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Topic: SN1 Product Help  (Read 6514 times)

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Offline Pt.Defiance

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SN1 Product Help
« on: July 12, 2011, 10:12:05 PM »
C3H6-CH2-OTs reacts with water to produce C4H8-OH and C3H6-CH2-OH.
I understand the mechanisms involved, but am having understanding a point my professor made. He said that in this particular SN2 reaction only 50% of the products are of the C4H8-OH variety. How can that be with the presence of a tertiary substrate and the lowered angle strain?

Offline AC Prabakar

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Re: SN1 Product Help
« Reply #1 on: July 13, 2011, 02:56:11 AM »
Could u bit elaborate this? I mean structure.

Offline Dan

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Re: SN1 Product Help
« Reply #2 on: July 13, 2011, 03:59:50 AM »
Quote
C4H8-OH and C3H6-CH2-OH...
...a tertiary substrate and the lowered angle strain?

Are you talking about n-butanol and s-butanol or something else? Without defining your structures, the post is nonsensical.
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Offline Pt.Defiance

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Re: SN1 Product Help
« Reply #3 on: July 13, 2011, 12:47:37 PM »
I'm sorry about that, we haven't learned how the IUPAC nomenclature for these molecules yet, so I'll try my best. 1-Methyl-1-(1-TDs)-cycloproprane reacts with water to produce Cyclobutanol and Hydroxymethylcyclopropane. I am wondering why cyclobutanol only constitutes 50% of the prodcuts when angle strain is lessened?

Offline Dan

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Re: SN1 Product Help
« Reply #4 on: July 13, 2011, 01:35:37 PM »
1-Methyl-1-(1-TDs)-cycloproprane reacts with water to produce Cyclobutanol and Hydroxymethylcyclopropane.

Do you mean:
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Offline Honclbrif

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Re: SN1 Product Help
« Reply #5 on: July 13, 2011, 02:04:27 PM »
That one's a bit of a trick question. Look up the cyclopropylcarbinyl carbocation.
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Offline Pt.Defiance

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Re: SN1 Product Help
« Reply #6 on: July 13, 2011, 03:10:00 PM »
1-Methyl-1-(1-TDs)-cycloproprane reacts with water to produce Cyclobutanol and Hydroxymethylcyclopropane.

Do you mean:

Yes that is what I mean, there is also a solvent present. Sorry for being so difficult, this is my first organic chemistry and I'm only a couple weeks into it. I'm guessing my nomenclature was way off. The problem that I am having is why cyclobutanol constitutes only 50% of the products when angle strain is much lower 49.5 v. 40 degrees. I am guessing that it has to do with the difference in structure, but I can't figure out why.

Offline Honclbrif

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Re: SN1 Product Help
« Reply #7 on: July 13, 2011, 03:18:19 PM »
After the leaving group leaves the starting material, is the cation primary, secondary, or tertiary?
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Offline Pt.Defiance

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Re: SN1 Product Help
« Reply #8 on: July 13, 2011, 03:21:27 PM »
After the leaving group leaves the starting material, is the cation primary, secondary, or tertiary?

Primary.

Offline Pt.Defiance

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Re: SN1 Product Help
« Reply #9 on: July 13, 2011, 03:35:17 PM »
Now I'm even more confused, if Sn1 reactions prefer tertiary to secondary to primary substrates why would the formation of cyclobutanol be identical to the other molecule. Angle strain is lessened and a secondary carbocation is formed leading me to believe that cyclobutanol would be produced in a much higher proportion.

Offline Honclbrif

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Re: SN1 Product Help
« Reply #10 on: July 13, 2011, 05:31:57 PM »
Yeah, I thought I could go a different way with that but in light of the equal product distribution it was an incorrect starting point. I'm going back to the "this is actually a trick question" statement. Your reasoning is sound, but there's something else happening here which is honestly beyond the first few weeks of an o-chem lecture.

Again, I recommend that you look up the cyclopropylmethyl carbocation and the cyclobutyl carbocation. They are different from other carbocations.
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Offline Pt.Defiance

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Re: SN1 Product Help
« Reply #11 on: July 13, 2011, 05:59:41 PM »
I really appreciate your help Honclbrif. It's probably pretty difficult explaining things to someone that has just started with OChem, but I think I understand now. Despite the secondary carbocation and lessened angle strain of cyclobutanol, the cyclopropylmethyl carbocation is more stable due to the ability of the bent bonds to donate electrons in to the vacant p orbital on the carbon and the resultant conjugation.

Offline Honclbrif

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Re: SN1 Product Help
« Reply #12 on: July 13, 2011, 06:13:41 PM »
"Despite the secondary carbocation and lessened angle strain of cyclobutanol, the cyclopropylmethyl carbocation is more stable due to the ability of the bent bonds to donate electrons in to the vacant p orbital on the carbon and the resultant conjugation."

There's actually more to it than that. I've been trying to skirt around giving a full on answer, but this falls under the heading of "nonclassical carbocations". They are really interesting, but are usually reserved for more advanced lectures.
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Offline BluePill

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Re: SN1 Product Help
« Reply #13 on: July 14, 2011, 02:32:04 AM »
+1 to non-classical carbocations.

There is also another explanation: "Neigbouring-Group Effects".

You see, the product does not really form the usual carbocation:



It then rearranges into cyclobutane and cyclopropane. The rate of formation of either of this is controlled kinetically.


Offline napoleon79

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Re: SN1 Product Help
« Reply #14 on: July 16, 2011, 12:36:35 PM »
I think that
If you do SN1, you must do sustainable form carbocation.

if you do practice with KOH or NaOH then the reaction will be SN2.

thanks  :)

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