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Hi guys! I have question regarding number 103 ochem section (2017 edition). I'm wondering why choice E gives Sn1 product? Why can't it produce alkene product via E1?

Also, for choice A, can't it also undergo Sn2 reaction? Is it saying that both E2 and Sn2 products form? and since the question asked for which anower choice can nproduce alkene product, so A is correct?

Thanks in advance!!
 
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themuffinman11

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I think this question is asking for which reaction would give the best yield of an alkene, or in other words, which reaction has the best conditions for and alkene to form. So right away you can eliminate B/D as they will definitely not give alkenes. Choice C can also be eliminated because I- is not a good base.

So now comparing A and E, A looks like the best to form an alkene because it has a strong base (negatively charged oxygen) with a secondary carbocation and polar protic solvent (both of which favour E2 over SN2). In E, it's a weak nucleophile and base, tertiary carbocation and a polar protic solvent (it's aqueous, so includes water), thus you probably get E1, but also a lot of SN1 products too.

Hope this helps!
 
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orgoman22

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Hi guys! I have question regarding number 103 ochem section (2017 edition), picture is attached as well. I'm wondering why choice E gives Sn1 product? Why can't it produce alkene product via E1?

Also, for choice A, can't it also undergo Sn2 reaction? Is it saying that both E2 and Sn2 products form? and since the question asked for which anower choice can nproduce alkene product, so A is correct?

Thanks in advance!!
You have a good eye,,,,,this tertiary halide does indeed compete with the E1 here, but since we have no heat......the major is by Sn1. If you heat this up.....E1 would predominate. A tertiary halide with reflux or heat will almost always give major by E1. For choice A...indeed Sn2 can compete,,,,,but a nice rule of thumb is this..... On a secondary or tertiary halide with methoxide or ethoxide,,,,,E2 is the major product. For a wonderful review of all this, I think the text by David Klein PhD is the clearest and best.

Hope this helps.

Dr. Romano
 
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May 5, 2017
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That makes sense! thank you so much!

I think this question is asking for which reaction would give the best yield of an alkene, or in other words, which reaction has the best conditions for and alkene to form. So right away you can eliminate B/D as they will definitely not give alkenes. Choice C can also be eliminated because I- is not a good base.

So now comparing A and E, A looks like the best to form an alkene because it has a strong base (negatively charged oxygen) with a secondary carbocation and polar protic solvent (both of which favour E2 over SN2). In E, it's a weak nucleophile and base, tertiary carbocation and a polar protic solvent (it's aqueous, so includes water), thus you probably get E1, but also a lot of SN1 products too.

Hope this helps!
 
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May 5, 2017
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You have a good eye,,,,,this tertiary halide does indeed compete with the E1 here, but since we have no heat......the major is by Sn1. If you heat this up.....E1 would predominate. A tertiary halide with reflux or heat will almost always give major by E1. For choice A...indeed Sn2 can compete,,,,,but a nice rule of thumb is this..... On a secondary or tertiary halide with methoxide or ethoxide,,,,,E2 is the major product. For a wonderful review of all this, I think the text by David Klein PhD is the clearest and best.

Hope this helps.

Dr. Romano

Thank you so much Dr. Romano! I really appreciate your help :)
 
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