[Ben Bob2]

Why is it necessary for the hydrogen to be removed from the oxonium ion intermediate when an alkyl halide an alcohol is treated with SOCl₂, but not when it is treated with PBr₃?

I answer by looking at electron densities based on electronegativity values:

• In the case of SOCl₂, there is a slight inductive effect due to the double-bonded oxygen on the sulfur, which makes the oxonium hydrogen more positive which reduces oxygen’s pull on the R group thereby slowing down the reaction. Once hydrogen is removed, the inductive effect works exclusively on the R group making it more attractive to the nucleophile.
• With PBr₃, there is no inductive effect because there aren't any groups as strong as or stronger than oxygen elsewhere in the molecule.  Therefore, removing the hydrogen would not be necessary.

My reasoning feels very weak but it’s all I can figure right now. It’s driving me crazy that I can’t explain this easily.
Thanks.
EDIT: corrected first sentence to alcohol, not alkyl halide...sorry

[Dan]

Why is it necessary for the hydrogen to be removed from the oxonium ion intermediate when an alcohol is treated with SOCl₂, but not when it is treated with PBr₃?

Who says it is?

[Ben Bob2]

My teacher and my text book both add pyridine to the reaction with SOCl₂ to take off the hydrogen, but this is not done with PBr₃.

[Dan]

You don't have to add pyridine with SOCl₂. If you run the reaction without pyridine it still works but you can get a different mechanism, which can change the stereochemical outcome in the case of secondary alcohols.

Some textbooks mention it, others don't. See what information you can find on the S_Ni mechanism - SOCl₂ + alcohol with and without pyridine is the classic example.

[Ben Bob2]

OK, thank you for your time.