Question

Imagine that you are given the mass spectra of these two compounds, but the spectra are missing the compound names. Which peaks occur in one isomer but not in the other isomer? That is, which peaks could be used to distinguish one isomer from the other?

Imagine that you are given the mass spectra of these two compounds, but the spectra are missing the compound names. Which peaks occur in one isomer but not in the other isomer? That is, which peaks could be used to distinguish one isomer from the other?

Answer 1

Concepts and reason

This problem is based on interpreting the mass spectra of organic compounds.

Mass spectroscopy is used to measure the mass of different charged fragments of a molecular ion of a particular compound which can be useful in order to understand the structure and chemical composition of the compound.

Fundamentals

When the positive molecular ion breaks, a positively charged fragment and a free radical is produced. The fragment ions can further break to form smaller ions. The positive ions detected with different mass to charge ratios is represented in a mass spectrum, as peaks, with their relative abundance. This method can be used to distinguish between two compounds.

The structure of first and second isomers are as follows:

These are isomers of a saturated alkane. There are eight carbon atoms. So, the molecular formula for saturated alkane is.

The molecular mass for this is calculated by the formula as follows:

$(w)8+(°W)8=W $

Substitute 12 for and 1 for .

So, the molecular ion peak of would be at .

The possible fragmentation for first isomer are as follows:

The molecular mass of is 15. Thus, isomer forms fragment peak for the positive ion at 15.

Similarly, molecular mass of , , , , and are 29, 43, 57, 71, 85 and 99 respectively. Thus, fragment peak for positive ion are at 29, 43, 57, 71, 85 and 99 respectively.

Now, write the possible fragmentation for second isomer as follows:

The molecular mass of , , , and are 15, 43, 57, 71 and 99 respectively. Thus, fragment peak for positive ion in isomer B are at 29, 43, 57, 71, 85 and 99 respectively.

The first isomer, A shows peak at 29 and 85 which are not present in second isomer B.

Thus, peaks at 29 and 85 can be used to distinguish the two isomers.

Ans:

The two isomers are distinguished by peaks at 29 and 85.