OCR Chemistry A H432 Carbonyl Compounds
Carbonyl Chemistry
Carbonyl compounds are those which contain >C=O
- aldehydes
- ketones
- carboxylic acids
- esters
You should recall how to name aldehydes and ketones:
O
- longest chain is 5 carbons, so "-pentan-" stem
CH3 CH2 C CH CH3 - ketones have "-one" ending
- C=O is located on 3rd carbon in chain
CH3
- has a methyl group
- methyl group on 2nd carbon (numbering for smallest)
2-methylpentan-3-one
O - longest chain is 5 carbons so "-pentan-" stem
- aldehydes have an "-al" ending
CH3 CH CH2 CH2 C
- numbering starts from the end with the aldehyde
CH3 H - methyl group is therefore on 4th carbon
4-methylpentanal
The simplest aromatic aldehyde and ketone are:
H O
C benzaldehyde
- colourless liquid (at RT) with almond-like aroma
- almond essence, gives flavour to marzipan
phenylethanone
O - used to create fragrances which resemble cherry,
H3C C strawberry, honeysuckle or jasmine
Chemical tests to distinguish carbonyl compounds
1: Detecting an aldehyde or ketone
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) to
form an orange or yellow precipitate. No precipitate is formed with other carbonyl
compounds such as carboxylic acids or esters. Brady's Reagent is a solution of 2,4-DNPH
in methanol and sulphuric acid.
A few drops of the carbonyl compound are put in a test tube with about 5cm3 of Brady's
reagent. The precipitate formed, referred to as a 2,4-dinitrophenylhydrazone derivative,
can be used to help identify the specific aldehyde or ketone, after purifying, by
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, OCR Chemistry A H432 Carbonyl Compounds
measuring its melting point. This works well because the different derivatives have
melting points that are many degrees apart.
e.g.
heptan-2-one b.p. = 151°C m.p of 2,4-DNP derivative = 90°C
cyclohexanone b.p. = 156°C m.p of 2,4-DNP derivative = 162°C
octan-2-one b.p. = 173°C m.p of 2,4-DNP derivative = 58°C
Identifying an aldehyde/ketone from the 2,4-DNPH derivative:
! The orange/yellow solid substance is purified by recrystallisation
! Melting point is determined
! Melting point is compared to a database of published melting points for these
derivatives
You will not be asked to write an equation for the formation of a 2,4-DNP derivative or
to recall who to draw the structure of a 2,4-dinitrophenylhydrazone derivative.
2: Telling an aldehyde from a ketone
A further test is necessary to distinguish an aldehyde from a ketone. Aldehydes can be
further oxidised to carboxylic acids, but ketones cannot.
Tollens' reagent is a weak oxidising agent containing silver nitrate in ammonia.
Aldehydes can be oxidised further whereas ketones are not oxidised. The oxidising
agent is the aqueous silver (I) ion, Ag+(aq). When warmed with Tollens reagent, the
aldehyde is oxidised to a carboxylic acid, and the silver ions in solution are reduced to
silver metal. A "silver mirror" is formed on the walls of the test tube (or sometimes just
a silver-grey solid is formed).
Oxidation of the aldehyde:
R-CHO + [O] " R-COOH
Reduction of the silver ions:
Ag+(aq) + e- " Ag(s)
How the carbonyl group reacts
Firstly we need to understand a little more about the carbonyl group:
1. Like a C=C double bond, it is comprised of a sigma bond and a pi bond formed by the
overlap of p-orbitals on the C and O atoms.
π-bond above and
C O à C O below plane of
other bonds
C C
σ-bond
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Carbonyl Chemistry
Carbonyl compounds are those which contain >C=O
- aldehydes
- ketones
- carboxylic acids
- esters
You should recall how to name aldehydes and ketones:
O
- longest chain is 5 carbons, so "-pentan-" stem
CH3 CH2 C CH CH3 - ketones have "-one" ending
- C=O is located on 3rd carbon in chain
CH3
- has a methyl group
- methyl group on 2nd carbon (numbering for smallest)
2-methylpentan-3-one
O - longest chain is 5 carbons so "-pentan-" stem
- aldehydes have an "-al" ending
CH3 CH CH2 CH2 C
- numbering starts from the end with the aldehyde
CH3 H - methyl group is therefore on 4th carbon
4-methylpentanal
The simplest aromatic aldehyde and ketone are:
H O
C benzaldehyde
- colourless liquid (at RT) with almond-like aroma
- almond essence, gives flavour to marzipan
phenylethanone
O - used to create fragrances which resemble cherry,
H3C C strawberry, honeysuckle or jasmine
Chemical tests to distinguish carbonyl compounds
1: Detecting an aldehyde or ketone
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) to
form an orange or yellow precipitate. No precipitate is formed with other carbonyl
compounds such as carboxylic acids or esters. Brady's Reagent is a solution of 2,4-DNPH
in methanol and sulphuric acid.
A few drops of the carbonyl compound are put in a test tube with about 5cm3 of Brady's
reagent. The precipitate formed, referred to as a 2,4-dinitrophenylhydrazone derivative,
can be used to help identify the specific aldehyde or ketone, after purifying, by
Page 1
, OCR Chemistry A H432 Carbonyl Compounds
measuring its melting point. This works well because the different derivatives have
melting points that are many degrees apart.
e.g.
heptan-2-one b.p. = 151°C m.p of 2,4-DNP derivative = 90°C
cyclohexanone b.p. = 156°C m.p of 2,4-DNP derivative = 162°C
octan-2-one b.p. = 173°C m.p of 2,4-DNP derivative = 58°C
Identifying an aldehyde/ketone from the 2,4-DNPH derivative:
! The orange/yellow solid substance is purified by recrystallisation
! Melting point is determined
! Melting point is compared to a database of published melting points for these
derivatives
You will not be asked to write an equation for the formation of a 2,4-DNP derivative or
to recall who to draw the structure of a 2,4-dinitrophenylhydrazone derivative.
2: Telling an aldehyde from a ketone
A further test is necessary to distinguish an aldehyde from a ketone. Aldehydes can be
further oxidised to carboxylic acids, but ketones cannot.
Tollens' reagent is a weak oxidising agent containing silver nitrate in ammonia.
Aldehydes can be oxidised further whereas ketones are not oxidised. The oxidising
agent is the aqueous silver (I) ion, Ag+(aq). When warmed with Tollens reagent, the
aldehyde is oxidised to a carboxylic acid, and the silver ions in solution are reduced to
silver metal. A "silver mirror" is formed on the walls of the test tube (or sometimes just
a silver-grey solid is formed).
Oxidation of the aldehyde:
R-CHO + [O] " R-COOH
Reduction of the silver ions:
Ag+(aq) + e- " Ag(s)
How the carbonyl group reacts
Firstly we need to understand a little more about the carbonyl group:
1. Like a C=C double bond, it is comprised of a sigma bond and a pi bond formed by the
overlap of p-orbitals on the C and O atoms.
π-bond above and
C O à C O below plane of
other bonds
C C
σ-bond
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