Our last example addresses a situation very common in chemistry—working out the structure of a product of a reaction. The situation is this: you have treated propenal (acrolein) with HBr in ethane-1,2-diol (or glycol) as solvent for 1 hour at room temperature. Distillation of the reaction mixture gives a colourless liquid, compound X. What is it?

The mass spectrum shows a molecular ion (181) much heavier than that of the starting material, C3H4O = 56. Indeed, it shows two molecular ions at 181 and 179, typical of a bromo compound, so it looks as if HBr has added to the aldehyde somehow. High resolution mass spectrometry reveals a formula of C5H9BrO2, and the five carbon atoms make it look as though the glycol has added in too. If we add everything together, we find that the unknown com pound is the result of the three reagents added together less one molecule of water.

Now, how many DBEs have we got? With a formula like this the safest bet is to draw something that has the right formula—it need not be what you expect the product to be. Here’s something in the margin—we just added atoms till we got there, and to do so we had to put in one double bond. C5H9BrO2 has one DBE. The next thing is to see what remains of the hydrocarbon skeleton of propenal by NMR. The 13C NMR spectrum of CH2 CH–CHO clearly shows one carbonyl group and two carbons on a double bond. These have all disappeared in the product and for the five carbon atoms we are left with four signals, two saturated, one next to oxygen, and one at 102.6 ppm, just creeping into the double bond region.

The IR spectrum gives us another puzzle—there appear to be no functional groups at all! No OH, no carbonyl, no alkene—what else can we have? The answer is an ether, or rather two ethers as there are two oxygen atoms. Now that we suspect an ether, we can look for the C–O single bond stretch in the IR spectrum and find it at 1128 cm−1.

Each ether oxygen must have a carbon atom on each side of it, but we seem to have only one carbon in the saturated C next to O region (50–100 ppm) in the 13C NMR. Of course, as you’ve already seen, these limits are arbitrary, and in fact the peak at 102 ppm is also a saturated C next to O. It is unlikely to be an alkene anyway as it takes two carbons to make an alkene. What would deshield a saturated C as much as this? The answer is two oxygen atoms. We can explain the 13C spectrum if we assume a symmetrical fragment C–O–C–O–C accounts for three of the five carbon atoms. So, where is our double bond equivalent? We know we haven’t got a double bond (no alkene and no C O) so the DBE must be a ring. You might feel uncomfortable with rings, but you must get used to them. Five-, six-, and seven-membered rings are very common. In fact, most known organic compounds have rings in them. We could draw many cyclic structures for the formula we have here, such as this one in the margin. But this won’t do as it would have five different carbon atoms. It is much more likely that the basic skeletons of the organic reagents are preserved, that is, that we have a two-carbon (from the ethylene glycol) and a three-carbon (from the propenal) fragment joined through oxygen atoms. This gives four possibilities, all containing the C–O–C–O–C fragment we deduced earlier (highlighted in black).

These are all quite reasonable, although we might prefer the third as it is easier to see how it derives from the reagents. The product is in fact this third possibility, and to be sure we would have to turn to the fi ne details of 1H NMR spectroscopy, which we return to in Chapter 13.