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Question: What happens when Bromine attacks \(C{H_2} = CH - C{H_2} - C \equiv CH\)...

What happens when Bromine attacks CH2=CH−CH2−C≡CHC{H_2} = CH - C{H_2} - C \equiv CH

Explanation

Solution

We should be well aware of the fact that in such a type of reaction the amount of the attacking component decides the fate of the reaction and the product formation. Also Bromine is considered as enough of a reducing agent, when amount is sufficient it will reduce the reactant into product and results in colour formation.

Complete step by step answer:
As we all know that bromine is a strong reducing agent when compared with chlorine and weak reducing agent when compared with iodine because it can easily give up its valence electrons due to the larger atomic radius which lessens the force of attraction of nucleus on the electron as compared to chlorine.
Another factor that we should keep in mind while answering such types of questions is the amount of the attacker. In this case bromine is the one to attack, so if it is present in sufficient amount or say excess amount then the possible result will be bromination of double and triple bonds. The reaction will be as shown:
CH2=CH−CH2−C≡CH→excess  Br2CH2Br−CHBr−CH2−CBr2−CHBr2C{H_2} = CH - C{H_2} - C \equiv CH\xrightarrow{{excess\;B{r_2}}}C{H_2}Br - CHBr - C{H_2} - CB{r_2} - CHB{r_2}
In this reaction colour given by bromide ions will disappear as bromine is present in excess, otherwise no change in colour will be observed.
But we also have to consider the situation when bromine is less or limited in amount, in such condition it will be able to reduce only the double bonds and not the triple bond, so the reaction will be as followed:
CH2=CH−CH2−C≡CH→less  Br2CH2Br−CHBr−CH2−C≡CHC{H_2} = CH - C{H_2} - C \equiv CH\xrightarrow{{less\;B{r_2}}}C{H_2}Br - CHBr - C{H_2} - C \equiv CH
In this reaction, as the bromide ions are less the colour produced by them will remain as such and will not disappear.
Therefore, these two reactions can be used to distinguish alkenes and alkanes.

Note:
Some points to remember:
1. Bromide will break the double bond first and then only the triple bonds will be broken if it is present in excess.
2. Bromine is non-metallic in nature and at ordinary conditions of temperature, it evaporates as red vapours with a strong smell of chlorine.