Identify the compound that gives the mass spectrum and infrared spectrum shown here.

High resolution mass spectrometry calculates exact mass of the fragments up to 5 ppm level.

It is given that M⁺=84

Consider the mass of compound is 84.

Thus, the given mass is even which means that the given compound contains zero or even number of nitrogen atoms, according to nitrogen rule.

According to rule of 13, divide 84 by 13.

$\frac{84}{13}=13684-786$

The remainder is 6. The basic C-H formula is:${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{n}+\mathrm{r}}$

Here, n=6 and r=6

(n= quotient, r=remainder)

${\mathrm{C}}_6{\mathrm{H}}_{(6+6)}={\mathrm{C}}_6{\mathrm{H}}_{12}$

It contains zero nitrogen atoms because mass number is even. So, according to nitrogen rule, zero or even nitrogen added to the compound .

Degree of unsaturation is also known as double bond equivalent (DBE). If molecular formula is given, plug in the number of C, H, N, and halogen into the formula, to get DBE

$\text{DBE=}\frac{\text{2C+2+N-X-H}}{2}$

Where, C is the number of Carbon.

N is the number of nitrogen.

H is the number of hydrogen.

X is the number of halogen.

For data-custom-editor="chemistry" ${\mathrm{C}}_6{\mathrm{H}}_{12}(\mathrm{m}/2=84)$

$\text{DBE}=\frac{2\left(6\right)+2+0-\text{X}-12}{2}$

There is only one unsaturation or ring present in the compound.

The absorption band at vicinity of 3000 indicate that the compound has hydrogen attached to both sp² and sp³ carbons. The absence of absorption bands at 1600-1800 and absence of broad absorptions between 2500 and 3650 rules the possibility of having C=O, N-H, and O-H groups. Even the lack of absorption at 1600 and 1500 indicate that compound does not have a benzene ring. So, the sp²hydrogens must be due to an alkene.

The lack of absorption at 1600 indicates that the compound must be an alkene with a relatively small dipole moment change when vibration starts.

The IR absorption band at 965 confirms that the compound is trans-alkene.

The molecular ion with m/z = 84 identifies that the compound has molecular formula${\text{C}}_6{\text{H}}_{12}$with one degree of unsaturation in it. The Mass spectra shows base peak at m/z=55 which indicates that molecular ion most easily loses an ethyl radical (84 – 29 = 55).

Because the molecular ion loses an ethyl radical easily, the ethyl radical group must be attached to allylic carbon. The IR absorption band at 965 confirms that the compound is trans-alkene. Therefore, the compound is trans-2-hexene.