To determine the number of monobromo derivatives formed by alkanes with the molecular formula \(C_5H_{12}\), we start by identifying the structural isomers of \(C_5H_{12}\). These are:
n-Pentane: A straight-chain alkane. It can form three distinct monobromo derivatives by substituting a hydrogen from each of the different types of carbon (primary, secondary, tertiary):
Substitution at one of the four primary carbons yields one derivative.
Substitution at the secondary carbon results in another derivative.
Isopentane (2-methylbutane): A branched isomer, giving three monobromo derivatives:
Substitution at one of the six equivalent primary carbon leads to one derivative.
Substitution at the secondary carbon adjacent to the terminal carbon results in another distinct derivative.
Substitution at the central tertiary carbon yields a third distinct derivative.
Neopentane (2,2-dimethylpropane): With a highly symmetrical structure, this isomer forms only one type of monobromo derivative due to all 12 hydrogens being equivalent.
In total, the sum of all derivatives is 8. Thus, the computed solution is:
