For the dehydration shown, use curved arrows to show the formation of the carbocation intermediate in the presence of sulfuric acid H2SO4H2SO4, then draw the structures of the minor and major products of the elimination. For the dehydration shown, use curved arrows to show the formation of the carbocation intermediate in the presence ofsulfuric acid H, SO,, then draw the structures of the minor and major products of the elimination.H2SO4Major and Minor productsНО:

We need to show the mechanism for dehydration reaction of given the molecule and predict the major and minor products. Alcohols undergo dehydration in the presence of sulfuric acid to give corresponding alkenes. Alcohols undergoes dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid such as concentrated H2SO4 or H3PO4 or catalyst such as anhydrous zinc chloride or alumina. The mechanism of dehydration of alcohol involves the following steps: Step 1: Formation of protonated alcohol. Step 2: Formation of carbocation. It is the slowest step and hence, the rate determining step of the reaction. Step 3: Formation of alkene by elimination of a proton. (Please note that the acid used in step 1 is released in step 3 and thus acts as a catalyst) For the given molecule, the nucleophilic lone pair of electrons on the O atom abstracts a proton from the acid, H2SO4 that leads to the formation of protonated alcohol. Water, being a good leaving gets removed from the pronated alcohol and the formation of stable carbocation intermediate takes place. This carbocation has two different type of H atoms; one highlighted in red (secondary hydrogen) and the other highlighted in green (tertiary hydrogen) and both can be eliminated to form alkene as shown:Depending upon the stability of alkene formed, the major and minor product is decided. The elimination of H highlighted in red results in major product while the minor product is achieved if elimination of H highlighted in green takes place. The elimination of H highlighted in red (in step-2) results in an alkene that has more hyper conjugative structures (alpha H) thereby contributes in major product (left). On the other hand, less hyper conjugative structures are observed from the alkene formed by the elimination of H highlighted in green (step-2). The alpha H are shown below: The overall reaction is represented as: