(a) Spectra A and B belong to isomers of C₆H₁₂O. Explain how you can tell which isomer goes with each spectrum.

(b) The intensity of theM + 1peak at m/z,101mustbe incorrect in both spectra. It is entirely missing in spectrum A and too intense (15.6% of intensity of M^+* ) in spectrum B. What should be the intensity of M + 1 relative to M^+* for the compositionC₆H₁₂O?

Oxepane is a heterocyclic chemical molecule with the formula C₆H₁₂O, which is a cycloheptane with one of its methylene groups replaced by oxygen.

(a)

It is necessary to explain the supplied spectrum.

It can deduce the following from the given spectrum:

The arrival of a strong peak at is the first difference between the two spectra. m/z = 72 It's possible that something in A isn't present in B.

The highest point is at m/z = 72 indicates that the molecular ion has lost a neutral molecule with an even mass of 28 Da.

3-Methyl-2-pentanone yields ethylene, but 3,3-Dimethyl-2-butanone yields no ethylene due to a lack of oxygen $\gamma$-CH group It's known as the McLafferty rearrangement, and it's represented like this:

3-methyl-2-pentanone should be used for spectrum A, while 3,3-dimethyl-2-butanone should be used for spectrum B.

(b)

What should the intensity of the given chemical be M + 1 in relation to M⁺ It must be determined.

The chemical in question is C₆H₁₂O .

The intensity with which M + 1 in relation to M⁺ The following calculation can be used to find this. Intensity

$=\underset{{}^{13}\mathrm{C}}{\underbrace{6\times 1.08\%}}+\underset{\mathrm{H}}{\underbrace{12\times 0.012\%}}+\underset{{}^{12}\mathrm{O}}{\underbrace{1\times 0.038\%}}=6.7\%{\mathrm{of}}^{+}{\mathrm{M}}^{+}$