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Chemical properties of Aldehydes and Ketones

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Chemical properties of Aldehydes and Ketones Reactions common to both Aldehydes and Ketones The difference between an Aldehyde and a Ketone is the position of the C=O (Carbonyl group). Aldehyde – always on a carbon at the end of the chain Ketone – Always on a carbon at middle carbon of the chain 1. BOTH Aldehydes and Ketones can be reduced Aldehyde reduced to Primary Alcohol Reducing agent = Sodium tetrahydroborate III (NaBH4) Ketone reduced to Secondary Alcohol Reduction Oxidisation Gain of Electrons Loss of Electrons H O H3C H + 2[H]  H3C C OH H True for all secondary alcohols Carbonyl + 2[H] ─(NaBH4) Alcohol CH3CH2CHO + 2[H]  CH3CH2CH2OH Propanal + NaBH4  Propan-1-ol CH3COCH3 +2[H]  CH3CH(OH)CH3 Propanone + NaBH4  Propan-2-ol 2. BOTH Aldehydes and Ketones undergo nucleophilic substitution using HCN Def. A nuclophile is an electron pair donor E.g. NH3 – 1 lone pair H2O – 2 lone pairs OH– – 3 lone pairs Cl– – 4 lone pairs Therefore nucleophiles attack E.g. C δ+=Oδ– 1 Due to a difference in electronegativity the O is electronegative than the C so attracts the electron pair more strongly C = C O NOTE: – A C=C will undergo electrophillic addition reactions and the reagents are added across the double bond – A C=O double bond undergo nucleophillic addition reactions and the reagents are added across O the double bond Mechanism of the reaction between Ethanal and HCN Conditions: K +CN– = catalyst Reagents: δ + O H3C C O H + HCN(g) δ + δ – H3C C H H CN O H N C KCN Ethanal  Intermediate O H3C C H CN  H3C C H CN 2 Hydroxylpropanenitrile Nitrile group Mechanism of the reaction between Propanone and HCN Conditions: K +CN– = catalyst δ + δ – H CN Reagents: O H3C C CH3 O H C C CH O H H C C CH 2 Propanone  3 3  3 3 CN CN N C KCN Intermediate 2 Hydroxyl 2 methylpropanenitrile C C C C C = C + 7 7 3. BOTH react with 2,4, Dinitrophenylhydrazine (2,4, DNPH)  2,4,DNPH is an orange solution  The reaction is used as a test for the C=O double bond  Reacts to give an orange/yellow precipitate H C C O H2N H3C C N + H2O 3 NH O H NH O H Ethanal + N + O O N 2,4, DNPH O + O O N Orange/Yellow O precipitate Example exam Q. Q. How can you distinguish between an Aldehyd and a Ketone? (4 marks) A.  Dissolve in 2,4 DNPH if precipitate is found the C=O bond present  Orange precipitate is purified by recrystalisation  Dry the crystals and heat them until their melting point is found  check the melting point found against data  Identify the exact Aldehyde or Ketone found Reactions for Aldeydes only 1. ONLY Aldeydes can be oxidised with Acidified Dichromate (H+ /Cr2O 2- ) O H3C H H O + [O]  H C C H OH 3 H + /Cr2O 2- Ethanal Acidified Dichromate Ethanoic Acid

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Chemical Properties Of Aldehydes And Ketones React
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Chemical properties of Aldehydes and Ketones

Reactions common to both Aldehydes and Ketones

The difference between an Aldehyde and a Ketone is the position of the C=O (Carbonyl group).

Aldehyde – always on a carbon at the end of the chain
Ketone – Always on a carbon at middle carbon of the chain

1. BOTH Aldehydes and Ketones can be reduced

Aldehyde reduced to Primary Alcohol
Reducing agent = Sodium tetrahydroborate III (NaBH4)
Ketone reduced to Secondary Alcohol

Reduction Oxidisation
Gain of Electrons Loss of Electrons

H
O

H3C C
+ 2[H]  H3C C OH

H H

True for all secondary alcohols
Carbonyl + 2[H] ─(NaBH4) Alcohol

CH3CH2CHO + 2[H]  CH3CH2CH2OH
Propanal + NaBH4  Propan-1-ol

CH3COCH3 +2[H]  CH3CH(OH)CH3
Propanone + NaBH4  Propan-2-ol

2. BOTH Aldehydes and Ketones undergo nucleophilic substitution using HCN

Def.
A nuclophile is an electron pair donor

E.g. NH3 – 1 lone pair
H2O – 2 lone pairs
OH– – 3 lone pairs
Cl– – 4 lone pairs
Therefore nucleophiles attack
E.g. Cδ+=Oδ– 1
Due to a difference in electronegativity the O is
electronegative than the C so attracts the electron
pair more strongly

, NOTE:

C C – A C=C will undergo
C C = electrophillic addition
reactions and the reagents
are added across the double
bond



– A C=O double bond
undergo nucleophillic
addition reactions and the
C reagents are added across
C O = O
the double bond




Mechanism of the reaction between Ethanal and HCN

Conditions: K+CN– = catalyst
Reagents: O
+ HCN(g)
H3C C H


δ+ O δ+ δ–
H3C C H CN
O H
H
Ethanal  Intermediate  HC C H
O 3
2 Hydroxylpropanenitrile
N C C CN
H3C H
KCN Nitrile group
CN


Mechanism of the reaction between Propanone and HCN
δ+ δ–
H CN
Conditions: K+CN– = catalyst
Reagents: O
O O H
2
H3C C CH3 HC C CH HC C CH
3 3 3 3
Propanone  
CN CN
N C 2 Hydroxyl 2 methylpropanenitrile
Intermediate
KCN

, 3. BOTH react with 2,4, Dinitrophenylhydrazine (2,4, DNPH)

 2,4,DNPH is an orange solution
 The reaction is used as a test for the C=O double bond
 Reacts to give an orange/yellow precipitate


H3C

H2N C N + H2O
HC C O
3
NH H NH
O O
H
Ethanal + +
 +
N N

+ O + O
O N O N Orange/Yellow
2,4, DNPH precipitate
O O



Example exam Q.
Q. How can you distinguish between an Aldehyd and a Ketone? (4 marks)

A.  Dissolve in 2,4 DNPH if precipitate is found the C=O bond present
 Orange precipitate is purified by recrystalisation
 Dry the crystals and heat them until their melting point is found
 check the melting point found against data
 Identify the exact Aldehyde or Ketone found


Reactions for Aldeydes only

1. ONLY Aldeydes can be oxidised with Acidified Dichromate (H+/Cr2O 72-)


Primary Alcohol  Aldehyde  Carboxylic Acid

Secondary Alcohol  Ketone

Tertiary Alcohol


O H O
H3C C + [O]  H C C
3
H H OH
H+/Cr2O72-
Ethanal Acidified Dichromate Ethanoic Acid

, 2. ONLY Aldehydes can react in the Silver mirror test (Tollens Test)

Reagents: Silver nitrate dissolved in Amonnia (Ag+ (NH 3)2 )
Conditions: Hot water bath

 If an Aldehyde is present the Silver ion (Ag+) changes to a silver atom (Ag)

Ag+ + 1e–  Ag(s)
ion – 46e– atom – 47e–

 As the Aldehyde was oxidised the Ag+ gained an electron and was reduced
 Aldehydes Carboxylic Acid KetonesNo further reaction




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