Tardigrade
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Tardigrade
Question
Chemistry
CH3COOH ->[NH3] A ->[Δ] B ->[P2O5] C. C will be
Q.
C
H
3
COO
H
NH
X
3
A
Δ
B
P
X
2
O
X
5
C
.
C
will be
7269
231
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A
C
H
3
N
H
2
B
C
H
3
C
H
2
N
H
2
C
C
H
3
CN
D
C
H
3
C
H
2
CN
Solution:
C
H
3
COO
H
NH
X
3
C
H
3
CO
O
−
N
H
4
+
Heat
C
H
3
CON
H
2
P
X
2
O
X
5
Methyl cyanide
C
H
3
CN