Question

How many isomers are possible in${\mathrm{C}}_4{\mathrm{H}}_8$?

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is B

$5$

The explanation for correct option:

Option (B):$5$

• The number of isomers possible for the compound having a formula ${\mathrm{C}}_4{\mathrm{H}}_8$ can be found out by using the DBE (Double bond equivalent) formula which is given by:
• $DBE=C+1-{\displaystyle \frac{H}{2}}-{\displaystyle \frac{X}{2}}+{\displaystyle \frac{N}{2}}$
  Where,
  $C$ = Number of carbon atoms present in organic molecules.
  $H$ = Number of hydrogen atoms present in organic molecules.
  $X$ = Number of halogen atoms present in organic molecules.
  $N$ = Number of nitrogen atoms present in organic molecules.
• Similarily DBE of $C_4{H}_8$ can be calculated as:
  $$\begin{gathered} \Rightarrow \mathrm{DBE}=4+1-\frac{8}{2}-0+0 \\ \Rightarrow \mathrm{DBE}=5-4 \\ \Rightarrow \mathrm{DBE}=1 \end{gathered}$$
• The DBE value of 1, suggests there must be present at least one one ring structure or double bond in the given structural formula.
• Hence, for $C_4{H}_8$there are 5 constitutional isomers possible.
• The five constitutional isomers included are given as: But-1-ene, But-2-ene, Cyclobutane, 2-Methylpropane, Methylcyclopropane.
• The possible five structural isomers of all the isomers of $C_4{H}_8$ can be drawn as:

Therefore, option (B) 5 is the correct answer.