What is the role of red phosphorus in hvz reaction? Initially I think that red phosphurus is nothing but a catalyst. But it is not a catalyst.
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$\begingroup$ chemistry.stackexchange.com/a/90666/102629 $\endgroup$– cngzz1Commented Jan 13, 2021 at 3:51
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TL;DR: When $\ce{PBr3/PCl3}$ is not available as catalyst, red phosphorus is used which is reacted to bromine or chlorine to form the catalyst in situ.
Long answer:
Hell–Volhard–Zelinsky halogenation reaction halogenates carboxylic acids at the α-carbon. The reaction is initiated by addition of a catalytic amount of $\ce{PBr3}$, after which one molar equivalent of $\ce{Br2}$ is added. $\ce{PBr3}$ replaces the carboxylic -OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with the $\ce{Br2}$ to brominate a second time at the α position.
What can be done if $\ce{PBr3}$ is not available?
You can use red phosphorus. It is reasonably air stable (whereas white phosphorus combust in air almost spontaneously). Phosphorus reacts with bromine to give phosphorus tribromide which now becomes the catalyst for the reaction.
$$\ce{2P_{(red)} + 3Br2 -> 2PBr3}$$
Why exactly $\ce{PBr3/PCl3}$ is needed and why an extra molar equivalent of $\ce{Br2/Cl2}$ is used?
When left out in the air, $\ce{PBr3}$ will readily combine with any water vapor present to form a new $\ce{P-O}$ bond along with $\ce{HBr}$. The strength of the $\ce{P-O}$ bond [about $\pu{140 kcal/mol}$] makes $\ce{PBr3}$ very useful for swapping out $\ce{C-O}$ bonds for $\ce{C-Br}$ bond and that's how carboxylic acids gets converted to acyl bromides.
However, $\ce{PBr3}$ by itself is not sufficient to perform the HVZ reaction. A molar excess of $\ce{Br2}$ is added to the already present $\ce{PBr3}$, which perform the halogenation of the alpha-carbon. That's why the catalyst is always mentioned as ($\ce{PBr3 + Br2}$).
References
- https://en.wikipedia.org/wiki/Hell%E2%80%93Volhard%E2%80%93Zelinsky_halogenation
- https://www.organic-chemistry.org/namedreactions/hell-volhard-zelinsky-reaction.shtm
That generates another question: Can we use phosphorous triiodide in Hell-Volhard-Zelinsky reaction?
answered Jan 13, 2021 at 5:48
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$\begingroup$ I see it written at some places like "catalytic amount of red phosphorous" or other words which suggest "catalyst". I know you've mentioned it as a reagent in your answer, but can you please confirm whether using the word catalyst in this case for red phosphorous is wrong or not? $\endgroup$– AshishCommented Jul 19, 2021 at 6:06
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$\begingroup$ @Ashish Yes, you are correct with the usage of term "catalytic amount". Any substance or mixture for use in chemical analysis and that causes chemical reaction is called reagent. That also includes catalyst. In this context, "catalyst" should be a better term because it is causing catalysis. I have edited appropriately. $\endgroup$ Commented Jul 19, 2021 at 6:26