A bromoalkane may be obtained.
Bromine reacts with an alkane to produce a colorless solution, due to the formation of a colorless alkyl halide. On the other hand, potassium permanganate (KMnO4) reacts with an alkane to form a brown precipitate of manganese dioxide.
Cl2 + 2LiBr -> 2LiCl + Br2
Any number of chemical moieties could react with alkanes to produce new compounds in a substitution reaction. For example, hydrohalic acids (HCl, HBr, HI) could react with an alkane to produce a haloalkane. Here, the halogen atom would replace one of the hydrogen atoms in the alkane. (HCl + ethane --> chloroethane) (HBr + propane --> bromopropane) This also works with other reactive species, such as: - nitric acid + alkane --> nitroalkane
When BR2 reacts with H2O, it undergoes oxidation to form HBr and HOBr. This reaction involves the transfer of electrons from BR2 to H2O, resulting in the formation of these products.
The reaction between Br2, NaOH, and H2O involves the formation of hypobromite ions (OBr-) and bromide ions (Br-) through a series of chemical reactions. Initially, Br2 reacts with NaOH to form NaOBr and NaBr. Then, NaOBr further reacts with water to produce hypobromite ions and hydroxide ions. Overall, the mechanism involves the oxidation of bromide ions to hypobromite ions in the presence of NaOH and water.
The chemical equation is:2 AlBr3 + 3 Cl2 = 2 AlCl3 + 3 Br2
The balanced equation for Cl2 + 2KBr -> 2Br2 + 2KCl is balanced as it conserves the number of atoms on both sides of the reaction. Two moles of KBr reacts with one mole of Cl2 to produce two moles each of Br2 and KCl.
An alkene can undergo halogenation when combined with chlorine or bromine in a halogenation reaction to form a dihalogenated alkane. This reaction involves the addition of a halogen atom across the double bond of the alkene.
The mole ratio of Cl2 to Br2 in the given reaction is 1:1. This means that for every 1 mole of Cl2 that reacts, 1 mole of Br2 is also involved in the reaction.
The chemical equation is:C2H2 + 2 Br2 = C2Br2H4
A methyl group can be added to an alkane through a process called alkylation, where a methyl halide, such as methyl iodide, reacts with the alkane in the presence of a strong base, such as sodium hydroxide. This reaction results in the substitution of a hydrogen atom in the alkane with a methyl group, forming a new alkylated compound.
They may be ethene, propene! ethyne, propyne and all the unsaturated hydrocarbons