Question

If 2-chloro 3-methylpentane is treated with ethanolic KOH solution. How many different alkenes would be formed via ${E_2}$ elimination reaction?

Answer 1

In this reaction, we will see the $\beta -$hydrohalogenation. This means that a HX group will undergo this reaction and leave and give us different alkenes as products.

Complete step by step answer:
-In the ${E_2}$ mechanism, a base abstracts a proton neighboring the leaving group, forcing the electrons down to make a double bond, and, in so doing, forcing off the leaving group.
-When numerous things happen simultaneously in a mechanism, such as the ${E_2}$ reaction, it is called a concerted step.
-Now let us see the particular compound given to us:

So, now with the reaction with alcoholic KOH, if favours $\beta -$dehydrohalogenation.
-So now let us understand what is $\beta -$dehydrohalogenation. The $\beta -$positions to the halide group in the compounds are those carbons just next to them. The carbon containing Cl is $\alpha$ carbon while the other two carbon on its left and right are $\beta -$carbons.
-Now, the elimination of -HX group should be from any of these $\beta -$ carbons.
But how to decide?
-Thus, we follow Saytzeff's rule here.
-It states that more substituted alkene will always favour hydrogen elimination from that particular $\beta -$ carbon which has a lesser number of hydrogens.
-Then, the products formed are:

-According to Satyzeff rule, we can see that in the major product the elimination occurred from the $\beta -$carbon containing a lesser number of hydrogens.
-Since, this is an ${E_2}$ reaction, so no question of formation of carbocation. The intermediates form products.

Therefore, the correct answer is 4.

Note: Always follow Saytzeff rule when in ${E_2}$ elimination. In ${E_2}$, the steps are all concerted while for ${E_1}$, the steps involving carbocation formations are more stepwise. It depends more on the stability of the carbocation but that is not the scene for ${E_2}$ eliminations.