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

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 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 :

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

1. What is the general formula of an aldehyde?

Ans. The general formula of an aldehyde is R-CHO, where R represents an alkyl or aryl group.

2. How can aldehydes be oxidized to form carboxylic acids?

Ans. Aldehydes can be oxidized to form carboxylic acids by using oxidizing agents such as potassium permanganate (KMnO4) or chromic acid (H2CrO4).

3. What is the difference between an aldehyde and a ketone?

Ans. The main difference between an aldehyde and a ketone is the presence of a carbonyl group (-C=O) attached to a hydrogen atom in an aldehyde, whereas in a ketone, the carbonyl group is attached to two alkyl or aryl groups.

4. How can ketones be prepared from aldehydes?

Ans. Ketones can be prepared from aldehydes by using a dehydrating agent such as acid anhydrides or phosphorus pentoxide (P2O5). This reaction is called the Aldol condensation.

5. What is the functional group present in carboxylic acids?

Ans. The functional group present in carboxylic acids is the carboxyl group (-COOH), consisting of a carbonyl group (-C=O) and a hydroxyl group (-OH) attached to the same carbon atom.

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