Question

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

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

Step-by-step solution

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

IR spectroscopy

This alkene show IR peaks between$675-1000{\mathrm{cm}}^{-1}$ 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 $675-750{\mathrm{cm}}^{-1}$.
2. If the product is trans or E isomer, it shows the IR peak $920-960{\mathrm{cm}}^{-1}$.
3. If the product is disubstituted, it shows the IR peak$860-895{\mathrm{cm}}^{-1}$ .

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