[xerthelias]

A little bit of background: This was a problem on the organic chemistry test that I recently took. We were given 1 mol of HBr and an unknown amount of acetylene gas and were told to predict the product of the two reactants. From what I know, acetylene has a triple bond and should undergo an addition reaction with the HBr. The first thing I did after reading this question was ask the teacher exactly how much acetylene gas was given. His reply was that you could take as much as you want. After hearing this, I took half a mole of acetylene (I wrote down how much I took on the side of the paper), reacted it with 1 mol of HBr and created C₂H₄Br₂. My exact answer was the following:

C₂H₂ + 2HBr C₂H₄Br₂

I got the question wrong; I don't know why. I tried asking the teacher but got very little useful information out. Apparently, since there was 1 mol of HBr, I was supposed to only break one of the bonds of Acetylene and add only one H and one Br. I don't get the logic behind this since the mols of acetylene is a variable. The correct answer was the following:

C₂H₂ + HBr C₂H₃Br

This was the reasoning behind my answer:

What we know is that we have 1 mol of HBr and an unknown amount of acetylene. I was told that that number mols of acetylene present was up for me to decide.

Moving on, what I thought was that if there are twice as many HBr molecules as there are acetylene molecules, we can break both of the pi bonds of acetylene and make it ethane. We could then add 2 bromines to one carbon and the remaining 2 hydrogens into the other to make it 1,1-dibromoethane or C₂H₄Br₂. So, if I make the ratio of HBr to C₂H₂ 2:1, I should be able to create what I wrote on the test. But this is only for if there's 0.5 moles of acetylene.

If, however, we had one mol of acetylene, the number of acetylene molecules would equal the number of HBr molecules. Each acetylene could only, therefore, take in one HBr molecule. In other words, we will not have enough HBr molecules to break both of the pi bonds of all the acetylene molecules; we will only have enough to break a single pi bond. If this were the case, I could definitely see why the product would be C₂H₃Br. The HBr would simply break a single pi bond and bond with the acetylene. At the end of the reaction, we'd get C₂H₃Br. Simple enough.

End of reasoning

I'm not sure what I'm doing wrong. I was told that if there were 2 moles of HBr (instead of one), that the answer would be something else. (I think this means that if there were two moles, I would be correct. I don't know, I could be wrong). Can someone please explain this reaction and point out exactly where I made the errors in my thinking process?

Tl;DR: n_HBr = 1.0 mols. What is the product when you react C₂H₂ with HBr? I wrote 1,1-dibromoethane on the test but but the question wrong. Right answer: 1-bromoethene. Why is the latter produced and not the former?

[BetaAmyloid]

Maybe it has something to do with Markovnikov selectivity. The original acetylene would allow bromine to attack either carbon 1 or 2; however, not only is there spatial interference in addition of another bromine, but I'm assuming that the reactivity is no longer high enough for HBr to be selective towards that same carbon. And then HBr doesn't attack the other carbon because the antimarkovnikov reaction is not energetically favorable enough to complete.

I'm not sure if this is exact, hopefully someone confirms or disproves what I said!

Otherwise, take this with a grain of salt until then.

[helenn]

The HBr is the limiting reagent, you only have 1 mole available to react. No matter how much acetylene you have, 1 mole HBr will react with 1 mole acetylene to give C2H3Br and the excess acetylene will not react. If you have 2 moles of HBr it can now react with 2 moles of acetylene to give C2H4Br2.

[Dan]

The HBr is the limiting reagent, you only have 1 mole available to react. No matter how much acetylene you have, 1 mole HBr will react with 1 mole acetylene to give C2H3Br and the excess acetylene will not react.

If HBr is the limiting reagent. The OP's point is that HBr was not specified to be the limiting reagent.

If you have 2 moles of HBr it can now react with 2 moles of acetylene to give C2H4Br2.

No, 2 mol HBr can react with 1 mol acetylene to give 1 mol 1,1-dibromoethane. Balance the equation.
Equally, 1 mol HBr can react with 0.5 mol acetylene to give 0.5 mol 1,1-dibromoethane.

Moving on, what I thought was that if there are twice as many HBr molecules as there are acetylene molecules, we can break both of the pi bonds of acetylene and make it ethane. We could then add 2 bromines to one carbon and the remaining 2 hydrogens into the other to make it 1,1-dibromoethane or C₂H₄Br₂. So, if I make the ratio of HBr to C₂H₂ 2:1, I should be able to create what I wrote on the test. But this is only for if there's 0.5 moles of acetylene.

If, however, we had one mol of acetylene, the number of acetylene molecules would equal the number of HBr molecules. Each acetylene could only, therefore, take in one HBr molecule. In other words, we will not have enough HBr molecules to break both of the pi bonds of all the acetylene molecules; we will only have enough to break a single pi bond. If this were the case, I could definitely see why the product would be C₂H₃Br. The HBr would simply break a single pi bond and bond with the acetylene. At the end of the reaction, we'd get C₂H₃Br. Simple enough.

End of reasoning

You are correct in your reasoning. In the presence of excess HBr, you can get 1,1-dibromoethane. If HBr is not in excess, you would expect bromoethene as the major product.

There are two possibilities:

1. You didn't read the question properly and it refers to 1 molar equivalent of HBr. In this case while your reasoning is sound, you didn't answer the question.
2. The question is poorly constructed, and ambiguous as a result. In this case both 1,1-dibromoethane and bromoethene could be correct answers. If you are allowed to choose any reagent stoichiometry you want then I don't see why they have specified the mols of HBr since it is irrelevant. I am puzzled as to why your teacher told you you could choose any stoichiometry and would then mark your answer wrong.

It sounds like #2 is the case, and what they meant was #1 (molar equivalents rather than moles). Your teacher probably did not personally write the question, did not realise the ambiguity, gave bad advice and will not admit the error. If you post the original question verbatim we can check.

[sjb]

Even if the ratio was 1:1, there are cases where the intermediate reacts faster than the starting material.

So you might have

2 acetylene + 2 HBr  1 vinyl bromide + 1 HBr + 1 acetylene; and the question is now if vinyl bromide + HBr  dibromoethane faster than acetylene + HBr  vinyl bromide. In this case I think it's likely that reaction of the alkyne is faster due to sterics if nothing else

[helenn]

Quote from: helenn

    If you have 2 moles of HBr it can now react with 2 moles of acetylene to give C2H4Br2.


No, 2 mol HBr can react with 1 mol acetylene to give 1 mol 1,1-dibromoethane. Balance the equation.
Equally, 1 mol HBr can react with 0.5 mol acetylene to give 0.5 mol 1,1-dibromoethane.

Apologies, I meant 2 moles HBr reacts with 1 mole acetylene, my fingers didn't type what my brain was thinking! And I clearly didn't read the OP thoroughly and wrote a rushed answer.

After re-reading the original post I assumed that when the teacher said you can have as much acetylene as you want he was taking it to be in excess.

Your answer as written is correct, your reasoning and understanding make sense. I suspect the teacher was looking for one HBr (1 equivalent) in the equation not 2 as you have written.

[BetaAmyloid]

The question is poorly constructed, and ambiguous as a result. In this case both 1,1-dibromoethane and bromoethene could be correct answers. If you are allowed to choose any reagent stoichiometry you want then I don't see why they have specified the mols of HBr since it is irrelevant. I am puzzled as to why your teacher told you you could choose any stoichiometry and would then mark your answer wrong.

I don't want to be offensive because I know your tag says organic chemist and by all means I've only had organic 1 and 2, so with saying that:

I thought 1,1-dibromoethane would not form. 1 equivalent of acetylene with 1 equivalent of HBr would form vinyl chloride, and then 1 equivalent of acetylene and 2 equivalents of HBr would then form 1,2-dibromoethane.

I say that because after the electrophilic attack of one set of pi electrons to the proton, and then that addition of the bromine, there would be a second electrophilic attack of the second set of pi electrons to a second equivalent proton. But between this step a bromonium ion would form, and either way cause the second positive charge to form on the second carbon, thus causing both bromines to be on opposite carbons.

So basically:
Acetylene + HBr -> 1-bromoethene
1-bromoethene + HBr -> 1,2-bromoethane (due to the formation of a bromonium ion, thus causing the addition of the second bromine to be more energetically favored in the trans form)

I'm not saying that the resonance structure with the positive charge on carbon 1 for the second addition of bromine would NEVER happen, but the bromonium resonance giving the rearrangement of attack would be more so favored and thus be the major product if not 100%.

Is that reasoning wrong?

[Dan]

I don't want to be offensive because I know your tag says organic chemist

It is not offensive to question!

In the protonation of vinyl bromide, you have two options.

1. Protonate at C1, leaving a carbocation at C2, which could be stabilised by bromonium formation.
2. Protonate at C2, leacing a carbocation at C1, which is stabilised by resonance.

It seems that #2 wins, and 1,1-dibromoethane is the major product.

1,2-dibromoethane is formed preferentially in the presence of peroxides (by a radical mechanism).

For a detailed classical study: http://pubs.acs.org/doi/pdf/10.1021/ja01333a048

[BetaAmyloid]

Ah...so the normal reaction is 1,1-dibromoethane but in the lab if vinyl chloride has been sitting around it would probably yield 1,2-dibromoethane.

Gotcha, difference between theoretical and reality I suppose.

Thanks for the paper Dan, I made a post called "The Peroxide Effect," if you don't mind checking it out.

[xerthelias]

You are correct in your reasoning. In the presence of excess HBr, you can get 1,1-dibromoethane. If HBr is not in excess, you would expect bromoethene as the major product.

There are two possibilities:

1. You didn't read the question properly and it refers to 1 molar equivalent of HBr. In this case while your reasoning is sound, you didn't answer the question.
2. The question is poorly constructed, and ambiguous as a result. In this case both 1,1-dibromoethane and bromoethene could be correct answers. If you are allowed to choose any reagent stoichiometry you want then I don't see why they have specified the mols of HBr since it is irrelevant. I am puzzled as to why your teacher told you you could choose any stoichiometry and would then mark your answer wrong.

It sounds like #2 is the case, and what they meant was #1 (molar equivalents rather than moles). Your teacher probably did not personally write the question, did not realise the ambiguity, gave bad advice and will not admit the error. If you post the original question verbatim we can check.

I know for a fact that I did not misread the question; it was what I described it to be. Recalling from memory, it was something along the lines of "Acetylene reacts with one Mol of HBr, predict the reaction and draw the structural diagrams." (My memory is kinda bad so I can't grantee 100% accuracy of that quote). It only gave one mol of HBr and the mols of ethyne were variable (I have the exact quote of what the teacher said written down on the paper as well as the assumption of 0.5 mols of acetylene). That being said, you're probably correct your second prediction of what likely happened (the question was definitely ambiguous). Now, here's the other problem: I've already argued with the teacher over this matter and he's already dismissed me by saying that I was incorrect and that he doesn't want to discuss this further. Because of this, I'm not sure if I should bring up the matter again (which I would need to do to get a picture of the exact question and the work I did on the test). Getting the marks for this question could raise my test grade from a 92% to a 94%. I'm not sure what to do. The teacher was fairly generous with the marking scheme so I do not want to annoy him too much with this two-mark question.

Furthermore, even if I do decide to bring the matter up with the teacher, I can't just say that a guy called Dan at an online chemistry forum told me that I could be correct. The chances that I would accomplish anything are slim to none; he could just as easily dismiss me again (because I won't be able to hold my side of the argument against him). After all, he's a chemistry teacher with over 20 years of teaching experience and I'm just some kid who only learned organic chemistry in the past month. :/ Lets see though, I'll try to talk to him tomorrow and maybe show him this thread. After that, I'll have to accept whatever mark I get.

[Dan]

You could show him some references, I provided one, there are others. The reaction of alkynes with excess HX to form geminal dihalides is in the textbooks after all.

However, this is not worth a fight. Let it go. There are plenty of teachers who will refuse to acknowledge their mistakes - unfortunately they are the same ones that will deliberately make your life harder and harder if you keep bringing it up. If you needed that extra 2% to pass then argue, but for the sake of 92 to 94 take the moral victory.

What you can take away is that if the moles of only one reagent are specified, you can probably assume the other reagent is 1 mol - i.e. they mean molar equivalents

[xerthelias]

You could show him some references, I provided one, there are others. The reaction of alkynes with excess HX to form geminal dihalides is in the textbooks after all.

However, this is not worth a fight. Let it go. There are plenty of teachers who will refuse to acknowledge their mistakes - unfortunately they are the same ones that will deliberately make your life harder and harder if you keep bringing it up. If you needed that extra 2% to pass then argue, but for the sake of 92 to 94 take the moral victory.

What you can take away is that if the moles of only one reagent are specified, you can probably assume the other reagent is 1 mol - i.e. they mean molar equivalents

You're right. I tried to approach him about this after but nothing good came out of it. I think I'll just let it go this time. Thanks for all your help.