Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Each C-O bond in an ether has a bond dipole of about . The two bonds meet at the oxygen atom at an angle of in methoxymethane. Find the net dipole of the two bonds. Then find the net dipole if the angle were , as in water. Then say what the answer for a straight (180) arrangement would be.
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 .
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. for each bond.
Step 1. At . .
Step 2. At . .
Step 3. Straight. .
Check. The measured value for methoxymethane is , close to the first result. The bent shape matters: a linear C-O-C would have no dipole, as in carbon dioxide.
Answer: , and .
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