Organic Chem- Protic and Aprotic solvent!!

[ammarm]

The question is as follows:

treatment of 2-bromoethanol with OH-/H2O results in:
a) HO-CH2-CH2-OH
b) Oxirane
c) Acetone
d) Oxalic acid
e) None of these

Can anyone figure the answer to this, and i was also wondering if OH-/H2O was used as a protic or aprotic solvent (in other words, is this reaction SN1 or SN2). I am a bit confused about it, so please help. Thanks

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[Maygyver]

Is it B?

 

[K Niner]

The question is as follows:

treatment of 2-bromoethanol with OH-/H2O results in:
a) HO-CH2-CH2-OH
b) Oxirane
c) Acetone
d) Oxalic acid
e) None of these

Can anyone figure the answer to this, and i was also wondering if OH-/H2O was used as a protic or aprotic solvent (in other words, is this reaction SN1 or SN2). I am a bit confused about it, so please help. Thanks

solvents don't act as either protic or aprotic. They are either protic or aprotic, its a characteristic. I'm on my phone, so it takes to long to exlain, but look up the difference. it will be on the dat.

 

[RCT PC CRN]

Isn't this stright Sn2? -Br = great leaving group -OH = good nucleophile

Sn2 is favored by Polar aprotic solvent. We have H2O = Polar protic solvent but would not make much difference cause we have great leaving group and good nuclephile. Top of that primary alcohol - no steric hindrance.

Answer: A

Am I correct?

 

[RCT PC CRN]

I just remembered the following conversation I had with my Orgo prof.

Polar protic or aprotic solvents add to the effect of Sn2 or Sn1 (i.e. Sn2 favored by polar aprotic and Sn1 favored by polar protic) but if you have good Sn2 going and you add polar protic solvent won't make it a Sn1 or vice-versa.

More thoughts !!!

 

[Maygyver]

Isn't this stright Sn2? -Br = great leaving group -OH = good nucleophile

Sn2 is favored by Polar aprotic solvent. We have H2O = Polar protic solvent but would not make much difference cause we have great leaving group and good nuclephile. Top of that primary alcohol - no steric hindrance.

Answer: A

Am I correct?

That makes sense to me. It seems weird having both an OH group and a Br group on there too though which made me think epoxide. Not sure though 🙁

 

[UndergradGuy7]

I didn't really review this stuff yet, but from what I remember I thought it would make the epoxide in answer B. This is an "internal Sn2" displacement reaction.

The mechanism the OH- takes the proton from the alcohol and then the O- from the alcohol attacks the carbon with the Br causing it to leave.


Anyway as far as the Sn2 stuff goes. Sn2 wants a strong nucleophile and a polar aprotic solvent (one that does not have hydrogens to give off). It also prefers primary.

Common protic solvents are H2O, alcohols, etc.
Common aprotic solvents are DMSO, DMF, etc.

It is also important to remember that in an aprotic solvent nucleophicity increases going up in the halogen group. So F>Cl>Br>I. But in an protic solvent, the solvent stabilizes the nucleophile so it does not react as well. The bigger atoms are harder to surround by the solvent and are less stabilized and more reactive in a protic solvent. So for a protic solvent the nucleophicity is I>Br>Cl>F.

So I guess like the person above said, if you were doing an Sn2 look if it is tertiary, secondary, primary, etc and if the reagent is a strong or weak nucleophile to decide if it undergos Sn2, Sn1, E2 or E1. Usually you can also use the solvents to decide.

Correct me if I am wrong.

 

[RCT PC CRN]

Overlooked oxirane, mainly cause I didn't know oxirane is an epoxide (Ethylene oxide). Just looked at "How to make epoxide" and came with the same reactant that we have in queston. Alkali hydoxide (-X n -OH next to each other) and Base (OH-) to grab H. Though don't know the roll of H2O. May be just solvent for NaOH or anyother OH- !!

Now I'm more towards Answer B.

Maygyver: Thanks for the comment otherwise I wouldn't have looked oxirane.

 

[UndergradGuy7]

I had to look up the structures too 🙂. I didn't memorize the naming/common names yet.

 

[ammarm]

well the answer is B, it says it is an intramolecular SN2..

now my first question is as follows:
since OH-/H2O is of a polar protic characteristic wouldn't it undergo SN1?

secondly,
the halide is the primary jhalide so this will support SN2.. which contradicts my first question..

so i am confused.. can anyone help me clear my problem out?

 

[ammarm]

what undergradguy7 is perfectly correct.. i just have a problem that i have explain in the post reply above this message.. does anyone know where i am going wrong?

 

[UndergradGuy7]

what undergradguy7 is perfectly correct.. i just have a problem that i have explain in the post reply above this message.. does anyone know where i am going wrong?

Well you are right that the protic should favor Sn1. In this case I think we should just look at the carbon that it is primary and that it is using a strong nucleophile. These 2 overlap the protic solvent and make it Sn2.

Also I think that aprotic solvents are preferred, but a protic solvent can still be used just not as effectively. I am not 100% on this, just a guess.

Maybe someone else can better explain why there is a protic solvent, but it is Sn2.

 

[Maygyver]

Well, Sn1 generally wont happen on a primary carbon. But either way, whether it is SN1 or SN2 doesnt really matter. What hinted it to me was that there was a leaving group as well as an OH.

 

[UCB05]

IMO the solvent is used to increase the efficiency of the reaction. Sure H2O is protic, but considering it's in basic solution and partially deprotonated in the from of -OH, it's not a very good one. What it DOES do is push the reaction in favor of the oxirane/ethane diol equilibrium. -OH acts as a good nucleophile to replace Br- In this solution the -OH can either come from solution or intramolecularly, favoring the latter by virtue of vicinity. Either way the Br- is getting bumped off to create the oxirane or ethane-1,2-diol, which exist in equilibrium. Oxirane gets attacked by -OH to form ethanediol; ethanediol undergoes internal Sn2 to reform the oxirane. Since it's under basic conditions, the oxirane is favored. Who cares if the Br- comes off by itself (Sn1) or gets bumped off by -OH (Sn2)? (though I would also add it's a primary halide, and not in a great "protic" solvent, so it's likely to go Sn2) It's always a mix of the two reactions in any case.

 

[sfoksn]

So.. wouldn't answer technically be both A and B?

 

[UCB05]

So.. wouldn't answer technically be both A and B?

Teeeechnically there's both, but oxirane is favored over diol in basic solution. There are many ochem questions to which there are technically two or more answers. Which is the major product is implied 😛

 

[RSD2014]

Hey UndergradGuy7,

How do you know when "internal SN2" would occur? Something bout the molecule being 2 Carbon long? My initial answer was A, so it's wrong. I thought OH would have gone E2 and form a Diol.

 

[UndergradGuy7]

I don't think it matters how long the chain is. The thing that matters is that you have a halohydrin (a halogen on 1 carbon and a OH on the carbon next to it).

This is an intramolecular Williamson ether synthesis. You just have to memorize this reaction.
It would also be important to know that the O- attacks the halogen from the opposite side causing inversion of configuration at the carbon that has the leaving group (like a typical Sn2 displacement reaction).

Here is the normal Williamson ether synthesis. http://en.wikipedia.org/wiki/Williamson_ether_synthesis

The one we had in this problem is pretty much the same except the nucleophile and the leaving group are on the same molecule (intramolecular).

 

[RSD2014]

I don't think it matters how long the chain is. The thing that matters is that you have a halohydrin (a halogen on 1 carbon and a OH on the carbon next to it).

This is an intramolecular Williamson ether synthesis. You just have to memorize this reaction. It would also be important to know that the O- attacks the halogen from the opposite side causing inversion of configuration at the carbon that has the leaving group (like a typical Sn2 displacement reaction).

Do you mean the O- attacks the Carbon bearing the LG? Which causes the LG to leave and yield and inverted oxirane(?)

 

[luv8724]

Do you mean the O- attacks the Carbon bearing the LG? Which causes the LG to leave and yield and inverted oxirane(?)

-OH attacks the the Hydrogen of OH which deprotonates it... and electronegativity of deprotonated O will attack backside of Br which makes the oxirane and kicking Br out of the place.
This is called intramolecular Sn2 reaction...