How many geometrical isomers are possible for the given compound?
$C H_{3} - C H = C H - C H = C H - C_{2} H_{5}$

Four

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Three

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Two

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Five

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A Four

Since the number of double bonds in the given compound is 2 and the ends are different $(i . e . C H_{3}$ and $C_{2} H_{5})$ , therefore the total number of geometrical isomers are $2^{2} = 4$

Hence, the correct option is $A$

Suggest Corrections

Similar questions

C H_{3} - C H = C H - C H = C H - C H_{3}

C H_{3} - C H = C H - C H = C H - C_{2} H_{5}

Join BYJU'S Learning Program