I want to ask a question about the hydrolysis of haloalkanes to form alcohols.
I have been presented with a flowchart diagram of possible reactions of haloalkanes below:
Now, I have been presented with the question below:
I have identified the reagent I relevant to the question as NaOH (aq)
First, I need to compare 1-bromobutane and 2-bromo-2-methylpropane.
I can see that there are 3 alkyl groups in 2-bromo-2-methylpropane and only 1 alkyl group donating electrons to the Carbon atom attached to the Bromine.
So, during the reaction, the carbocation intermediate formed by reaction of NaOH with 2-bromo-2-methylpropane will form be more stable than the carbocation intermediate formed with 1-bromopropane (application of Markownikoff's rule) as alkyl groups are electron-donating groups and the three methyl groups will stabilise the positive charge on the carbon atom to a greater extent in 2-bromo-2-methylpropane.
However, I am slightly confused.
Why does this mean that the rate of reaction with 2-bromo-2-methylpropane is faster than the reaction with 1-bromopropane?