Which of the underlined protons in each pair absorbs farther downfield: (a) CH,CH,CH, or CH;OCH3; (b) CH;OCH3 or CH3SCH3?

(a) Compound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning. (b) Compound B has molecular formula C8H6O2. The IR, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning. (c) Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for C and explain your reasoning.

Choose the structure that best fits this proton NMR spectrum. CH3(CO)OCH₂CH₂CH3 CH3CH₂(CO)OCH₂CH3 CH3O(CO)CH₂CH₂CH3 CH3(CO)OCH₂(CH3)2 CH,(CO)CH,CH,CH, PPM ta 0

2. Draw structures for compounds that have the proton nmrs described: a) C2H6O; one singlet b) CAH&Cl2O; two triplets c) CAHBO2; one singlet, one triplet and one quartet

( find structure with detail explanation

Given the following 13C NMR signals, construct a structure for the unknown compounds.  A. Molecular formula: C6H14O DEPT-135 (positive): 14.1 δ DEPT-135 (negative): 22.7 δ, 25.3 δ, 31.8 δ, 32.2 δ, 62.8 δ B. Molecular formula: C7H12O2 Broadband decoupled: 19.1 δ, 28.0 δ, 70.5 δ, 129.0 δ, 129.8 δ, 165.8 δ DEPT-90: 28.0 δ, 129.8 δ DEPT-135 (positive): 19.1 δ, 28.0 δ, 129.8 δ DEPT-135 (negative): 70.5 δ, 129.0 δ

(b) The 1H-NMR spectrum of compound B, C5 H10 O2, consists of the following signals: 8 1.2 (d, 6H), 2.0 (s, 3H), and 5 (septet, 1H); contains an ester. Draw the structural formula of compound B. • You do not have to consider stereochemistry. • Explicitly draw all H atoms. ୫) C n ? ChemDoodle ® Show Hint

A molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm1. Provide the structures of the starting material (1 pt) and product (2 pts) NaOH H20 C5H1,Br 6H triplet 4H quintet 1H quintet 10 8 7 6. 4 3 1 HSP-06-347 ppm

Kk.290.