adding H-X to alcohol

[joonkimdds]

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Hi~

If we add H-X, such as H-Cl to 3-methyl-butanol, H from HCl and OH will form H2O and leave, and then there will be an rearrangement of carbocation to more stable carbocation, and then Cl will be attached.

so that was an example with 3-methyl-butanol.


I looked at another problem.
CH3CH(OH)CH(CH3)2 + catalytic amount of HCl

and the answer was 2-methyl-2-butene.

If I follow the pattern from the first example with 3-methyl-butanol,
OH and H will form H2O and leave, there will be rearrangement of carbocation and Cl should attack that carbocation, so how come the answer is 2-methyl-2butene instead of 2-chloro-2-methyl-butene?

does catalytic amount have something to do with it?

 

[Flipper405]

The mechanism isn't quite right. The OH will spontaneously leave first to form the tertiary carbocation, then the H will be pulled off by a base (Cl- or H2O for example) to form the alkene. This is a classic example of an E1 mechanism. The more stable alkene will form in both cases due to Zaitsev's rule.

The catalytic amount is important because if there were a significant amount of H-X, the SN1 substitution would be more likely to occur.

 

[joonkimdds]

The mechanism isn't quite right. The OH will spontaneously leave first to form the tertiary carbocation, then the H will be pulled off by a base (Cl- or H2O for example) to form the alkene. This is a classic example of an E1 mechanism. The more stable alkene will form in both cases due to Zaitsev's rule.

The catalytic amount is important because if there were a significant amount of H-X, the SN1 substitution would be more likely to occur.

ah~ so the first example was substitution and second example was elimination?

does catalytic amount = significant amount?
the example with catalytic amount of HCl was elimination reaction, not substitution.

btw, are there any other rules besides catalytic amount that decides whether it will undergo Elimination or Substitution?

 

[Flipper405]

Well, based on the answers you gave, they're both eliminations. The Cl- CAN attach to the carbocation, but it is not likely to happen. There is not very much in there; catalytic amount means very little of it.

There are some other things that will make it tend to go one way or the other. However, especially with SN1/E1, they will be competing. Both SN1 and E1 products will be formed most of the time.

 

[dontwakeme]

The hydroxyl group is a poor leaving group. Thus, it is protonated by the HCl to form water, which is a good leaving group. OH will never leave by itself.

Like he said above, catalytic amount just means a very small amount to get the reaction going i.e. protonate the OH. Thus, you're probably not going to be seeing much substitution. The major product would probably come from elimination. This question is pretty ambiguous. You would really need to go in the lab and actually do the experiment to see what products would really be formed because both SN1 and E1 are competitive.

 

[joonkimdds]

How do we know if this is E1 or E2?


can it be
step 1 = H-Cl bond is broken, the H forms a bond with OH, making a good leaving group of H2O.
step 2 = before H2O leaves, Cl pulls H from beta carbon, form double bond, and H2O leaves at the same time [transition state]

or does H2O have to leave first and how do we know?


p.s. Flipper, when r u gonna take DAT? U've been at SDN for forever.

 

[Flipper405]

Yeah I have lol...

I'm was going to take it in January, but I realized pretty soon after I started studying that I need a lot more time because my bio knowledge sucks. At this point I'm shooting for March. I'm taking both biochemistry and microbiology next semester so that should help with the bio section some hopefully.

dontwakeme is correct; I was incorrect in saying that OH leaves, it is actually H2O that leaves after the strong acid H-X protonates the OH.

I think the best way to tell about E1 vs E2 is probably the base strength. You don't have a strong base to pull off the H immediately. Also, because the carbon bearing the H2O is tertiary, it is relatively easy to form a carbocation. These indicate E1.

 

[dontwakeme]

1st order reactions (Sn1,E1) occur when 1) there is no good nucl/base 2) OH is protonated to form water as LG 3) a stable carbocation can be formed. So, clearly this qualifies as first order.

As for whether it is Sn1 or E1, it's tough to tell. I don't really see why this would necessarily be E1. It could just as easily be Sn1. Cl- could be a good nucleophile and do Sn1. It's not possible to predict unless you actually do the experiment. I doubt they give a question like this without atleast some more detail.