Final answer:
The best compound for radical halogenation with minimal complications due to isomeric by-products would be a symmetric molecule, such as Boron trihalides including BF3, BCl3, BBr3, and BI3. These compounds involve a direct reaction between boron and the halogens, resulting in fewer by-products. Additionally, sodium chloride and preparation of hydrogen halides using suitable acids can minimize by-product formation.
Step-by-step explanation:
The question focuses on which compound can be synthesized via radical halogenation with minimal complications from isomeric by-product formation. Radical halogenation is a reaction where a hydrogen atom in an alkane is replaced by a halogen atom through a chain reaction mechanism, often leading to a mixture of isomeric products. In general, to minimize by-products and complications, one would choose an alkane with a symmetric structure or lacking secondary or tertiary hydrogen atoms, which would otherwise lead to various isomeric halogenated products.
Boron trihalides could be a good example here. Compounds like BF3, BCl3, BBr3, and BI3 can be prepared by the direct reaction of boron with the respective halogen. Since these are symmetrical molecules due to the sp² hybridization and trigonal planar molecular geometry of boron, they are less prone to forming by-products through radical halogenation. It is also worth noting that halides like hydrogen chloride and hydrogen bromide can be directly synthesized from halogens, and haloalkanes can be produced from simpler alkanes with a high concentration of alkane and low concentration of halogen to reduce side reactions.
Sodium chloride is often the chloride of choice in many reactions due to its low cost and availability. While hydrogen bromide and hydrogen iodide typically can't be prepared with sulfuric acid, phosphoric acid can be used due to its lower oxidizing potential, leading to fewer by-products.