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This reaction will be both an E2 and SN2, where the E2 will be major and the SN2 minor (If I have done it correctly). I was drawing out the mechanism as I noticed that I was making my nucleophile attack the wrong hydrogen, the one attached to the alpha-carbon. I know that the nucleophile is supposed to attack the hydrogen on the beta-carbons however how do I determine if the $MeO^-$ attacks the hydrogen on the second carbon to the left or to the right of the alpha-carbon? Or does it attack both and since the sturcture of the compound formed is the same either way, it doesn´t matter which hydrogen it removes?


Well done to narrow the options down to E2 and SN2 and recognize that E2 is likely favored over SN2.

As the comments suggest, in this case, it doesn't matter which one is deprotonated. The reason is that deprotonation at either beta-position gives the same product. If you are unsure, draw out both options.

In future similar examples, you may come across reactants with multiple possible reactive positions that are not equivalent. For example, consider 3-chlorohexane (instead of 3-chloropentane, as in the original problem). In this case, again there are two possible sites to be deprotonated (Ha and Hb). Looking at the two possible products, both are disubstituted alkenes, so they will be close in stability. There isn't much difference in the groups nearby Ha and Hb, so the kinetics of the deprotonation are probably similar. So which is the "correct" answer? In reality, both are correct. They may not be formed in exactly equal amounts, but they will probably both be formed in substantial amounts.

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There are a couple of key lessons here for those studying organic chemistry:

1) Draw out all of the possible products that come from mechanisms that you think make sense. This might be lots of drawing! Nevertheless, it is much easier to realize the differences if everything is drawn out in front of you.

2) Organic chemistry is messy. Often times multiple mechanistic pathways will be viable and multiple products will form. Much of the creativity in organic chemistry is finding ways to ensure that only one mechanism can occur or one (desired) compound is formed as the major product.

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  • $\begingroup$ Re: "In reality, both are correct." I was taught small nucleophiles favor Satyzeff alkene, while bulky nucleophiles (like tert-butoxide) favor Hoffmann alkenes. So, how do you justify this statement? Thank you! $\endgroup$ – Gaurang Tandon Feb 16 '18 at 1:43
  • $\begingroup$ @GaurangTandon in this case, which is which? They are both di-substituted. And to answer a question with a question, what does "favor" mean? Is the product distribution 100:0, 75:25, 51:49? Those all have a "major product" but they vary greatly in how much each is favored. $\endgroup$ – jerepierre Feb 16 '18 at 15:25
  • $\begingroup$ "in this case, which is which?" i'd say the one on the right is the Saytzeff alkene, since it is more stable by an extra hyperconjugative structure; "what does "favor" mean?" Well I was just asking a question based on the knowledge I got at high school ^_^ I haven't done the experiments myself, so I really don't know what ratio it is exactly; but in this case the alkenes are so close in stability I guess you are right in that "both be formed in substantial amounts." $\endgroup$ – Gaurang Tandon Feb 16 '18 at 15:30

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