1559T_ch11_199-21911/3/0516:02Pa9e206 EQA 206.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY Because the highfield signals are triplets,these can only be CH groups:three of them Combining CHa=C with three CHa's can only give CH2= (f)H.=14+2 16:degrees of unsaturation =(16-10)/2 =3,or 1 T bond and 2 rings Again,be careful.No ow ther are four signals,bu 1352).Be an this signal must represent two equivalent alkene CH groups: -CH-CH-.So you have at least wo CH 's.an alk e CH.and CH-CH-.for a total to groups aready identificd n order to keep the NMR spectrum as simples it is.In her words ⊕ -CH2- -CH2- -CH- c-- -CH=CH- Each isa reasonable possibility (the secondone..is actually correct) 32.Hat =10+2=12:degrees of unsaturation=(12-10)/2 =1. (a)The ony way for five carbons to be equivalent is to make a ring:is the answer. (b)Three CH,'s,and a-CH=C: CH;-CH=( (c)Two CH3's.one CH2.and-CH=CH-:CH3-CH2-CH=CH-CH3 is the answer (stereochemistry is ambiguous). Because the highfield signals are triplets, these can only be CH2 groups: three of them. Combining with three CH2’s can only give (f) Hsat
14 2
16; degrees of unsaturation
(16 10)/2
3, or 1  bond and 2 rings. Again, be careful. Now there are four signals, but seven carbons in the molecule. Upfield, there are two different kinds of CH2’s (
25.2 and 48.5) and one kind of CH (
41.9). There is one kind of alkene carbon (
135.2). Because a double bond must connect two alkene carbons, this signal must represent two equivalent alkene CH groups: OCHPCHO. So you have at least two CH2’s, an alkane CH, and OCHPCHO, for a total of C5H7. So two carbons and three H’s are still required: One more CH2 and one more CH would do, and these each must be equivalent to groups already identified in order to keep the NMR spectrum as simple as it is. In other words, A here are the pieces you have for the molecule: 2 equivalent OCH2O’s, 2 equivalent OCH’s, a A single unique OCH2O, and the OCHPCHO group, for a total of C7H10. How do you put this all together? Remembering that symmetry can make groups equivalent, you can write these groups in symmetrical arrangements and connect them in a trial-and-error manner. Each is a reasonable possibility (the second one, norbornene, is actually correct). 32. Hsat
10 2
12; degrees of unsaturation
(12 10)/2
1. (a) The only way for five carbons to be equivalent is to make a ring: is the answer. (b) Three CH3’s, and a (c) Two CH3’s, one CH2, and OCHPCHO: CH3OCH2OCHPCHOCH3 is the answer (stereochemistry is ambiguous). CH3 CH3 CH3 CH C CH C : CH2 CH2 CH2 CH CH HC CH or or CH2 CH2 The 
31.1 signal accounts for the C two equivalent CH2 groups (circled). CH2 CH2 CH2 C 206 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 1559T_ch11_199-219 11/3/05 16:02 Page 206