Chem 213 Extra Problems Set 2 Solutions Fall 2013 DBE = 4 (aromatic) [EASY] Problem 62 C10H14 1
H and 13C indicate symmetry: one type of aromatic H (2H integration) and two types of CH3 (6H each) Left hand structure has two types of CH3 but right has only one so left is correct: _____________________________________________________________________________________ DBE = 3 [EASY] Problem 68 C5H7NO2 IR: 2266 = nitrile CN 1749 = ester, non‐conjugated –CO2‐ These account for all DBE 13
C says we have CH2, CH2 and CH3 which accounts for all H 1
H shows downfield q (2H) and a t (3H) consistent with –OCH2CH3; other CH2 is isolated (singlet): Only options is: _____________________________________________________________________________________ DBE = 6 (aromatic + 2) [EASY] Problem 110 C9H10N2 IR: 2209 = nitrile CN → accounts for all DBE so other N MUST be a tertiary amine (no NH stretch) 1
H NMR shows two aromatic doublets (2H each) and with IR band at about 820 cm‐1 plus 13C pattern, this strongly suggests a para‐disubstituted aromatic ring. CH3
H3C
N
Other 1H NMR signal is a 6H singlet at about δ 3 indicating two equivalent CH3 on N, so the structure can only be: C
N
_____________________________________________________________________________________ Problem 214 C8H5O3Cl3 DBE = 5 (aromatic + 1) [MODERATE to HARD] This problem is only hard if you pay too close attention to the IR stretch and the shifts predicted by the yellow sheets! I’ll show the best answer and then comment on what the issues are with it… acid OH stretch, although not as strong as often seen; confirmed by 1H at 11.8 IR: 3400‐2400 1745 not great for an acid but definitely not conjugated MS: not usually that useful but a distinguishing point from other options in this case. Note the loss of 45 (CO2H) to 209, then loss of a CH2CO2 to 196 and finally loss of OCH2CO2 to 179. Note also that 179 corresponds to C6H2Cl3 indicating quite convincingly that all three Cl are on the aromatic ring (isotope pattern is still present in the 179 peak as well). 1
H NMR shows two aromatic s (isolated from each other) and a CH2 s far downfield in the 1H and 13C consistent with an –OCH2‐ meaning we must have an –OCH2CO2H unit as one ring substituent with the rest Cl: 13
C shifts for the aromatic CH carbons are much more consistent with the left hand structure because this one has only one CH carbon that is shielded while the right hand structure would have two. Okay, so what are the issues with this structure from the data provided?  IR C=O stretch is not great for an acid (too high) and the OH stretch isn’t as pronounced as usual  1H NMR chemical shifts do not agree well at all based on the electron donating property of O and do not match predictions from the yellow sheet very well (they do however match with prediction software I use) I also considered the possibility that this could be a conjugated acid chloride (to get the IR C=O stretch low enough) with a OCH2OH group on the ring. The structure below actually matches the 1H NMR aromatic shifts reasonably well based on yellow sheet predictions and the downfield OH could be okay with H‐bonding to the carbonyl. However, this simply cannot match the fragmentation pattern in the MS as it cannot lead to a 179 peak with isotopes plus, the 13C shifts aren’t a good match and neither is the OCH2O shift in the 1H NMR. So, despite some unusual pieces of info, the best fit is the one above. _____________________________________________________________________________________ Problem 262 C7H14O2 DBE = 1 [EASY to MODERATE] IR is not useful here 13
C shows =CH2 (118) and =CH (136) so we have a mono‐substituted alkene ‐CH=CH2 as the DBE 1
H says the δ 5.73 ddd has one next door H (otherwise it would be a dd due to cis and trans coupling) and this CH must be the resonance at δ 4.77 as it is the only other resonance of integration 1. The shift of this CH is far downfield in the 1H but it is even more exceptional in the 13C since it has to be the one at δ 102 ppm. This fact and the obvious symmetry observed in the 13C plus the fact the O have to be ethers, tells us we have to identical O attached to this CH. All that remains are a CH2 at δ 61 (on O) and a CH3 at δ 16. The 1H NMR shows that the methyls are identical a triplets so they must be connected to the CH2 as two identical ethyl groups. So, why are there two sets of multiplets for the CH2 groups? This is because the ‘inside’ and ‘outside’ H on each methylene C are inequivalent (and couple to one another as well as the methyl group) but the two ‘outer’ H (dashed boxes) on different C are equivalent, as are the two ‘inner’ H (circles) on the different C. _____________________________________________________________________________________ Problem 308 C10H18O DBE = 2 [HARD] IR: 3327 br indicates an alcohol and 1001 suggests it is primary, ‐CH2OH, confirmed by 13C DEPT (‐ at δ 59) and 1H (2H, δ 4.05) Need some DBE and 13C says we likely have an alkene or two. Looks like 2 x =CH and 2 x =C but a quick check of the COSY shows that the CH are NOT coupled to one another so we have: 2 x ‐CH=C Rest of the 13C says we have: 2 x CH2 = 4H 3 x CH3 = 9H + CH2OH + 2 x CH=C = 5H 18H which is correct Now the COSY can help: ‐CH2‐ groups that are NOT on O overlap but show coupling to ONE alkene H at δ 5.07 ?
5.07 t H
(partially resolved)
H2C
?
The –CH2OH group shows coupling to the other alkenyl proton at δ 5.4 (making it a triplet): Although it is not very strong, there does appear to be a weak COSY between the alkenyl H at δ 5.4 and one CH3 while the other alkenyl H at δ 5.07 shows similar coupling to TWO CH3 suggesting we have the following units (note these are 4J couplings and they can sometimes be seen in COSY with alkenes but they are usually weak): We are still missing one CH2 so we connect the two units through that group. The NOESY suggests that the CH2OH is trans to the methyl group because no NOESY correlation is seen between the CH2 H and any CH3 protons. CH3
H3C
CH OH
2
H2
C
H
H
C
H2
CH3