Here we have to get the major eleminated product in the reaction of 1-bromobutane with potassium tert-butoxide in tert-butanol.
In the reaction of 1-bromobutane with potassium tert-butoxide in tert-butanol gives 1-butene.
Here, in the reaction of 1-bromobutane with potassium tert-butoxide, elemination reaction takes place.
Potassium tert-butoxide is bulky in size, it eliminates hydrogen from β-position and β-elimination reaction takes place. Hence, the product formed is 1-butene.
Answer:
1-butene
Explanation:
For this case we will have a very voluminous and strong base (the ter-butoxide), so the elimination would go by a E2 mechanism. In this case the attack of the base would be to the less hindered carbon. Then the Br would go at the same time to form the double bond between carbons 1 and 2. To form the 1-butene molecule (see figure).
