Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Hex-2-en-4-yne is
Q. Hex-2-en-4-yne is
2616
223
JIPMER
JIPMER 2012
Organic Chemistry – Some Basic Principles and Techniques
Report Error
A
C
H
3
−
C
H
2
−
C
≡
C
−
C
H
=
C
H
2
10%
B
C
H
3
−
C
≡
C
−
C
H
=
C
H
−
C
H
3
78%
C
C
H
3
−
C
H
2
−
C
H
=
C
H
−
C
≡
C
H
4%
D
C
H
3
−
C
≡
C
−
C
H
2
−
C
H
=
C
H
2
8%
Solution:
Correct answer is (b)
C
H
3
−
C
≡
C
−
C
H
=
C
H
−
C
H
3