Question

• For IR, annotate the major peaks used in structure determination. Indicate what is causing the particular absorption.

• Annotate MS spectra

• For NMR, label each signal ( a, b, c, ...) starting from the right. Draw the structure of the compound and label the carbons and protons using separate structures. Calculate the index of unsaturation (U). Draw individual structures for each of the NMR spectra (1HNMR and 13C) and annotate each.

C,H,OBr U= 200 180 160 140 120 100 30 60 4 0 20

Answer 1

IR SPECTROSCOPY ANALYSIS

1680: C=O stretching of conjugated ketone

1560: C=C stretching

1400:C-H stretching of alkanes

Rules for finding degree of unsaturation

There are 3 basic steps in calculating the degree of unsaturation:

Step 1 – take the molecular formula and replace all halogens by hydrogens

Step 2 – omit all of the sulfur or oxygen atoms

Step 3 – for each nitrogen, omit the nitrogen and omit one hydrogen

After these 3 steps, the molecular formula is reduced to CnHm and the degree of unsaturation is given by:

U = m - (n/2) + 1

The given formula C8H7OBr will become C8H8, m=8, n=8

U = 8- (8/2) + 1 = 8 - 4 + 1 = 5

So it contains an aromatic ring (4 unsaturation) and a ketone (one unsaturation)

The structure matched with 4-bormoacetophenone.

MS SPECTRA

In the MS. the m/z will come at 197.9

Chemical Formula: CH Bro Exact Mass: 197.96803

1H NMR SPECTRA ANALYSIS

7.95 7.78 2.50 Br V 7.95 7.78

Singlet proton at 2.5ppm (CH3)

Doublet at 7.95 (two CH protons)

Doublet at 7.78 (two CH protons)

13C NMR SPECTRA ANALYSIS

129 131.5 338 | 197.0 Br 1298 135.7 135.7 26.6 127.5|| 26.6 127.5 131.5

CH3 carbon at 26.6 ppm

Carbon attached to bromine at 127.5 ppm

Two carbons ortho to bromine at 131.5 ppm

Two carbons meta to bromine at 129.8 ppm

Carbon attached to carbonyl at 135.7 ppm

Carbonyl carbon at 197.0 ppm

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