Elimination Reaction: OH & H Removal Explained

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In the elimination reaction of ethanol, the process specifically involves the removal of the hydroxyl (OH) group and a hydrogen (H) atom from adjacent carbons due to the E1 mechanism. The hydroxyl group is protonated by sulfuric acid, transforming it into a good leaving group that departs, forming a carbocation. Subsequently, a water molecule deprotonates a neighboring carbon, facilitating the formation of a double bond between the two carbons. This mechanism explains why only the OH and H are removed, rather than other combinations of hydrogen atoms. The focus on this specific pathway highlights the importance of the reaction's mechanism in determining the outcome.
DespicableMe
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For the elimination reaction of ethanol:
elimination1.jpg


How come only the OH and H get taken out? I understand that you have to take out one atom from each carbon, so how come we can't take out, for example, one H from the left carbon and one H from the right carbon to get?

Or is it mandatory to take out the hydroxyl from an alcohol?
 
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There's probably some reaction that can convert ethanol to vinyl alcohol (ethenol), which would subsequently tautomerize to acetaldehyde... BUT that's not what is happening here. The answer lies in the mechanism.

This is an E1 elimination. First step is the hydroxyl oxygen gets protonated by the rather strong acid present (sulfuric), which is then a pretty darn good leaving group and like a good leaving group does, it leaves (and forms a carbocation)! Next, a water molecule comes in and deprotonates the neighboring carbon, the electrons migrate between the two carbons, and whammo: double bond.
 
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