Lab Conclusion: Dehydrohalogenation of 2-Bromoheptane (Elimination)

Lab Report: Dehydrohalogenation of 2-Bromoheptane (Elimination)

Total word count: 405
Time spent researching and writing: 2 hrs


Brandon Skenandore

CHEM 308-004




Answered questions from Conclusion:

  1. See attached full spectrum
  2. See attached full spectrum
  3. See attached expanded spectrum
  4. See attached expanded spectrum
  5. See attached expanded spectrum
  6. See attached expanded spectrum
  7. See emailed excel spreadsheet
  8. See attached full spectrum and emailed excel spreadsheet
  9. Ratios of products formed from different alcohols:
    Example: 1-heptene:trans-2-heptene:cis-2-heptene
    (1)Methanol: 1.2:2.4:1
    (2)Absolute Ethanol: 1.7:2.6:1
    (3)Isopropyl Alcohol: 2.2:1.7:1
    (4)2-butanol: 4:2.6:1
    (5)Isobutyl Alcohol: 2.6:2.1:1
    (6)Potassium tert-butoxide: 6:1.3:1
    As seen from the data on the spreadsheet, it can be proven that a more bulky base hinders the reaction sterically and the base removes the hydrogen that is the most accessible. Looking at the patterns of the ratios of the products formed, the reaction would follow one of two pathways. The first pathway would favor a ratio where 1-heptene was the major product, and the two isomers of 2-heptene would be formed secondary, with trans-2-heptene in greater quantity.

Alcohols that favored these product ratios were isopropyl alcohol, 2-butanol, isobutyl alcohol, and potassium tert-butoxide. The second pathway would favor trans-2-heptene, with 1-heptene in about half the quantity and the cis-2-heptene just less than half. Alcohols that did this were methanol and absolute ethanol. The methanol that was used experimentally favored this second formation of products explained. The experiment manual refers to Zitsev’s rule, where the more substituted alkene will be formed by removing the proton from the b-carbon bonded to the fewest number of hydrogen. This was the case when it came to the second scenario discussed above. However, alcohols that produced bulky bases (such as methanol and ethanol) went to the second scenario, where the trans-2-heptene was the major product. That makes sense because this is the less stable alkene that bulkier bases can produce.

  1. Most of the students reported different amounts of Sn2 products for the same alcohols used, and because of this it was necessary to go back and examine the original spectra for all the alcohols. After doing so, it appears that the bulkier the base, the less Sn2 products. For instance, methanol and ethanol had the greatest amount of Sn2 products, whereas other bases like isopropyl alcohol and isobutyl alcohol had lower amounts of Sn2 products.
  2. No, there were large amounts of starting material left. It would also stand to reason that more time would be necessary to carry out the full reaction.


Brandon Skenandore