15597ch11199-21911/02/0521:43Page209 EQA Solutions o Problems09 OH CH,CH-CH2CH +CH OH mkz 73 CH,CH2CH2+CH￥CH OH CHCH CH CH,CH, Baheseiosgecuiosbladbyencefonanoygnlaet.isk,nf6ae cThem.os of water doen't rally help much,ept toue ou (CH)CCHOH.which has n herefore cannot dehydrate.That leaves three possibilities for A CHs CHCH2CH2CH2CH2OH CH CH2CHCH2OH CH,CHCH-CH2OH omer or -pm时 41.(a)CH1 H=2(7)+2 16;degrees of unsaturation =(16-14)/2=1 (b)C:H CI Ham =2(3)+2-1(for the Cl)=7;degrees of unsaturation =(7-5)/2=1 e:2m+2-16女of=06-22=2 (d)CsHc Ht =2(5)+2=12:degrees of unsaturation =(12-6)2=3 (6)++1(for the N)=15;degrees of unsaturation =(15-11)/2=2 CH.O 2(5)+=12;degrees of unsaturation =(12-8)/2=2 (g)CsHo 2(5)+=12;degrees of unsaturation =(12-10)/2=1 (h)C1oH4 Ha =2(10)+2=22:degrees of unsaturation=(22-14y2=4 42.(a)Ha =2(7)+=16;degrees of unsaturation =(16-12)/2=2 (b)H=2(8)+2+I (for the N)=19:degrees of unsaturation =(19-7)2 =6 (e)=(6)+-6(for the CI's)=8:degrees of unsaturation=(8-0)2=4 (d)H=2(10)+2=22:degrees of unsaturation =(22-22)/2=0 (e)Har=2(6)+2=14:degrees of unsaturation =(14-10)2=2 (f)Ha =2(18)+=38;degrees of unsaturation =(38-28)2=5Both fragmentations give cations stabilized by resonance from an oxygen lone pair. This is, in fact, the correct answer. Isomer A does not lose CH3 or CH3CH2 (no peaks at m/z 73 or 59). That pretty much rules out any tertiary or secondary alcohol structure as a possibility. (Any example that you can write should show those fragmentations.) How about possible primary alcohol structures? Look again to intense fragment peaks for clues. The m/z 70, loss of water, doesn’t really help much, except to rule out (CH3)3CCH2OH, which has no -hydrogens and therefore cannot dehydrate. That leaves three possibilities for A: CH3 CH3 A A CH3CH2CH2CH2CH2OH CH3CH2CHCH2OH CH3CHCH2CH2OH The data that you have are in fact quite consistent with either of the first two (the third is difficult to maneuver into a fragment with m/z 42). If you got this far, you are doing well! (The actual spectrum of isomer A is that of 1-pentanol, by the way.) 41. (a) C7H14 Hsat  2(7)  2  16; degrees of unsaturation  (16 14)/2  1 (b) C3H5Cl Hsat  2(3)  2 1 (for the Cl)  7; degrees of unsaturation  (7 5)/2  1 (c) C7H12 Hsat  2(7)  2  16; degrees of unsaturation  (16 12)/2  2 (d) C5H6 Hsat  2(5)  2  12; degrees of unsaturation  (12 6)/2  3 (e) C6H11N Hsat  2(6)  2  1 (for the N)  15; degrees of unsaturation  (15 11)/2  2 (f ) C5H8O Hsat  2(5)  2  12; degrees of unsaturation  (12 8)/2  2 (g) C5H10O Hsat  2(5)  2  12; degrees of unsaturation  (12 10)/2  1 (h) C10H14 Hsat  2(10)  2  22; degrees of unsaturation  (22 14)/2  4 42. (a) Hsat  2(7)  2  16; degrees of unsaturation  (16 12)/2  2 (b) Hsat  2(8)  2  1 (for the N)  19; degrees of unsaturation  (19 7)/2  6 (c) Hsat  2(6)  2 6 (for the Cl’s)  8; degrees of unsaturation  (8 0)/2  4 (d) Hsat  2(10)  2  22; degrees of unsaturation  (22 22)/2  0 (e) Hsat  2(6)  2  14; degrees of unsaturation  (14 10)/2  2 (f ) Hsat  2(18)  2  38; degrees of unsaturation  (38 28)/2  5 CH3CH2CH2 CH CH3 OH OH CH3CH2CH2CH m/z 73  CH3   OH CH3CH m/z 45  CH3CH2CH2  b a a b Solutions to Problems • 209 1559T_ch11_199-219 11/02/05 21:43 Page 209