Ocimene, a compound isolated from the herb basil, has three double bonds and the IUPAC name 3,7-dimethyl-1, 3-6-octatriene.
b. Draw the structure of the compound formed if enough HBr is added to react with all the double bonds in ocimene.

Question

Ocimene, a compound isolated from the herb basil, has three double bonds and the IUPAC name 3,7-dimethyl-1, 3-6-octatriene.

b. Draw the structure of the compound formed if enough HBr is added to react with all the double bonds in ocimene.

Accepted Answer

Hi, everyone. Welcome back. Our next problem says pharmacy is the compound responsible for the characteristic smells of green apples. Its Iupac name is 3 7-Eleven tri methyl Dodek 136 10 tetra predict the major product formed when pharmacy reacts with excess HBR. And we have four different answer choices labeled a through d of different major products that could be formed in this circumstance. So we're talking here about the halogen nation A an alkene. So the double bond or double bonds I should say will be broken with a bromine added to one of the carbons participating in the double bonds. So first, our first step needs to be to draw the original structure of the compound that's going to react to that H pr. So let's look at our Iupac name to be able to draw that. So we have that prefix Dodek meaning 12. So our main chain is a 12 carbon chain. So let's draw that not that I tried to start my zigzag for lack of a better word in the same direction as say my molecule and choice A where it shows carbon one being up and carbon tubing down. I want to be able to compare my two structures uh when I'm trying to find the right answer. So I just drew it that way for convenience to be able to match it a little easier to the answers. It's easy in these long chains to lose track of my carbon. So it's always a good idea to number them, especially as we're going to be adding groups and double bonds onto them. So I'm just going to number my 12 carbons here, make sure I did draw them and look, I'm actually too short. So that is a good example of why we always want to numb those number those carbons because it's really easy with a skeletal structure when the carbons are just corners to lose some somewhere. So I've only got 10 carbons. I need to add 1112, 2 more. So right, there are a good illustration of why you want to number your carbons. So I have my 12 carbon main chain. Then I see that my name continues to say it's a tetra E. So it's an alkene with four double bonds and I use the numbers to show me where they go. So it says 136, 10 tetra. So my first double bond is between carbons, one and two. Put that on there, the second double bond between three and four, the next double bond between six and seven and the last one between 10 and 11. So I've got my double bonds drawn on here. And then finally, I see I have three methyl groups, my 3, 7-Eleven triethyl go on carbon three on uh carbon seven and on carbon 11. So this is why I drew my carbon numbers in red just so that I didn't get them sort of mixed up as I drew my structure in blue. So now with the original structure, I can think about where bromine would be added. And when I talk about the halogen generation of alkenes, I actually have a rule to tell me we're going to break a double bond, which carbon from the double bond does the B roaming get added to? And this rule is called Markov Niko's rule. And it says when HXX being halogen is added to an alkene, the major product is formed when H hydrogen attaches to the carbon with more hydrogen atoms, while X attaches to the carbon with the lesser number, and that's the lesser number of hydrogen atoms. So with that in mind, let's work our way through our double bonds here. So our first old bond is between carbons. One and two, carbon, one has three hydrogens. It's the end of the chain while carbon two being internal has two hydrogens still. So our bromine will be added to draw that a little more clearly to carbon two in this case. So now we need to look at our answer choices and see which one shows the correct addition here So when I look at choice A, I see, I look at carbons one and two and I see bromine is indeed added correctly to carbon number two. When I look at Choice B, the bromine's been added to carbon one. So I'm going to go ahead and cross that out and I can eliminate Choice B as my answer choice. Then let's look at Choice C looking at carbon one again, this is a little tricky to see which is which here, I've got B roaming here attached to see. This looks like I, I was tempted to go to. This is carbon one at first because the structure has gotten a little confusing. Here's my bromine, the end of my chain. So on the left, there's carbon one and carbon two. So the bromine is correctly added to carbon to. So let's move on to choice D. When I look at answer choice D and look for the bromine, I see that the bromine, in this case, there is no bromine added to either carbon one or two. That double bond has mysteriously been broken. But no, bro means been added to choice D definitely incorrect. So Ive eliminated two of my answer choices. Let's move on to my next double bond. My next little bond is between three and four. Carbon three has a methyl group on it. So there's actually no hydrogens on carbon three while carbon four still has one hydrogen. So that means my bromine will be added to carbon three. So lets look at choices A and B again, when I look at choice A and look at carbons three and four. I see the bromine was added to carbon four. So that makes choice a incorrect when I look. However, at choice C I see the bromine has been correctly added to carbon three. So by process of elimination, I've gotten down to choice C is my correct answer. But to be thorough, let's just keep going and look at our other double bonds really quickly. Our next double bond between six and seven carbon six as one hydrogen, carbon seven as a methyl group and no hydrogens. So we would expect our bromine to be added to carbon seven, double check our molecule in choice C and we see that the bromine was indeed added to carbon seven. So finally, we move on to our last double bond between 10 and 11. Carbon 10 has one hydrogen, carbon 11 with a methyl group has none. So our bromine would be added to carbon 11. So once again, when I look at my answer, choice, C I see that bromine has been added to carbon 11. So the major product that I would predict to be formed when my molecule, so we have our correct answer for the major project products. Excuse me, we expect with this reaction in choice C where we have our bromine's added to our carbons 236 and 11 in our molecule. See you in the next video.

Ocimene, a compound isolated from the herb basil, has three double bonds and the IUPAC name 3,7-dimethyl-1, 3-6-octatriene.
b. Draw the structure of the compound formed if enough HBr is added to react with all the double bonds in ocimene.

Verified video answer for a similar problem:

Key Concepts

Hydrohalogenation

Alkene Reactivity

IUPAC Nomenclature

Ocimene, a compound isolated from the herb basil, has three double bonds and the IUPAC name 3,7-dimethyl-1, 3-6-octatriene.b. Draw the structure of the compound formed if enough HBr is added to react with all the double bonds in ocimene.

Verified video answer for a similar problem:

Key Concepts

Hydrohalogenation

Alkene Reactivity

IUPAC Nomenclature

Ocimene, a compound isolated from the herb basil, has three double bonds and the IUPAC name 3,7-dimethyl-1, 3-6-octatriene.
b. Draw the structure of the compound formed if enough HBr is added to react with all the double bonds in ocimene.