How many geometrical isomers are possible for the given compound? CH3 — CH = CH — CH = CH — C2H5

Question

How many geometrical isomers are possible for the given compound? CH3 — CH = CH — CH = CH — C2H5 (A) Four (B) Three (C) Two (D) Five

Tardigrade Admin · Accepted Answer

Since the number of double bonds in the given compound is 2 and the ends are different (i.e. CH 3 and C 2 H 5 ), therefore the total number of geometrical isomers are 22=4 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/47ad345fed143de41351024c1fa70a77-.png" />

Q. 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

Three

Two

Five

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 $2^{2} = 4$

Q. How many geometrical isomers are possible for the given compound? CH3​—CH=CH—CH=CH—C2​H5​

Four

Three

Two

Five

Since the number of double bonds in the given compound is 2 and the ends are different (i.e. CH3​ and C2​H5​ ), therefore the total number of geometrical isomers are 22=4

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

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 $2^{2} = 4$