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Tardigrade
Question
Chemistry
(C H3)2C=CHCOCH3 can be oxidized to (C H3)2C=CHCOOH by
Q.
(
C
H
3
)
2
C
=
C
H
COC
H
3
can be oxidized to
(
C
H
3
)
2
C
=
C
H
COO
H
by
2434
246
NTA Abhyas
NTA Abhyas 2020
Aldehydes Ketones and Carboxylic Acids
Report Error
A
chromic acid
7%
B
NaOI, followed by acidification
35%
C
Cu at 573 K
8%
D
K
M
n
O
4
+
H
2
S
O
4
49%
Solution:
(
C
H
3
)
2
C
=
C
H
COC
H
3
is a methyl ketone and therefore, it can be easily oxidized to
(
C
H
3
)
2
CC
H
COO
H
by haloform or iodoform reaction (NaOI).
(
C
H
3
)
2
C
=
C
H
COC
H
3
⟶
N
a
O
I
(
C
H
3
)
2
C
=
C
H
COC
I
3
⟶
N
a
O
H
C
H
I
3
+
(
C
H
3
)
2
C
=
C
H
COON
a
→
(
C
H
3
)
2
C
=
C
H
COO
H
Chromic acid and
K
M
n
O
4
, being strong oxidizing agents, will also attack the multiple bond and cause the cleavage giving the mixture of carboxylic acids.