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Question
The correct decreasing order of stability of alkene formed from the following alkyl halides according to the β− elimination reaction with alcoholic KOH will be:
(i) 
(ii) CH3−CH2−Br
(iii) CH3−CH2−CH2−Br
The correct decreasing order of stability of alkene formed from the following alkyl halides according to the β− elimination reaction with alcoholic KOH will be:
(i)
(ii) CH3−CH2−Br
(iii) CH3−CH2−CH2−Br
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Solution
The correct option is D (i) > (iii) > (ii)
There are two types of elimination reaction,
1. α− Elimination reaction
2. β− Elimination reaction
In α− Elimination reaction, the leaving group and the proton are removed from the same α− carbon which leads to reactive carbene intermediate.
In β− Elimination reaction, the leaving group and the proton are removed from the adjacent atoms i.e., leaving group from α− carbon and proton from β− carbon. It leads to the formation of alkene compounds.
According to saytzeff rule, more substituted alkenes are more stable.
β− Elimination reaction gives more substituted alkenes as the major product.
In given alkyl halides, all will undergo β− Elimination to give alkene compounds.
Alkene formed from compound (i) has 2 methyl substituted group, (iii) has 1 methyl group and (ii) has 0 methyl substitutes.
Thus, the correct order of stability is,
(i) > (iii)> (ii).
There are two types of elimination reaction,
1. α− Elimination reaction
2. β− Elimination reaction
In α− Elimination reaction, the leaving group and the proton are removed from the same α− carbon which leads to reactive carbene intermediate.
In β− Elimination reaction, the leaving group and the proton are removed from the adjacent atoms i.e., leaving group from α− carbon and proton from β− carbon. It leads to the formation of alkene compounds.
According to saytzeff rule, more substituted alkenes are more stable.
β− Elimination reaction gives more substituted alkenes as the major product.
In given alkyl halides, all will undergo β− Elimination to give alkene compounds.
Alkene formed from compound (i) has 2 methyl substituted group, (iii) has 1 methyl group and (ii) has 0 methyl substitutes.
Thus, the correct order of stability is,
(i) > (iii)> (ii).
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