[ih]

I need to know how much reactant to use. I have the following reaction:
2CH₃I + 2CH₃CH₂CHO^-CH₂CH₃ 2CH₃CH₂CH(OCH₃)CH₂CH₃ + I₂
I need 10g of the organic product.
10g/(102g/mol)=0.098mol
for 50% loss would need 0.196mol
.196mol x 142g/mol ≈28g
therefore with density of 2.28g/mL ≈ 13mL

Is how I have done this correct?

[Hunter2]

Sounds good, if you mean the methyl iodide. But why is your yield only 50%.

[magician4]

2CH₃I + 2CH₃CH₂CHO^-CH₂CH₃ 2CH₃CH₂CH(OCH₃)CH₂CH₃ + I₂

sure?


regards

Ingo

[Hunter2]

Iodine is wrong. Iodide looks better.

[ih]

Yeah I did mean the methyl iodide. I'm doing 50% yield because I have to recover it from an ether and other impurities from a previous reaction so I'm accounting for a big loss in this. I'll change it to Iodide too.

Thanks guys