Question

thanol in CTO 20 crum expected for dicthyl ether(CH.CHOCH.CH 12. Propose a structure to go with the proton NMR spectrum and the following mas po [Mj - 136 (100), [M+11 - 137(3.3). [M+25 - 138 (973) 13. PPM Construct an explanation for why the po at an explanation for why the peak in the spectrum above found at 4.5 ppm is outside the ypical range for a hydrogen alpha to a carbonyl (2-3) and the typical range for a hydrogen bound to the same carbon as a halogen (3-4). 14. Draw the structure of an unknown to go with this proton NMR and the following mass spectral data: (M) - 166 (50), (M+1 -167 (5.5). [M+2] - 168 (16.1) 2 2 987 PPM Copyright Houghton Mifflin Harcourt Publishing Company. All rights res

Answer 1

(12) @ mass spectrum data mlz = [mt] = 136 (100). (mt2j= 138097.3) # molecular weight of compound is 136. the relahve intensity of Mand m+2 peak is 14:1 This indicate there is presence of Bromine atom in the compound. (b) molecular Loomula - Using rule of 13. # mass indicate compound have one boomine atom and Peale al 9.80 indlicate presence of aldehyde up. These compound have one Br and one batom. Carbonyl group. Thus compound have 136 ja = n= 10,8= $ CnHntre Clo H16 (Raleols 13) But com pound have one Br and one'o then remove Cy Hil Isrom rule of 13, then molecular Loomula of Compound is - C3H5 Bro Degree os Unsaturahºcm) = (2C+2-H4+) = 2(3)+2-5*] 2 ( 2C+2-HX) = L = 1 - 2 (e) 1 H NMR data § 1.95( 3H, d) 5 Cutçu- 4.50 (141, 6) petencie qunktion 36 9.80 (81H, d) = Haff- +

structure си — сH- Hit 9.80(d) - loasld) 4-5( quinlet) 0+ m (13) The Proten containing this carbon attached to the # two electoco-coithdrawing group ie. Brand-çti Thus -I ektect of -36 and -G- ceshield the this Pooten and peak of this protess is outside the typical range Lor a hycksogen alpha to carbuoyl and Typical range of hydrogen boound to the same carbon as halogen. the fcht a -Iekbect Br -I effect