(Solved): 1) Draw the curved arrow mechanism for the 1st step of the reaction. 2) Carbocation intermediate #1 ...

Questions: 1) Draw the curved arrow mechanism for the $1^{\text{st }}$ step of the reaction. 2) Carbocation intermediate $\#1$ is ( methyl $/1^{?}/2^{?}/3^{?}$ ) and (alkyl / allyl / benzyl / vinyl). 3) Carbocation intermediate \#1 will rearrange by a ( hydride / methyl) shift to form a more stable carbocation $\$2$. 4) Carbocation intermediate $\#2$ is ( methyl / $1^{?}/2^{?}/3^{?}$ ) and (alkyl / allyl / benzyl / vinyl). 5) Draw the curved arrow mechanism for the rearrangement that forms the more stable carbocation intermediate \#2. 6) Draw the curved arrow mechanism for the last step of the mechanism and also draw the final major product(s). 7) The final Product is ( chiral / achiral ) and will be optically ( active / inactive ). Note: Even though the bromide does not end up on the more substituted vinylic carbon of the initial alkene, this regiochemistry is still considered to be Markvonikov, since the hydrogen ended up the less substituted vinylic carbon!!!!