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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Pt.Defiance on July 12, 2011, 10:12:05 PM

Title: SN1 Product Help
Post by: Pt.Defiance 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?
Title: Re: SN1 Product Help
Post by: AC Prabakar on July 13, 2011, 02:56:11 AM
Could u bit elaborate this? I mean structure.
Title: Re: SN1 Product Help
Post by: Dan 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.
Title: Re: SN1 Product Help
Post by: Pt.Defiance 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?
Title: Re: SN1 Product Help
Post by: Dan 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:
Title: Re: SN1 Product Help
Post by: Honclbrif on July 13, 2011, 02:04:27 PM
That one's a bit of a trick question. Look up the cyclopropylcarbinyl carbocation.
Title: Re: SN1 Product Help
Post by: Pt.Defiance 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.
Title: Re: SN1 Product Help
Post by: Honclbrif on July 13, 2011, 03:18:19 PM
After the leaving group leaves the starting material, is the cation primary, secondary, or tertiary?
Title: Re: SN1 Product Help
Post by: Pt.Defiance on July 13, 2011, 03:21:27 PM
After the leaving group leaves the starting material, is the cation primary, secondary, or tertiary?

Primary.
Title: Re: SN1 Product Help
Post by: Pt.Defiance 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.
Title: Re: SN1 Product Help
Post by: Honclbrif 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.
Title: Re: SN1 Product Help
Post by: Pt.Defiance 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.
Title: Re: SN1 Product Help
Post by: Honclbrif 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.
Title: Re: SN1 Product Help
Post by: BluePill 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:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi56.tinypic.com%2F34iqz2g.jpg&hash=f4a9745642d3ee3027945b783831253bc35a5abe)


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

Title: Re: SN1 Product Help
Post by: napoleon79 on July 16, 2011, 12:36:35 PM
I think that
If you do SN1, you must do sustainable form carbocation. (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi1201.photobucket.com%2Falbums%2Fbb344%2Fquocvu1986%2F1.jpg&hash=21d0860f17e4bcd4ab4a8a0ea6683246230c9632)

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

thanks  :)