In the dehydrohalogenation of 2-bromopentane using a bulky, sterically hindered base such as potassium tert-butoxide (), what is the major alkene formed?
Answer and explanation
Correct answer: A. Pent-1-ene (Hofmann product)
- Step 1: Steric bulk in the base dictates the regiochemical pathway of eliminations.
- Step 2: Potassium tert-butoxide is an extremely bulky base whose access to the more hindered internal methylene protons at C3 is sterically restricted.
- Step 3: Consequently, it preferentially abstracts a proton from the sterically accessible terminal methyl group at C1, yielding the less substituted alkene, pent-1-ene (Hofmann product), as the major product.
Answer
Pent-1-ene (Hofmann product).
Why not the others: Pent-2-ene is the Saytzeff product obtained using unhindered bases like ethoxide or hydroxide; Cyclopentane requires an intramolecular coupling; 2-Methylbut-2-ene involves a skeletal rearrangement not seen in .
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