Video Transcript
During the halogenation of ethyne, ethyne gas is passed through bromine water dissolved in carbon tetrachloride. The orange color of the bromine is lost as the addition reaction occurs. What is the final product of this reaction?
To answer this question, we need to determine the product of the halogenation of ethyne using bromine water dissolved in carbon tetrachloride. The reaction which occurs is an addition reaction. An addition reaction is a type of chemical reaction where one or more molecules combine to form one larger molecule without any byproducts forming. Halogenation is a specific type of addition reaction which involves the addition of one or more halogens.
Two halogens are mentioned in the question. Bromine water consists of diatomic bromine dissolved in water. And carbon tetrachloride is a colorless liquid. Looking at the answer choices, we can see that all of the structures contain bromine atoms but none of the structures contain chlorine atoms. So this halogenation involves the addition of bromine atoms. So we can conclude that diatomic bromine is a reactant and carbon tetrachloride is an unreactive solvent.
Now that we can recognize that diatomic bromine is the reactant and bromine atoms will be added during this reaction, let’s take a look at the other reactant ethyne. Ethyne is a two-carbon alkyne. So the two carbon atoms must be joined by a carbon-carbon triple bond. The carbon-carbon triple bond is fairly reactive because of the two 𝜋 bonds it contains. During an addition reaction, one of the 𝜋 bonds can break, allowing for the addition of two new substituents.
During the reaction of ethyne with diatomic bromine, one of the 𝜋 bonds between the two carbon atoms along with the bond between the two bromine atoms will break, allowing for two new bonds to form between the carbon atoms and the bromine atoms. This produces 1,2-dibromoethene. But this molecule contains a 𝜋 bond which can react with diatomic bromine. Once again, a 𝜋 bond between the two carbon atoms will break along with the bond between the two bromine atoms. This allows for the formation of two new carbon-bromine bonds. This produces the compound 1,1,2,2-tetrabromoethane.
Another point we are told in the question is that the orange color of the bromine is lost as the addition reaction occurs. Bromine water has a characteristic orange color. But 1,1,2,2-tetrabromoethane ranges in color from colorless to pale yellow. So when ethyne is bubbled through the bromine water, the bromine reacts with the ethyne producing a colorless substance. So the orange color of the solution will become lighter and lighter as the reaction proceeds. And when all of the bromine has reacted, the orange color will be completely lost.
Returning to the question, we have determined that the product of the halogenation of ethyne is 1,1,2,2-tetrabromoethane. This compound contains four bromine atoms. As such, we can eliminate answer choices (C), (D), and (E), as these structures do not contain the correct number of bromine atoms. In the product, the two carbon atoms are joined by a single bond. As such, we can eliminate answer choice (B), which shows the two carbon atoms joined by a double bond. So answer choice (A) must be the correct skeletal structure of 1,1,2,2-tetrabromoethane. Therefore, the halogenation of ethyne with bromine water dissolved in carbon tetrachloride produces the product shown in answer choice (A).
