Show how you might synthesize the following compounds, using a...

Question

Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why.
d. 4-methylhex-2-yne
e. 5-methylhex-2-yne
f. cyclodecyne

d. 4-methylhex-2-yne

e. 5-methylhex-2-yne

f. cyclodecyne

Accepted Answer

All right. Hello everyone. So this question is asking us to give a step by step process of synthesizing the compounds below using acetylene and an O hide. If the compound cannot be synthesized using the method or the mentioned starting materials explain why. Report one, we have 55 dimethyl ox three I. For part two, we have seven methyl oct three I. And for part three, we have Seko Aine recall first and foremost that acetylene otherwise referred to as ethane is an example of a terminal all kind and terminal, all kinds are unique in that their outermost carbon containing a hydrogen is relatively acidic or specifically that proton is relatively acidic. So a very common reaction is one in which that terminal all kind is derotated and then reacting reacted with a primary ocular Halide to extend the carbon chain in an S and two reaction in which the conjugate base of the terminal all kind referred to as the acetyl ion behaves as the nucleo. So to understand which Ocu halides we're going to use for this synthesis, we have to consider what our groups are branching off of the triple bond in each intended product. So starting off here with 55 dimethyl oct three I. So that is a five carbon sorry eight carbon parent chain with a triple bond in between carbons, three and four and two methyl substituents both on carbon five. So now to this structure, I'm just adding the remaining hydrogens necessary to fulfill each carbon's four bond requirement. So the triple wand itself is what stems from acetylene. So the ocho halides necessary must include the R groups branching on either side of that triple bond. But there is an issue when we consider the R group on the right side because notice how carbon five that's directly attaching to are all kind carbons here from Celine would be tertiary once it's disconnected from the all kind itself. And thats poses an issue, right? Because if we imagine a leaving group like a halogen attached to this R group that becomes a tertiary ocular hide. And if it's a tertiary Ocu Halide, it's going to be too sternly hindered to participate in an S and two reaction, the nucleo file, the acetyl ion would have a difficult time approaching that tertiary carbon. So because of that 55 dimethyl oct three I cannot be produced using this method. So here I went ahead and paused the recording temporarily to write out our answer for part one, which is that the compound cannot be synthesized using this method since it requires a hindered starting material. So now with this in mind. Let's go ahead and scroll down somewhat to make space for part two. Part two is severed metal three. I, so that's the same seven carbon or sorry eight carbon parent chain with a triple bond in between carbons three and four. But this time with a methyl substituent on carbon number seven. So this time around, we can in fact use primary halides to extend the carbon chain on either side of the triple bond. If we consider the R group to the left of the all kind, that is a two carbon chain with our leaving group, say a halogen on the first carbon here that is directly connecting to carbon three of the parent chain from the all kind. So that turned out to be Chloro Ethane to give an example here. And on the other hand, if we consider the R group to the right, that's an additional four carbons with a methyl group on the third position. And our halogen, let's say chlorine again on carbon one. So that's one Chloro three methyl butane. And from here, we have our synthesis. So starting off with acetylene or ey first, we must deproteinate that terminal all kind using a strong base like sodium amide or nanh two followed by Chloro Ein. This extends the carbon chain on one side of the triple bond by two carbons leaving us with one more terminal all kind proton that can be derotated once more. So we have sodium amide once again, this time with one chloral three methyl butane. And I recognize now that I am a little limited on space towards the right side of the screen. So I'm going to have this last arrow point upwards and I'm going to move the structure of the intended product that I went ahead and drew previously. There we go. So this is the synthesis that is going to result in seven methyl three iron. So last but not least, lets talk about Cyclo Aine. Now, cyclo octane is an eight carbon ring with a triple bond in between two carbons of that ring. So for part three, thats a triple bond inside of a ring, an eight carbon ring. So once again, I'm going to need to add hydrogens to each of the remaining carbons here to make sure they fulfill their for bond requirement. And once more, the triple bond stems from acetylene. But because our final product is a ring, an intra molecular sn two reaction must have happened at some point. This means that the ideal cul Halide must actually have been a die Halide because by having two halogens present in the same molecule, there are two sites for nucleic attack. So using the remaining six carbons of this eight carbon ring, by adding our halogen, let's say chlorine to either end of this chain, we can close out a six member ring or excuse me, an eight member ring with the two carbons already coming from acetylene. So now let's go ahead and write out that synthesis. So here we have acetylene and our first step once again is going to be de protonation with sodium amide followed by SN two with one of the carbons bearing a halogen. So this is going to extend the carbon chain tr six from the die, he lied, followed by two more from acetylene. But now in the second step, instead of adding another cul Halide, we can simply dep proton the remaining terminal all kind proton and make way for our intra molecular sn two reaction. Now, once again, for the sake of space, I'm going to go ahead and erase the dial that I drew previously because I'm going to have to show our product above this intermediate here. So this will give us cyclo octin and there you have it. Now, I did have to get a little creative with the formatting and I actually almost forgot to add hydrogens to my intermediate here. But now we have our completed synthesis for part three and that concludes our final answer to this question, brief reminder that for part one, the compound cannot be synthesized using this method because it requires a hindered starting material. However, here is our synthesis four, part two and also for part three and there you have it. So if you watch this video all the way through, thank you so very much. And I hope you found this helpful.

Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why.
d. 4-methylhex-2-yne
e. 5-methylhex-2-yne
f. cyclodecyne

Key Concepts

Alkyne Synthesis

Regioselectivity in Reactions

Cycloalkyne Formation

Watch next

Show how you might synthesize the following compounds, using acetylene and any ­suitable alkyl halides as your starting materials. If the compound given cannot be ­synthesized by this method, explain why.d. 4-methylhex-2-ynee. 5-methylhex-2-ynef. cyclodecyne

Key Concepts

Alkyne Synthesis

Regioselectivity in Reactions

Cycloalkyne Formation

Watch next

Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why.
d. 4-methylhex-2-yne
e. 5-methylhex-2-yne
f. cyclodecyne