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.

Short Answer

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Answer

The first spectrum represents alkenes.

The second spectrum represents alkanes.

The third spectrum represents aromatic hydrocarbon

Step by step solution

01

IR spectrum

The functional group present in a molecule can be determined by using its IR spectrum.

For the compounds with bond, the characteristic peak is observed at 1700cm-1 .

For alkanes, the characteristic stretch is observed at 3000cm-1 .

For alkenes the characteristic stretching peak is observed at 1500cm-1 .

02

First spectrum:Alkenes

A thin absorption at about 1700 cm-1is observed pointing downward. It is considered as carbonyl group (C=O). Alkenes (C=C) stretch is observed around 1500 cm-1. To be sure about alkene, there is unsaturated hydrocarbon at 3000 cm-1(C=C-H).

03

Second Spectrum:Alkanes

Alkanes have stretches at about 3000 cm-1 (C-H), C-H scissoring at about 1470 cm-1, methyl rock at about and long chain methyl rock at about.

04

Third Spectrum: Aromatic Hydrocarbon

Aromatic ring absorption spectra consist of series of bumps (1650 cm-1to 2000 cm-1). Number of bumps depends upon substitution of benzene ring.

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Most popular questions from this chapter

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.

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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(b) base peak at 72

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

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