Question

6. Give equations for the following reactions:
(a)2- iodohexane with sodium methoxide​

Answer 1

Answer:

This is an example of either alkene synthesis or the Williamson ether synthesis. It is likely that two types of products will form, one being an ether and the other an alkene (two possibilities). The likely products are shown in the equation below.

Regarding the alkene products there are two competing factors so predicting which will be major is a little problemmatic. To form the alkene under base conditions, the mechanism is likely an E2 so a proton has to be abstracted by the base (methoxide). Sterically, the most accessible proton is the one on the terminal methyl group leading to hex-1-ene. However, the more highly substituted alkene is more stable, the base is relatively small, and the methylene protons (3 position) are not very hindered so the hex-2-ene product may form to a greater extent.

Now, if the desire is to make the Williamson product, a modest change will avoid the formation of the alkene products. In particular, use the sodium alkoxide of 2-hexanol and sodium iodide.

Elimination is just not an option for Methyl Iodide.