CHEM 213 FALL 2018 MIDTERM EXAM 2 - VERSION A

Transcription

1 CEM 213 FALL 2018 MIDTERM EXAM 2 - VERSIN A Answer multiple choice questions on the green computer sheet provided with a PENCIL. Be sure to encode both your NAME and Registration Number (V#). You will draw the structures of Molecules A and B on the Separate Double-Sided STRUCTURE ANSWER SEET. You will also enter your selection for Molecule Z (bonus question) on that sheet. Be sure to put your NAME on that sheet as well. There should be 6 pages (double-sided) in this examination AND you should have a green Scantron form, PLUS a STRUCTURE ANSWER SEET (double-sided). There are 10 graded multiple choice questions worth 20 marks and two structure solutions worth 25 marks for a TTAL of 45 MARKS. There is also a BNUS QUESTIN (molecule Z) on the last page worth 5 extra marks. There are 50 minutes available for this examination so ration your time accordingly. YU MAY NLY USE A Non-Programmable CALCULATR. Substituent constants for 13 C and 1 chemical shift prediction: δ aromatic C = Δ ipso + Δ ortho + Δ meta + Δ para δ alkene = Δ gem + Δ cis + Δ trans R Δ ipso Δ ortho Δ meta Δ para Δ gem Δ cis Δ trans Br C= (non-conj) C= (conj) N C=C

2 CEM 213 Midterm 2 Version A November 20, 2018 Page 2 of 9 1. This is exam VERSIN A, mark A on the computer sheet as the answer to question 1 Consider Molecule T when answering Questions 2-6; hydrogens are numbered and carbons are lettered. 2 N 1 2 A B 3 T C J 7 8 L 9 Br G I K F E 6 5 D N ow many unique 13 C signals will Molecule T show in the 13 C DEPT-135 spectrum? (b) (a) 3 (b) 4 (c) 5 (d) 6 (e) 7 (f) 8 (g) 9 (h) 10 (i) 11 (j) Which of the following best matches the multiplicity and chemical shift of the hydrogen-2 in the 1 spectrum of T at room temperature (use the data on page 1 if necessary)? (j) (a) 6.96 dt (b) 7.17 dt (c) 6.05 dt (d) 5.05 dt (e) 5.05 d (f) 7.17 d (g) 6.05 q (h) 6.96 q (i) 7.17 q (j) 6.96 d 4. What is the predicted chemical shift of carbon-l in the 13 C NMR of T (using the data on page 1)? (b) (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) What pattern is possible for 1 in the low temperature 1 NMR spectrum of T assuming you can see coupling up to 4 J (assume there is free rotation around the C-N bond)? (d) (a) s (b) d (c) t (d) dd (e) ddd 6. If the Br in T was replaced with a 19 F and you could see up to 4 J couplings, what would the spin system of the aromatic system (only) be? Assume this is run on a 500 Mz spectrometer. (e) (a) ABCX (b) AA BX (c) AA MX (d) A 2 X (e) A 2 MX

3 CEM 213 Midterm 2 Version A November 20, 2018 Page 3 of 9 Consider Molecule U when answering Questions Use the following coupling constants when answering this question: 1 J Pt-P (cis to Cl) 1600 z 1 J Pt-P (trans to Cl) 900 z 2 J P-P 65 z 3 J Pt-F 25 z NTE: Molecule U is planar but there is 3 J -F 10 z rotation about the Pt-C and Pt-P bonds 4 J Pt- 5 z 4 J F-P 2 z (to both P) REMEMBER: { 1 } means proton decoupled! 5 J -P 0 z 7. ow many lines in total would you expect to observe in the 195 Pt{ 1 } spectrum of U? (note, for example that a doublet is 2 lines in total while two doublets would be 4 lines in total). (f) (a) 1 (b) 2 (c) 4 (d) 8 (e) 9 (f) 12 (g) 16 (h) 36 (i) 72 (j) What is the total span in z (distance in z between the outermost lines) on the 19 F{ 1 } spectrum of molecule U? (f) (a) 2 (b) 4 (c) 10 (d) 25 (e) 27 (f) 29 (g) 35 (h) 39 (i) 52 (j) Which of these statements is true regarding the relationship between 1 and 2? (b) (a) chemically and magnetically equivalent (b) chemically equivalent but magnetically inequivalent (c) chemically inequivalent but magnetically equivalent (d) chemically and magnetically inequivalent 10. What pattern would you expect for the 31 P resonance cis to the Cl? (g) (a) d with satellites (b) t with satellites (c) dd (d) dt (e) ddt (f) dd with satellites (g) dt with satellites (h) ddt with satellites (i) 1:4:1 pattern (j) 1:4:1:1:4:1 pattern

4 CEM 213 Midterm 2 Version A November 20, 2018 Page 4 of What pattern would you expect to see for the C 2 groups in V at low temperature? (d) (a) s (b) d (c) t (d) q (e) pentet (f) sextet (g) septet (h) dd (i) ddq (j) ddt For EAC STRUCTURE below, DRAW your answer on the separate structure sheet. Compound A: C No IR data available [10 pts] 1 NMR (300 Mz): 13 C{ 1 } NMR with DEPT-135 phasings: δ 199.2(x), 163.0(x), 133.2(x), 129.6(+), 114.1(+), 55.2(+), 33.2(-), 12.0(+) ppm

5 CEM 213 Midterm 2 Version A November 20, 2018 Page 5 of 9 Compound B: C N 2 [15 pts] IR (cm -1 ): 3410, 1698, NMR (300 Mz): 13 C{ 1 } and DEPT-135 NMR: Don t forget to enter your structures for A and B on the separate structure sheets!

6 CEM 213 Midterm 2 Version A November 20, 2018 Page 6 of 9 TIS IS A BNUS QUESTIN: Molecule Z is a highly toxic substance with the formula C 4 10 P 2 F containing a pentavalent P atom of the general type X 3 P= (shown at right). Use the coupling constant data, 13 C{ 1 }, 19 F{ 1 } and 31 P{ 1 } spectra (REMEMBER { 1 } means 1 -decoupled) to deduce the structure of Molecule Z. Draw your structure for Molecule Z on the attached sheet in the space provided. [5 bonus pts] 31 P{ 1 } NMR: δ 35.0 (d, J = 180 z) 19 F{ 1 } NMR: δ (d, J = 180 z) 13 C{ 1 } NMR: δ 65.1 (dd, J = 40, 15 z), 21.0 (d, J = 9 z), 9.1 (dd, J = 175, 45 z). All three 13 C resonances have a positive phase in the DEPT-135 spectrum. Don t forget to draw your structures for A, B and Z on the separate structure sheet! END

7 CEM 213 Midterm 2 Version A November 20, 2018 Page 7 of 9 Structure solutions Molecule A C10122 DBE = 5 (Arom +1) Look at 13 C: 199.2x is a ketone so this cannot be an ester; other must be an ether 1 NMR clearly shows the two doublet pattern of a para-disubstituted ring and the 13 C resonances are consistent with this: p-c64 1 also shows 3, s at 3.75 ppm consistent with an C3 group and confirmed by the in the 13 C Remaining 1 resonances are a 2, q and a 3, t which must be coupled to one another and therefore a C2C3 group. So, we have everything: p-c64, C= ketone, -C3 and C2C3. The ethyl group must be on the ketone and this must also be on the aromatic ring. The C3 cannot be on the C= or it would be an ester. Final structure must therefore be:

8 CEM 213 Midterm 2 Version A November 20, 2018 Page 8 of 9 Molecule B C1215N2 DBE = 6 (Aromatic + 2) IR: 3410 ( or N stretch), 1698 (conjugated ketone; 13 C says ketone at 206 ppm), 1220 (usually an sp 2 C- stretch) The aromatic ring is tri-substituted with a s, d, d pattern: suggests 1,2,4-substitution This means all the C are inequivalent and realizing we can only have one ketone (one 13 C ketone signal) means we do not have any further unsaturation but as we are 1 DBE short, we must have a ring! The 2, q in the 1 is past 4 ppm so this is likely to be an -C2C3 unit and that also means that the 3410 stretch must be due to an N (matches the 1, s at 3.6 ppm). There is a 3, triplet at 1.3 ppm so this is all consistent with a terminal -C2C3 unit. The remainder of the unassigned 1 resonances are two 2, triplets (3.2, 2.7 ppm) and a 2 pentet (2. ppm) consistent with a C2-C2-C2- unit. The two C2 triplets are downfield quite a bit so are likely attached to the ketone (which must be conjugated so on the ring) and the N- so this gives us a N- C2C2C2C(=)- bifunctional unit which must be attached to the aromatic ring while the C2C3 is attached to the third ring position. It is really only reasonable to have the two chain ends connected ortho to one another as it is very difficult for a chain of this length to span meta-sites. We therefore have two possible structural options: owever, these are distinguishable by their chemical shifts for the aromatic region. The observed 1 NMR has the aromatic singlet furthest downfield. This is inconsistent with the left hand structure because this would put the s ortho to BT donors (Et and N). In the right hand structure the singlet is ortho to one donor (Et) and an electron withdrawing group (C=) so a normal shift makes some sense. The correct structure is therefore the right hand one. d d s N d d N s

9 CEM 213 Midterm 2 Version A November 20, 2018 Page 9 of 9 Molecule Z (Bonus) Use the magnitude of the coupling constants to assess what is connected to what. The 19 F and 31 P both show 180 z couplings which means (a) they are coupled to each other and (b) this is a 1 bond magnitude so they are directly attached to one another. In the 13 C{ 1 } NMR, there is a resonance at 9.1 ppm that is a dd meaning it must be coupled to the 19 F and the 31 P. ne coupling is large (175 z) consistent with a 1 J coupling and the other is still fairly large (45 z) and consistent with a 2 J coupling. This must be a C3 on P (and 2 bonds away from F). Note the C is + in DEPT phase. All that remains is an and C37. There is a 13 C at 65.1 ppm consistent with a C on and it phases + in the DEPT so it must be a C (note it cannot be a C3 or we would have completed the chain and have no place left to place the other two C). It is a dd with 40 and 15 z couplings to nuclei other than, consistent with a 2 J coupling to 31 P and a 3 J coupling to 19 F. The remaining resonance at 21.0 ppm must be two identical remaining C3 and these show 3 J coupling to 31 P. The molecule is therefore: This deceptively simple molecule is Sarin and it is responsible for one of the bestknown terrorist attacks using a chemical agent in peacetime the 1995 Tokyo subway attacks: