When ethyl alcohol, CH3CH2OH, dissolves in water, how many hydrog...

Question

When ethyl alcohol, CH3CH2OH, dissolves in water, how many hydrogen bonds are formed between one ethyl alcohol molecule and surrounding water molecules? Sketch the hydro- gen bonding interactions. (Hint: Add lone pairs of electrons to the structure before drawing hydrogen bonds.)

Accepted Answer

Hello. Everyone in this video we're talking about methanol and properly molecules and we're seeing how the interactions of hydrant bonding works here. So we go ahead and schedule those hydrogen bonding interactions and we are given a hit in this problem that before drawing any hydrant bonds will have to go ahead and add the lone pairs on our auction atoms. So first let's go ahead and drop the structure for methanol. So methanol has our method group as well as our alcohol group. So we have the prefix of meth meaning we have just one carbon and then our suffix of O. So O. L. And that's just because we have an alcohol group. So then we have our methyl group here. So H three C connected to an auction and hydrogen that's our alcohol. And because auction likes to have two bonds and two lone pairs we already have the two bonds. So adding those two lone pairs. That's methanol here now for propranolol. So we have the prefix of probe that's going to mean that we have three carbon. So we have H three C. Ch two ch here, those three carbons and our alcohol groups of H. Again we have two bonds. So adding the necessary two lone pairs. So as we call the hydrant bonding means that we're dealing with are hydrogen atoms bonding between a very electro negative atom such as nitrogen often and flooring. So when it has those bonds that's what we call a hydrogen bonding. So we said earlier that there's going to be too long pairs on the oxygen atom of propranolol. And because of that it means that the molecules can go ahead and set to bonds just from those two lone pairs. We all see for O. H. That we have a hydrogen bond to our auction. So because we see that O. H bond there's going to be a total of three pro bono molecules that can form. So now start shooting out our interactions because we already dropped the lone pairs on each of our molecules here, we'll go ahead and do this in green. So starting off with our proposal, we have H three C CH two, CH 2 and then our oxygen and hydrogen of course with our long pairs here now for the methanol groups, I go ahead to create this interaction. We'll do we'll do it in blue and as well as showing the hydrogen bonding interactions with a dotted blue line. So the auction is going to form a bond with our hydrogen of our methanol and then we'll have comparison oxygen and the methyl group. Same thing with the lone pair in the bottom this dotted line with our hygiene and then we'll have our auction number two lone pairs and another bond to our method group. Now for the auction instead of dealing with another hygiene were being bonded with the auction now so then we'll have our dash line here with the auction and its lone pair, two lone pairs. And then we have our methyl group and the other end being a hydrogen. So let's see here that we have had made three interactions total.

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