Original post by BBeyond
When doing an elimination reaction, for example 2-bromo-3-methylbutane reacting with ethanolic potassium hydroxide, how do we know where the double bond will form? Surely it could form either side of the bromine bond?
Depends on the stability of the intermediate ...
Original post by charco
Depends on the stability of the intermediate ...
Ah so the double bond is formed with the most stable carbon? (e.g. the carbon bonded to the most carbons?)
Original post by BBeyond
Ah so the double bond is formed with the most stable carbon? (e.g. the carbon bonded to the most carbons?)
Both possible products will be formed but the major product is the one with the most stable intermediate (usually)
In this mechanism the intermediate is a carbanion and hence DE-stabilised by +I groups such as alkyl.
You'll got both products, the yield of each product will depend on the stability of the carbocation intermediate.
Original post by BenLynch9
You'll got both products, the yield of each product will depend on the stability of the carbocation intermediate.
It's not a carbocation, see post above ...
In the example you gave, 2-bromo-3-methylbutane, the major product would be 3-methylbut-1-ene, the Seytzeff product.
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