Including stereoisomers, how many products are theoretically possible from the following reaction?

Question

The starting material is 3-bromo hexane, NaOCH2CH3

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HOCH2CH3

Answer 1

This content would be covered in your textbook section on SN2 and E2 reactions. If you haven't gone there yet to figure it out on your own, I strongly recommend doing so before reading on. Building the skill to solve your own problems with the resources you already have is very important for most classes, but perhaps especially so for organic chemistry....

Given the conditions, this would be a competition between an E2 elimination and an SN2 substitution. The ethoxide (CH3CH2O-, from dissociation of sodium ethoxide in the ethanol solvent) is both an excellent nucleophile and a strong base, so it could cause either one of those reaction mechanisms. With a protic solvent (ethanol), the rate of SN2 is decreased by hydrogen bonding of the solvent on the anionic nucleophile. This is especially true for the secondary substrate (3-bromohexane) -- primary halides will often do SN2 reaction well even in protic solvents, but secondary halides usually do not. As such, I would suspect the major products would be E2 products -- but I would expect some (a minor amount) of SN2 substitution to occur along with it.

If SN2 occurs, the CH3CH2O- nucleophile would simply replace the -Br, making 3-ethoxyhexane. Although you were not given stereochemical configuration of the reactant (C3 is a chiral center), the SN2 would go through inversion, so if you had started with a single enantiomer, your product would be the single enantiomer formed through an inversion of configuration. Without configuration of the reactant specified, one would assume it is racemic, so both enantiomers of the product would be formed.

For elimination, there are two pairs of beta-hydrogen atoms -- the two on C2, and the two on C4. (E2 reactions occur when a hydrogen is removed from a carbon ADJACENT to the carbon bearing the leaving group.) Removing one of the H's on C2 to drive the elimination would make 2-hexene; removing a hydrogen from C4 would form 3-hexene. Both of these isomers are equivalently substituted (both are disubstituted) so similar amounts of each constitutional isomer would be expected to form. (If one were more substituted than the other, that constitutional isomer would be expected to be major, as the base is not "big and bulky" so Zaitsev's rule would be followed.)

But both of those alkene isomers could be formed as either the cis or trans stereoisomers. There would be less cis than trans in each case, for steric reasons, but I would expect both stereoisomers to form.

In total there would be 5 (or six, depending on how you count the enantiomers of the ether product) different products formed. Trans-2-hexene and trans-3-hexene would be formed in similar amounts, and would be the two major products. Cis-2-hexene and cis-3-hexene would also be formed, but in smaller amounts. And I would expect the product formed in the smallest amount would be 3-ethoxyhexane (both enantiomers, assuming a racemic reactant). It's not impossible that the ethoxyhexane MIGHT be formed in a higher amount than the cis- isomers of hexene, depending on reaction conditions, but I doubt it would be the case in ethanol solvent.