Q 13 P - Organic Chemistry (Mcmurry)

Question

Question: Show the structures of the fragments you would expect in the mass spectra of the following molecules:

Step-by-Step Solution

Answer

Structure of fragments of molecule:

a)

b)

Step 1: Fragmentation

Fragmentation occurs when high-energy cation radical flies apart by spontaneous cleavage of the bond. One fragment is carbocation while the other fragment is a neutral radical. 

The positive charge often remains with a more stable carbocation that forms. Carbocations are electron-deficient species.

Step 2: Alpha-cleavage

Alpha-cleavage ($\mathrm{\alpha }$-cleavage) is the breaking of a carbon-carbon bond adjacent to the carbon bearing a specified functional group. Here, cleavage forms one carbocation and one radical.

Cleavage can occur on either side of the molecule but practically occurs on the side which gives more stable carbocation. So, this gives two products if asymmetric molecule.

Step 3: Mc-Lafferty rearrangement

In mass spectrometry, this process involves the β-cleavage of carbonyl-containing radical cation to produce an enol radical cation and an alkene.

For Mc-Lafferty rearrangement $\mathrm{\gamma }$ hydrogen is needed. 

Step 4: Methods to form fragments from the parent ion

• Mc-Lafferty rearrangement

• Alpha-cleavage ($\mathrm{\alpha }$-cleavage)

Step 5: Mc-Lafferty rearrangement of the given molecule (a)

Here it can be seen two fragments are formed on β-cleavage one is alkene and other is enol radical cation.

Step 6: Alpha-cleavage of given molecule (a)

As given molecule is asymmetric and has carbon on both side of carbonyl group so here cleavage on both side can occur resulting in forming two carbocation and two radical molecules. 

Step 7: Mc-Lafferty rearrangement of the given molecule (b)

As here no $\mathrm{\gamma }$ hydrogen is present so this reaction of the Mc-Lafferty rearrangement is not possible.

Step 8: Alpha-cleavage of given molecule (b)

As given molecule is asymmetric and has carbon on both side of carbonyl group so here cleavage on both side can occur. One cleavage will cleave methyl radical and the other will result in cleavage of the cyclic ring giving carbocation radical molecule.