Question

How many different isomers of an alkyl bromide, upon treatment with ${C_2}{H_5}ONa$ in ${C_2}{H_5}OH$ and give $3 - methyl - 2 - hexene$ as one of the elimination product by $E2$ mechanism? Correct answer is ‘ $6$ ’. Can you explain this answer?

Answer 1

Hint: here in the above question it is given that there are different isomers of alkyl bromide which we have to react with ${C_2}{H_5}ONa$ And in the presence of ${C_2}{H_5}OH$ By this reaction we get the product as $3 - methyl - 2 - hexene$ . So, in this question we have to find the number of isomers used in alkyl bromide to form $3 - methyl - 2 - hexene$ .

Complete Step by Step Solution:
Firstly we write all things which is given in the question,
Reactants, first reactant = alkyl bromide
Second reactant = ${C_2}{H_5}ONa$
In presence with the reaction got completed = ${C_2}{H_5}OH$
As a product we get, = $3 - methyl - 2 - hexene$
As by getting full data the reaction will be,
\[R - Br + {C_2}{H_5}ONa\xrightarrow{{{C_2}{H_5}OH}}3 - methyl - 2 - hexene\]
As in the above reaction we know that we have to use an elimination reaction, then we have to perform an elimination of Halogen with the hydrogen atom which is present on $\beta - Carbon$ . Hence, from which we get that there will be 6 isomers of alkyl bromide is there because other halogen are eliminated and hydrogen are bonded over the last and a double bond can form and to balance the reaction we get to add more to the alkyl bromide.
Hence, we can say that the number of isomers of alkyl bromide will be ‘ $6$ ’.

Note: As we know that this is a tricky question used mainly in many competition exams as from which, in these types of questions we only have to focus on balancing the reactions by which we can easily get the correct answer as per also the correct bonding positions too.