North American termite soldiers, when encountering enemy insec...

Question

North American termite soldiers, when encountering enemy insects, contract their mandibular muscles, expelling a mixture of chemicals that essentially trap their enemies in a glue-like substance. This weapon, built into the face of the termite, is called the fontanellar gun. It releases a mixture of pinene (62%), myrcene (27%), and limonene (11%).

(b) Suggest an acid-catalyzed mechanism by which pinene could be produced from limonene.

(b) Suggest an acid-catalyzed mechanism by which pinene could be produced from limonene.

Accepted Answer

Hi, everyone. Welcome back. Our next problem says, Mercy, which is a mono Turpin and a fragrant oily liquid. Can I soar into Liman, Osam and Alo Osam just an acid catalyze mechanism by which Liman could be produced from Merc. And we have the structure of Merc drawn here. It's reacting in H 30 plus. So this acidic environment has plenty of hydro ions and producing limine. And when we look at these two structures, we see that mercein has three double bonds and an open ring, it has sort of five carbons in a ring that doesn't have a last peace to close it. There are three double bonds, it's carbon carb Neel bond on top, a double bond on the side next to the open space and on the bottom. So let's number these carbons to be able to refer to them. I'll number the carbons in the ring starting at the bottom since that has this isopro pal group with a double bond attaching it to the ring. So that will be carbon one at the bottom and then we'll go around. It's not a ring actually. So it's a chain. So going to our left 23, carbon four is at the top of the molecule with a double bond to another carbon. And then carbon five on the top right of the molecule. And then carbon six is the very last one. Next to this open space, there's a double bond between carbons five and six. So we'll just use those numbers to be able to refer to the different carbons here. So we see that we've got those three double bonds in mercy in the open space. And in Liman, in the open space has been closed to make a closed six carbon ring. And we've dropped down from three double bonds overall just to two double bonds. We'll keep the numbering of the carbon the same. We have a double bond going from carbon 4 to 5. So it's now within the ring instead of on top and down at the bottom, instead of having a double bond going from carbon one out to the first carbon on the branch, the double bond now goes from on the last, the terminal end of that branch to one of the carbons that were a methyl group. So I'm going to redraw my hydro ion as hoh two plus just to be able to show my electron movement here. And I'm going to think about what I need to do to convert this, I have an acidic environment. So I can generate a Carbocaine intermediate by adding a proton to a double bond this will also bring me down one double bond, which is what I'm looking for. And once we're that Carbocaine, we have the possibility of movement of electrons from double bonds shifting over and getting our sort of rearranged situation that we have in Liman. So we'll start by looking at this double bond at the top of the molecule going from carbon four out to this little branch. If we take those pi electrons and grab a proton from our hydron ion, the electrons that bound that hydrogen to the oxygen go back to the oxygen generating water. And so now we've generated a stable tertiary Carbocaine intermediate. So we now just have a single bond methyl group at the top. Our ring is still open with a double bond between carmens five and six. And then down at the bottom that double bond from carbon one out to its substituent, but we now have a positive charge on carbon four. So now we have this Carbocaine intermediate. So we can start pushing things around and moving our double bonds. So we can have the pi electrons from that double bond on the bottom that carbon one can move on over and bond a carbon six, thereby closing the gap and making a ring. Now there was a double bond between five and six. So those pi electrons can move on up and form a new double bond between carbons four and five that we see in limine. So we've done two things we want to see in Liman, we've turned the double bond at the bottom into a single bond, close the ring and moved our double bond up to between carbons four and five and we just have a single bond on the very top. So now what's happened to our positive charge? Well, that's now down below on the carbon that's attached to carbon one down at the bottom. So we're almost there, just draw what we have here. All we have left is a spare positive charge down on that carbon that will eventually be part of a double bond at the very end of that branch there. So if we draw in the hydrogen, that's at the very end there, we we'll have a water molecule that was generated in the first step. So let's draw 02 with its two long pairs, we can go ahead and deproteinate this terminal carbon here, thereby regenerating our acid. The electrons from that bond to the hydrogen go over to form that final double bond that also removes the positive charge. So from there, we've gone back and gen generated our laminin molecule. So once again, we had this addition of a proton to a double bond at the top generating carbo cion, we were able to have all these py electrons shifting around closing up the ring, moving the double bonds to where we're desired with a final dep protonation step and regeneration of our hydro num ion. So there's our mechanism, we'll see you in the next video.

Key Concepts

Acid-Catalyzed Reactions

Isomerization

Terpene Structure and Reactivity

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