Transcript The carbonyl functional group Formation of the C=O group π

•
Recognise and name aldehydes and ketones.
The carbonyl functional group
Formation of the C=O group π-bond
Structures of benzaldehyde and phenylethanone
•
Describe the reduction of carbonyl compounds to form alcohols.
•
Outline the mechanism for nucleophilic addition reactions of aldehydes and
ketones with hydrides.
Reduction of an aldehyde produces a primary alcohol
Lithium
tetrahydridoaluminate
LiAlH4
Source of HReacts specifically with polar
π bonds
The primary alcohol propan-1-ol and the aldehyde propanal
Reduction of a ketone produces a secondary alcohol
Lithium
tetrahydridoaluminate
LiAlH4
Source of HReacts specifically with polar
π bonds
Steps
1. Addition of H- ions to the δ+ C
2. Reagents must be kept dry, carrried out
in ether
step2
1. Addition of aqueous acid solution,
protonates the O-
Week 3
Reduction of an aldehyde by nucleophilic addition
Both carbonyl compounds and alkenes can
be reduced using H2 & Pt catalyst
Addition Elimination Reactions
•
Describe the use of 2,4-dinitrophenylhydrazine to detect a carbonyl group
and to identify a carbonyl compound.
•
Describe the use of Tollens’ reagent to detect the presence of an aldehyde
group.
Carbonyl compounds react with compounds containing H2N- gp. LP on N acts
as a nucleophile and forms a bond with δ+ C in C=O.
Instead of an H+ adding, it loses a water (elimination) and C=N is formed
C=O + H2N-X  C=N-X + H2O
Racemic mixture of products are formed
Test for carbonyl group
Reaction of propanal with 2,4-dinitrophenylhydrazine
© Pearson Education Ltd 2009
This document may have been altered from the original
2,4 DNP (Brady's reagent)
• Product is insoluble, used as test for
carbonyl group
• Simple aldehydes & ketones = yellow ppt
• Aromatic = orange ppt
Identify the product
• React carbonyl with 2,4 DNP
• Filter ppt
• Recrystallise using minimum of hot
ethanol
• Dry & measure melting point
• Compare to data tables to identify the
carbonyl
•
Describe the oxidation of primary alcohols to form aldehydes and carboxylic
acids.
•
Describe the oxidation of secondary alcohols to form ketones.
•
Describe the oxidation of aldehydes to form carboxylic acids.
© Pearson Education Ltd 2009
This document may have been altered from the original
Ethanol oxidised to ethanal, and finally to ethanoic acid
Propan-2-ol can be oxidised to propanone
Oxidation of an aldehyde to a carboxylic acid (acid or neutral conditions)
Potassium
Dichromate
(K2Cr2O7)/ H2S04
KMnO4 in neutral –
brown ppt
© Pearson Education Ltd 2009
This document may have been altered from the original
Distinguish between an aldehyde & Ketone
Oxidation of an aldehyde using Tollens’ reagent (alkaline conditions – NaOH
added to silver nitrate)
+ H2O
Salt of
Fehlings reagent (copper(II)sulphate + sodium potassium copper (I) oxide),
red ppt when added to an aldehyde
Iodoform Reaction
• Reaction with ethanal and methyl
ketones & iodine in alkaline (sodium
hydroxide) solutions
• H atoms of CH3C=O are replaced by I
• Alkali breaks C-C bond to form pale yellow
ppt of iodoform CHI3 is formed
• Test if substance is ethanol or ethanal
• Sodium hydroxide solution is added to
iodine solution to form iodate ions (IO-)
• These substitute into the CH3 gp next to
C=0. IO- & O are electron withdrawing
thus breaking the bond between the 2 C.
• Result formation of iodoform
The reaction
• I2 + OH-  IO- + I• CH3COR  CI3COR  CHI3 + RCOO• Overall
• CH3COR + 3I2 + 4NaOH  CHI3 +
RCOONa + 3H20
conditions
Solutions is warmed with a mixture of iodine
and sodium hydroxide
Or
Potassium iodide in sodium chlorate
Summary - Aldehydes
+ HCN(pH 8) = RCH(OH)CN
+ [H] = RCH2OH
+ 2,4DNP = yellow orange ppt
+ K2Cr2O7/H+ = RCOOH
+ Fehlings/tollens = RCOO+ I2/NaOH = CHI3 + HCOO-
Summary Ketones
+ HCN(pH 8) = RC(OH)R’CN
+ [H] = RCH(OH)R’
+ 2,4DNP = yellow orange ppt
+ K2Cr2O7/H+ = no reaction
+ Fehlings/tollens = no reaction
+ I2/NaOH = CHI3 + R’COO-
HCN
• Adding HCN to carbonyl
• Need to add KCN as this is easiest to
break the bond and form CN• Need to add SULPHURIC acid, help to
amplify the polarisation of C=O
• Oxygen is very electronegative and
therefore makes the carbonyl bond very
polar
• Why is KCN added?
• To increase the number of CN ions
present, to speed up the reaction
• Why is sulphuric acid added?
• Increases the polarity of the C=O bond
Optical isomers – ethanal &
HCN
• Optical isomerism occurs in compounds
which have four different groups attached
to a single carbon atom. In this case, the
product molecule contains a CH3, a CN,
an H and an OH all attached to the central
carbon atom.
• Ethanal is a planar molecule, and attack
by a cyanide ion will either be from above
the plane of the molecule, or from below.
There is an equal chance of either
happening.
• the existing groups get forced down away
from the approaching cyanide ion
• Attack from below forces the existing
groups upwards.
• Now compare that with the molecule
formed by attack from above.
• This argument applies to all aldehydes,
and to ketones as long as they are
unsymmetrical - with a different alkyl group
either side of the carbonyl group.
• A symmetric ketone like propanone,
CH3COCH3, will only produce a single
product - not a mixture of isomers. The
product doesn't have four different groups
around the central carbon atom, and so
won't have optical isomers