Bromine water is Br2 / H2O, and when one brominates an alkene with bromine water, you are correct in that you obtain a halohydrin. To obtain a dibromide product, you have to use Br2 in an inert solvent such as CCl4 or CH2Cl2.

2021年12月16日 · In the presence of aprotic solvent, the product is a vicinal dihalide, as shown here for the addition of chlorine to propene. The reaction between C=C double bond and bromine (Br 2) can be used as a test for the presence of alkene in an unknown sample. The bromine reagent is in reddish color, and the product vicinal dibromide is colorless.

2023年1月23日 · Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom becomes attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed.

2013年3月15日 · When alkenes (also known as olefins) are treated with bromine (Br 2) or chlorine (Cl 2) in an inert solvent [Note 1] such as carbon tetrachloride (CCl 4) or dichloromethane (CH 2 Cl 2), they are converted into dihalides (specifically, ‘vicinal’ dihalides since the C-halogen bonds are on adjacent carbons).

2023年8月8日 · Under UV light, BrX2 B r X 2 undergoes homolytic splitting to generate Br⋅ B r ⋅ radicals: BrX2−→hv 2Br⋅ B r X 2 → h v 2 B r ⋅. The formation of 3-bromocyclohexene is an example of substitution of alkanes, which require the free-radical mechanism:

2023年10月13日 · When suspended in tetrachloride (CCl 4), NBS reacts very rapidly with the HBr formed during the reaction mechanism to provide bromine (Br 2) which is required for the reaction to continue. Under the correct conditions, NBS provides a constant but very low concentration of Br 2 in the reaction mixture.

R Br2 R Br C C R Br R Reaction is first order in alkene R At low concentrations of Br2, rxn is also first order in Br2 At higher concentrations of Br2 in nonpolar solvents rxn is second order in Br2. Oxymercuration Pasto, JACS 1970, 92, 7480 The basic process: 反式加成 H C R C H H X–Hg–X HgX ROH H R C C H H NaBH4 OR H R H C C H H OR

2020年6月30日 · The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide. The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.

2010年4月6日 · Normally if you have a lot of Br2, you will get 1,2 dibromo butane (both on the first carbon) but in this case the answer lists only 1 Br is added. But to answer your question, the difference is in the conc of the Halogen. Low = formation of allylic radical (double bond exists) while the high = formation of vicinal halogens (no double bond lefT)

When treated with bromine (Br 2) and light (hν) alkanes are converted into alkyl bromides. Bromination of tertiary carbons is selective when in absence of any double bonds. The bromination mechanism is the same as for any other free radical halogenation and consists of three stages: initiation, propagation, and termination.