Question

$(E) - 3 - $bromo$ - 3 - $hexene when treated with $C{H_3}{O^ - }$ in $C{H_3}OH$ gives:
A. $3 - $hexyne
B. $2 - $hexyne
C. $2,3 - $ hexadiene
D. $2,4 - $hexadiene

Answer 1

Hint : We can get the product of the given reaction with the help of Saytzeff’s rule which is used in analysis of elimination reactions of organic chemistry. Elimination reaction of any halides and alcohol produces different alkenes. In such a type of reaction Saytzeff’s rule is used to predict major products. According to this rule, stable alkene is formed if the removal of hydrogen from $\beta - $ carbon which has a low number of substituents. This rule is applicable for dehydrohalogenation reactions. This problem is also based on dehydrobromination of the reactant so the product will be formed with the help of Saytzeff's rule.

Complete step by step solution:
So we can say that according to the Satzev’s rule $(E) - 3 - $bromo$ - 3 - $hexene when treated with $C{H_3}{O^ - }$ in $C{H_3}OH$ it produce $3 - $hexyne. It undergoes dehydrobromination where a molecule of hydrobromide gets eliminated from $(E) - 3 - $bromo$ - 3 - $hexene. A dehydrobromination reaction is a kind of elimination reaction where elimination of hydrogen bromide from a substrate takes place. This reaction is mainly associated with the synthesis of alkenes. The reaction of $(E) - 3 - $bromo$ - 3 - $hexene when treated with $C{H_3}{O^ - }$ in $C{H_3}OH$ is given below.


Hence option A is correct that is $(E) - 3 - $bromo$ - 3 - $hexene when treated with $C{H_3}{O^ - }$ in $C{H_3}OH$ gives $3 - $hexyne.

Note : We have approached this reaction with the help of saytzeff rule which favours the alkene with less number of hydrogen on double bonded carbon atoms.As the reaction is dehydrobromination so saytzeff rule is applicable here. We know that during the elimination reaction proton is removed from that carbon atom which have less number of substituents.Therefore $(E) - 3 - $bromo$ - 3 - $hexene when treated with $C{H_3}{O^ - }$ in $C{H_3}OH$ gives $3 - $hexyne.