What would distinguish the mass spectrum of 2,2-dimethylpropane from the mass sp

Question

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

Check all that apply.

Check all that apply.

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 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 less stable 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 tertiary carbocation, whereas loss of a methyl radical from pentane forms a less stable primary 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 tertiary carbocation, whereas loss of a methyl radical from 2-methylbutane forms a more stable secondary 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=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. 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

Explanation / Answer

TRUE: 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. (Reason: 2,2-dimethylpropane cannot form ethyl radical and penatne has less stable primary carbocation)

TRUE: The peak at m/z=57 will be more intense for 2,2-dimethylpropane than for 2-methylbutane or pentane. (Reason: 2,2-dimethylpentane has tertiary carbocations which is most stable than secondary and primary carbocations)

TRUE: 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.

TRUE: 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.

All the above statements are TRUE.

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