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Assume that you are carrying out the base-induced dehydrobromination of 3-Bromo-3-methyl pentane (Section 11-7) to yield an alkene. How could you use IR spectroscopy to tell which of three possible elimination products is formed, if one includes E/Z isomers?

Short Answer

Expert verified

We can distinguish between monosubstituted alkene (3-methyl pentane), E, and Z-alkenes based on IR peaks.

Step by step solution

01

Base induced reaction

Base induced dehydrobromination of 3-Bromo-3-methyl pentane gives 3-methylpentane, (E)-3- methylpent-2-ene, and (z)-3-methylpent-2-ene which is represented as:

Base induced dehydrobromination

02

IR spectroscopy

This alkene show IR peaks between675−1000 cm−1 due to bending of =C-H, but it depends on the substituents on the double bonds as follows:

  1. If the product is cis or Z isomer, it shows the IR peak 675−750 cm−1.
  2. If the product is trans or E isomer, it shows the IR peak 920−960 cm−1.
  3. If the product is disubstituted, it shows the IR peak860−895 cm−1 .

Therefore, we can distinguish between monosubstituted alkene (3-methyl pentane), E, and Z-alkenes based on IR peaks.

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