What is the major product of the elimination reaction of 2-bromobutane with a st

What is the major product of the elimination reaction of 2-bromobutane with a strong base?

What is the major product of the elimination reaction of 2-bromobutane with a strong base?

Step 1: Identify the starting material, which is 2-bromobutane. This molecule has a bromine atom attached to the second carbon of a four-carbon chain.
Step 2: Recognize that a strong base will be used in the reaction. Common strong bases include sodium hydroxide (NaOH) or potassium tert-butoxide (KOtBu).
Step 3: Understand that the elimination reaction (E2 mechanism) will occur, where the strong base will remove a hydrogen atom from a carbon adjacent to the carbon with the bromine atom.
Step 4: Determine which hydrogen atom will be removed. In 2-bromobutane, there are two possible adjacent carbons (C1 and C3) that have hydrogen atoms.
Step 5: When the base removes a hydrogen from C3, the bromine leaves, forming a double bond between C2 and C3, resulting in 2-butene.
Step 6: Consider the stability of the alkene formed. 2-butene is more stable than 1-butene because it is a more substituted alkene (the double bond is between two carbons that each have more substituents).
Step 7: Conclude that the major product of the elimination reaction of 2-bromobutane with a strong base is 2-butene.

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