orgo question

[polarmolar]

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If you have an:

R-OH and you add to it HCl

does the OH become protonated and leave via E1 or does it get substituted with the Cl through SN1?

 

[Electrons]

If you have an:

R-OH and you add to it HCl

does the OH become protonated and leave via E1 or does it get substituted with the Cl through SN1?

I think it depends on the R. Anti periplanar position needed also? Usually you need to add heat to get into E1. I think you would get SN reaction rather in this case.

Anyone, correct me if I'm wrong.

 

[polarmolar]

So:

R-OH -----Strong Acid + heat-----> E1

 

[doc3232]

Tell us what the R is. We can go from there.

Can someone explain the antiplanar thing, I really don't get that.

 

[polarmolar]

lets assume all three, that R is primary, secondary or tertiary Carbon...how would each react.

 

[artanis]

lets assume all three, that R is primary, secondary or tertiary Carbon...how would each react.

You have to take a look at the solvent as well.

If its Polar Aprotic, its gonna be SN2 or E2, i.e. DMSO, DMF, Acetonitrile
Polar Protic, its gonna be SN1 or E1

If its tertiary, its definately gonna be a SN1 unless there is a BASE present.

If its secondary, thats where the gray area is. you gotta look at the solvent and if there is base.

If its Allylic and Secondary, or Benzylic and Secondary, then its gonna be most likely SN1.

I hope this helps.

 

[artanis]

Tell us what the R is. We can go from there.

Can someone explain the antiplanar thing, I really don't get that.

K there is 2 kinds of "periplanars". There is SYN periplanar, in which you the hydrogen you are attacking as well as the leaving group sit in SYN position together. Or, there is Anti periplanar in which the hydrogen and the leaving group are Trans to one another. Now if this is put into a cyclohexane molecule which it is commonly represented, the condition for an E2 is that the groups are Anti periplanar which is equivalent to "trans diaxial". For drawing a chair confirmation, your leaving group and the adjacent hydrogen being attacked must both be in a trans di-axial positioning in order for it to take place. Remember, ring flips are possible, so if given both in equatorial position, you can do a ring flip. I hope this helps..

 

[Electrons]

Anti peri planar...kinda like antiplanar (opposite almost 180 but tilt a little bit to the side)

Checkout the pix in the this link about antiperiplanar

http://en.wikipedia.org/wiki/Alkane_stereochemistry.

 

[harrygt]

There is no need to ask about a solvent as long as HCl and alcohols are missible. In such cases we don't use any solvent.
Answer: The major reaction will be an Sn1 substitution of Cl-.
The alcohol gets protonated by HCl, leading to R-OH2+. Water is a good leaving group, thus it leaves the R group, giving us a carbocation [R+]. Cl- then reacts with the carbocation to give us R-Cl. E1 can also occur, but it's much less likely. The reason would be the weak basicity of Cl- which does not make it able to detach a proton from the R group easilly.

This reaction is more feasible when the alcohol is tertiary or secondary. There would not be any noticeable reaction if the R group was primary, becuase 1) the primary carbocation would not be stable 2) Cl- is a weaker nucleaphile compared to OH-. Sn2 wouldn't occur in this case.

 

[polarmolar]

Thanks for that harrygt, thats what I thought!

One more thing, I remember reading that R-OH would undergo E1 in a strongly acidic solution and the example given was H2SO4 (strong acid). Isn't HCl a strong acid and would react the same way?

 

[harrygt]

Thanks for that harrygt, thats what I thought!

One more thing, I remember reading that R-OH would undergo E1 in a strongly acidic solution and the example given was H2SO4 (strong acid). Isn't HCl a strong acid and would react the same way?

This is a bit tricky. The mechanisms are the same. Heating the alcohol with H2SO4 would lead to the alkene [E1 happens], while heating it with HCl does not give you that much of alkenes.
Here is the reason: HSO4- is not a nucleophile, so you will never get R - HSO4. Therefore, the only product left after hours of heating the mixuture would be the alkene which is still hard to produce.
As mentioned in my previous post, Cl- has nucleophilic characteristics and can react with the carbocation, leading to the formation of R - Cl. The alkene would be the minor product in this case.

 

[polarmolar]

This is a bit tricky. The mechanisms are the same. Heating the alcohol with H2SO4 would lead to the alkene [E1 happens], while heating it with HCl does not give you that much of alkenes.
Here is the reason: HSO4- is not a nucleophile, so you will never get R - HSO4. Therefore, the only product left after hours of heating the mixuture would be the alkene which is still hard to produce.
As mentioned in my previous post, Cl- has nucleophilic characteristics and can react with the carbocation, leading to the formation of R - Cl. The alkene would be the minor product in this case.

gotcha!