Question

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl phosphine would you need to prepare 220 mg of benzyltriphenylphosphonium chloride?

Answer 1

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,

`(353420 mg of Phosphonium is formed by reacting)/(220 mg of Phosphonium will be formed by)` =
`(126580 g of benzyl chloride)/(X)`

Solving for X,
X =
`(220 mg . 126580 mg)/(353420 mg)`
X = 78.79 mg

For PPh₃:

`(353420 mg of Phosphonium is formed by reacting)/(220 mg of Phosphonium will be formed by)` =
`(262290 g of PPh3)/(X)`

Solving for X,
X =
`(220 mg . 262290 mg)/(353420 mg)`
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when
`(95 percent)/(100 percent)` =
`(220 g)/(X)`
Solving for X,

X =
`(220 . 100)/(95)` = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when
`(220 mg phosphonium required)/(231.57 mg require)` =
`(78.79 g of benzyl chloride)/(X)`
Solving for ,

X =
`(231.57 . 78.79)/(220)` = 82.93 mg of Benzyl chloride

when
`(220 mg phosphonium required)/(231.57 mg require)` =
`(163.27 g of PPh3)/(X)`
Solving for ,

X =
`(231.57 . 163.27)/(220)` = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.