Draw a condensed and a skeletal structure for each of the followi...

Question

Draw a condensed and a skeletal structure for each of the following: (a) 3,4-diethyl-2-methylheptane

Accepted Answer

Hey, everyone, let's do this problem and says, draw the appropriate skeletal and condensed structures for the following compound four ethyl, 23 dimethyl hep tane. So before we can draw the skeletal or condensed structures, we want to first get an idea of the Lewis structure, right? Just seeing the general connectivity of the compound. So first we're going to draw the Lewis structure then from that, once we see all of the atoms in the compound and their connectivity, we can draw the condensed structure. So for the condensed structure, we list the carbon atoms, the longest continuous carbon chain. And then we add the atoms that are born into each carbon, writing them next to it. So Adams directly connected two carbons, we write those atoms Usually after the carbon that it's connected to, but that's not 100%. Always the case. If we have a double bond or triple bond, we make sure that we show the carbons if it's carbon atoms that are being double bonded or triple bond into each other. And lastly, we'll go back revisit our Lewis structure and get our skeletal structure by drawing the compound in a zigzag shape and then removing all the carbons and the hydrogen. So remember that carbons are all the corners and the ends of the lines. So let's look at our compound and break this down to see how we can draw our Lewis structure. So I like to break down the name at the root. So the root is Hep tane, which we know has seven carbons, then we have dimethyl. So two ch, three groups, two methyl groups on carbons two and three. And then we have an ethyl group which is CH two ch three, right two carbon chain On Carbon four. Alright. So drawing our Lewis structure 1st Lewis structure, we're gonna draw our seven carbon chain 1234567. And let's say carbon number one is on the right hand side, we number 12, we have a methyl on carbon to three methyl on carbon three and ethel on carbon four, ch two ch three. And now we just add our hydrogen is to satisfy the carbons octet. So our end carbons on the chain will get three hydrogen, our carbons in the chain that don't have another substitution group get to hydrogen and the carbons that already have a substitution group get one hydrogen. Okay. So now let's convert this to our condensed structure. We see all the atoms that we have and how they are connected. So our longest carbon chain is the seven carbons, right? 1234567. So starting from the left hand side of our chain, we can just write the carbon and then any atoms that are attached to it. So I guess the opposite order from when we numbered in our Lewis structure, right, we started numbering right to left. Now we'll just when we do the condensed structure, we usually read it left to right. So our seventh carbon has three hydrogen, six carbon has two hydrogen, fifth carbon has two hydrogen, fourth carbon has a hydrogen and an ethyl group. So first you want to put the hydrogen and then in parentheses, include all of the atoms in the ethyl group. So ch two ch three showing that all of those atoms are a single group connected to the previous carbon. Oops I missed a hydrogen here on carbon three. So then we have carbon three with one hydrogen parentheses, ch three, close parentheses. And then carbon two also has a metal. So we have the hydrogen open parentheses, ch three, close parentheses. And then our first carbon has three hydrogen, but we can make this a little bit more condensed when we have a CH three carbon on the end. And the previous carbon also has a methyl group, we can simply add this CH three group to the tally The subscript outside of the previous method group. So we simply remove the CH three and add that subscript to outside of the methyl group on the previous carbon. So that is the condensed structure for this compound. Now, going back to the skeletal structure, oops, we want to revisit the Lewis structure, writing skeletal structure. So let's revisit our Lewis structure and draw this in a zigzag shape. So from left to right carbon is down, then we go up, down, up, down, up 1234567, all right. And then on carbon one, no substitutions, carbon two methyl group, carbon three methyl group, carbon four ethyl group making our substitue ints zigzag as well. So we want to alternate the sides that are larger substitution groups are on because of starik hindrance, right? So then our hydrogen will be on the other side that gives the substitution groups more room less interaction with each other. And notice that also applies to the direction of the ethyl group, right? We're pointing that ch three and the ethyl group towards the left, which is away from this methyl group, right? So always thinking about our stare IX. So now I'm not going to draw all the hydrogen for these because we're just going to remove them. So we just want the overall shape. We already know how many are supposed to be there. How many hydra Jin's? So drawing our zig zag 1234567, adding our methyl groups on 23 in our ethyl group on four. So that is our skeletal structure which makes a the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching.

Draw a condensed and a skeletal structure for each of the following: (a) 3,4-diethyl-2-methylheptane

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