With this goal in mind we can see that in the following steps (shown below), that this creates a stereogenic center which leads to the question, what stereospecific reaction will be necessary to form this center, and how can we arrange our molecule to set us up for this reaction. In present day chemistry, this is often where some transition metal catalyst from some far out land comes into play, and quite honestly I can't stand when people do this. After all, it's not like I have that catalyst just lying around the lab. What I love about Woodward, was in ingenuity in forming these stereocenters. In this reaction we will see that a common base (pyridine) is used to form this stereocenter.
I will only cover the first step today (1 to 2), the treatment with methyl iodide.
Overview:
Let's first start with what methyl iodide is, and how it acts. Methyl iodide is a very potent electrophile. The iodine (as you may remember from electronegativity trends) is pretty electronegative and pulls lots of the electron density from the carbon. This polarization makes it a good leaving group for attack by a nucleophile. If we look at our reaction scheme above the numbered compound 1 we see only one nucleophilic nitrogen. Recall that the nitrogen on the indole ring is not nucleophilic as its electron pair is involved in the conjugated aromatic system between both fused rings (it is also conjugated to the dimethoxy benzene substituent too!). One thing to note about nitrogens is that they are very nucleophilic. I remember first learning organic chemistry and noting that nitrogens and oxygens were both nucleophilic, and just accepting it as so. In some of my current research, actually I have come to learn that nitrogens can displace a halide in an SN2 reaction in the presence of an alcohol group. I was surprised as well to see some acetylation reactions of nitrogens using acetic anhydride done with methanol as the solvent with good yields. It just goes to show that Orgo I and II are really just scraping the surface of the actual practice of organic chemistry.
Back to methyl iodide. Since the tertiary nitrogen is the only nitrogen allowed to attack the methyl iodide we see displacement of the iodine to form the tetraalkyl amine salt. You may be wondering why a tertiary and thus hindered amine is allowed to form such a salt, well, it's because the methyl iodide is is such a great electrophile, and also because even tertiary amines are good nucleophiles for a reaction like this. I don't believe that doing this reaction in an alcoholic solvent would be a good idea as I am sure that the alcohols would react to form methyl ethers (unlike the anhydride reactions). Since our compound is now positively charged, what we will actually isolate will be its salt. Since the reaction is performed in benzene (nowadays we would not use such a solvent, but perhaps toluene, or THF would be better) the only counteranion is the iodine leftover from the methyl iodide.
Another thing you may or may not be asking yourself is why methyl bromide or methyl chloride couldn't be used. In fact, they could be used, however, methyl iodide is a liquid, while the bromide and chloride are gases. Not only are gases more difficult to handle, they are also potent electrophiles which in a chemists mind should instantly hint at carcinogens. They will react with logs of nucleophiles in the body and cause a ton of dammage to tissues and the like. In gaseous form, these are not fun to deal with.
Workup:
The work up for this reaction is actually quite easy. Because the reaction is performed in a non-polar solvent at ice bath temperatures (0C). What is nice about reactions like this, where the starting material is non-polar (relatively) and the products are a salt, is that often they will just precipitate from the reaction. This is the case with this reaction. The crystals will slowly precipitate from the reaction, and can simply be vacuum filtered away. Washing with a volatile non-polar solvent (diethyl ether) washes away any benzene or unreacted material away leaving the pure, pink, crystals in this case in great yield (92%)! This is because methyl iodide is such a great electrophile, the reaction goes well toward the right, giving excellent yields, which are typical of these methylations with methyl iodide.
Looking Ahead:
The next step is to actually remove this group we just made to add another carbon into the chain giving a primary amine. The positively charged tetraalkyl ammonium ion is a great leaving group for the cyanide ion. We will cover this step (and the precautions to take while using cyanide) in the next post.
Take care!
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