Chapter_15-2 - Nuclear Magne,c Resonance Spectroscopy...

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Unformatted text preview: Nuclear Magne,c Resonance Spectroscopy Miscellaneous and 1H and 13C “Known & unknown” spectra interpreted and explained Coupling to Nuclei With Very Similar Chemical ShiEs 60 MHz Coupling to Nuclei With Very Similar Chemical ShiEs 300 MHz 1-Bromopropane 1 ­Bromopropane The methyl group (0.9 ppm) has two 1H neighbors and is a triplet; the CH2Br group (3.5 ppm, deshielded) has two 1H neighbors and is a triplet; the central CH2 group (1.8 ppm, somewhat deshielded) has 3 + 2 1H neighbors; here the coupling constants are iden,cal, so that the group shows a sextet. 1 ­Iodopropane 1 ­Iodoropropane Equivalent Nuclei Coupling to Non ­ The methyl group has two neighbors and is a triplet (as expected) J = 7.3 Hz The CH2I group has two neighbors and is a triplet (as expected) J = 6.8 Hz The central CHCl group is more complicated Carbon ­13 Nuclear Magne,c Resonance A triplet of quartets A quartet of triplets Broad ­Band 1H Decoupling simplifies 13C spectra by elimina,ng 1H coupling all Methylcyclohexane 5 types of carbon atoms – 5 signals Three Dichloropropanes Because of the wide range of 13C chemical shiEs, nonequivalent carbons are more likely to be resolved than nonequivalent 1H nuclei; the number of signals is more likely to equal the number of nonequivalent carbons Chemical ShiE Type of Carbon___ 5 – 20 20 – 30 30 – 50 30 – 45 20 – 40 20 – 40 25 – 50 50 – 90 65 – 95 100 – 160 170 – 175 190 – 210 Primary Alkyl Secondary Alkyl Ter,ary Alkyl Quaternary Alkyl Allylic Alkyl–Br Alkyl–Cl Alkyl–OH or  ­OR Alkyne Alene, Aroma,c Ester Aldehyde, Ketone HH Cl Cl H four 1H signals CH3 1:1:1:3 3.75, 3.55, 4.15, 1.6 ppm Unknown Spectrum   0.9 ppm 9H t (=2n) shielded   1.5 ppm 6H q (=3n) slightly deshielded   The combina,on of t (=2n) and q (=3n) is always  ­CH₂ ­CH₃   1.3 ppm 1H s (=0n) Because the signal @ 1.3 cannot be less than 1H, there must be three  ­CH₂ ­CH₃ groups. The signal @ 1.3 is not coupled: OH The compound is (CH₃CH₂)₃C ­OH 13C spectrum (75 ppm, C ­O) confirms Unknown Spectrum   4.3 ppm 1H   2.4 ppm 1H   2.2 ppm 1H t (=2n) strongly shielded t (=2n) deshielded qui (=4n) deshielded 3 nuclei (1+1+1) cannot explain the observed mul,plets: we try 2+2+2 This works if the nuclei @ 4.3 AND 2.4 ppm HH interact with those @ 2.2 ppm: H Y ­CH₂ ­CH₂ ­CH₂ ­X H 2.4 2.2 4.2 O 13C spectrum confirms H O H Unknown Spectrum   1.1 ppm d(?), 6 two possibili,es: 2 equiv. CH₃ with 1 neighbor or two nonequiv. CH₃ w/o neighbor   2.8 ppm sep., 1 (=6n) requires that the 6H have 1n. This gives us isopropyl. 2.8 ppm is deshielded, but not enough for a halogen: how and to what is isopropyl aqached? The answer comes from the 13C spectrum. The ,ny signal @ 220 ppm signifies a carbonyl. Our compound is diisopropyl ketone. 13C spectrum confirms Unknown Spectrum   1.2 ppm 3H t (=2n) shielded   3.5 ppm 2H q (=3n) strongly deshielded   The combina,on of t (=2n) and q (=3n) is again  ­CH₂ ­CH₃   4.0 ppm 2H d (=1n) The signals @ 3.5 and 4.4 ppm could be adjacent to O (signals @ 62 and 66 ppm are typical for C ­O), giving us a CH₂ ­O ­ CH₂ ­ CH₃ unit. [The remaining signals belong to a vinyl group, a mono ­subs,tuted alkene group: (see Chapter 11).] Unknown Spectrum 1.0 t 3H CH₃ 2n (CH₂) 1.5 sex 2H CH₂ 5n (CH₃, CH₂) 1.8 qui 2H CH₂ 4n (CH₂, CH₂) 4.3 t 2H CH₂ 2n (CH₂) deshielded by Eneg group Eneg cannot be a Hal because a Hal would “terminate” the molecule; we need O (or S or N, etc) to aqach the rest of the molecule. This adds up to a –O–CH₂–CH₂–CH₂–CH₃ group. The rest of the molecule is a phenyl group, aqached to the O via a C=O; O ­C=O is an ester group (167 ppm); you learn about that in 308. Unknown Spectrum 1.0 t 3H CH₃ 2n (CH₂) 1.4 sex 2H CH₂ 5n (CH₃, CH₂) 1.6 qui 2H CH₂ 4n (CH₂, CH₂) (this requires one more CH₂; there are three, @ 3.5, 3.6, 3.8, all t 2H CH₂ 2n all deshielded by Eneg group: 2 O ? This adds up to a –O–CH₂–CH₂–CH₂–CH₃ group and a –O–CH₂–CH₂–O – One O has an OH (2.4, s, 1H) Molecule: H–O–CH₂–CH₂–O –O–CH₂–CH₂–CH₂–CH₃ ...
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