Question

Please help. I need a step by step solution to find the compound. I'm not too confident with what m/z means and how to read the IR graph yet. What does quin (5) have to do with 2.36 ppm in the H NMR? A description about those would be great too. Thank you so much.

ea (e) ()-соон, ON COOH, tr COOH COOH, CH3o 31. Propose a structure for a compound that displays the spectro- scopic data that follow. The molecular ion in the mass spec- trum appears at mlz 116. IR: 1710 (s), 3000 (s, broad) cm-1; 1H NMR: δ 0.94 (t, .-7.0 Hz, 6 H), 1.59 (m, 4 H), 2.36 (quin, J = 7.0 Hz, 1 H), 12.04 (broad s, 1 H) ppm; 13C NMR: δ 11.7, 24.7, 48.7, 183.0 ppm 1. Hg(OCCHs)2, H20 2. NaBH CrO.. H C&H10 acetone, 6112 Section 12-7 Section 13C NMR:6-22,1 "C NMR: 1 1 .7, 24.7, 48.7, 1 83.0 ppm. 5.5 acetone, 0C 2 → ChHnO section 8-5 CH12O Section 12-3 100 "C NMR:δ 22.1 24.5 126.2 ppm 25.9 35.5 HNMR-B 69.5 ppm 1. BHs, THF chi-P(CJA-h→ 2. HOT, H,0 -Section 12-8 C7H12 C6H100 Section & 13C NMR: 23.8 26.5 IR-E IR-F H NMR-F 40,4 208.5 ppm IR IR-D IR-D 100 100 1704 IR 3040 IR 4000 3500 3000 2500 2000 4000 1500 1000 600 cm Wavenumber IR-A IR-E

Answer 1

IR analysis

The band at 1710 cm-1 Is for carbonyl group

The broad band at 3000 cm-1 is for hydroxyl group of carboxylic acid

Therefor compound has –COOH group

1H NMR analysis

The triplet at 0.94 ppm is for methyl proton (CH3) attached to methylene group

The multiplet at 1.59 ppm is for methylene group proton (CH2) attached to methyl and CH group

The quintet at 2.36 ppm is for CH proton attached to carboxylic acid and two methylene group, since CH proton couple with two methylene group it has quintet signal (n+1) where n = 4, and it is downfielded due to attachment of carboxylic acid

The broad signal at 12.04 ppm is for hydroxyl group of carboxylic acid

13C NMR analysis

The peak at 183 ppm is for carbonyl carbon

The peak at 48 ppm is for CH carbon

The peak at 24.7 ppm is for CH2 carbon

The peak at 11 ppm is for CH3 carbon

Based on the above observation structure of the compound is

но CH3 H3C

m/z = 116