Question Video: Identifying the Product of the Reaction of Ethyne Gas with Bromine Water | Nagwa Question Video: Identifying the Product of the Reaction of Ethyne Gas with Bromine Water | Nagwa

Question Video: Identifying the Product of the Reaction of Ethyne Gas with Bromine Water Chemistry • Third Year of Secondary School

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What is the major product of the reaction when ethyne gas is bubbled through a solution of bromine water, which is decolorized in the process?

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Video Transcript

What is the major product of the reaction when ethyne gas is bubbled through a solution of bromine water, which is decolorized in the process? (A) 1,1-Dibromoethene, (B) 1,2-dibromoethene, (C) 1,1,2-tribromoethane, (D) 1,1,1,2-tetrabromoethane, or (E) 1,1,2,2-tetrabromoethane.

To answer this question, we need to determine the name of the major product of the reaction of ethyne gas with bromine water. The ending Y-N-E tells us that ethyne gas is an alkyne, a hydrocarbon that contains a carbon–carbon triple bond. The prefix eth- tells us that ethyne only contains two carbon atoms. We can use this information to draw the displayed formula of ethyne.

Bromine water is a common reagent that consists of diatomic bromine dissolved in water. This solution has a characteristic brownish-orange color. When alkynes react with diatomic bromine, a bromination reaction occurs. A bromination reaction is a type of addition reaction where one or more bromine atoms are added to a compound. During the bromination of an alkyne, one of the 𝜋 bonds in the alkyne breaks, as does the bond between the halogen atoms. This allows for the formation of two new carbon–bromine single bonds. So we might expect the product of this reaction to look something like this.

But alkynes, like ethyne, have two 𝜋 bonds. So the second 𝜋 bond could also break, allowing for the addition of two more bromine atoms. This produces a product which contains four bromine atoms. In this reaction, this is indeed the product that will be produced. This compound is classified as a haloalkane as it contains at least one halogen bonded to a saturated carbon chain. The alkane portion is the two-carbon alkane ethane.

There are four bromine atoms bonded to the carbon chain, two on carbon number one and two on carbon number two. Therefore, this compound is called 1,1,2,2-tetrabromoethane. This compound is a liquid that ranges in color from colorless to yellow. As the bromine is used up in the reaction to produce the colorless haloalkane, the brownish-orange color will disappear. In other words, the bromine water will be decolorized.

After examining the reaction between ethyne gas and bromine water, we have determined that the major product of the reaction is 1,1,2,2-tetrabromoethane, answer choice (E).

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