# Is it possible to form an 8 carbon ring using a Diels-Alder reaction?

I need to convert the starting material to the product on the right but I was wondering if it was possible to use a Diels-Alder reaction and use $\ce{KMnO4}$ to cleave the ring.

• A better way would be the tetramerization of ethyne to cyclooctatetraene, then hydrogenation to cyclooctene followed by ozonolysis. – aventurin Apr 21 '18 at 23:25
• what is tetramerization? – Chem Apr 22 '18 at 1:56
• The combination of 4 molecules (monomers, here ethyne) to a single molecule (here cyclooctatetraene). – aventurin Apr 22 '18 at 8:31

You can obviously use Diels-Alder Reaction for the synthesis of the desired product but not so directly. But Diels-Alder will give an optimal yield when the ring formed is Six-Membered. After the formation of that Six-Membered product, you have to increase the no. of carbon atoms in the carbon-chain by some way, and the yield of your product will be also better if you no. of steps as less as possible.

I will not suggest you to directly add 1,3-butadiene to acetylene because there will be a chance of aromatisation, and then you cannot do anything, as to break the aromatic rings are highly unfeasible. Instead I can suggest you the following steps:

1. First you partially reduce the acetylene to ethene, and then perform Diels-Alder reaction with butadiene. You will get cyclohexene.
2. In the second step, you perform ozonolysis, and then to the resulting compound, perform Wittig reaction with excess reagents. You will get octa-1,7-diene.
3. Lastly, you can perform hydroboration, oxidation to get less substituted diol, and then oxidise that to acid by any strong oxidising agent like $\ce{KMnO4}$ or $\ce{K2Cr2O7}$ in slightly acidic medium.

Thus, you will get your desired product. The path of reaction is shown below:

• is it possible to use h2 pd/c instead of the witting reaction to reduce the oxygen? My professor never taught that reaction so I don't know the mechanism. – Chem Apr 21 '18 at 19:31
• NO! You cannot. The reduction by $H_2/Pd$ will reduce the double bond too and you can't proceed further. In fact this reagent can reduce almost everything to saturated forms , which are not always desired. That 's why different reducing agents are made which can reduce to certain extent according to our choice. – Soumik Das Apr 22 '18 at 4:34
• Please also observe that by Wittig reaction, I am increasing the no of Carbons in the chain. You cannot do it by using a reducing agent.. – Soumik Das Apr 22 '18 at 4:38