Name all unbranched ether and alcohol isomers with formula C 5 H 12 O and write their structural formulas.

Hi, everyone. Welcome back. Our next question says, draw the line formula and name all unbranched ether and alcohol structural isomers with a molecular formula of C 4h 10 0. So we're looking at both ethers and alcohols and looking for unbranched structural isomers. So we have four carbons. So let's think about possible ether arrangements. This could have recalling the ether is in the format of R zero R. So our oxygen is in the middle of that carbon chain. So we can have our oxygen after carbon one. So COCCC, so let's think about whether this has the right amount of hydrogens for our formula. So we have four carbons and we would have three hydrogens on each of our terminal carbons and two hydrogens, two hydrogens on each of the internal ones. And this will give us a total of six plus 410. So that is fine. We could also have our oxygen being in the middle. So CCOCC, that'll give us the same number of hydrogen. So we're good there. So let's draw these line structures. So we would have for a first one, we have our carbon leading to a bond to oxygen and then three more carbons count to make sure we're correct. 1123, no other groups, substituent groups or anything like that. So that's all we need for our line structure. For our second possible arrangement, we have 12 carbons and then our oxygen and then two more. So let's think about the names of these two ethers. We're called up to name an ether. We name each of the groups on either side, putting these names with spaces in between them and then the word ether. So in number one, we have methyl prop ether and number two are the oxygens in the middle. This is two ethyl groups. So we have die ethel either. So we've got our names and our line structures of our possible ethers, the unbranched ethers. But we need to go on and do this for the alcohol. So we recall that our alcohol as a format of roh so our oxygen has to be a terminal oxygen. So we have two possible arrangements here. We have our four carbons and then an an oh group at the end or we might have our oxygen as a substituent group. So four carbons with an oxygen as a substituent group on carbon two or three, it wouldn't be one or four because then that would be identical to the structure. We just drew, let's draw line structures for both of those arrangements. For our first arrangement, we have one s 1234 carbons, 1234. And then our bond too, an alcohol group at the end. And then for our second structure, we have 1231234 carbons with an oxygen group coming up from carbon two or an alcohol group I should say. So let's give them their names. In this case, the first case, the, the alcohol groups at the end. So we have one butanol because it has four carbons. And even though we drew the ox, the alcohol group at the end, we know that we start with numbering from the carbon that gives our substituent group its lowest number. So one butanol in our second molecule, that al that alcohol group is on the second carbon. So it will be two butanol. So those are our two possible arrangements for alcohols. And so we have all unbranched ether and alcohol structural isomers with this merle formula of C 4h 10 0. We have methyl copy ether, diethyl ether, one butanol and two butanol. See you in the next video.