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A particular compound’s mass spectrum is indicated as a bar graph comprising m/z values on the x axis and the intensity or relative abundance of the m/z values on the y axis.The tallest peak in a mass spectra is termed the base peak.

Wavelengths in the IR region are mentioned in micrometers and the frequencies are indicated in hertz.The useful IR region ranges from 4000 to $\mathbf{400}\mathbf{}{\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$ .The full interpretation of IR is not possible as all organic molecules have several stretching and bending motions.

The given compound shows 70 and 55 in the mass spectrum and the IR spectrum shows$1660{\mathrm{cm}}^{-1}$ and $3000{\mathrm{cm}}^{-1}$.

The given compound shows molecular ion peak at${\mathrm{M}}^{+}=70$ . Hence, the molecular formula of the compound is ${\mathrm{C}}_5{\mathrm{H}}_{10}$.

The formula for identifying the degree of unsaturation is:

$$\begin{gathered} \mathrm{Degree}\mathrm{of}\mathrm{unsaturation}=\frac{2\mathrm{C}+\mathrm{N}-\mathrm{H}-\mathrm{X}}{2} \\ =\frac{2×5+2+0-10-0}{2} \\ =1 \end{gathered}$$

The molecule has one degree of unsaturation, hence it has one double bond or a ring.

The molecular fragmentation is 55 indicates the cleavage of methyl group. Hence, the molecule has a terminal methyl group.

IR spectrum shows at $1660{\mathrm{cm}}^{-1}$and$3000{\mathrm{cm}}^{-1}$ , which shows the compound has a carbon-carbon double bond. The IR spectrum does not show peaks at$890-900{\mathrm{cm}}^{-1}$ which shows that the compound has a terminal alkene.

The possible structure can be given as:

The correct structure is: