Give the product of the structure formed when 2-methyl-2-butene reacts with bromine in water
*structure of the product formed
To determine the product formed when 2-methyl-2-butene reacts with bromine in water, we need to know the reaction that takes place.
When an alkene (such as 2-methyl-2-butene) reacts with bromine in water, it undergoes an addition reaction called bromination. In this reaction, bromine adds across the double bond of the alkene, resulting in the formation of a bromoalkane.
Let's consider the reaction between 2-methyl-2-butene and bromine:
2-methyl-2-butene (C₆H₁₂) + Br₂ → Product
To determine the product, we'll add one bromine atom to each carbon atom of the double bond. In this case, the double bond is between the second and third carbon atoms in 2-methyl-2-butene.
The bromine atom adds in an anti-Markovnikov manner, meaning that it adds to the carbon atom with fewer hydrogen atoms attached. The product will be a bromoalkane with the bromine atom attached to the carbon adjacent to the methyl group.
The structure of the product will be:
CH₃-C(CH₃)CH₂Br
This is 1-bromo-2-methylpropane.
So, when 2-methyl-2-butene reacts with bromine in water, it forms 1-bromo-2-methylpropane as the product.