Chemical equation for the reaction of ethyne with bromine in carbon tetrachloride?

Answer 1

The question is not very specific, so there is more than just one answer, but I'm assuming you are referring to a radical bromination of an alkane (ethane) versus an electrophilic bromination of an alkene (ethene).

Br2 in the presence of a radical initiator (such as light or heat) will add to ethane to form 2-bromoethane as the major product in a radical mechanism. This goes through an initiation step (forming 2 bromine radicals), followed by propagation to the alkane (forming a secondary ethyl radical), followed by a termination step. The termination step leading to the product is one where another bromine radical joins with the ethyl radical.

In the absence of light or heat, bromine cannot react with an alkane, but it can react as an electrophile with an alkene. In this type of reaction (electrophilic addition to an alkene), the ∏-bond (double bond) on ethene attacks a bromine atom (from Br2) and kicks out a bromide (Br-). The bromine that was just added forms two bonds (one on each carbon of the double bond), giving a three-membered C-Br-C ring called a bromonium ion (since the bromine atom now has a positive charge). The bromide that left before can now attack the backside of the bromonium ion, opening the 3-membered ring, and adding anti to form a dibromoalkane (1,2-dibromoethane in this example). This reaction is stereospecific because in the major product the bromine atoms will always add anti (to the opposite side) on the alkene.

Answer 2

The chemical reaction is:
C2H2 + Br2 = C2H2Br2

Answer 3

The chemical equation for the reaction of ethyne (C2H2) with bromine (Br2) in carbon tetrachloride (CCL4) is: C2H2 + Br2 → C2H2Br2