
H (o)
CH3 C OH
b) Oxidation of ketone
Ketones are not easily oxidised. Under drastic condition or with powerful oxidising agent like Con.HNO3, H+/KMnO4, H+/K2Cr2O7, cleavage of carbon-carbon bond takes place to give a mixture of carboxylic acids having less number of carbon atom than the parent ketone.
HCOOH + CH3COOH O
(O) Con HNO3
Formic acid Acetic acid
CH3 C CH3
The oxidation of unsymmetrical ketones is governed by Popoff ’s rule. It states that during the oxidation of an unsymmetrical ketone, a (C–CO) bond is cleaved in such a way that the keto group stays with the smaller alkyl group.
O
CH CH3 3 CH3CH2 CH2CH2 CH3C COOH COOH (O)
Con HNO3
ethanoic acidPropanoic acidpentan – 2 – one
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C) Reduction reactions
(i) Reduction to alcohols
We have already learnt that aldehydes and ketones can be easily reduced to primary and secondary alcohols respectively. The most commonly used reducing agents are Lithium Aluminium hydride (LiAlH4), and Sodium borohydride (NaBH4).
a) Aldehyde are reduced to primary alcohols. Example
CH3 C + 2 (H)
O
LiAlH4
H
Acetaldehyde Ethylalcohol (1o)
CH3 CH2 OH
b) Ketone are reduced to Secondary alcohols. Example
CH3 C CH3 + 2(H)
O
NaBH4
Acetone OH
Isopropyl alcohol (2o)
CH3 3CH CH
The above reactions can also be carried out with hydrogen in the presence of metal catalyst like Pt, Pd, or Ni. LiAlH4 and NaBH4 do not reduce isolated carbon – carbon double bonds and double bond of benzene rings. In case of α, β unsaturated aldehyde and ketones, LiAlH4 reduces only C = O group leaving C = C bond as such.
ii) Reduction to hydrocarbon
The carbonyl group of aldehydes and ketones can be reduced to methylene group using suitable reducing agents to give hydrocarbons.
O
4(H) C CH2 + H2O Reducing
agent
a) Clemmensen reduction Aldehydes and Ketones when heated with zinc amalgam and concentrated hydrochloric
acid gives hydrocarbons. Example
CH3 C H + 4(H)
O
Zn - Hg
Acetaldehyde
Con HCl Ethane
CH3 CH3 + H2O
CH3
C
CH3 +
4(H)
O
Zn - Hg
Acetone
CH3CH2CH3 +
H2O Con
HCl
Propane
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b) Wolf Kishner reduction Aldehydes and Ketones when heated with hydrazine (NH2NH2) and sodium ethoxide, hydrocarbons are formed Hydrazine acts as a reducing agent and sodium ethoxide as a catalyst.
Example
CH3 C H + 4(H)
O
NH2 NH2
Acetaldehyde
CH3 CH3 + H2O + N2 C2H5ONa
Ethane
CH3
C
CH3 +
4 (H)
O
NH2 NH2
Acetone
CH3CH2CH3 +
H2O
N2
C2H5ONa
Aldehyde (or) ketones is first converted to its hydrazone which on heating with strong base gives hydrocarbons.
(iii) Reduction to pinacols: Ketones, on reduction with magnesium amalgam and water, are reduced to symmetrical diols known as pinacol.
CH3 - C
=
O
O
=
C
- CH3 +
2(H) Mg
Hg
H2O CH3 CH3
CH3 - C - C -
CH3
CH3 CH3
OH OH Acetone Acetone 2,3 dimethyl butane 2,3 - diol
(pinacol)
D) Haloform reaction
Acetaldehyde and methyl ketones, containing
O
CH3 C group, when treated with
halogen and alkali give the corresponding haloform. This is known as Haloform reaction.
CH3–C–CH3
O
CCl3–C–CH3
O
CHCl3 + CH3–C–ONa
O
NaOH NaOH
3Cl2
E) Reaction involving alkylgroup
i) Aldol condensation
The carbon attached to carbonyl carbon is called α - carbon and the hydrogen atom attached to α - carbon is called α - hydrogen.
In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α - hydrogen add together to give β- hydroxyl aldehyde (aldol) or β - hydroxyl ketone (ketol). The reaction is called aldol condensation reaction. The aldol or ketol readily loses water to give α,β – unsaturated compounds which are aldol condensation products.
Propane
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162
a) Acetaldehyde when warmed with dil NaOH gives β - hydroxyl butyraldehyde (acetaldol) H
C + H CH2 CHO dil.NaOH
O OH Acetaldehyde Acetaldol
(3 - Hydroxy butanal)
CH3 CH CH2 CHOCH3
Mechanism
The mechanism of aldol condensation of acetaldehyde takes place in three steps.
Step 1 :
The carbanion is formed as the α - hydrogen atom is removed as a proton by the base.
HO - + H - CH2 - CHO CH2 - CHO + H2O
Step 2 :
The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion. H
CH3 - C
CH2
- CHO CH3 - CH- CH2 - CHO
O O
Step 3 :
The alkoxide ion formed is protonated by water to form aldol.
CH3 -
O OH
+ OH -CH - CH2 - CHO CH3 - CH- CH2 - CHOH - OH
3-Hydroxy butanal
The aldol rapidly undergoes dehydration on heating with acid to form α - β unsaturated aldehyde.
CH3 CH = CH CHO + H2OH+
OH H Crotonaldehyde (But - 2- enal)
CH CH CHOCH3
ii) Crossed aldol condensation
Aldol condensation can also take place between two different aldehydes or ketones or between one aldehyde and one ketone such an aldol condensation is called crossed or mixed aldol condensation. This reaction is not very useful as the product is usually a mixture of all possible condensation products and cannot be separated easily.
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Example :
HCHO + CH3CHO dil.NaOH
HO - CH2 - CH2 - CHO formaldehyde acetaldehyde 3-hydroxy propanal
HCHO + CH3 - C - CH3
dil.NaOH HO - CH2 - CH2 - C - CH3
O O formaldehyde acetone 4-hydroxy butan-2-one
F) Some important reactions of benzaldehyde
i) Claisen – Schmidt Condensation
Benzaldehye condenses with aliphatic aldehyde or methyl ketone in the presence of dil. alkali at room temperature to form unsaturated aldehyde or ketone. This type of reaction is called Claisen – Schmidt condensation.
Example
Benzaldehyde
C6H5 CH = O + H2 CH CHO C6H5 CH = CH CHO + H2O
Cinnamaldehyde
dil NaOH
Acetaldehyde
Benzaldehyde
C6H5 CH = O + H2 CH C CH3 C6H5 CH = CH C CH3 + H2O
Benzylidene acetone (Benzal acetone)
dil NaOH
Acetone O O
ii) Cannizaro reaction
In the presence of concentrated aqueous or alcoholic alkali, aldehydes which do not have α - hydrogen atom undergo self oxidation and reduction (disproportionation) to give a mixture of alcohol and a salt of carboxylic acid. This reaction is called Cannizaro reaction.
Benzaldehyde on treatment with concentrated NaOH (50%) gives benzyl alcohol and sodium benzoate.
Benzaldehyde
C6H5CHO
C6H5CHO + 50% NaOH
C6H5CH2OH
C6H5COONa +
Sodiumbenzoate
Benzylalcohol
This reaction is an example disproportionation reaction
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Mechanism of Cannizaro reaction Cannizaro reaction involves three steps.
Step 1 : Attack of OH- on the carbonyl carbon.
C6H5 C H
O
C6H5 C H
OH
O
+ OH fast
Step 2 : Hydride ion transfer
C6H5 C H
O
C6H5 C + C6H5CH2O
OH Benzaldehyde
O
OH
+ C6H5 - C - H
O slow
Step 3 : Acid – base reaction.
benzoate
C 6 H
5 C OH + C
6 H
5 CH
2 O C
6 H
5 C
- O + C
6 H
5 CH
2 OH
O O
Benzyl alcohol
Proton
exchange
Cannizaro reaction is a characteristic of aldehyde having no α – hydrogen.
Crossed Cannizaro reaction
When Cannizaro reaction takes place between two different aldehydes (neither containing an α hydrogen atom), the reaction is called as crossed cannizaro reaction.
C6H5 HCHO NaOH C6H5CH2OH
Benzaldehyde Formaldehyde Benzyl alcohol
CHO +
sodium formate
HCOONa+
In crossed cannizaro reaction more reactive aldehyde is oxidized and less reactive aldehyde is reduced.
3) Benzoin condensation
The Benzoin condensation involves the treatment of an aromatic aldehyde with aqueous alcoholic KCN. The products are a hydroxy ketone.
Example Benzaldehyde reacts with alcoholic KCN to form benzoin
Benzaldehyde
C6H5 C + H C C6H5
Benzoin
alcKCN
O O
H
OHO
C6H5 CH C C6H5
2-hydroxy – 1, 2 – diphenyl ethanone
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4) Perkins’ reaction
When an aromatic aldehyde is heated with an aliphatic acid anhydride in the presence of the sodium salt of the acid corresponding to the anhydride, condensation takes place and an α, β unsaturated acid is obtained. This reaction is known as Perkin’s reaction.
Example:
Benzaldehyde
C6H5 C = O + H2CH C
O
H O C6H5 CH = CH C
O
O
O
C6H5CH = CH COOH + CH3COOH CH3COONa
- H2O
H2O
Acetic anhydride
Cinnamic acid Acetic acidCH3 C
CH3 C
O
5) Knoevenagal reaction
Benzaldehyde
C6H5 CH = O + H2 C Pyridine
- H2O
COOH
COOH C6H5 CH = C
COOH
COOH - CO2
Malonic acid Cinnamic acid
C6H5 CH = CH COOH
Benzaldehyde condenses with malonic acid in presence of pyridine forming cinnamic acid, Pyridine act as the basic catalyst.
6) Reaction with amine
Aromatic aldehydes react with primary amines (aliphatic or aromatic) in the presence of an acid to form schiff ’s base.
Example
Benzaldehyde
C6H5 CH = O + H2 N C6H5 H C6H5 CH= N C6H5 + H2O
Aniline Benzal aniline (Schiff’s base)
7) Condensation with tertiary aromatic amines
Benzaldehyde condenses with tertiary aromatic amines like N, N – dimethyl aniline in the presence of strong acids to form triphenyl methane dye.
Benzaldehyde N, N - Dimethyl aniline Malachite green dye
C = O +
H H
H
N (CH3)2
N (CH3)2
con H2SO4 - H2O
H
C
N (CH3)2
N (CH3)2
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|——|——|
166
8) Electrophilic substitution reactions of benzaldehyde
CHO
conc H2SO4
CHO
NO2
+ H2O
m - Nitrobenzaldehyde
m - Benzaldehyde sulphonic acid
CHO
SO3H
+ H2O
CHO
Cl
+ HCl
Conc. HNO3
Conc. H2SO4
Conc. FeCl3
No Catalyst
m - chlorobenzaldehyde
C - Cl +HCl
O
Benzoyl chloride
Cl2
Cl2
Electrophilic substitution reaction of acetophenone Acetophenone reacts with Nitrating mixture
to form m – nitroacetophenone.