Question
Asked Aug 1, 2019
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Identify the structures of isomers H and I (molecular formula C&HuN).
a. Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm
2H
5 H
2H
2H
0
2
3
4
5
6
7
ppm
b. Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm1
3 H
5 H
1 H
2 H
3
2
8
5
4
7
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Identify the structures of isomers H and I (molecular formula C&HuN). a. Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm 2H 5 H 2H 2H 0 2 3 4 5 6 7 ppm b. Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm1 3 H 5 H 1 H 2 H 3 2 8 5 4 7

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Expert Answer

Step 1

(a)

Analyzing the IR Spectrum:

The absorption peak at 3365 cm-1 and 3284 cm-1 corresponds to N–H stretching. The absorption peak at 3026 cm-1 and 2932 cm-1 indicates C=C stretching frequency. The absorption peak at 1603 cm-1 and 1497 cm-1 indicates the presence of benzene ring.

Step 2

Analyzing the NMR Spectrum:

There are four signals in the spectrum. This implies that the compound has five different non-equivalent proton environments. The multiplet about 7.3 ppm indicates a monosubstituted benzene ring. A triplet about 1.0 ppm comes from amine having two protons. A triplet about 2.7 ppm indicates the methyl group. One triplet about 3.0 ppm indicates the methyl group near to benzene ring.

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(t, 3H) H2 C NH2 (s, 2H) C H2 (t, 3H) (m, 5H)

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Step 3

(b)

Analyzing the IR Spectrum:

The absorption peak at 3367 cm-1 and 3286 cm-1 corresponds to N–H stretching. The absorption peak at 3027 cm-1 and 2962 cm-1 indicates C=C stret...

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