A chemist heats 7.41 g of diethyl ether with excess hydrogen iodide until both ethyl groups are…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A chemist heats of diethyl ether with excess hydrogen iodide until both ethyl groups are converted to iodoethane. The yield is . Use for the ether, for iodoethane and for hydrogen iodide. Find the mass of iodoethane and the least mass of hydrogen iodide that the reaction needs.
Need a reminder? The key ideaShow
HI cleaves the C-O bond. Iodide attacks the less crowded carbon (or the tertiary one). An aryl C-O bond stays, so an aryl alkyl ether gives a phenol and an alkyl iodide. Excess HI turns an alkyl ether into two alkyl iodides. The net dipole of two bond dipoles at angle is .
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. of ether, yield.
Step 1. Ether. .
Step 2. Iodoethane. Two for each ether: . At : .
Step 3. Hydrogen iodide. Two for each ether: .
Check. The iodine is the heavy part of the product, which is why the mass of iodoethane is so much larger than the mass of the ether.
Answer: of iodoethane and at least of hydrogen iodide.
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