1. A2 Chemistry Aldehydes and Ketones
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2. ALDEHYDES and KETONES Aldehydes CH3CH2CHO Propanal CH3CHO Ethanal
Collectively known as CARBONYL compounds C H O
Aldehydes
General functional group is
Abbreviated –CHO
Named using (name of corresponding alkane….al) C H O
CH3CH2CHO
Propanal C H O
CH3CHO
Ethanal C H O
CH3CH2CH2CHO
Butanal

3. Ketones General functional group is Abbreviated C-CO-C
Named using (name of corresponding alkane….one) C O H
CH3COCH3
Propanone C O H
CH3COCH2CH2CH3
Pentan-2-one C O H
CH3COCH2CH3
Butanone

4. PHYSICAL PROPERTIES
1. Boiling points HIGHER than similar Mr hydrocarbons
because stronger dipole-dipole forces are present between aldehyde and ketone molecules.
2. Boiling points LOWER than similar Mr alcohols / acids
because stronger hydrogen bonds are present between alcohol and acid molecules.
3. Provided Mr not too large, aldehydes and ketones are relatively soluble in water
because the O atoms can hydrogen bond to H in water.

5. = first oxidation products of PRIMARY alcohols
PREPARING ALDEHYDES
= first oxidation products of PRIMARY alcohols
R-C-O-H + [O]  R-C H2O H O
[O] provided by acidified dichromate(VI), Cr2O72-.
Reaction mixture heated and IMMEDIATELY distilled
to avoid further oxidation to the carboxylic acid.
R-C [O]  R-C H O OH
eg Naming the organic products, write equations for oxidations of :
(a) ethanol (b) 2-methylpropan-1-ol

6. CH3-C-O-H + [O]  CH3-C + H2O H
eg Naming the organic products, write equations for oxidations of :
(a) ethanol (b) 2-methylpropan-1-ol
(a)
(b)
CH3-C-O-H + [O]  CH3-C H2O H O
Ethanal
CH3-C [O]  CH3-C H O OH
Ethanoic acid
CH3-C C-O-H + [O]  CH3-C C H2O H O
H3C
CH3
2-Methylpropanal
CH3-C C [O]  CH3-C C H O
CH3 OH
2-Methylpropanoic acid

7. = oxidation products of SECONDARY alcohols
PREPARING KETONES
= oxidation products of SECONDARY alcohols
R1-C-O-H + [O]  R1-C-R H2O H R2 O
[O] provided by acidified dichromate(VI), Cr2O72-.
Reaction mixture heated under reflux
since further oxidation does not occur.
 ketones are not reductants since not oxidisable.
Used in chemical tests to distinguish aldehydes from ketones, since aldehydes are reductants.
eg Naming products, write equations for oxidations of :
(a) propan-2-ol (b) 3-methylbutan-2-ol

8. CH3-C-O-H + [O]  CH3-C-CH3 + H2O H
eg Naming products, write equations for the oxidations of :
(a) propan-2-ol (b) 3-methylbutan-2-ol
(a)
(b)
CH3-C-O-H + [O]  CH3-C-CH H2O H
CH3 O
Propanone
CH3-C-O-H + [O]  CH3-C-CH(CH3)CH3 + H2O H
CH(CH3)CH3 O
3-Methylbutan-2-one

9. DETECTING & IDENTIFYING ALDEHYDES & KETONES
Unknown organic compound
+ 2,4-dintrophenylhydrazine dissolved in conc. H2SO4 (Brady’s reagent)
NO2
NH-NH2
+ O=C R1 R2
Orange ppt. if compound is an aldehyde or ketone
Ppt. is the HYDRAZONE derivative of the aldehyde or ketone
NO2
NH-N=C R1 R2
+ H2O
Isolate the hydrazone derivative, purify by recrystallisation from ethanol and measure its melting point
Identify the particular aldehyde or ketone by matching melting point to databook values

10. DISTINGUISHING ALDEHYDES FROM KETONES Dichromate test Fehling’s Test
Silver Mirror Test
Heat with ORANGE acidified Cr2O72-
Heat with BLUE alkaline Cu2+ ( + NaOH)
Heat with COLOURLESS alkaline Ag+ ( + NH3)
For Aldehyde :
ORANGE  GREEN SOLN [Cr3+(aq)]
BLUE SOLN  BRICK-RED PPT [Cu2O(s)]
COLOURLESS SOLN  SILVER MIRROR [Ag(s)]
ALL ARE REDUCTIONS OF METAL ION BY ALDEHYDE
Cr(+6)  (+3)
Cu(+2)  (+1)
Ag(+1)  (0)
ALDEHYDE IS OXIDISED TO THE CAROXYLIC ACID
For Ketone :
Remains ORANGE
Remains BLUE
Remains COLOURLESS

11. REDUCTION OF ALDEHYDES & KETONES
= reverse of oxidation of alcohols
NB (2) preferred to reduce aldehydes & ketones – (1) and (3) also reduce acids etc
Reducing agents =
(1) H2 gas with Ni catalyst
or (2) NaBH4 followed by water
or (3) LiAlH4 in dry ether
all represented by [H]
Applicable to both aldehydes and ketones
C=O R1 R2
+ 2[H]  C R1 R2 H OH
Acid  ALDEHYDE  primary alcohol
KETONE  secondary alcohol

12. Predict the reduction product(s) of :
1. Propanal 2. Butan-2-one
3. Butan-2,3-dione 4. Hexanoic acid
5. Cyclohexanone
CH3 CH2 C H OH
CH3 CH CH3 C OH H
Propan-1-ol
Butan-2-ol
CH3 C CH3 C OH H
CH3CH2CH2CH2CH2CHO
(hexanal)
then CH3CH2CH2CH2CH2CH2OH
(hexan-1-ol)
Butan-2,3-diol H OH
Cyclohexanol

13. = NUCLEOPHILIC ADDITION
MECHANISM OF REDUCTION
Stage 1 : Nucleophile (:H-) attacks +C of C=O
:H- C
O- +
from NaBH4 H C
:O-
Stage 2 : Protonation of O- by water H C
:O- H C OH
Primary or secondary alcohol formed
+ :OH-
H OH
= NUCLEOPHILIC ADDITION

14. - NUCLEOPHILIC ADDITION OF HCN  common to both aldehydes and ketones
:CN = the nucleophile, attacks +ve C of +C=O -
C=O R1 R2
+ HCN  C R1 R2 CN OH
2-hydroxynitriles
Note : HCN provided by mixing K+CN- with dil. H2SO4 because HCN is so weak that the concentration of :CN- would be very low.

15. Predict the product(s) of HCN reacting with:
Methanal Propanone
3. Ethanal Cyclohexanone C H CN OH C
CH3 CN OH
2-hydroxyethanenitrile
2-hydroxy-2-methylpropanenitrile CN OH C H
CH3 CN OH
2-hydroxypropanenitrile
1-cyanocyclohexanol

16. MECHANISM OF NUCLEOPHILIC ADDITION OF HCN
Stage 1 : Nucleophile (:CN-) attacks +C of C=O
:CN- C
O- + CN C
:O-
Stage 2 : Protonation of O- by H2SO4 CN C
:O- CN C OH
Note: Exactly parallel to reduction
+ HSO4-
H OSO3H

17. OPTICAL ACTIVITY OF 2-HYDROXYNITRILES
If R1 and R2 not the same, CN C OH R2 R1
2- hydroxynitrile will be optically active because
it contains C bonded to 4 different groups
= a CHIRAL or
ASYMMETRIC carbon atom
 stereoisomers which are non-superimposable mirror images of eachother
 rotate plane-polarised light EQUALLY but OPPOSITE
Measured by polarimeter (see “isomerism” notes)

18. Optical isomers (“enantiomers”) of 2-hydroxynitrilesOH R1 CN R2
mirror HO R1 NC R2 C
A 50:50 mixture of such optical isomers is called a RACEMATE.
This is OPTICALLY INACTIVE because rotational effects CANCEL.

19. C O R1 R2 C OH R1 CN R2 C NC OH R1 R2 “Back” gives: “Front” gives:
Why does aldehyde / ketone + HCN reaction produce a racemate? C O R1 R2
Aldehydes and ketones are PLANAR (flat) around the C=O carbon
 CN- nucleophile may (50:50 chance) attack from “front” or “back” of this plane
 50:50 mixture of 2 different optical isomers C OH R1 CN R2
“Back” gives: C NC OH R1 R2
“Front” gives:
NB Racemisation provides evidence to support the proposed nucleophilic addition mechanism.

20. CHI3 = triiodomethane = “iodoform” = bright yellow precipitate
THE IODOFORM REACTION
Test for methyl aldehydes and ketones.
Heat with I2 dissolved in aq. NaOH
CH3 C R O
Also applies to alcohols of the type:
R = H (aldehyde) or CnH2n+1
CH3 CH R OH
+ 3I OH-
i.e. ethanol or any secondary methyl alcohol
These are oxidised to the corresponding methyl aldehyde or ketone by the alkaline I2 -O C R O
CHI3 +
+ 3I H2O
CH3-CH(OH)-R + I OH-
CHI3 = triiodomethane = “iodoform” = bright yellow precipitate
CH3-CO-R + 2I H2O

21. A. Propanal B. 2-Methylpropanal C. Pentan-3-one
SAMPLE QUESTIONS
A. CH3CH2CHO B. (CH3)2CHCHO C. CH3CH2COCH2CH3
Name compounds A – C
By selecting from A – C, select all the compounds which satisfy the following requirements.
Could be reduced to a secondary alcohol.
Could be oxidised to a carboxylic acid.
C be reduced to a primary alcohol.
(iii) Would produce a precipitate when treated Brady’s reagent.
A. Propanal B. 2-Methylpropanal C. Pentan-3-one
 ketone  C only
 aldehyde  A and B only
 aldehyde  A and B only
 Carbonyl compound  A, B and C

22. (a) 2-hydroxybutanenitrile (b) 2-hydroxyethanenitrile
Name the compound formed when (a) propanal (b) methanal is treated with HCN.
State the conditions for carrying out these reactions.
Write an equation for each reaction.
Are the products optically active? Explain.
(a) 2-hydroxybutanenitrile
(b) 2-hydroxyethanenitrile
+ KCN and dil. H2SO4 at room temperature
(a) CH3CH2CHO + HCN  C H
CH3 CH2 CN OH
(b) HCHO + HCN  C H CN OH
Neither is optically active because:
(a) racemisation will occur
(b) the product does not contain a chiral C (R1 = R2 = H)

23. In dry ether under reflux without heat
Aldehydes and ketones can be reduced to alcohols by lithium tetrahydridoaluminate(III), LiAlH4.
State the conditions under which this reagent is used.
P. CH3CH(OH)COOH Q. CH3COCOOH
P may be produced from Q by reduction, but this cannot be achieved with lithium tetrahydridoaluminate(III).
What will be the product if this reagent is used?
How would you obtain P from Q?
In dry ether under reflux without heat
C=O and COOH in Q will both be reduced
propan-1,2-diol (CH3CH(OH)CH2OH)
Use a milder reducing agent
 NaBH4 in water

24. The End