wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

How many geometrical isomers are possible in the following two alkenes?
(i) CH3CH=CHCH=CHCH3
(ii)CH3CH=CHCH=CHCl

A
4 and 4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4 and 3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3 and 3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3 and 4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 3 and 4
1. Compound (i) is a symmetrical compound. It has same methyl terminal groups.
The formula used to determinne the number of geometrical isomers in compound having same terminal groups with even number of double is, 2(n1)+2[(n/2)1]
where, n is the number of double bond present in the molecule.
In the given molecule we have two double bond so,
No. of geometrical isomers = 2(n1)+2[(n/2)1]
= 2(21)+2[(2/2)1]
=21+20
=2+1
=3

2. Compound (ii) is a unsymmetrical compound. It has different terminal groups (methyl and chlorine).
The formula used to determinne the number of geometrical isomers in compound having different terminal groups is, 2n, where n is the number of dounle bond present in the molecule.
In the given compound we have 2 double bond so n = 2
Therefore,
No. of geometrical isomers = 2n
= 22
= 4.
Therefore, the correct option is (d).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Geometrical Isomerism
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon