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What is the major product of the following elimination reaction and what is the mechanism involved? CH3CH2CH2Br + KOH ------>

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The major product of this elimination reaction is an alkene, specifically, propene  CH3CH=CH2 . The mechanism involved is E2  bimolecular elimination  mechanism.Here's a step-by-step explanation of the E2 mechanism for this reaction:1. KOH is a strong base, and it will take a proton  H+  from the carbon next to the carbon attached to the bromine atom  Br . In this case, it will take a proton from the second carbon in the chain  CH3CH2CH2Br .2. The electrons from the carbon-hydrogen bond will move to form a double bond between the first and second carbons in the chain.3. At the same time, the carbon-bromine bond breaks, and the bromine atom leaves as a bromide ion  Br- .4. The final product is propene  CH3CH=CH2  and the by-products are water  H2O  and potassium bromide  KBr .The overall reaction can be written as:CH3CH2CH2Br + KOH  CH3CH=CH2 + H2O + KBr
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