sternman318
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Questions about synthesis of tetrahydrothiophene
So I am in first year organic chemistry and we are learning about S<sub>N</sub>2 reactions, and one example of an intramolecular reaction
confused me.
The example was 4-bromobutane-1-thiol in the prescence of a strong base ( NaOH) produces tetrahydrothiophene. One would expect the OH- to attack at the bromine, but instead the base deprotonates the thiol and the deprotonated thiol
'twists' back and reacts at the bromine instead. What confuses me is:
1. I understand that the thiol is a stronger nucleophile than OH-, but would the OH- not react first, before the thiol is able to twist over itself to
react at the bromine? I understand that molecules undergo very rapid, constant conformational changes, but is that not quite a large movement?
2. What is to stop the thiol of another molecule from reacting at the bromine instead, essentially creating a polymer?
This is the only reference to the reaction I could find, though I did not extensively research it: http://www.chemicalforums.com/index.php?topic=26998.0
And I apologize for incorrect phrasing or other errors, we just started learning about it 
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spong
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The thiophene would be the main product as intramolecular reactions are favoured over intermolecular reactions. I think you'd get some polymer as a
byproduct however you'd want the bromothiol to be added slowly to the base with stirring to stop polymer formation.
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Nicodem
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Quote: Originally posted by sternman318  | | 1. I understand that the thiol is a stronger nucleophile than OH-, but would the OH- not react first, before the thiol is able to twist over itself to
react at the bromine? I understand that molecules undergo very rapid, constant conformational changes, but is that not quite a large movement?
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You are comparing a proton transfer (these are just about the faster known type of reactions) with an intermolecular SN2 substitution of bromide with
hydroxide (generally a very slow reaction, certainly >9 orders of magnitude slower than proton transfer). So, the answer is in the kinetic control,
though organic chemists generally don't even consider proton transfers as reactions to be considered as potentially kinetically relevant as long as
the pKa difference is less than 7 magnitudes (they are simply so fast that the proton transfer is taken as a given fact before taking other
mechanistic steps into consideration).
There are many exceptions though. For example, if you have a reaction that is about 7 magnitudes slower than proton transfers (which is still mighty
fast!) and a proton transfer irreversibly gives you a side product, then you can consider the proton transfer rate as a highly relevant factor (e.g.,
such cases can be enolizations during grignard reagent additions on some acidic aldehydes/ketones, particularly on dicarbonyl compounds, etc.).
| Quote: | | 2. What is to stop the thiol of another molecule from reacting at the bromine instead, essentially creating a polymer? |
Again kinetics. The rate of an intermolecular attack is negligible if you compare it to the intramolecular attack. You need a bit of playing with a
molecular model and understanding of the SN2 transition state geometry to understand why is this so. For the sake of simplicity you can use a rule of
thumb which says the kinetics are fastest in the SN2 based cyclisations to 3, 5 and 6-membered rings (particularly the 3 membered rings form readily).
4-Membered rings are extremely difficult to form via SN2, if at all (they will much more likely be subject to intermolecular attack). Note that the
readiness of formation of cyclic compounds via an SN2 does not correlate with thermodynamic stability (3-membered cycles are the less
thermodynamically stable!). For cyclizations based on polar mechanisms you can use the Baldwin'r rules to understand which of these proceed readily and which don't.
Note also that the product tetrahydrothiophene is still a relatively nucleophilic compound and can thus easily react with electrophiles like alkyl
bromides to give the corresponding sulfonium salt. As a curiosity, if it would for some reason react with 4-bromobutane-1-thiol it would form a
sulfonium salt which itself can cyclize into two molecules of tetrahydrothiophene in the presence of a base. You can draw down such a pathway for an
exercise.
The starting compound itself does not look stable at all, so all this looks like a pretty much "made up exercise" for students. Thiols react with
alkyl bromides even in the absence of a base. And NaH and NaOH are total overkill. Even NaHCO3 would suffice.
Please open homework or pedagogic threads in the Beginnings section where I'm moving this.
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Nicodem
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Thread Moved
19-10-2011 at 07:40 |
fledarmus
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5 and 6 membered rings are fairly easy to form by intramolecular reactions. The bonds are all free to rotate, and if you play with a set of models for
a while, you will see how easy it is to get the sulfur anion to line up perfectly with the back of the carbon bonded to Br, a perfect geometry for Sn2
reactions. 4 membered rings are very difficult to form in this manner because it requires a lot of torsional strain in the bonds to get the sulfur to
line up in the proper geometry. Larger rings are less difficult than the 4 membered ring but harder than 5 and 6, because there are so many other ways
to arrange all the atoms that getting the perfect geometry for backside attack isn't as strongly favored and intermolecular reactions predominate. If
you do the reaction in a very dilute solution, however, you can reduce the chances that two separate molecules will run into each other, minimizing
the intermolecular reactions, and allowing enough time for the two ends of each molecule to get into the right position to allow the intramolecular
reactions.
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SmashGlass
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All of the above being true. As well as thiols being more acidic than their corresponding alcohol
counterparts make them far more reactive in basic solutions and in the case of this 5 membered ring
intermolecular SN2 reactions.
Thiols are famous for forming thio-ethers when reacting an alkyl halide with a soft metal thiol, such as NaSH.
While the corresponding NaOH with an alkyl halide would primarily give the alcohol. (In stiochiometric amounts).
If it ain't broke don't fix it....
Now where are my screwdrivers?  |
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sternman318
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Thank you very much everyone for your responses!! Definately cleared up all of my questions.
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