Important Aldehyde, Ketones & Carboxylic Acid Formulas for JEE and NEET

 Page 1


  
?  Points to remember in Aldehyde & ketone
Aldol condensation : 
Carbonyl compounds having acidic sp
3
 ?-H shows this reaction in presence
of dil. NaOH or dil. acid.
O C CH 2
|
H
3
? ?
Dil
NaOH
????
H
|
CHO CH C CH
|
OH
2 3
? ? ?
O H
, H
2
?
?
? ? ? ? ?
?
CHCHO CH CH
3
?
Crossed aldol condensation
(i) CH
3
CHO + HCHO ? ? ? ? ? ? ?
NaOH . Dil
 HOCH
2
?CH
2
?CHO 
?
? ? ? ? ? ?
?
O H / H
2
 CH
2
=CH?CHO
(ii) CH
3
COCH
3
 + HCHO ? ? ? ? ? ? ?
NaOH . Dil
CH
3
CO?CH
2
CH
2
OH 
?
? ? ? ? ? ?
?
O H / H
2
CH
3
CO?CH=CH
2
Cannizzaro reaction :
Carbonyl compounds not having sp
3
?-H shows following disproportion reaction
%) 50 (
NaOH H C H 2
||
O
? ? ?   ?? ?
OH CH
3
?
 +  
HCOONa
CHO H C 2
5 6
 +  
%) 50 (
NaOH ?? ? OH CH H C
2 5 6
  +  
benzoate . Sol
COONa H C
5 6
Crossed Cannizzaro reaction :
+ HCHO +
%) 50 (
NaOH ?? ? +
formate . Sod
HCOONa
Formation of hydrzones and azines
C=O + NH NH
2 2
C
NHNH
2
OH
?H O
2
C= NNH
2
CO
C= N?N = C
Perkin reaction :
When benzaldehyde (or any other aromatic aldehyde) is heated with the
Page 2


  
?  Points to remember in Aldehyde & ketone
Aldol condensation : 
Carbonyl compounds having acidic sp
3
 ?-H shows this reaction in presence
of dil. NaOH or dil. acid.
O C CH 2
|
H
3
? ?
Dil
NaOH
????
H
|
CHO CH C CH
|
OH
2 3
? ? ?
O H
, H
2
?
?
? ? ? ? ?
?
CHCHO CH CH
3
?
Crossed aldol condensation
(i) CH
3
CHO + HCHO ? ? ? ? ? ? ?
NaOH . Dil
 HOCH
2
?CH
2
?CHO 
?
? ? ? ? ? ?
?
O H / H
2
 CH
2
=CH?CHO
(ii) CH
3
COCH
3
 + HCHO ? ? ? ? ? ? ?
NaOH . Dil
CH
3
CO?CH
2
CH
2
OH 
?
? ? ? ? ? ?
?
O H / H
2
CH
3
CO?CH=CH
2
Cannizzaro reaction :
Carbonyl compounds not having sp
3
?-H shows following disproportion reaction
%) 50 (
NaOH H C H 2
||
O
? ? ?   ?? ?
OH CH
3
?
 +  
HCOONa
CHO H C 2
5 6
 +  
%) 50 (
NaOH ?? ? OH CH H C
2 5 6
  +  
benzoate . Sol
COONa H C
5 6
Crossed Cannizzaro reaction :
+ HCHO +
%) 50 (
NaOH ?? ? +
formate . Sod
HCOONa
Formation of hydrzones and azines
C=O + NH NH
2 2
C
NHNH
2
OH
?H O
2
C= NNH
2
CO
C= N?N = C
Perkin reaction :
When benzaldehyde (or any other aromatic aldehyde) is heated with the
  
anhydride of an aliphatic acid (containing two ?-hydrogen atoms) in the
presence of its sodium salt, condensation takes place to form a
?-arylacrylic acid ; e.g., with acetic anhydride and sodium acetate, cinnamic
acid is formed.
C
6
H
5
CHO + (CH
3
CO)
2
O 
? ? ? ? ? ?
Na CO CH
2 3 C
6
H
5
CH = CHCO
2
H
Mechanism :
CH
3
COOCOCH
3
 + CH
3
CO
2
?
  H CO CH COOCOCH H C
2 3 3 2
?
?
C H C + CH COOCOCH
6 5 2 3
O
H
  
C H CCH COOCOCH
6 5 2 3
O
H
 
H
+
C H CCH COOCOCH
6 5 2 3
OH
H
 ? ? ? ? ?
O H ?
2
 C
6
H
5
CH=CHCOOCOCH
3
 ? ? ? ?
O H
2
C
6
H
5
CH = CHCO
2
H + CH
3
CO
2
H
Haloform reaction :
Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl
2
, Br
2
 or
I
2
) in the presence of alkali to give haloform and acid salt.
3
CH C R
||
O
? ?
 
? ? ? ? ? ? ?
NaOH / Br
2
  
3
CHBr ONa C R
||
O
? ? ?
  (Bromoform)
In this reaction ? CH
3
 of ? ? C CH
||
O
3
 group is converted into haloform as it
contains acidic hydrogen atom and rest-part of alkyl methyl ketone give acid
salt having carbon atom corresponding to alkyl ketone.
Preparation of haloform from methylketone involves two steps.
(a) Halogenation
3
CH C R
||
O
? ?
  
? ? ? ?
2
Br
  
3
CBr C R
||
O
? ?
  (Halogenation)
Page 3


  
?  Points to remember in Aldehyde & ketone
Aldol condensation : 
Carbonyl compounds having acidic sp
3
 ?-H shows this reaction in presence
of dil. NaOH or dil. acid.
O C CH 2
|
H
3
? ?
Dil
NaOH
????
H
|
CHO CH C CH
|
OH
2 3
? ? ?
O H
, H
2
?
?
? ? ? ? ?
?
CHCHO CH CH
3
?
Crossed aldol condensation
(i) CH
3
CHO + HCHO ? ? ? ? ? ? ?
NaOH . Dil
 HOCH
2
?CH
2
?CHO 
?
? ? ? ? ? ?
?
O H / H
2
 CH
2
=CH?CHO
(ii) CH
3
COCH
3
 + HCHO ? ? ? ? ? ? ?
NaOH . Dil
CH
3
CO?CH
2
CH
2
OH 
?
? ? ? ? ? ?
?
O H / H
2
CH
3
CO?CH=CH
2
Cannizzaro reaction :
Carbonyl compounds not having sp
3
?-H shows following disproportion reaction
%) 50 (
NaOH H C H 2
||
O
? ? ?   ?? ?
OH CH
3
?
 +  
HCOONa
CHO H C 2
5 6
 +  
%) 50 (
NaOH ?? ? OH CH H C
2 5 6
  +  
benzoate . Sol
COONa H C
5 6
Crossed Cannizzaro reaction :
+ HCHO +
%) 50 (
NaOH ?? ? +
formate . Sod
HCOONa
Formation of hydrzones and azines
C=O + NH NH
2 2
C
NHNH
2
OH
?H O
2
C= NNH
2
CO
C= N?N = C
Perkin reaction :
When benzaldehyde (or any other aromatic aldehyde) is heated with the
  
anhydride of an aliphatic acid (containing two ?-hydrogen atoms) in the
presence of its sodium salt, condensation takes place to form a
?-arylacrylic acid ; e.g., with acetic anhydride and sodium acetate, cinnamic
acid is formed.
C
6
H
5
CHO + (CH
3
CO)
2
O 
? ? ? ? ? ?
Na CO CH
2 3 C
6
H
5
CH = CHCO
2
H
Mechanism :
CH
3
COOCOCH
3
 + CH
3
CO
2
?
  H CO CH COOCOCH H C
2 3 3 2
?
?
C H C + CH COOCOCH
6 5 2 3
O
H
  
C H CCH COOCOCH
6 5 2 3
O
H
 
H
+
C H CCH COOCOCH
6 5 2 3
OH
H
 ? ? ? ? ?
O H ?
2
 C
6
H
5
CH=CHCOOCOCH
3
 ? ? ? ?
O H
2
C
6
H
5
CH = CHCO
2
H + CH
3
CO
2
H
Haloform reaction :
Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl
2
, Br
2
 or
I
2
) in the presence of alkali to give haloform and acid salt.
3
CH C R
||
O
? ?
 
? ? ? ? ? ? ?
NaOH / Br
2
  
3
CHBr ONa C R
||
O
? ? ?
  (Bromoform)
In this reaction ? CH
3
 of ? ? C CH
||
O
3
 group is converted into haloform as it
contains acidic hydrogen atom and rest-part of alkyl methyl ketone give acid
salt having carbon atom corresponding to alkyl ketone.
Preparation of haloform from methylketone involves two steps.
(a) Halogenation
3
CH C R
||
O
? ?
  
? ? ? ?
2
Br
  
3
CBr C R
||
O
? ?
  (Halogenation)
  
(b) Alkalihydrolysis
3
CBr C R
||
O
? ? ? ? ? ? ?
NaOH
  CHBr
3
 +  
ONa C R
||
O
? ?
 (Alkalihydrolysis)
Note : This reaction is used to distinguish the presence of ? ? C CH
||
O
3
 group.
Other reactions :
(1)
(2)
(3)
Page 4


  
?  Points to remember in Aldehyde & ketone
Aldol condensation : 
Carbonyl compounds having acidic sp
3
 ?-H shows this reaction in presence
of dil. NaOH or dil. acid.
O C CH 2
|
H
3
? ?
Dil
NaOH
????
H
|
CHO CH C CH
|
OH
2 3
? ? ?
O H
, H
2
?
?
? ? ? ? ?
?
CHCHO CH CH
3
?
Crossed aldol condensation
(i) CH
3
CHO + HCHO ? ? ? ? ? ? ?
NaOH . Dil
 HOCH
2
?CH
2
?CHO 
?
? ? ? ? ? ?
?
O H / H
2
 CH
2
=CH?CHO
(ii) CH
3
COCH
3
 + HCHO ? ? ? ? ? ? ?
NaOH . Dil
CH
3
CO?CH
2
CH
2
OH 
?
? ? ? ? ? ?
?
O H / H
2
CH
3
CO?CH=CH
2
Cannizzaro reaction :
Carbonyl compounds not having sp
3
?-H shows following disproportion reaction
%) 50 (
NaOH H C H 2
||
O
? ? ?   ?? ?
OH CH
3
?
 +  
HCOONa
CHO H C 2
5 6
 +  
%) 50 (
NaOH ?? ? OH CH H C
2 5 6
  +  
benzoate . Sol
COONa H C
5 6
Crossed Cannizzaro reaction :
+ HCHO +
%) 50 (
NaOH ?? ? +
formate . Sod
HCOONa
Formation of hydrzones and azines
C=O + NH NH
2 2
C
NHNH
2
OH
?H O
2
C= NNH
2
CO
C= N?N = C
Perkin reaction :
When benzaldehyde (or any other aromatic aldehyde) is heated with the
  
anhydride of an aliphatic acid (containing two ?-hydrogen atoms) in the
presence of its sodium salt, condensation takes place to form a
?-arylacrylic acid ; e.g., with acetic anhydride and sodium acetate, cinnamic
acid is formed.
C
6
H
5
CHO + (CH
3
CO)
2
O 
? ? ? ? ? ?
Na CO CH
2 3 C
6
H
5
CH = CHCO
2
H
Mechanism :
CH
3
COOCOCH
3
 + CH
3
CO
2
?
  H CO CH COOCOCH H C
2 3 3 2
?
?
C H C + CH COOCOCH
6 5 2 3
O
H
  
C H CCH COOCOCH
6 5 2 3
O
H
 
H
+
C H CCH COOCOCH
6 5 2 3
OH
H
 ? ? ? ? ?
O H ?
2
 C
6
H
5
CH=CHCOOCOCH
3
 ? ? ? ?
O H
2
C
6
H
5
CH = CHCO
2
H + CH
3
CO
2
H
Haloform reaction :
Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl
2
, Br
2
 or
I
2
) in the presence of alkali to give haloform and acid salt.
3
CH C R
||
O
? ?
 
? ? ? ? ? ? ?
NaOH / Br
2
  
3
CHBr ONa C R
||
O
? ? ?
  (Bromoform)
In this reaction ? CH
3
 of ? ? C CH
||
O
3
 group is converted into haloform as it
contains acidic hydrogen atom and rest-part of alkyl methyl ketone give acid
salt having carbon atom corresponding to alkyl ketone.
Preparation of haloform from methylketone involves two steps.
(a) Halogenation
3
CH C R
||
O
? ?
  
? ? ? ?
2
Br
  
3
CBr C R
||
O
? ?
  (Halogenation)
  
(b) Alkalihydrolysis
3
CBr C R
||
O
? ? ? ? ? ? ?
NaOH
  CHBr
3
 +  
ONa C R
||
O
? ?
 (Alkalihydrolysis)
Note : This reaction is used to distinguish the presence of ? ? C CH
||
O
3
 group.
Other reactions :
(1)
(2)
(3)
  
(4) 
(5)
(6)
Page 5


  
?  Points to remember in Aldehyde & ketone
Aldol condensation : 
Carbonyl compounds having acidic sp
3
 ?-H shows this reaction in presence
of dil. NaOH or dil. acid.
O C CH 2
|
H
3
? ?
Dil
NaOH
????
H
|
CHO CH C CH
|
OH
2 3
? ? ?
O H
, H
2
?
?
? ? ? ? ?
?
CHCHO CH CH
3
?
Crossed aldol condensation
(i) CH
3
CHO + HCHO ? ? ? ? ? ? ?
NaOH . Dil
 HOCH
2
?CH
2
?CHO 
?
? ? ? ? ? ?
?
O H / H
2
 CH
2
=CH?CHO
(ii) CH
3
COCH
3
 + HCHO ? ? ? ? ? ? ?
NaOH . Dil
CH
3
CO?CH
2
CH
2
OH 
?
? ? ? ? ? ?
?
O H / H
2
CH
3
CO?CH=CH
2
Cannizzaro reaction :
Carbonyl compounds not having sp
3
?-H shows following disproportion reaction
%) 50 (
NaOH H C H 2
||
O
? ? ?   ?? ?
OH CH
3
?
 +  
HCOONa
CHO H C 2
5 6
 +  
%) 50 (
NaOH ?? ? OH CH H C
2 5 6
  +  
benzoate . Sol
COONa H C
5 6
Crossed Cannizzaro reaction :
+ HCHO +
%) 50 (
NaOH ?? ? +
formate . Sod
HCOONa
Formation of hydrzones and azines
C=O + NH NH
2 2
C
NHNH
2
OH
?H O
2
C= NNH
2
CO
C= N?N = C
Perkin reaction :
When benzaldehyde (or any other aromatic aldehyde) is heated with the
  
anhydride of an aliphatic acid (containing two ?-hydrogen atoms) in the
presence of its sodium salt, condensation takes place to form a
?-arylacrylic acid ; e.g., with acetic anhydride and sodium acetate, cinnamic
acid is formed.
C
6
H
5
CHO + (CH
3
CO)
2
O 
? ? ? ? ? ?
Na CO CH
2 3 C
6
H
5
CH = CHCO
2
H
Mechanism :
CH
3
COOCOCH
3
 + CH
3
CO
2
?
  H CO CH COOCOCH H C
2 3 3 2
?
?
C H C + CH COOCOCH
6 5 2 3
O
H
  
C H CCH COOCOCH
6 5 2 3
O
H
 
H
+
C H CCH COOCOCH
6 5 2 3
OH
H
 ? ? ? ? ?
O H ?
2
 C
6
H
5
CH=CHCOOCOCH
3
 ? ? ? ?
O H
2
C
6
H
5
CH = CHCO
2
H + CH
3
CO
2
H
Haloform reaction :
Acetaldehyde and methylalkyl ketones react rapidly with halogen (Cl
2
, Br
2
 or
I
2
) in the presence of alkali to give haloform and acid salt.
3
CH C R
||
O
? ?
 
? ? ? ? ? ? ?
NaOH / Br
2
  
3
CHBr ONa C R
||
O
? ? ?
  (Bromoform)
In this reaction ? CH
3
 of ? ? C CH
||
O
3
 group is converted into haloform as it
contains acidic hydrogen atom and rest-part of alkyl methyl ketone give acid
salt having carbon atom corresponding to alkyl ketone.
Preparation of haloform from methylketone involves two steps.
(a) Halogenation
3
CH C R
||
O
? ?
  
? ? ? ?
2
Br
  
3
CBr C R
||
O
? ?
  (Halogenation)
  
(b) Alkalihydrolysis
3
CBr C R
||
O
? ? ? ? ? ? ?
NaOH
  CHBr
3
 +  
ONa C R
||
O
? ?
 (Alkalihydrolysis)
Note : This reaction is used to distinguish the presence of ? ? C CH
||
O
3
 group.
Other reactions :
(1)
(2)
(3)
  
(4) 
(5)
(6)
  
?  Points to remember in Carboxylic acid & Derivative
Summary of reactions of carboxylic acids :