91Ó°ÊÓ

This is an elimination reaction (E1). In the presence of acid the OH group is protonated to form${\text{OH}}_{\text{2}}^{\text{+}}$ which is a better leaving group. When${\text{OH}}_{\text{2}}^{}$ leaves a secondary carbocation is formed. In the next step double bond is formed when the proton leaves from the adjacent carbon.

The first product is formed by elimination reaction. The benzylic OH group is protonated to form ${\text{OH}}_{\text{2}}^{\text{+}}$which is a better leaving group. The reaction proceeds with the elimination of the leaving group and loss of proton to give .the double bond. The second product is formed when a proton is lost from the adjacent carbon. The third product is formed through ring expansion as shown in the mechanism below.

This is an elimination reaction. The OH group is protonated and gets converted to a better leaving group. Now, the proton can be lost from two positions giving two different alkenes as given in the mechanism below.

The first product is formed through ${\text{S}}_{\text{N}}\text{1}$ mechanism where bromine is substituted by the methoxy$\left({\text{-OCH}}_{\text{3}}\right)$ group. For the second product the bond in the cyclobutane cleaves leading to ring expansion and formation of a carbocation. The methoxy anion attacks at the carbocation to form the required product. Formation of the third product occurs when elimination reaction takes place. The proton is abstracted by a base and$\left({\text{-OCH}}_{\text{3}}\right)$ group is eliminated to give the required alkene.