12.7 Elimination Reactions of Alcohols | Organic Chemistry

[@dhruvshah6461]

Thank you so much for uploading this and all you your other videos! Because of you I am able to survive college chem AND GET AN A! THANK YOU!!!!!!!! YOU TRULY HAVE A GIFT OF TEACHING!!

[@ChadsPrep]

You are welcome, dhruv. Thank you for your comment.

[@Viona730]

11:35 should the back side attack alcohol be propanol instead of ethanol in this case? Because the product has 6 carbons in the chain. If the chain with 3 carbons is added with a two-carbon chain, then the product will have 5 carbons only. I really like your videos btw :)

[@nilsnickname4455]

Hello Chad, there is a fault in your argumentation at about 12:18. It is not very important, but I want to tell you. You said "Elimination gets favoured at higher temperature, because it's more entropically favoured...". Than you tell that in the elimination would have same number of reactants like the number of products. And that in the substitution there would be more reactants than products. When you write the equations and eliminate the parts, which you find on the left and also on the right side of the reaction equation arrow, you get the following result: Elimination: one alcohol molecule reacts. One alkene molecule and one water molecule are built. So the number of reactants and products aren't the same. One molecule becomes two molecules. Substitution: two molecules of alcohol react to give one ether molecule and one water molecule. So the number of molecules doesn't change. But, you are the best! Noone explains as good as you do! I only repeat things that I learned years before. I want to understand every single word and rewind the video if there is only a little bit that I don't understand. Thats the only reason I find some "errors". By the way - your wrong counting of molecules of the both reactions cancel each other, so that your interpretation lead to the correct overall result. Best regards Nils

[@ChadsPrep]

Hey Nils! Not a mistake - and actually a universal trend in elimination versus substitution products. The basic elimination reaction forms the end product (1) plus the conjugate acid (1) plus the leaving group = 3 whereas the substitution gives the substituted product molecule and the leaving group only = 2. I'm not sure which equation you are referring to in the video but if you look around the 6:30 mark with the E2 reaction we get the alkene + water + H2SO4 = 3 molescules

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[@ChadsPrep]

Excellent!

[@tzvifriedman4986]

Greetings! The bimolecular sn2 reaction: how did the alcohol work if it is a WEAK nucleophile?

[@ChadsPrep]

Hello tzvi...Great question! This is actually a bit unexpected for exactly the reason you claim...alcohols are weak nucleophiles. We actually might have predicted that this reaction wouldn't work for primary alcohols (or methanol) since they wouldn't be able to undergo SN1 reactions (due to instability of the carbocation intermediate), but it turns out that these reactions do occur. And on top of that there is evidence to suggest that they are proceeding via the SN2 mechanism. So we don't really have a "How" this happens as weak nucleophiles typically can't do SN2, but just an observation that they do in this case which is a little unexpected. Hope this helps!

[@tzvifriedman4986]

@@ChadsPrep Thank you! You the man!

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