Write the product formed when (CH3)3C-CHO reacts whith the CH3COCH3in the presence of diluteNaOH
Answers
Answer:
Acetaldehyde (CH
3
CHO) reacts with hydrogen cyaide HCN to give 2-hydroxypropapanenitrile as product.
Acctaldehyde
CH
3
−CHO
+
Cyanide
Hydrogen
HCN
⟶
CH
3
H
∣
2−hydroxypropanenitrile
CN
∣
−C−
OH
(ii) Acetaldehyde (CH
3
CHO) reacts with Hydroxylamine (NH
2
OH) to give acetaldoxime as a product.
Acetaldyde
CH
3
CHO
+
Hydroxylamine
NH
2
OH
⟶
Acetaldoxime
CH
3
−CH=NOH+H
2
O
(iii) Reaction of acetaldehyde with acetaldehyde in the presence of dilute NaOH, this is the kind of Aldol reaction by which obtained 3-hydroxybutanal as product. Further proceed reaction when using heat in the reaction, its gives aldol condensation product which is But-2-enal.
Acetaldehyde
CH
3
CHO
+
Acetaldehyde
CH
3
CHO
⇌
NaOH
dilute
3−Hydroxybutanal(aldol)
CH
3
−
OH
∣
C
H−CH
2
−CHO
Δ
−H
2
O
(Aldol condensation product)
But−2−enal
CH
3
−CH=CH−CHO
(b) Chemical tests to distinguish the following compounds:
(i) Propanal and propanone : These two are distinguished by the iodoform test when it reacts with I
2
in the presence of NaOH while propanone give iodoform test when reacts with I
2
in the presence of NaOH.
CH
3
COCH
3
+3NaOI→CHI
3
+CH
3
COONa+2NaOH Yellow ppt
CH
3
CH
2
CHO+NaOl→ No ppt of CHI
3
formed
Aldol condensation takes place!
(CH3)3C-CHO does not have alpha hydrogen but CH3COCH3 has alpha H
the product formed is
5,5- dimethyl hex-3-en-2-one
