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Complete the following conversion table:

v(cm-1)

4000

1800

1670

1620

400

λ(μm)

2.50

3.40

3.13

4.87

25.0

Short Answer

Expert verified

v(cm-1)

4000

2941.17

3095.97

2053.38

1800

1670

1620

400

λ(μm)

2.50

3.40

3.13

4.87

5.55

5.98

6.17

25.0

Step by step solution

01

Relation between wavelength and wavenumber

Wavenumber is calculated as,

v(cm-1)=1wavelength(λ)incm

The units of wavenumber are cm-1.

The units of wavelength are µm .

02

Conversion of  wavenumber to wavelength and vice-versa

Since, 1 cm =10,000μm

Wavenumber can be calculated by dividing 10,000 by the wavelength in microns.

03

Conversion of 3.40 µm wavelength to wavenumber

An absorption of wavelength 3.40 µm corresponds to a wavenumber that is calculated as:

vcm- 1=10,000μm/cm3.40μm=2941.17cm-1

04

Conversion of 3.13 µm wavelength to wavenumber

An absorption of wavelength 3.13 µm corresponds to a wavenumber that is calculated as:

v=10,000μm/cm3.13μm=3095.97cm-1

05

Conversion of 4.87 µm wavelength to wavenumber

An absorption of wavelength 4.87 µm corresponds to a wavenumber that is calculated as:

v=10,000μm/cm4.87μm=2053.38cm-1

06

Conversion of 1800 cm-1 wavenumber to wavelength

A wavenumber of 1800cm-1 absorbs a wavelength that is calculated as:

λ=10,000μm/cm1800cm-1=5.55μm

07

Conversion of 1670 cm-1  wavenumber to wavelength

A wavenumber of 1670 cm-1 absorbs a wavelength that is calculated as:

λ=10,000μm/cm1670cm-1=5.98μm

08

Conversion of 1620 cm-1 wavenumber to wavelength

A wavenumber of 1620cm-1 absorbs the wavelength that is calculated as:

λ=10,000μm/cm1620cm-1=6.17μm

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Most popular questions from this chapter

Question: A C-D (carbon–deuterium) bond is electronically much like a C-H bond, and it has a similar stiffness, measured by the spring constant, k. The deuterium atom has twice the mass (m) of a hydrogen atom, however.

(a) The infrared absorption frequency is approximately proportional tokm , when one of the bonded atoms is much heavier than the other, and m is the lighter of the two atoms (H or D in this case). Use this relationship to calculate the IR absorption frequency of a typical C-D bond. Use as a typical C-H absorption frequency.

(b) A chemist dissolves a sample in deuterochloroform (CDCl3) and then decides to take the IR spectrum and simply evaporates most of theCDCl3 . What functional group will appear to be present in this IR spectrum as a result of theCDCl3 impurity?

Question: (a) Use the mass spectrum to propose a molecular formula. How many elements of unsaturation are there? (b) Use the IR spectrum to determine the functional group(s), if any. (c) Propose one or more structures for this compound. What parts of the structure are uncertain? If you knew that hydrogenation of the compound gives n-octane, would the structure still be uncertain? (d) Propose structures for the major fragments at 39, 67, 81, and 95 in the mass spectrum.

Chapter-9 covered a synthesis of alkynes by a double dehydrohalogenation of dihalides. A student tried to convert trans-2,5-dimethylhex-3-yne by adding bromine across the double bond and then doing a double elimination. The infrared and mass spectra of the major product are shown here.

  1. Do the spectra confirm the right product? If not, what is it?
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For each hydrocarbon spectrum, determine whether the compound is an alkane, an alkene, an alkyne, or an aromatic hydrocarbon, and assign the major peaks above (to the left of) 1600 cm-1 . More than one unsaturated group may be present.

Question: A common lab experiment is the dehydration of cyclopentanol tocyclopentene.

a. Explain how you could tell from the IR spectrum whether your product was pure cyclopentene, pure cyclopentanol or a mixture of cyclopentene and cyclopentanol. Give approximate frequencies for distinctive peaks.

b. Explain why mass spectrometry might not be a good way to distinguish cyclopentene from cyclopentanol.

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