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Convert the following infrared wavenumbers into wavelengths.

(a) 1600cm-1, typical for an aromatic

(b) 3000cm-1, typical for a saturated bond

(c) 1715cm-1, typical for a ketone carbonyl

(d) 1750cm-1, typical for an ester carbonyl

(e) 2220cm-1, typical for a nitrile

(f) 3300cm-1, typical for an alcohol

Short Answer

Expert verified

The number of waves that are present in the unit distance is termed a wave number. The wave number and wavelength are related as:

wavenumber=1wavelength

Step by step solution

01

Wave number

The number of waves that are present in the unit distance is termed a wave number. The wave number and wavelength are related as:

wavenumber=1wavelength

02

Subpart (a)

The given wave number of a radiation is 1600cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=11600cm-1=6.2510-4cm=62500A

Hence, the wavelength of the given radiation is 62500 Ao.

03

Subpart (b)

The given wave number of a radiation is 3000 cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=13000cm-1=3.3310-4cm=33300A

Hence, the wavelength of the given radiation is 33300A.

04

Subpart (c)

The given wave number of a radiation is 1715 cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=11715cm-1=5.8310-4cm=58300A

Hence, the wavelength of the given radiation is 58300 Ao.

05

Subpart (d)

The given wave number of a radiation is 1750 cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=11750cm-1=5.7110-4cm=57100A

Hence, the wavelength of the given radiation is 57100 Ao.

06

Subpart (e)

The given wave number of a radiation is 2220 cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=12200cm-1=4.5410-4cm=45400A

Hence, the wavelength of the given radiation is 45400 Ao.

07

Subpart (f)

The given wave number of a radiation is 3300 cm-1.

The wavelength of the radiation can be determined by using the following formula:

wavelength=1wavenumber=13300cm-1=3.0310-4cm=30300A

Hence, the wavelength of the given radiation is 30300 Ao.

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Most popular questions from this chapter

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: (Another true story.) A student who was checking into her lab desk found an unlabeled sample from a previous student. She was asked to identify the sample. She did an IR spectrum and declared, 鈥淚t looks like an alkane.鈥 But it seemed too reactive to be an alkane, so she did a GC鈥揗S. The mass spectrum is shown next. Identify the compound as far as you can, and state what part of your identification is uncertain. Propose fragments corresponding to the numbered peaks.

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

Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds

a. 2-methylpentane

b. 3-methylhex-2-ene

c. 4-methylpentane-2-ol

d. 2-methyl-1-phenylpropane

e. Cyclohexylisopropy; ether [cyclohexyl--O-CH(CH3)2 ]

f. CH3CH2CH2NHC(CH3)2

g.

h.3-bromo-2-methylpentane


Question: Four infrared spectra are shown, corresponding to four of the following compounds. For each spectrum, determine the structure and explain how the peaks in the spectrum correspond to the structure you have chosen

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