Question:The mass spectrum (a) and the infrared spectrum (b) of an unknown hydrocarbon 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$\left(7.8\times {10}^{-7}\right)m$ to${10}^{-4}\mathrm{m}$ . But the wavelength in the range from$\left(2.5\times {10}^{-6}\right)$ to$\left(2.5\times {10}^{-5}\right)m$ is used by organic chemists.

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

The given compound has a molecular ion peak at ${\mathrm{M}}^{+}=67$but the hydrocarbon has even number of molecular mass. Therefore the molecular ion peak is$\mathrm{m}/\mathrm{z}=68$. The molecular formula of the compound is${\mathrm{C}}_5{\mathrm{H}}_8$.

The formula for identifying the degree of unsaturation is:

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

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

The possible structures can be given as:

The correct structure can be given as: