1559T_ch11_199-21911/02/0521:43Pa9e205 ⊕ EQA Solutions to Problems.205 31.Use the carbon types together with the chemical shifts to choose between alterative possibilities. ees of unsaturation =(10-612=2T bonds or rings resent eco一c今s一G年G6鸣。 CH=CH H2C-CH2 Cyclobutene (1 bond and I ring) and therefore,must be a-CH group because it is a doublet.The answer is therefore 0 -H(2 bonds).You do not have enough information about C NMR to (c)Ha =8+2=10;degrees of unsaturation =(10-8)/2=1. CH一CH2-CH=CH d)H=10+2=12: s of u ration =(12 1012 =1.This one has two CH ough (=58.8)to be attached to the and alkene carbons,of which only one (=125.7)has an H on it. The pieces:(2x)CH--CHa---CH-C.There is one H unlocated.Because it is not attached toone of the carbons,it must be on the oxygen.So the possible answers are CH e-CH:-oH CI:-OH G H0-CH、CH C=C You do not have the information to tell which of the three is the actual compound. (e)Notice that ther an alkene CH2.whereas =149.2 is an alkene C lacking hydrogens. What do you have so far?The molecule has the piece CHC 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