Calculation of the yield of an organic reaction

[Neon]

I was doing the iodination of phenylalanine. I weighed out 20 g of phenylalanine, 12.3 g of iodine (0.4 equivalents) and 5.9 g of KIO₃ (0.23 equivalents). I got 4.8 g of monoiodophenylalanine. What is the yield of the reaction? I am confused about what the mechanism is and what the limiting reactant is.

[ChenBeier]

1 mol Phenyalalanine correspond to 1 mol Monoiodophenylalanine.
20 g MW = 165,19g/mol is equal 0,12 mol so theroretical the same amount of the iodine derivate should be achieved, but you get only 4,8 g.
4,8 g MW = 291,086 g/mol is equal 0,016 mol
The yeald is 0,016/0,12 = 0,137 = 13,7%

But the reactants are strange. Iodine and potassium iodate together make no sense. Better would be potassiumiodide and potassiumiodate
Because iodate can oxidised iodide to iodine.

[Neon]

But shouldn't iodine be the limiting reactant?

[ChenBeier]

The question is the mechanism and the reaction equation.

Simple Phenyĺalanine + I₂ => Phenylalanine-I + HI

Each mol phenylalanine requires the same amount I₂.

12,3 g I₂ is 0,048 mol

You have 0,12 mol Phenylalanine available.

Yes in this case iodine is the limiting reactant.

But it was enough to get 4.8 g = 0,016 mol product.

[Neon]

In this case, should we consider that 1 mole of I₂ can produce 2 moles of I, or can we say that the ratio is 1:1. Because if I consider 1:2 then I get an unusually low yield, but if I consider 1:1 the yield is close to that in the literature. Are there any recommendations on how to calculate the yield in the iodination of an aromatic ring? Are both iodine atoms from I₂ really available for iodination?

[ChenBeier]

Simple
C₆H₆ + I₂ => C₆H₅I + HI

1 mol Aromat (Benzene) + 1 mol I₂ gives 1 mol Aromatiodide and 1 mol Hydrogen Iodide.
Ratio 1:1
But only 0.5 mol I₂ = 1 mol I is used to make the product. The other half is consumed in creating HI.

But what is the problem. The yield is calculated by the mass (moles) obtained divided by the mass( moles) theoretical amount.
Both masses reactant and product are given. No need to calculate with the iodine.
In my first post the calculation is already given.