1559T_ch11_199-21911/02/0521:43Pa9e204 ⊕ EQA 204.Chapter 11 ALKENES:INFRARED SPECTROSCOPY AND MASS SPECTROMETRY olecule everything is accounted for,and the two CH H2C=CH-CH2-CH2OH (e)Ha=6+2 unsaturation=(6-4)/2=1bond or ring 1.7 (doublet.3 H):CHa,next to CH =5.7 (quartet,1 H):alkene CH,next to CH3 29.The multi t8=231s rtet for the 0 CH se of the three neighbors,the CH s additi 心8hg in th TPi6mtpiae In other words.lines 5 and6overlap to give one tall line in the center of the pattem.and lines7and overlap to give the oth 30.(a)Yes.1-Butene>trans-2-butene(which should be zero) (b)No (e)Yes.cistrans(which,again,is zero)
As in (c), with only four carbons in the molecule everything is accounted for, and the two CH2 groups must be attached to each other; thus, H2CPCHOCH2OCH2OH (e) Hsat 6 2 2 6; degrees unsaturation (6 4)/2 1 bond or ring. 1.7 (doublet, 3 H): CH3, next to CH 5.7 (quartet, 1 H): alkene CH, next to CH3 All that’s left is one C and two Cl’s, so the pieces, and two Cl’s, combine to give the answer, CH3OCHPCCl2. 29. The multiplet at 2.3 is a quartet for the C2 CH2 group, because of the three neighbors, the CH2 of C1 and the CH of C3. The signal shows additional very fine splitting, indicating that there is a small coupling to the alkene hydrogens on C4 (J 1 Hz). The multiplet for the hydrogen signal at 5.7 can be interpreted in detail. It corresponds to the CH at C3, which is split by the alkane CH2 on one side and the alkene CH2 on the other. The result is a pattern that can be described as a doublet (for trans alkene coupling) of doublets (for cis alkene coupling) of triplets (for coupling to the alkane CH2), 12 lines total, of which 10 are seen. If we construct a plausible splitting diagram we find that the two “missing” lines are actually “hidden” under the tallest lines in the center of the pattern: In other words, lines 5 and 6 overlap to give one tall line in the center of the pattern, and lines 7 and 8 overlap to give the other. 30. (a) Yes. 1-Butene trans-2-butene (which should be zero) (b) No (c) Yes. cis trans (which, again, is zero) 1 2 3 4 5 6 7 8 9 10 11 12 Triplet splitting J 7 Hz Cis doublet splitting J 10 Hz Trans doublet splitting J 16 Hz CH3 CH C 204 • Chapter 11 ALKENES; INFRARED SPECTROSCOPY AND MASS SPECTROMETRY 1559T_ch11_199-219 11/02/05 21:43 Page 204
15597ch11199-21911/02/0521:43Pag0205 EQA Solutions o Problems205 31.Use the carbon types together with the chemical shifts to choose between alterative possibilities. hee must be two of each.ToCH plusCH-H-.which can only combin CH=CH H2C-CH Cyclobutene (1 bond and 1 ring) and therefore.mu CHa-CH=CH-C-H(2 bonds).You do not have enough information aboutC NMR to determine the stereochemistry. (e)H.=8+2=10:degrees of unsaturation =(10-8)/2=1. 1362581390112 CH3-CH2-CH=CH2 two alkene carbons.of which only one(=125.7)has an H on it. The pieces:(2xCH--CH2---CH-C.There is one H unlocated.Because it is not attached to one of the carbons,it must be on the oxygen.So,the possible answers are CH: CH、 CH2-OH C=C H CH2-OH H HO-CH2 C-C45 H You do not have the information to tell which of the three is the actual compound. ()ly(d an alkene CH2,whereas=149.2 is an alkene C lacking hydrogens. What do you have so far?The molecule has the picce CH2=C.leaving three C's and six H's to make up the formula.which must still contain one more element of unsaturation (a ring?)
31. Use the carbon types together with the chemical shifts to choose between alternative possibilities. (a) Hsat 8 2 10; degrees of unsaturation (10 6)/2 2 bonds or rings present. 30.2 is a CH2 group; 136.0 is an alkene CH. Because those alone only add up to C2H3, there must be two of each. Two OCH2O’s, plus OCHPCHO, which can only combine to make (b) Hsat 8 2 10; degrees of unsaturation (10 6)/2 2 again. 18.2 is a CH3, not attached to the oxygen; 134.9 and 153.7 are alkene CH’s; 193.4 is in the CPO region, O B and therefore, must be a OCH group because it is a doublet. The answer is therefore O B CH3OCHPCHOCOH (2 bonds). You do not have enough information about 13C NMR to determine the stereochemistry. (c) Hsat 8 2 10; degrees of unsaturation (10 8)/2 1. (d) Hsat 10 2 12; degrees of unsaturation (12 10)/2 1. This one has two CH3 groups ( 17.6 and 25.4), a CH2 downfield enough ( 58.8) to be attached to the O, and two alkene carbons, of which only one ( 125.7) has an H on it. The pieces: (2 )CH3O, OCH2OOO, There is one H unlocated. Because it is not attached to one of the carbons, it must be on the oxygen. So, the possible answers are 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. Hsat 10 2 12; degrees of unsaturation (12 8)/2 2 now. 15.8 and 31.1 are CH2 groups; 103.9 is an alkene CH2, 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?). CH2 C , CH3 CH3 CH2 C C H OH CH3 CH2 CH3 C C OH H CH2 CH3 CH3 C C HO H CH C . CH3 CH2 CH CH2 Answer directly available from the carbon types. 13.6 25.8 139.0 112.1 H2C CH2 CH CH Cyclobutene (1 bond and 1 ring) Solutions to Problems • 205 1559T_ch11_199-219 11/02/05 21:43 Page 205