Question

The five compounds below all have a molecular weight of 148 g/mol. Start by drawing in all implied hydrogens on the printed structures and then add up the chemical formula and molecular weight for each to make sure you have everything correct. The spectral data (EI-MS, IR, 1H NMR, and 13C NMR) for each compound are grouped together (1 compound per page). For 1H NMR, numerals underneath the spectrum represent integral values.
OH
**Compound A has spectral data 2, Compound B has spectral data 3, Compound C has spectral data 5, Compound D has spectral data 4, Compound E has spectral data 1.

**Looking at 1H NMR spectra for each compond, labelling the multiplicities (corresponding to CH2 groups only for question 8) and signals asking in question 9 on the graph.

Answer 1

HO H H н он O. но 7.90 .51 1.51 O H H 098 7.40 H NMR 0.88 2.96 2.96 0.09 7,87 tripletsextet 7.40 Two methylene groups (CH2) a and b a is sextet (since adjacent H s are 5, (n+1) multiplicity rule (5+16) b is triplet 11 10 875 4 32 3.45 3.45 2.45 НН 121 7.24 7.27 1.21 singlet 7.22 7.24 1.21 H NMR 7.27 Two methylene groups (CH2) a and b a is quartet (since adjacent H s are 3, (n+1) multiplicity rule (3+1-4) b is singlet, since no H s present on the adjacent carbons quartet 11 10 9876 532 6.78 6.31 2.25 0.79 467 407 543 0.90 2.25 H NMR (1.0-7.5 ppm) One methylene group (CH2) a with doublet multiplicity (since adjacent H s are 1, (n+1) multiplicity rule (1+1-2) doublet

6.82 5.32 он 1.83 8.92 4.95 H 1.88 aH 1.55 8.92 H 1.65 284 each proton on a.b and c are sextets 6.02 2.87 Three methylene groups (CH2) a, b and c with each proton on a,b and c are sextets H NMR 2.48 2.48 но 248 H 7.21 2.30 24810.38 There is no CH2 group present in compound E 7.21 2.30 2.48 2.48 Locate and label the signals of the following C-H groups (2 points each). I. Aldehyde Peak a is aldehydic proton peak II. Singlet representing 2 equivalent aromatic C-H groups Peak b represents 2 equivalent aromatic C-H proton groups 0 2.48 721 10.30 2.30 7.21 H NMR 6 3 12 11 10 9876 5 432