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228 Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY CH₃ CH₃ CH₂-OH H H CH₃ HO-CH, CH₃ H CH₃ You do not have the information to tell which of the three is the actual compound. (e) Notice that there are only four signals. but there are five carbons. Be careful. = 10 + 2 = 12; degree of unsaturation = (12 8)/2 = 2 now. δ = 15.8 and 31.1 are CH₂ groups; δ = 103.9 is an alkene whereas = 149.2 is an alkene C lacking hydrogens. What do you have so far? The molecule has the piece leaving three C's and six H's to make up the formula. which must still contain one more element of unsaturation (a ring?). Because the highfield signals are triplets. these can only be CH₂ groups: three of them. Combining with three CH₂'s can only give The 8 = 31.1 signal accounts for the CH₂=C CH₂ two equivalent CH₂ groups (circled). CH₂ (f) = 14 + 2 = 16; degree of unsaturation = (16 10)/2 = 3, or 1 bond and 2 rings. Again. be careful. Now there are four but seven carbons in the molecule. Upfield. there are two different kinds of CH₂'s (δ = 25.2 and 48.5) and one kind of CH (δ = 41.9). There is one kind of alkene carbon (δ = 135.2). Because a double bond must connect two alkene carbons, this signal must represent two equivalent alkene CH groups: So you have at least two CH₂'s, an alkane CH. and 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, here are the pieces you have for the molecule: 2 equivalent 2 equivalent -CH's, a single unique and the -CH=CH- group. for a total of 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. or or HC- -CH -CH=CH- Each is a reasonable possibility (the second one, is actually correct).