How many products are formed in the monosubstitution reaction of butane and bromine?
The compound butane is an alkane that consists of a four-carbon chain. Bromine is a halogen that exists as Br2. And when alkanes react with elemental halogens in the presence of light, one or more hydrogen atoms from the alkane are substituted with halogen atoms. And this reaction forms what is known as a haloalkane, also sometimes called a halogenoalkane. During this substitution process, the hydrogen atom that was substituted from the alkane becomes bonded to the second halogen atom, creating a hydrogen halide.
To be more specific with this example, let’s assume that one of the hydrogens attached to the terminal carbon or the one at the end is being substituted by bromine. This means that our terminal carbon will retain two bonds to hydrogen atoms as well as a carbon-carbon bond to the rest of the alkyl chain but will also have formed a new bond with bromine. And since the question is asking how many products are formed in the monosubstitution reaction, we are only representing one halogenoalkane as a substitution product, though it’s important to know that we could have represented the same reaction with the substitution taking place on a nonterminal carbon.
However, as we’ve already discussed, during this process, we also form a hydrogen halide molecule. And the hydrogen halide molecule consists of the hydrogen that was extracted from the alkane from the starting materials as well as the second bromine from the Br2. And this gives us the hydrogen halide product HBr. And with this information, we can revisit the question “How many products are formed in the monosubstitution reaction of butane and bromine?” And since we’ve already found that the products are a bromobutane molecule and a hydrogen bromide molecule, we can correctly say that the answer is two.