Why is iodine always Used in a solution containing excess I2
1 answer:
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Answer:
See explanation below
Explanation:
In this case, we are having a reaction between an anion and alkyl halide. The carbon 1 of the anion will act as nucleophile and will attack the electrophile, which is 5 carbon chain with the bromine in the third carbon.
Now, the nucleophyle is an alkyne of 3 carbon. According to the description, it should be:
CH₃ - C ≡ C⁻
And the alkyl halide is:
CH₃ - CH₂ CH(Br) - CH₂ - CH₃
And the final product after the reaction would be the following:
CH₃ - C ≡ C - CH - (CH₂CH₃)₂
However, in the attached picture you can see this better and the mechanism of reaction.
Hope this helps
