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Tardigrade
Question
Chemistry
Hot and conc. KMnO4 at 373-383 K reacts with pent-2-ene to form
Q. Hot and conc.
K
M
n
O
4
at 373-383 K reacts with pent-2-ene to form
1434
210
NTA Abhyas
NTA Abhyas 2020
Hydrocarbons
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A
propanoic acid only
12%
B
ethanoic acid only
18%
C
a mixture of propanoic acid and ethanoic acid
61%
D
a mixture of butanoic acid and formic acid
8%
Solution:
Hot and conc.
K
M
n
O
4
at 373-383 K reacts with pent-2-ene to form a mixture of propanoic acid and ethanoic acid.
C
H
3
C
H
2
C
H
=
C
H
C
H
3
→
K
M
n
O
4
P
ro
p
an
o
i
c
a
c
i
d
C
H
3
C
H
2
COO
H
+
Et
han
o
i
c
a
c
i
d
H
OOCC
H
3