In the presence of sulfuric acid, this alcohol is dehydrated to form an alkene through an E1 mechanism. In the box, draw the maj
or alkene product of this reaction.
1 answer:
Answer:
figure is attached
Explanation:
When we treat alcohol with H₂SO₄ we get elimination as the major product.
As we can see in the given reaction that in step 1 the lone pair of electrons of oxygen attached to the alcohol make a bond with the hydrogen of H₂SO₄.
In the 2nd step H₂O gets detached from the parent ring which generated a positive charge on the ring.
In the 3rd step elimination of hydrogen from the carbon next to the carbonium carbon results into formation of an alkene.
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