
If 2-chloro 3-methylpentane is treated with ethanolic KOH solution. How many different alkenes would be formed via \[{E_2}\] elimination reaction?
Answer
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Hint: 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.
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.
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