Write IUPAC name of all the products formed by cross aldol condensation of butanone and Ethanal.(Hydroxy forms)
Answers
Answer:
When an aldehyde (or a ketone) having alpha hydrogen atom is treated with dilute base NaOH, KOH or sodium carbonate, two molecules of aldehyde ( or ketone) add together to form β hydroxy aldehyde (aldol) or β hydroxy ketone (ketol). New C−C bond is formed in this reaction.
When aldol or ketol is heated in presence of an acid, a molecule of water is lost and α−β unsaturated aldehyde or α−β unsaturated ketone is formed.
Aldol condensation of ethanal:
2CH
3
−CHO
dil NaOH
CH
3
−CHOH−CH
2
−CHO
H
+
Δ
CH
3
−CH=CH−CHO+H
2
O
Aldol condensation of propanone:
2(CH
3
)
2
CO
dil NaOH
(CH
3
)
2
C−OH−CH
2
−CO−CH
3
H
+
Δ
(CH
3
)
2
C=CH−CO−CH
3
+H
2
O
Explanation:
When an aldehyde (or a ketone) having alpha hydrogen atom is treated with dilute base NaOH, KOH or sodium carbonate, two molecules of aldehyde ( or ketone) add together to form β hydroxy aldehyde (aldol) or β hydroxy ketone (ketol). New C−C bond is formed in this reaction.
When aldol or ketol is heated in presence of an acid, a molecule of water is lost and α−β unsaturated aldehyde or α−β unsaturated ketone is formed.
Aldol condensation of ethanal:
2CH
3
−CHO
dil NaOH
CH
3
−CHOH−CH
2
−CHO
H
+
Δ
CH
3
−CH=CH−CHO+H
2
O
Aldol condensation of propanone:
2(CH
3
)
2
CO
dil NaOH
(CH
3
)
2
C−OH−CH
2
−CO−CH
3
H
+
Δ
(CH
3
)
2
C=CH−CO−CH
3
+H
2
O