The mass spectrum (a) and the infrared spectrum (b) of an unknownhydrocarbon are shown. Propose as many structures as you can.

Mass spectrometry constitutes a technique used for computing the molecular weight of a particular molecule. The instrument used in this technique is the mass spectrometer. The different parts of a mass spectrometer include an ionization source, mass analyzer and a detector.

The functional groups in a particular compound can be understood using infrared spectroscopy.The infrared region varies from just above visible$\left(7.8\times {10}^{-7}\right)\mathrm{m}$ to ${\text{10}}^{\text{- 4}}\text{m}$. But the wavelength in the range from$\left(2.5\times {10}^{-6}\right)\mathrm{m}$ to$\left(2.5\times {10}^{-5}\right)\mathrm{m}$ is used by organic chemists.

The given compound shows 67, 53,40, 39 and 27 in the mass spectrum and the IR spectrum shows ${\text{2130cm}}^{\text{- 1}}$and ${\text{3320cm}}^{\text{- 1}}$.

The given compound has a molecular ion peak at $M^{+}=67$but the hydrocarbon has even number of molecular mass. Therefore the molecular ion peak is $m/z=68$. The molecular formula of the compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{8}}$.

The formula for identifying the degree of unsaturation is:

$$\begin{gathered} \mathrm{Degree}\mathrm{of}\mathrm{unsaturation}=\frac{2C+2+N-H-X}{2} \\ =\frac{2\times 5+2+0-8-0}{2} \\ =2 \end{gathered}$$

The molecule has two degrees of unsaturation, hence it has two double bond or a triple bond.

The molecular fragmentation is 67, 53, 40, 39 and 27. The molecular fragmentation of 67 indicates the loss of a hydrogen atom. The fragmentation at 53 indicates the cleavage of methyl group. Hence, the molecule has a terminal methyl group.

The IR spectrum at${\text{2130cm}}^{\text{- 1}}$ shows the compound has a carbon-carbon triple bond and the IR spectrum at${\text{3320cm}}^{\text{- 1}}$ indicates that the compound has a carbon-hydrogen stretching frequency of a terminal alkyne.

The possible structures can be given as:

Possible structures of the compound

The correct structure can be given as:

Correct structure of the compound