Textbook: Vollhardt 6th Ed. (2010)

Chapter 11: Alkenes and Infrared Spectroscopy

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Individual Problems

Individual Problems

Problem # 657

The mass spec of chlorocyclohexane shows a peak at m/z = 83.

Use curved arrows to show the heterolytic cleavage that accounts for this fragment.

Solution

Heteroatoms (atoms that are not C or H) are always the most likely atoms to lose an electron during mass spec, and this will be no exception. Mass spec will blow off an electron from the chlorine, leaving behind the molecular ion (M⁺). But the molecular ion will then fragment.

The problem told us it was a heterolytic cleavage, in which both electrons form the bond go towards the positive charge. Because two electrons are involved, we use curved arrows as usual.

This results in a neutral chlorine radical and a carbocation with formula C₆H₁₁, for a total mass of 83 (6 x 12 + 11 x 1). Note that the chlorine radical doesn't give an MS peak because it is neutral.

Problem Details

MendelSet practice problem # 657 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Mass Spectrometry (MS)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): mass spec, heterolytic cleavage

Problem # 658

The mass spec of methyl ethyl ether shows peaks at m/z = 45 and 59.

Use hooks to show the alpha cleavages that result in these two fragments.

Solution

MS ionization will knock off an electron from the heteroatom (atom that's not C or H), in this case, the oxygen, leaving behind a positively charge compound. This is the molecular ion (M⁺).

Oxygen usually undergoes homolytic cleavage- the bond splits and each atom gets one electron. Since only one electorn is involed, we use hooks instead of the usual curved arrows.

If your professor is into mass spec cleavage mechanisms, I suggest you practice this mechanism, called an alpha cleavage (see image below).

Problem Details

MendelSet practice problem # 658 submitted by Matt on July 17, 2011.

Subject: Organic Chemistry

Subject Topic(s): Mass Spectrometry (MS)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): mass spec, homolytic cleavage, alpha cleavage

Problem # 659

The mass spec of 4-nonanone shows peaks at m/z = 58, 71, 86, 99.

Using curved arrows or hooks, show how each of these fragments can form via alpha cleavage or the McLafferty rearrangement. (and draw the structure of the indicated species in the appropriate box).

Solution

Note that I drew the McLafferty rearrangements using arrows (2 electrons moving at once). Some textbooks use hooks instead, but the results are the same.

In most ungraduate organic chemistry courses, being able to draw an alpha cleavage is much more important than a being able to draw a McLafferty rearrangement (which tends to only show up on bonus problems).