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Tardigrade
Question
Chemistry
Sum of X , Y and Z is :- (a) C 4 H 8 O No. of aldehydes = X (b) C 5 H 12 No. of chain isomers = Y (c) C 4 H 8 Maximum no. of alkenes = Z
Q. Sum of
X
,
Y
and
Z
is :-
(a)
C
4
H
8
O
No. of aldehydes
=
X
(b)
C
5
H
12
No. of chain isomers
=
Y
(c)
C
4
H
8
Maximum no. of alkenes
=
Z
1729
208
Organic Chemistry – Some Basic Principles and Techniques
Report Error
A
10
B
7
C
9
D
8
Solution:
(a)
C
4
H
8
O
:
H
7
C
3
−
C
H
O
⇒
2
aldehyde (monovalent radical of
C
3
H
8
=
(
X
)
2
)
(b)
C
5
H
1
: No. of chain isomers
=
(
Y
)
3
C
H
3
C
H
2
C
H
2
C
H
2
C
H
3
;
C
H
3
C
H
2
C
∣
H
3
C
H
C
H
3
C
H
3
−
C
∣
H
3
C
∣
C
H
3
−
C
H
3
(c)
C
4
H
8
:- Maximum no. of alkenes
⇒
4
(
Z
)
(i)
C
H
3
−
C
H
2
−
C
H
=
C
H
2
Hence
X
+
Y
+
Z
⇒
2
+
3
+
4
=
9
