Final answer:
Converting an alcohol to an alkyl halide using PBr3 involves an SN2 reaction where a bromide ion replaces the OH group of the alcohol, usually with inversion of stereochemistry. The reaction is regioselective, favoring the more stable product, and stereospecific, preserving the stereochemistry of the reactant.
Step-by-step explanation:
The mechanism of converting an alcohol to an alkyl halide using PBr3 involves a substitution reaction where the bromide ion replaces the hydroxyl group of the alcohol, often with inversion of configuration at the carbon atom due to the SN2 reaction mechanism. This reaction is stereospecific and typically exhibits regioselectivity, meaning it favors the formation of the most stable carbocation intermediate during the reaction process, where applicable.
In this mechanism, the lone pair of electrons on the oxygen of the hydroxyl (OH) group of the alcohol first attacks the phosphorus atom of PBr3, displacing a bromide ion and forming a phosphite ester intermediate. This intermediate is then attacked by a bromide ion in a back-side attack, ejecting the leaving group and forming the alkyl bromide. The entire process occurs in two main steps: formation of the phosphite ester and nucleophilic substitution by bromide ion.
The process is stereospecific because it converts a chiral alcohol into an inverted chiral alkyl halide, assuming the original alcohol had a configuration to invert. Additionally, the reaction tends to be regioselective, choosing the most stable carbocation where applicable, although in the case of PBr3, the reaction proceeds via an SN2 mechanism and therefore proceeds without forming a carbocation.
Regarding elimination reactions, while they may sometimes compete with substitution reactions, the conditions under which PBr3 operates are generally favorable for substitution over elimination.