The SN1 solvolysis of one of the following alkenyl bromides is in MeOH approximately 100000 times faster than the other one. Which is the more reactive one? Draw the mechanism including the transition state, determine the rate determining step and draw the reaction product of each compound.
Now what is apparently important here is the rotation of the substituent in each relative to the aromatic ring. In the second compound (with two methyl groups), the rotational profile shows destabilized molecule except when 90 degrees. That means the rotation in the second molecule helps with the leaving of Br and this (rate determining) step is much faster than when there are no methyl groups (first compound). Is that a complete explanation or I am missing something important? Also is there anything special about the mechanism of the two reactions and the products in each case?