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Chapter 11 Suggested solutions for Chapter 11 73 Purpose of the problem show that it is helpful to predict the NMR spectrum of an expected product provided that the is rejected if the spectrum is 'wrong'. Suggested solution The spectrum is all wrong. There are only three aromatic Hs (7.0-7.5 p.p.m.), not the four expected. There are two exchanging hydrogens, presumably and not the one expected. The only thing as expected is the chain of three CH₂ groups. You could not be expected to find the right which is actually an amide with a new carbocyclic ring. CONH₂ This surprising result was reported HON by B. Amit and A. Hassner, Synthesis, H 1978, 932. The expected reaction was the Beckmann rearrangement (Chapter 37) but what actually happened was a Beckmann fragmentation (Chapter 38) followed by an intramolecular Friedel-Crafts Now that you know what the product is, you might like to assign the spectrum and convince alkylation (Chapter 22). that this does indeed fit the NMR spectrum. Problem 7 Assign the 400 MHz NMR spectrum of this enynone as far as possible, justifying both o chemical shifts and coupling patterns. Purpose of the problem at interpretation of high-field NMR spectra. More complex problems can be solved easily. Suggested solution First measure the spectrum and list the data. The expansions make it much easier to see the but even so we are going to have to call the signal at 5.6 p.p.m. a multiplet. For the rest of signals, you should have measured the J values. Coupling is measured in Hz and at 400 MHz chemical shift unit of 1 p.p.m. is 400 Hz, so each subunit of 0.1 p.p.m. is 40 Hz. Out with the or the dividers and get measuring! shift (δ, p.p.m.) Integration Multiplicity Coupling, J, Hz Comments 1H m ? alkene region 1H di with fine splitting 16.3 alkene region 1H with fine splitting 10.4 alkene region 2H t with fine splitting 6.5 next to or C=C? 2H t with fine splitting 6.5 next to or C=C? 2H q with fine splitting 6.5 next to C=O or C=C? 2H t with fine splitting 6.5 next to or C=C? 1H broad S - alkyne? 2H q 6.5 not next to anything That gives us three protons in the alkene region, five CH₂ groups, and one solitary proton, which be only the alkyne proton. If you were surprised by its small chemical shift, check the tables and right. In the alkene region, the multiplet must be H² which couples to the CH₂ at C3 and On C1, has a large trans coupling (16 Hz) to H² while H¹ᵇ has a smaller cis coupling Hz). The coupling between and H¹ᵇ is very small. H 0 7 3 10 1 9 5 8 6 4 2 H