CH908, Problem set 4
1. Spectra A and B are from α-Ionone and β-ionone:
Which is which? Assign as many peaks as possible, draw mechanisms for their formation.
A.
Answer: α-Ionone.
Notes: there’s a •CH3 loss, and a solid alkane/alkene series. The peak at 136 is an odd electron fragment, indicating a rearrangement product. It’s the retro Diels-alder rearrangement, with a series of alkane losses (peaks 129, 109, 93) from there. 77 and 91 are the usual benzyl/toluyl peaks, which must involve multiple methyl transfers.
B.
Answer: β-Ionone.
Notes: retro diels alder is suppressed due to presence of extensive resonance stabilization. •CH3 loss is amplified, probably from loss of one of the methyls at the 2-position of the ring – this would put a radical on a tertiary carbon that is resonantly stabilized with the pi-system – a very stable site.
2. Interpret the following spectrum. Assign as many peaks as possible, and draw probable mechanisms.
Notes: alkane series suggests a 4-carbon alkyl group – it goes up to 56 (butyl) and no higher – probably linear n-butyl because no particularly intense peak and lack of a -15 Da loss from the precursor. 77 indicates a benzyl group. Complete lack of 91 indicates that benzyl group has no alkyl derivatives. 105 suggests a carbonyl attached to the benzyl (+28 Da). 123 is 55 below 178, suggesting loss of the whole butyl group (as •C4H7) – this is a pseudo complementary ion to 56. 123-105 = 18, suggesting a bridging ether.
Answer: n-butyl benzoate
3. Interpret the following spectrum. Assign as many peaks as possible, and draw probable mechanisms.
Notes: very strong alkane series, up to at least hexane or octane, probably an n-alkane because of lack of particular abundant peaks. Exception is 112, which is anomalously abundant. 35, 47, 61, 75, 89 indicate a sulfide attached to an alkane. 146 and 89 peak clearly have one sulfur (from A+2 isotope abundance), but 112 does not.
Answer: n-octane thiol