Draw a condensed structure for each of the following:
g....

Question

Draw a condensed structure for each of the following:

g. 1-bromo-1-pentyne

h. 5-methyl-2-cyclohexenol

Accepted Answer

Hi, everyone. And welcome back. Our next question says, what is the condensed structure of the given compounds? Number one is one, Chloro four, methyl pent one I and number two is 23 dimethyl cyclone one all. So we're given the Iupac name, we need to generate a condensed structure. Well, when doing this, it's a lot easier to start with a more detailed structure that shows all the bonds so that you can then figure out how to bring it back down to a condensed structure. So you can easily either drew sort of draw sort of a partially condensed structure with the bonds or I personally prefer to start with a skeletal structure. So number one, let's start with our main chain. We have a pent one I as the main chain. So it's five carbons with a triple bond from carbon one to carbon two. So triple bonds are a little tricky to draw on skeletal structures because of their linear geometry. So I like to use little dots to help me keep track. So about carbon one, carbon two, carbon three all in a straight line and then turn C one to C two into a triple bond there and then I have five carbons in the chain. So from C three, then I have 45. So there's a main chain with the triple bond of carbon one. And then in terms of substi substituents, I have a chlorine on carbon one, that's my triple bond. So just extend that line linearly to chlorine and then I have a methyl group on carbon four. So we'll stick a methyl group on there. So that's my skeletal structure showing me how everything is related. Now, let's convert that to a condensed structure. So what do we need to remember about condensed structures? No bonds are shown. So that triple bond, no lines are going to be there. You need to infer from the number of bonds. So the number of substituents and the number of hydrogens that are present will tell you what kind of bond it is. Um We show all atoms sort of in a continuous line but continuous line. And to do that, we use parentheses for both substituents, putting the substituents in parenthesis next to the atom, it's bo to and for C for carbon hydrogen groups that are repeating. So repeating ch two units, we can put in parentheses with a subscript. So let's look at what we've got here. And number one, so we have carbon one to carbon two and we've got that chlorine first. So we'll put cl and then we've got two carbons in a row that have no hydrogens. So a triple bond, they each have a bond to one other atom. So we just write CC so that's carbon one and two. And then for carbon three, that's ach two group, the next one has a methyl substituent. So there's no repeating CH two units. So I write CH two, then we have carbon four has the methyl group. So write CH and then in parenthesis CH three and then our final carbon five CH three. So we can just number it to make sure we kept track here 345 carbons in the main chain, triple bond between one and two. So no hydrogens on either of those carbons, chlorine on carbon one methyl group on carbon four. So there's our condensed structure for molecule number one. Now let's move on to molecule number two. So this one has cyclops one all as its base chain. So we have a cyclone with an alcohol substituent. So draw my little Pentagon. Now again, I'm starting with a skeletal structure, the alcohol groups on carbon one. So I'll just put it up top again. You could start at any carbon in the circle. I'm just making number one at the top one. So you can call whichever one you want number one. And then we have a methyl group on carbon two and carbon three. So just draw those methyl groups there. So we go to do the condensed structure. Well, here's an exception we can't draw this all in one line. There's no way to indicate a cyclic ring in the type of structure we drew above. So we do need to have those bonds indicated. So for cyclic ring, we'll say cycle bonds are shown. So in this case, but we don't show the bonds between the substituents. So for our top carbon number one, I'll write ch it's only got one hydrogen because it has a substituent. Oh, all is one group and then draw a line from the carbon down to the next carbon. And then there I'll put, it's got a methyl group. So it's CH and then CH three, I don't need parentheses because there's no other carbon here, it could be attached to since we do do show the ring bonds and then line down to next carbon, scroll up a little. So we don't run out of space. This one also is methyl group. So chch three for carbon three straight across to carbon four, which is just CH two, line up to carbon five, which is also CH two and then back around to carbon one. So we see that we have our 500 ring indicated alcohol group on carbon one methyl groups on carbons two and three, no parentheses needed because since we do have to indicate those ring bonds, it's clear which carbon each of the substituents belongs to. So there we go, we have our condensed structures for both of our molecules here by means of starting with a skeletal structure. But you can start. However, you like just usually you do need an intermediate step along the way before you generate that condensed structure. See you in the next video.

Draw a condensed structure for each of the following:
g. 1-bromo-1-pentyne
h. 5-methyl-2-cyclohexenol

Key Concepts

Condensed Structural Formula

Nomenclature of Organic Compounds

Functional Groups

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