0ximene and myrcene, two hydrocarbons isolated from alfalfa...

0ximene and myrcene, two hydrocarbons isolated from alfalfa that have the molecular formula $C_{10}H_16$, both yield 2,6-dimethyloctane when treated with $H_2$ and a Pd catalyst. 0zonolysis of oximene forms ($CH_3)_2C$ = 0, $CH_2$ = 0 , $CH_2(CH0)_2$, and $CH_3C0CH0$. 0zonolysis of myrcene yields ($CH_3)_2$C = 0, $CH_2$ = 0 (two equiv), and $HC0CH_2CH_2C0CH0$. Identify the structures of oximene and myrcene.

0ximene and myrcene, two hydrocarbons isolated from alfalfa that have the molecular formula $C_{10}H_16$, both yield
2,6-dimethyloctane when treated with $H_2$ and a Pd catalyst. 0zonolysis of oximene forms ($CH_3)_2C$ = 0, $CH_2$ = 0 , $CH_2(CH0)_2$, and
$CH_3C0CH0$. 0zonolysis of myrcene yields ($CH_3)_2$C = 0, $CH_2$ = 0 (two equiv), and $HC0CH_2CH_2C0CH0$. Identify the structures
of oximene and myrcene.

Transcript

00:02 This is the answer to chapter 12, problem number 55, from the smith organic chemistry textbook.

00:10 And in this problem, we are told that oxamine and myersene, myrosine, two hydrocarbons isolated from alfalfa.

00:23 We're told that they both have the molecular formula c10h16.

00:27 They both give 2 -6 dimethylactane when treated with hydrogen gas and palladium.

00:33 And then we're told about the oz analysis products of each of them.

00:39 And so we are asked to identify the structures of oxymine and merced.

00:46 And so this is a good problem because this gives us lots of pieces of information, certainly enough information to determine these two structures.

00:58 And we just need to know basically what to do with the information that we've been given.

01:04 And so the very first thing to do, i think, when we're given a formula and asked to find a structure, the very first thing to do is always to determine the hydrogen deficiency index, the hdi.

01:20 And so that's going to tell us how many double bonds or rings there are.

01:24 In these molecules.

01:26 And we can see that there's not going to be any rings, right? because when the molecules are hydrogenated, they both give 2 -6 dimethyl octane.

01:36 And so we know it's going to be a matter of how many double bonds are in these molecules.

01:42 And so that's what the hdi is going to tell us.

01:45 And so remember, hdi is, well, for a molecule where it's just carbons and hydrogens, it's going to be two times the number of carbons plus two minus the number of hydrogens, and then all of that is going to be over two.

02:05 And so here we have 22 minus 16 is going to be six, over two is going to be three.

02:14 So each of these molecules is going to have three double bonds in it.

02:21 Okay.

02:25 So that's helpful.

02:28 Now we just need to determine where these three double bonds are going to be.

02:33 And so we can start with oxamine.

02:40 And so for oxamine, we're given the products of an oz analysis of oxymine.

02:48 And so what that tells us is exactly where our double bonds are.

02:58 So we just need to write these oz analysis products out.

03:02 And then line them up so that carbonyl groups are facing one another.

03:11 And then we just need to link all of that up.

03:16 And so i will demonstrate what i mean.

03:20 So let's see.

03:23 So we have acetone.

03:27 And because we start, well, because we know that we have something, something based on the structure of 2 -6 dimethyl octane, i think acetone is a good piece to start with because it's going to be the left section of our molecule here.

03:45 So here's our acetone.

03:49 Okay.

03:49 And so let's see what else we have.

03:55 We have a formaldehyde.

03:59 We have a ch2, so a three -carbon dialdehyde.

04:09 Okay.

04:10 And then we have a ketone and an aldehyde together.

04:17 Okay.

04:18 So i'm going to say that our three carbon dialdehyde is the next piece.

04:37 Okay.

04:39 Okay.

04:42 Okay.