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91影视

Complete the following conversion table:

v(cm-1)

4000

1800

1670

1620

400

(渭尘)

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

(渭尘)

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()incm

The units of wavenumber are cm-1.

The units of wavelength are 碌m .

02

Conversion of  wavenumber to wavelength and vice-versa

Since, 1 cm =10,000m

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 碌m corresponds to a wavenumber that is calculated as:

vcm- 1=10,000m/cm3.40m=2941.17cm-1

04

Conversion of 3.13 µm wavelength to wavenumber

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

v=10,000m/cm3.13m=3095.97cm-1

05

Conversion of 4.87 µm wavelength to wavenumber

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

v=10,000m/cm4.87m=2053.38cm-1

06

Conversion of 1800 cm-1 wavenumber to wavelength

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

=10,000m/cm1800cm-1=5.55m

07

Conversion of 1670 cm-1  wavenumber to wavelength

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

=10,000m/cm1670cm-1=5.98m

08

Conversion of 1620 cm-1 wavenumber to wavelength

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

=10,000渭尘/cm1620cm-1=6.17m

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Most popular questions from this chapter

Question: (A true story.) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp 100鈥105 掳C), marked only 鈥淪TUDENT PREP.鈥 She ran a quick mass spectrum, which is printed below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, 鈥淚 know what it is.鈥

(a) What compound is the 鈥渟tudent prep鈥? Any uncertainty in the structure?

(b) Suggest structures for the fragments at 136, 107, and 93. Why is the base peak (at m/z 57) so strong?

The mass spectrum of n-octane shows a prominent molecular ion peak (m/z 114). There is also a large peak at m/z 57, but it is not the base peak. The mass spectrum of 3,4-dimethylhexane shows a smaller molecular ion, and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57 and predict the intensities of the peaks at masses 57 and 114 in the spectrum of 2,2,3,3-tetramethylbutane.

These five structures all have distinguishing absorption in the IR. Match each structure with its characteristic absorption.

Structure 1

Structure 2

Structure 3

Structure 4

Structure 5

(a)sharp, 2254 cm-1 (b) very broad, centered about3330cm-1 (c) strong, slightly broadened, 1645cm-1 (d) broad with spikes at 3367cm-1and 3392cm-1(e) strong, sharp 1717cm-1

Question: Show the fragmentation that accounts for the cation at m/z 57 in the mass spectrum of 2-methylpentane. Explain why this ion is less abundant than those at m/z 71 and 43.

Question: Consider the following four structures, followed by mass spectral data. Match each structure with its characteristic molecular ion or fragment. In each case, give a likely structure of the ion responsible for the base peak

(a) base peak at 105

(b) base peak at 72

(c) doublet at 198 and 200, base peak at 91

(d) base peak at 91, large peak at 43

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