Question

A fellow research chemist. Dr. Al Kaloid purchased a compound for his group's latest synthesis elect Unfortunately, the label on the bottle was badly damaged during shipping. Dr. Kaloid was able to obtain the MS data of m - 150 well as the IR. "C NMR and H NMR data provided below. You will Interpret all of the data provided on the spectre below, as well as the questions on the next page), to correctly elucidate the structure of the compound in the bottle. FT-IR (NEAT *C NMR 62.3 MHz Goose we can solvert CDCI NO 1 0 na 700 30 49 200 250 MHz CDC 3.1 Hz/mm TH NMR 250 MHZ CDCIE HHH 2H LLLLLLLLL பாNVITIN

Answer 1

Tv) 12 & Me Quartet at 10 ppm due to -CH3 group. Quartet cet 20ppm due to - cH3 group Triplet cet 30ppm due to – CH2- group. Doublet cet 35 ppm due to _&H group. Doublet cat 110 ppm, 120 ppm, 128 pom. 130 ppm due to ip-substituted benzene ring. singlet cet 135 ppm and 154 ppm due to quaternosy carber of benzene. due to CH 3 -CH2- signal at 2.0 ppm due to signal at 1.5ppm due to – CH3 signal at 3.2 ppm due to – CH - signal cat 6.5ppm due to -OH group signal at 7.0ppm to 7.5 ppm. due to benzene sing. HNMLL signal cet 1.1ppm sit

VD probable ny- structure may be OH (1) e (b) (0) L (6) T CH 3 0) -CH2 - ca) - CH (9) if (d) CH3 (c) 'HNMR - - с — Т. 1 22 т . Зн b = c= d = 2.0ppm, 2 lt 1.5ppm, 3H 3.2ppm, IH NMR spectee ia = 10ppm, Q i b= 30 ppm, T. c = 20 ppm, Q. d = 35ppm, D @=110ppm D f = 128 ppm, D I = 120ppm, D h = 120ppm, D K = 152ppm ,s j = 135ppm, s e=7.0ppm f = 7.30pm I= 7.4ppm h = 7.2 ppm i = 6.0 ppm