Question

How many structural isomer are possible with the molecular $C_8{H}_{10}$?

Answer 1

Double bond equivalent $=C+1-\frac{H}{2}$

in formula $C_8{H}_{10},C=8,H=10$

$D.B.E=8+1-\frac{10}{2}=4$

there is either $4$ double bond or $1$ ring- $3$ double bond possible or $2$ double bond, $2$ ring or $2$ triple bond etc

$C_8{H}_{10}$ See Image $1$

This is the compound having $4$ double bond equivalent

Thus possible structure $C_8{H}_{10}$is

See image 2

Thus $5$ structural isomers are possible.