Chem 213 Final 2012 Detailed Solution Key for Structures A H

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1 Chem 213 Final 2012 Detailed Solution Key for Structures A H COMPOUND A on Exam Version A (B on Exam Version B) C 8 H 6 Cl 2 O 2 DBE = 5 (aromatic + 1) IR: 1808 cm 1 suggests an acid chloride since we have Cl and O (can t be an anhydride with only two O) but to be this high it doesn t look conjugated 1221 cm 1 suggests an aryl C O stretch NMR: 7.24 and 6.82 ppm 2H doublets (J = 8Hz) suggests a para disubstituted aromatic ring C 6 H ppm 2H strongly suggests the CH 2 is on oxygen and something else that is deshielding it. From this it appears we have two options: The left option is a much better fit to the data because it accounts for the band at 1221 cm 1 (an aliphatic C O would be at a lower frequency) and the OCH 2 is probably too far downfield to just have an aromatic ring as the additional deshielding group. An alternate structure with a CH 2 Cl group would also be insufficient to explain the additional chemical shift of this CH 2.

2 COMPOUND B on Exam Version A (A on Exam Version B) First use MS to determine the formula (there was an error on the exam that was discovered late in the exam this was taken into account during grading): M+ m/e = 176 rel. abundance 100% M+1 4% this tells us there are 4 C since 4 x 1.1% = 4.4% chance of having one 13 C M+2 98% one Br since this is the only element with a 1:1 M+ and M+2 ratio M+3 4% C4HxBrOy since fragment at 97 (for loss of Br) contains only C, H and O: Since 4C = 4 x 12 = (Br) = 127, this means the H and O total mass is = 49. Oxygen weighs 16 so we must have 3 O and 1 H to equal 49. Note it cannot have 2 O because it is impossible for only 4 carbons to have 17 H. Therefore the formula must be C 4 HBrO 3 which gives DBE = 4. IR: 1795, 1750 cm 1 suggestive of an anhydride since we have 3 O (two DBE) An anhydride is C 2 O 3 so this only leaves C 2 HBr and we still need two DBE: Options are a ring and an alkene OR an alkyne as shown below: The alkynes can be ruled out generally because there is no alkyne CC stretch. The bromo alkyne in the middle would have a aldehyde like C H stretch at 2750 cm 1 and the alkyne above right would have a strong terminal alkyne C H stretch at 3300 cm 1. Note: the structure at right cannot be ruled out from the data given and was also given full credit

3 COMPOUND C on Exam Version A (C on Exam Version B) The NOESY shows TWO interactions, each of which are between an aromatic doublet and an aromatic singlet. This immediately rules out any structure with a substituent adjacent to the ring fusion point (since there MUST be aromatic H there): Ruled out: 1,2,3,4,5,6,7,8,10,11,14 on Version A (1,2,3,4,5,6,7,8,10,12,13 on Version B) There are only three unique isomers with substituents ONLY in the beta position but 13 on Version A (11 on B) can be ruled out because it has one NOESY between two aromatic singlets and another between two aromatic doublets. You can distinguish between 9 and 12 on Version A (9 and 14 on Version B) without even using the COSY simply by noting that 9 on either exam would have some aromatic triplets and since there are NONE, the correct structure must be 12 on Version A (14 on Version B). You could also establish the same thing by COSY noting that there are just two isolated groups of COSY interactions between aromatic doublets ruling out three H in the 1,2 and 3 positions.

4 COMPOUND D on Exam Version A (E on Exam Version B) C 9 H 14 O 5 DBE = 3 IR: br cm 1 suggests a carboxylic acid (consistent with 13 C resonance at 174.1) 1718, 1709 cm 1 non conjugated acid and/or ketone (double check: shows a 13 C resonance at that disappears in the DEPT indicating there is definitely also a ketone) NMR: first observations too few peaks in the 13 C and simple 1 H tells us we have a symmetrical molecule. 1 H NMR shows an acid OH at 12.0 ppm but it integrates to 2H so there must be two acid groups. Does that work? Yes, since we have 5 oxygens, we can have 2 x CO 2 H + a ketone note this accounts for all 3 DBE. 13 C NMR: 40.8 (CH 2 ), 32.7 (CH 2 ) and 18.6 (CH 2 ). Note none of these can be on O since we have two acids and a ketone (no O left: two have H and one is a keto O) but two are fairly far downfield while one is not. 1 H NMR: 3 x 4H resonances indicates (again) that this is a symmetrical molecule. triplet pentet triplet pattern is only consistent with a CH 2 CH 2 CH 2 chain and there must be two identical ones. So, we have two CO 2 H which are identical and terminal groups and we have two CH 2 CH 2 CH 2 chains and a keto (C=O) group that are each bifunctional. Only way this can fit together and be symmetrical is if the keto group is in the middle of the chain with CH 2 CH 2 CH 2 CO 2 H groups on either side:

5 COMPOUND E on Exam Version A (D on Exam Version B) C 12 H 14 O 3 DBE = 6 (aromatic + 2) IR: 1710 cm 1 look like a non conjugated carbonyl and 13 C indicates a ketone (210.5) 1 H NMR: only 2 singlets in the aromatic region suggest a tetra substituted aromatic ring two singlets of 3H at almost 4 ppm and the two quartets in the 13 C at 56.1 and 56.0 ppm indicates that we have two OCH 3 groups which accounts for all the O (other is the ketone) What else is there? three 2H resonances: s at 3.5 and two triplets at 2.5 and 3.0 ppm. Note the triplets MUST be coupled to each other as there are no other multiplets at all, so we have a CH 2 CH 2 fragment. 13 C NMR confirms the three CH 2 with triplets in the coupled spectrum at 44.2, 38.6 and 28.1 ppm. So, to summarize so far, we have: 3 x CH 2 as an isolated CH 2 and a CH 2 CH 2 unit a keto group (C=O, 1 DBE) a tetra substituted aromatic ring C 6 H 2 (4 DBE) 2 x OCH 3 (terminal groups) Which totals to C 12 H 14 O 3, so we have everything: this MUST mean the remaining DBE is a RING. Connect CH 2 CH 2 to keto group and then the other CH 2 to ensure the keto group is non conjugated and one CH 2 is isolated from the others; OCH 3 are terminal groups and must go on the aromatic ring. The CH 2 CH 2 C(=O)CH 2 chain then connects to the aromatic to make it tetra substituted and it must connect ortho to make chemical sense. This leaves two three possible structures recognizing that the aromatic resonances are singlets: These can be distinguished most easily by recognizing that ONLY the structure in the box shows a long range HMBC between BOTH aromatic singlets and a CH 2 carbon (the other two only show one such HMBC correlation).

6 COMPOUND F on Exam Version A (G on Exam Version B) MS shows peaks at 160, 141 ( 19) and 122 ( 38) indicating sequential loss of two mass 19 fragments, consistent with loss of two F. The M+1 peak is 9% indicating we have 8C (8 x 1.1% = 8.8%) and the fragment after F loss at 122 has only C, H and one O in it: C 8 H x O so x is (12) = 10 and the formula for intact COMPOUND F is C 8 H 10 F 2 O DBE = 3 IR: 2205 cm 1 must be an alkyne CC stretch (no N so not a nitrile); not terminal since no 3300 cm 1 C H stretch 1690 cm 1 looks like conjugated carbonyl and 13 C confirms a ketone (191.0 ppm) 13 C NMR: ketone but shows coupling to two F and the J value suggests 2 J (2 bond) triplet in the 1 H decoupled spectrum and large J indicate this carbon bears two F; note in the DEPT it shows as + indicating it ALSO has one H on it so we have a CHF 2 group. Downfield shift is okay because two F are extremely deshielding alkyne C but this one shows a triplet with 11 Hz coupling suggesting it is 3 J coupled to the two F 88.0 other alkyne C but further away from F so not coupled 24.3, 23.4, 21.0 CH 2 groups from DEPT not near F since normal shifts 14.1 CH 3 group from DEPT 1 H NMR: Note the CHF 2 is isolated but all the other groups (3 x CH 2 and CH 3 ) show coupling. The triplet pentet sextet triplet pattern is only consistent with a butyl chain: CH 2 CH 2 CH 2 CH 3 which is a terminal group. Thus we have an alkyne, a conjugated ketone and two end groups: a butyl chain and a CHF 2 group. This places the ketone next to the alkyne, but which way around are the end groups? O CHF 2 CHF 2 O The observation of 2 J CF coupling between the keto C and the F places them close together and only supports the right hand structure.

7 COMPOUND G on Exam Version A (F on Exam Version B) MS indicates this compound contains one Cl (M+2 = 32% rel. abundance) and this is backed up by the loss of 35 from the molecular ion to give a fragment at m/e = 148. The M+1 on this fragment or the molecular ion is 11% suggesting there are 10 C (10 x 1.1 % 13 C) in the compound. High resolution MS: (exact mass of 35 Cl) = so range of possible masses within error are: to and there are only two masses in the table that match this fragment mass: C 7 H 16 O 3 ( ) and C 10 H 14 N ( ). The latter fits the 11% M+1 size and the 13 C NMR that shows 10 C. Thus the formula is confirmed as C 10 H 14 ClN. IR: 3430, 3355 cm 1 br s suggests a primary amine since there is a N but no O 1 H NMR: two aromatic doublets, coupled to each other from COSY, of integration 1H suggests a tetra substituted aromatic with 1,2,3,4 substitution on the ring. tetra substitution is also confirmed by the 13 C DEPT. 3.7 broad singlet of integration 2H is consistent with a primary amine NH 2 2.4, 2.7 two CH 2 quartets indicating they are both next to 3H two CH 3 triplets indicating they are both next to 2H COSY confirms that each CH 2 is coupled to a CH 3 so we have two CH 2 CH 3 groups Which isomer do we have? That is easily determined from the NOESY spectrum: since BOTH CH 2 CH 3 groups show a NOESY to the NH 2, the amine MUST be between the two CH 2 CH 3 groups. Then since we have two aromatic doublets, the H on the ring MUST be next to one another meaning that the Cl is next to one of the CH 2 CH 3 groups:

8 COMPOUND H on Exam Version A (H on Exam Version B) C 7 H 6 N 2 O DBE = 6 NOT aromatic IR: 2230, 2218 cm 1 very strong and sharp suggests two nitriles (we do have two N); an alkyne could be a possibility but there are no C consistent with this (70 90 ppm). 1 H NMR: 7.1 and 5.7 d (J = 10 Hz) consistent with a cis alkene (must be coupled to one another since there are no other multiplets). 4.8 s probably an alkene H since there is nothing else in this compound deshielding enough to make a H this far downfield except O but that is part of a OCH 3 groups (see below). Must be on an alkene bearing no other H. 3.9 s (3H) consistent with a OCH 3 group and confirmed by the quartet at 55.1 in the 13 C 13 C NMR: Other than the OCH 3 quartet, everything else is between 100 and ppm, consistent with both alkene and nitrile C. Of these, three are doublets in the coupled spectrum, consistent with two alkenes: RCH=CHR (cis from 1 H) and RCH=CR2 Summary of what we have: 2 x CN (terminal groups), OCH 3 (terminal group) and the two alkenes (CH=CH and CH=C) Is this everything? C 7 H 6 N 2 O YES DBE = 2 x 2 (CN) + 2 x 1 (alkene) = 6 CHECKS MUST connect the alkenes so we have: Establishing which isomer we have is fairly tricky. First you need to establish which carbons form the alkene backbone and which are nitrile C: 7.07 d and 5.68 d are on the C at and ppm. As a first guess you might assume that since that the ppm C is the one bearing a CN since the C attached to the nitrile will be shielded (this will be shown to be correct) s is on the C at ppm. Again, this shielded value suggests that Z is a nitrile.

9 The other C must bear a OCH 3 group and will not show in the short range HMQC; it will however show up in the long range HMQC between the CH 3 protons and the alkene C. This confirms that the C at ppm is the one bearing the OCH 3 group. Now that we have established the C of the alkene skeleton, we know the other two at and ppm are the nitrile C. NOTE The alkene singlet at 4.83 ppm is critical in determining the isomer since we know unequivocally which H this is and where on the alkene backbone it must lie: The 4.83 ppm alkene H shows a long range HMQC to only one C at ppm which is one of the alkene backbone C (bearing a proton that is part of the cis doublet): This means that Y CANNOT be a nitrile (or there would be a long range HMQC to a nitrile C as well) and MUST be the OCH 3. Once that is known, the structure is established with X and Z being nitriles. We can check the structure by making sure all of the other long range HMQC work: the 7.07 d shows a long range HMQC to a nitrile (115.0) and an alkene C (100.2) while the 5.68 d shows a long range HMQC to only the alkene C (155.1). A complete assignment of the 1 H and 13 C shifts is shown below with the structure. Note, the other isomer with Y and Z trans rather than cis, is equally acceptable without further information. 5.68d 7.07d H H 108.2d 142.9d H 4.83s NC 155.1s 100.2d 115.0s H 3 CO CN q 105.1s long range HMQC END