Question

In this problem you have to elucidate the structure of product B. A substrate A (structure shown in scheme 1) which was subjected to two consecutive reactions as shown in scheme 1 below to give the product B of molecular formula CsH14Br2. Provide the structure for Product B based on the NMR data provided (attachment 2: Problem set 2). (The 1H NMR and 13C NMR of starting substrate A has been provided for reference purpose only) 1. MnO2, CHCI3, rt, 24 h OH CgH14Br2 Product B 2. CBr4, PPH3, DCM, 2 h, rt Substrate A Scheme 1 The spectra for product B was recorded on 400 MHz instrument. Elucidate the structure based on the NMR spectra provided: 1H, 13C, DEPT 135, DEPT 90. In your answer include: 1H spectral details (Chemical shift values, Coupling pattern and constants with unit, Number of proton integration and assignment of peaks as aromatic/alkene/...etc and any other remarks) 13C spectral details (Chemical shift values, type [C, CH2, CH3] and assignment) Draw the proposed structure and assign 1H and 13C chemical shift values with brief justification ii) Spectra key 1. Reference 1H NMR and 13C NMR spectra of starting material, substrate A. This is only for reference purpose and may be used to elucidate the structure of product B. No discussion on this spectra is required The reaction and substrate A is again shown here for reference 1. MnO2, CHCI3, rt, 24 h OH CoH14Br2 Product B 2. CBr4, PPh3, DCM, 2 h, rt Substrate A 2. 1H NMR of Product B (whole spectra) 3. 1H NMR expanded (to reveal splitting, 5.4 ppm to 5.9 ppm) 4. 1H NMR expanded (to reveal splitting, 1.6 ppm to 2.1 ppm) 5. 1H NMR expanded (to reveal splitting, 0.8 to 1.15 ppm) 6. 1C NMR 7. DEPT 135 8. DEPT 90 Reference 1H NMR and 13C NMR spectra of starting material, substrate A OH 12 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2,0 1.5 1.0 O.5 ppm 8919 OH 12 220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 O ppm 20 10 69 - 99- "SET - 2. 1H NMR of Product B (whole spectra) sha NR55.001.00888 201287107 2.99299 0.93 0.95 TT 35 Chemical Shift (ppm) 30 4.5 4.0 7.0 6.5 6.0 5.5 5.0 2.5 2.0 1.5 1.0 0.5 26 0 99- 98 1 961- 661 933 2. 1H NMR of Product B expanded (to reveal splitting, 5.4 ppm to 5.9 ppm) _sha NG 1R55.001.001.1 r.esp 0.95 5,70 5.60 5.90 5.85 5.75 5.65 Chemic al Shift (pp 5.55 5.50 5.40 5.80 5.45 5.695 -5.691 5.676 999 9- 3. 1H NMR of Product B expanded (to reveal splitting, 1.6 ppm to 2.1 ppm) _sha NG1R55.001.001.1r.esp 2.01 2.87 1.07 1.90 1.85 Chemical Shift (ppm) 2,10 2.05 2.00 1.95 1.80 1.75 1.70 1.65 1.60 E91- 099- 949 970 686 66 4. 1H NMR of Product B expanded (to reveal splitting, 0.8 to 1.15 ppm) shaNG1R55.001.001.1r.esp 2.99 2.99 1.15 1.05 1.00 Chemical Shift (ppm) 0.85 0.80 1.10 0.95 0.90 9L6 0- S860- 5, 13C NMR of Product B P_sha NG1R55.002.001.1r.esp T TT 180 160 140 120 100 80 60 40 20 Chemical Shift (ppm) -141.04 132.02 E9S8- 28.53 6. DEPT 135 of Product B P_sha NG1R55.004.001.1r.esp 180 160 100 Chemical Shift (ppm) 80 60 40 20 140 120 -37.24 -28.54 22.43 -15.34 7. DEPT 90 of Product B _sha.NG1R55.004.001.1r.esp 100 40 180 160 140 120 80 60 20 Chemical Shift (ppm)

Answer 1

In last structure there should be a double bond...

H NMe Seeetial deteuils BA tmme, Cucl, St, 24h 2EB4, PPh3,Dcm 2h, ht CayH 1a BAZ Substrute-A Praduct-B Br chemicad shibt Sppm Assjrmert Constamt Prcten a,b CHGRCH Derulblet J-4H2 No.9 Mulliple piletadHI N 1.85 Smytet 3H (-CH3) Maitplet N1.ys-1.99 2H C-CH) mylHple S.es-S.59 TH (-CH) 1H C-CH LJ ct 13c NmR Br chemicl shift Sppm Type fugunrehset IS.34 CH3 bb 22.43 2xcH 28-53 C isoprepyt-CH 37.23 -cH2 e -CH -CH 132.02 134.47 -CH CH 141. 04 Frem the CH2 grp Confiamea Hhe DEPT Ne Confirms 3 Cenben -CH DEPT-g0 Staucture es be fon predriet-B is Br Hemce the BA Paedact- B