Question

Predict the product in the following reaction.
$H - C \equiv C - H + 2B{r_2}\xrightarrow{{CC{l_4}}} Product$

Answer 1

Hint: Bromination is a chemical reaction that involves the addition of one or more halogens to a compound or material. The pathway and stoichiometry of bromination depends on the structural features and functional groups of the organic substrate, as well as on the bromine atom. Inorganic compounds such as metals also undergo bromination.

Complete step by step answer:
The bromines add to opposite faces of the triple bond (“anti addition”). Sometimes the solvent is mentioned in this reaction. A common solvent is carbon tetrachloride (\[CC{l_4}\] ). \[CC{l_4}\] actually has no effect on the reaction, it’s just to distinguish this from the reaction where the solvent is \[{H_2}O\] , in which case a bromohydrin is formed. In the above equation, there is an ethyne as the substrate and it is reacted with two moles of bromine in the presence of a non- polar solvent, carbon tetrachloride ($CC{l_4}$ ). There is an opposite face addition of the two bromine atoms in the reaction. The reaction takes place as follows:
$H - C \equiv C - H + 2B{r_2}\xrightarrow{{CC{l_4}}}H - C{(Br)_2} - C{(Br)_2} - H$
As there is a breakage of the multiple bonds in the substrate, this reaction is an exothermic reaction which causes an immense release of heat when the bromination of ethyne takes place. The triple bond has an additional stability because of its linear geometry and bond order.

Note:
Several pathways exist for the halogenation of organic compounds, including free radical halogenation, ketone halogenation, electrophilic halogenation, and halogen addition reaction. The structure of the substrate is one factor that determines the pathway. There is an anti addition in the above reaction in order to minimize the hindrances in the substrate. The two moles of bromine molecules break down two moles of pi-bonds in the substrate.