$C_{4} H_{8}$ can have total of ----- isomers (geometrical not included)

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Solution

The correct option is C
5

1 ) $C H_{2} = C H - C H_{2} - C H_{3}$ 1 - butene

2) $C H_{3} - C H = C H - C H_{3}$ 2 - butene

3) $C H_{3} - C | C H_{3} = C H_{2}$ Iso butene,

4) methyl cyclopropane

5) cyclobutane

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