Question

draw the structure for the spectroscopy and show steps and label each part

Mass Spectrum (not shown): [M] = 302 (75%), 304 (100%), 306 (25%) m/z (all listed are strong (s) unless otherwise IR Spectrum (not shown): 3090, 2985, 2850, 2750, 1715, 1610, 1500 cm indicated) 'H NMR Spectrum (400 MHz, CDCI, 25 °C) t(J = 2 Hz) t(J= 2 Hz) +(J= 2 Hz) 2H 6H 3H 1H 1H 1H PPM HC NMR Spectrum (with DEPT), proton-decoupled (125 MHz, CDCI, 25 °C) (CH) (CC)(CH) (CH)(CH)(C) (CH) (CH2)(C) (CH) 220 . 200 . 180 . 160 . 140 120 100 80 60 40 20 Q6.R

Answer 1

Mass Specbrum : [m] = 302 (75%) , 304 (100%) 306 (25% miz intensity pattem is [M ] - 75 %. [M+2] = 100 % LM+47 = 25% Indicates compound containing halogens. one Bromine and one chlorine present in compound therefore compound shows mass spectrum like this. -oor and och present Rule of thirteen = Molecular ion] 13 - 302 = 23 + 3 13 13 molecular formula = C23H26. But one Br and a present in compound. therefore cotty replaces by Br and C2H11 replaced by 35 cl en M.F= H8 Brel IR specbum: 3ogo, 2985 2850, 2750 1715, 1610, 1500 cmt.

. 3ogo cant 2985 & 2850 cm 2750 cmt → 1715 cost 1610, 1500 cont & Aromatic cat streething Alkane ch streething Aldehyde (H strecthing. Aldehyder- carbonyl group Aromatic cac shetching. 2750 &1715 indicates aldehyde functional group present in compound, ei CH4 Replaced by to & M.F - C14H4 Obra. t2 proton reptarted by Cic' replaced by 12 proton, ei mir - CIB H16 o brol. 'H NMR spectrum: 0.7 ppm, 34, og ppm 6H 102 PPM, 2 . 3.4 ppm, 14 7.5 ppm, 14, 7.6 ppm, 14, 7.8 ppm, 4. go ppm, 10 t. - CH3 7 a t 5, 24-CH3 – CH2-CH3 q . - CH2 d, och attach to EWG group tin t. Aromatic proton J = 2H2 t, I-value indicates mete coupling d, aldehyde proton ENG- Electron withdrawing group,

13C NMR Shows spechum, 12 different types of carbon (CH) 200 ppm & con CH) 68 ppm Attach to ENG ACHCH Ar. (CH3) 9 ppm (012) 34 ppm (CH3)24 ppm – - CH3 T – CH2 - CH2-CH3 shong peak Indicates +шо - енд Troup cs go me chemical errorment tebraubstituted curabon. ke) 32 PPR (c) 124 ppm --Brehatubst substituled cromatic carbon. (CH) 127 ppm - 9 (CH) 129 ppm & Aromatic carbon (CH) (134) ppm ) ) (c) 136 ppm –& substituted cromatic calabon (c) 145 ppm - R group substituted cromatic carbon

From the above the structure of data we conclude that compound is im spor 1715 2750 cnt o pipo og.oppio [200ppm] (v.y ppm). (14.5 ppos H [ 24 ppm] (7,5 ppm) FAR (H3 [32 ppm 7 at CH2 124 ppm 1 - H -сH3 - (o.1 үр т) [9 ppm] é (7.8 ppm) (7.6 ppm) 3ogo cml [136 ppon on ppm) [84 ppm (3.4 ppm) [68 ppm) CH3 Mir= C13H160 Bra. 6.14 - 05 ( - H-NMR E I - 13-NMR