Answer :
in the presence of sulfuric acid, this alcohol is dehydrated to form a pair of alkenes through an e1 mechanism.hence we treat alcohol with H2SO, the main product is elimination. As we can see in the given reaction, the lone pair of oxygen electrons attached to the alcohol form a bond with the hydrogen of H2SO4 in step 1.
In the second step, H2O separates from the parent ring, resulting in a positive charge on the ring. In the third step, hydrogen is removed from the carbon next to the carbonium carbon, resulting in the formation of an alkene. An alkene is a hydrocarbon with a carbon-carbon double bond in organic chemistry. Alkene is frequently used as a synonym for olefin, which is any hydrocarbon with one or more double bonds. Monoalkenes are classified into two types: terminal and internal. The E1 mechanism, also known as unimolecular elimination, typically involves two steps: ionisation and deprotonation. As an intermediate during ionisation, a carbocation is formed. The carbocation loses a proton during deprotonation.
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