A. How could their mass spectra be used to distinguish between the following com

Question

A. How could their mass spectra be used to distinguish between the following compounds?

3-methyl-2-butanone, 3-pentanone, 2-pentanone

2-Pentanone will have a base peak at m/z=57, whereas the other two ketones will have base peaks at m/z=43.

B. What would distinguish the mass spectrum of 2,2-dimethylpropane from the mass spectra of pentane and 2-methylbutane?

3-Methyl-2-butanone will have a peak at m/z=58 due to a McLafferty rearrangement. 3-Pentanone will have a peak at m/z=58 due to a McLafferty rearrangement. 3-Methyl-2-butanone will have a base peak at m/z=57, whereas the other two ketones will have base peaks at m/z=43. 2-Pentanone will have a peak at m/z=58 due to a McLafferty rearrangement. 3-Methyl-2-butanone does not have any ?-hydrogens. Therefore, it cannot undergo a McLafferty rearrangement, so it will not have a peak at m/z=58. 3-Pentanone will have a base peak at m/z=57, whereas the other two ketones will have base peaks at m/z=43. 3-Pentanone has  ?-hydrogens. Therefore, it cannot undergo a McLafferty rearrangement, so it will not have a peak at m/z=58.

2-Pentanone will have a base peak at m/z=57, whereas the other two ketones will have base peaks at m/z=43.

B. What would distinguish the mass spectrum of 2,2-dimethylpropane from the mass spectra of pentane and 2-methylbutane?

The peak at m/z=57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms a tertiary carbocation, whereas loss of a methyl radical from 2-methylbutane forms a less stable secondary carbocation. The peak at m/z=57 will be more intense for 2,2-dimethylpropane than for 2-methylbutane or pentane. The peak at m/z=57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms a tertiary carbocation, whereas loss of a methyl radical from 2-methylbutane forms a more stable secondary carbocation. The peak at m/z=43 will be most intense for 2-methylbutane because such a peak is due to loss of an ethyl radical, which forms a secondary carbocation. The peak at m/z=57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms a secondary carbocation, whereas loss of a methyl radical from pentane forms a more stable primary carbocation. The peak at m/z=57 will be more intense for pentane than for 2-methylbutane or 2,2-dimethylpropane. The peak at m/z=57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms a tertiary carbocation, whereas loss of a methyl radical from pentane forms a less stable primary carbocation. The peak at m/z=43 will be most intense for 2,2-dimethylpropane because such a peak is due to loss of an ethyl radical, which forms a secondary carbocation.

Explanation / Answer

a) The molecules can be distinguished with the help of their base peaks.

3methyl-2butanone and 3-pentanone donot have gamma-hydrogens. So, they will not give Mc lafferty rearrangement.

But 2-pentanone gives a peak at m/z = 58 for Mc Lafferty rearrangement.

Hence, 2-pentanone has the base peak at m/z = 58

3-pentanone has a base peak at m/z = 57 for CH3CH2CO+ fragment.

3-methyl-2-butanone will have a base peak at m/z = 43 for isopropyl cation.

(b) The molecules can be distinguished with the help of their base peaks.

2,2-dimethylpropane has the base peak at m/z = 57 due to tert-butyl cation fragment, (CH3)3C+but pentane and 2-methylbutane has the base peak at m/z = 43 due to propyl cation and iso-propyl cation fragments respectively.

Note:

The peak at m/z=57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms a tertiary carbocation, whereas loss of a methyl radical from 2-methylbutane forms a less stable secondary carbocation.

The peak at m/z=57 will be more intense for 2,2-dimethylpropane than for 2-methylbutane or pentane.

The peak at m/z=43 will be most intense for 2-methylbutane because such a peak is due to loss of an ethyl radical, which forms a secondary carbocation.

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