Why Alkenes React with HBR in the Cold while not with other Hydrogen Halides?

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GMAIL USER
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Kindly, answer this if you can.
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pj80602
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I don't know why but you must be trying to say "Hydrogen Halides" I guess.
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GMAIL USER
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Yeah, I mistyped in a hurry. I should correct it now.
Thank you.
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David Getling
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I'd not heard previously of this being the case, but if it is then the reason is as follows.

You have two competing effects. The polarity of the HX and the strength of the HX bond. A more polar molecule is better at attracting the alkene's pi bond electrons. In decreasing order the polarity is HF, HCl, HBr and HI. However this is also the bond strength order, and we need to break the HX bond, so weaker is better. Hence HBr represents the compromise between HX having enough polarity but also a bond that isn't too hard to break.
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GMAIL USER
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It makes sense for me to understand what you are trying to say but how about the HCL reaction with Alkenes? HCL must be having polarity enough to attract one of the pi electrons in the double bond of alkenes as well and must be having a bond strength to get broken and react with alkenes without having to give a warm environment to the reaction or heating it up but the cold condition does not satisfy for this.

Thank you for the response though. Appreciated it.
(Original post by David Getling)
I'd not heard previously of this being the case, but if it is then the reason is as follows.

You have two competing effects. The polarity of the HX and the strength of the HX bond. A more polar molecule is better at attracting the alkene's pi bond electrons. In decreasing order the polarity is HF, HCl, HBr and HI. However this is also the bond strength order, and we need to break the HX bond, so weaker is better. Hence HBr represents the compromise between HX having enough polarity but also a bond that isn't too hard to break
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MexicanKeith
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(Original post by GMAIL USER)
It makes sense for me to understand what you are trying to say but how about the HCL reaction with Alkenes? HCL must be having polarity enough to attract one of the pi electrons in the double bond of alkenes as well and must be having a bond strength to get broken and react with alkenes without having to give a warm environment to the reaction or heating it up but the cold condition does not satisfy for this.

Thank you for the response though. Appreciated it.
The arguments to do with polarity and attracting the alkene are irrelevant, both hydrogen halide and alkene are overall neutral and so any interactions between them will be interactions of permanent dipole with induced dipole which drop off very rapidly with separation between the molecules. Besides the whole reaction is happening in a solvent which will be shielding the two reagents from eachother influence anyway (until they are right next to each other).
It's also worth pointing out that all the reactions with the different hydrogen halides with be exothermic and have similar entropy changes associated with them and so we dont need to worry about thermodynamic considerations.
This therefore is a question of rates (or kinetics) so we have to think about the energy activation barrier, ie, how much energy we have to put in to break bonds in order to make the reaction go.
In this reaction the first step involves the breaking of the C-C pi bond, the formation of a C-H bond and the breaking of the H-X bond. The first two of these processes arent important for the relative rates because they dont involve H-X, so all we need to worry about is the strength of the H-X bond
strongest ---------> weakest
H-F > H-Cl > H-Br > H-I

so the rate is faster for the heavier hydrogen halides, because their bonds are most easily broken. This means that at high temperature all reactions will take place (but with H-Br reacting quicker than H-Cl for example) and as we lower temperature they will all get slower. H-Br will now react slower than before but still fast enough that the reaction happens, H-Cl on the other hand will have such a slow rate that you dont see reaction (unless you leave the reaction for a very long time).
(Bonus If you know about Arrhenius' equation) This is the same as saying Ea in the Arrhenius equation is smaller for the reactions with heavier hydrogen halides. If the temperature is high then the exponential term in the equation tends to one and the rate doesnt depend very strongly on temperature. As the temperature gets lower however, the exponential term gets smaller and smaller until the rate becomes negligible and the reaction effectively stop. The temperature at which any given reaction stops depends on Ea, for small Ea the reaction will only stop at very low temperatures, for big Ea the temperature will stop at less cold temperatures.

For the reaction with H-Br Ea is small enough that the reaction still happens at lowish temperatures, for the reaction with H-Cl Ea is bigger so the reaction does not proceed at an appreciable rate at the same low temperatures.

sorry for the essay, hope it makes sense
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GMAIL USER
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#7
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It definitely helped me more than I needed. Appreciated your efforts.


Thank you for the response.
(Original post by MexicanKeith)
The arguments to do with polarity and attracting the alkene are irrelevant, both hydrogen halide and alkene are overall neutral and so any interactions between them will be interactions of permanent dipole with induced dipole which drop off very rapidly with separation between the molecules. Besides the whole reaction is happening in a solvent which will be shielding the two reagents from eachother influence anyway (until they are right next to each other).
It's also worth pointing out that all the reactions with the different hydrogen halides with be exothermic and have similar entropy changes associated with them and so we dont need to worry about thermodynamic considerations.
This therefore is a question of rates (or kinetics) so we have to think about the energy activation barrier, ie, how much energy we have to put in to break bonds in order to make the reaction go.
In this reaction the first step involves the breaking of the C-C pi bond, the formation of a C-H bond and the breaking of the H-X bond. The first two of these processes arent important for the relative rates because they dont involve H-X, so all we need to worry about is the strength of the H-X bond
strongest ---------> weakest
H-F > H-Cl > H-Br > H-I

so the rate is faster for the heavier hydrogen halides, because their bonds are most easily broken. This means that at high temperature all reactions will take place (but with H-Br reacting quicker than H-Cl for example) and as we lower temperature they will all get slower. H-Br will now react slower than before but still fast enough that the reaction happens, H-Cl on the other hand will have such a slow rate that you dont see reaction (unless you leave the reaction for a very long time).
(Bonus If you know about Arrhenius' equation) This is the same as saying Ea in the Arrhenius equation is smaller for the reactions with heavier hydrogen halides. If the temperature is high then the exponential term in the equation tends to one and the rate doesnt depend very strongly on temperature. As the temperature gets lower however, the exponential term gets smaller and smaller until the rate becomes negligible and the reaction effectively stop. The temperature at which any given reaction stops depends on Ea, for small Ea the reaction will only stop at very low temperatures, for big Ea the temperature will stop at less cold temperatures.

For the reaction with H-Br Ea is small enough that the reaction still happens at lowish temperatures, for the reaction with H-Cl Ea is bigger so the reaction does not proceed at an appreciable rate at the same low temperatures.

sorry for the essay, hope it makes sense
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David Getling
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(Original post by MexicanKeith)
The arguments to do with polarity and attracting the alkene are irrelevant, both hydrogen halide and alkene are overall neutral and so any interactions between them will be interactions of permanent dipole with induced dipole which drop off very rapidly with separation between the molecules. Besides the whole reaction is happening in a solvent which will be shielding the two reagents from eachother influence anyway (until they are right next to each other).
It's also worth pointing out that all the reactions with the different hydrogen halides with be exothermic and have similar entropy changes associated with them and so we dont need to worry about thermodynamic considerations.
This therefore is a question of rates (or kinetics) so we have to think about the energy activation barrier, ie, how much energy we have to put in to break bonds in order to make the reaction go.
In this reaction the first step involves the breaking of the C-C pi bond, the formation of a C-H bond and the breaking of the H-X bond. The first two of these processes arent important for the relative rates because they dont involve H-X, so all we need to worry about is the strength of the H-X bond
strongest ---------> weakest
H-F > H-Cl > H-Br > H-I

so the rate is faster for the heavier hydrogen halides, because their bonds are most easily broken. This means that at high temperature all reactions will take place (but with H-Br reacting quicker than H-Cl for example) and as we lower temperature they will all get slower. H-Br will now react slower than before but still fast enough that the reaction happens, H-Cl on the other hand will have such a slow rate that you dont see reaction (unless you leave the reaction for a very long time).
(Bonus If you know about Arrhenius' equation) This is the same as saying Ea in the Arrhenius equation is smaller for the reactions with heavier hydrogen halides. If the temperature is high then the exponential term in the equation tends to one and the rate doesnt depend very strongly on temperature. As the temperature gets lower however, the exponential term gets smaller and smaller until the rate becomes negligible and the reaction effectively stop. The temperature at which any given reaction stops depends on Ea, for small Ea the reaction will only stop at very low temperatures, for big Ea the temperature will stop at less cold temperatures.

For the reaction with H-Br Ea is small enough that the reaction still happens at lowish temperatures, for the reaction with H-Cl Ea is bigger so the reaction does not proceed at an appreciable rate at the same low temperatures.

sorry for the essay, hope it makes sense
WRONG! If the polarity didn't matter then we would expect the reaction to be fastest with HI. Also, there's absolutely no reason why the reaction should be in a solvent. For the first few alkenes one would do it in the gas phase, and for subsequent ones the alkene itself is a liquid.
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GMAIL USER
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#9
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Can you provide me with a reference to what you are saying so I can verify this being the case, please?


Thank you for the response in pointing out if the preceding one was wrong.
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MexicanKeith
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(Original post by David Getling)
WRONG! If the polarity didn't matter then we would expect the reaction to be fastest with HI. Also, there's absolutely no reason why the reaction should be in a solvent. For the first few alkenes one would do it in the gas phase, and for subsequent ones the alkene itself is a liquid.
I believe the reaction with HI is the fastest. Ofcourse you're correct that solvents are strictly necessary, although in practise one would almost always use a solvent to avoid having to deal with gaseous H-X.

A little search of the literature has suggested that, at least for some small alkenes, the "gas phase" reaction is in fact heterogeneously catalysed by the reaction vessel and so reaction actually takes place at the surface, interesting!

either way, dipole dipole interactions are not strong enough to alter the activation barriers of these reactions in any significant way, and so as far as I am aware they follow the trend of H-X bond strengths.

I'd be interested to hear from GMAIL USER as to where this question comes from and if the source makes any allusion to H-I.
I expect it must also react in the cold, because the reaction involving H-I must have the smallest activation barrier.
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GMAIL USER
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Sure!

Here is the reference to my question.
https://www.chemguide.co.uk/mechanis...dd/symhbr.html

On the 1st Line in the 2nd paragraph of the Page.
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MexicanKeith
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(Original post by GMAIL USER)
Sure!

Here is the reference to my question.
https://www.chemguide.co.uk/mechanis...dd/symhbr.html

On the 1st Line in the 2nd paragraph of the Page.
Thanks!
Nothing on that page suggests that H-I wouldn't also react in the cold, in fact it makes it clear that H-I reacts fastest, it must therefore have the smallest activation barrier and hence must also react in the cold!

I have to say chemguide is a very good place to start if you're doing A-levels!
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GMAIL USER
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(Original post by MexicanKeith)
Thanks!
Nothing on that page suggests that H-I wouldn't also react in the cold, in fact it makes it clear that H-I reacts fastest, it must therefore have the smallest activation barrier and hence must also react in the cold!

I have to say chemguide is a very good place to start if you're doing A-levels!


Okay. It means HCl, HBR and HI can easily react in the cold as they have low to lowest activation energies and does not require much temperature for their bonds to get broken and hence participate in a reaction with alkenes.
Why can't they react in warmer environment as well; will their electrons get loose out from their shell if we give them higher temperatures and will get more in a reactive state and form their diatoms?

On the other hand, HF being one of the highest electronegative elements require tenperature enough to get its bond broken so it must not react in the cold. Right?

Correct me if I am wrong, please. I would highly appreciate it.


Thank you!
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MexicanKeith
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(Original post by GMAIL USER)
Okay. It means HCl, HBR and HI can easily react in the cold as they have low to lowest activation energies and does not require much temoerature for their bonds to get broken and hence participate in a reaction with alkenes.
Why can't they react in warmer environment; will their electrons get loose out from their shell if we give them higher temperatures?

On the other hand, HF being one of the highest electronegative elements require tenperature enough to get its bond broken so it must not react in the cold. Right?

Correct me if I am wrong. I would highly appreciate it.


Thank you!
All of them will react fast when its very hot and the rate will decrease as it gets colder.

At a given temperature the rates are in the order H-F < H-Cl < H-Br < H-I

So, when its very hot all the reactions will take place (with rates in the order above)

a bit colder and the reactions will all slow down until at some point H-F (the slowest reactant because it has the strongest bond) is reacting so slowly that its basically not reacting at all. H-Cl, H-Br and H-I will still be reacting with rates still in the order H-Cl < H-Br < H-I.

Cool down more (to roughly room temperature) and all the reactions slow down again, now the next slowest reaction (with H-Cl) has slowed to a stand still. The reactions with H-Br and H-I are slower than before, but they're still taking place and their rates are still in the order H-Br < H-I.

Cooling down even more and the next slowest reaction (with H-Br) will become so slow it effectively stops. now only the reaction with H-I will still be happening.

If we kept getting colder and colder, the reaction with H-I would get slower and slower and eventualy that would effectively stop aswell.


SO:
weaker bond -> less energy needed to react
less energy needed to react -> faster rate of reaction
faster rate of reaction -> lower temperature needed before the reaction 'stops'
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GMAIL USER
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(Original post by MexicanKeith)
All of them will react fast when its very hot and the rate will decrease as it gets colder.

At a given temperature the rates are in the order H-F < H-Cl < H-Br < H-I

So, when its very hot all the reactions will take place (with rates in the order above)

a bit colder and the reactions will all slow down until at some point H-F (the slowest reactant because it has the strongest bond) is reacting so slowly that its basically not reacting at all. H-Cl, H-Br and H-I will still be reacting with rates still in the order H-Cl < H-Br < H-I.

Cool down more (to roughly room temperature) and all the reactions slow down again, now the next slowest reaction (with H-Cl) has slowed to a stand still. The reactions with H-Br and H-I are slower than before, but they're still taking place and their rates are still in the order H-Br < H-I.

Cooling down even more and the next slowest reaction (with H-Br) will become so slow it effectively stops. now only the reaction with H-I will still be happening.

If we kept getting colder and colder, the reaction with H-I would get slower and slower and eventualy that would effectively stop aswell.


SO:
weaker bond -> less energy needed to react
less energy needed to react -> faster rate of reaction
faster rate of reaction -> lower temperature needed before the reaction 'stops'
I have got my concepts cleared now.
I highly appreciated it.


Thank you.
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David Getling
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(Original post by MexicanKeith)
All of them will react fast when its very hot and the rate will decrease as it gets colder.

At a given temperature the rates are in the order H-F < H-Cl < H-Br < H-I

So, when its very hot all the reactions will take place (with rates in the order above)

a bit colder and the reactions will all slow down until at some point H-F (the slowest reactant because it has the strongest bond) is reacting so slowly that its basically not reacting at all. H-Cl, H-Br and H-I will still be reacting with rates still in the order H-Cl < H-Br < H-I.

Cool down more (to roughly room temperature) and all the reactions slow down again, now the next slowest reaction (with H-Cl) has slowed to a stand still. The reactions with H-Br and H-I are slower than before, but they're still taking place and their rates are still in the order H-Br < H-I.

Cooling down even more and the next slowest reaction (with H-Br) will become so slow it effectively stops. now only the reaction with H-I will still be happening.

If we kept getting colder and colder, the reaction with H-I would get slower and slower and eventualy that would effectively stop aswell.


SO:
weaker bond -> less energy needed to react
less energy needed to react -> faster rate of reaction
faster rate of reaction -> lower temperature needed before the reaction 'stops'
Take a look at the very first sentence of my first reply. Yes my university text (not some noddy school one) says that generally the order is HI > HBr > HCl > HF. This is clearly due to the bond strengths. However IF the question given in the title isn't wrong (maybe it is) then the only explanation I can think of for HBr reacting faster than HI is down to polarity. Yes, the colliding molecules need at least the activation energy, but they also need to be lined up suitably, and maybe, at cold (whatever is meant by that) temperatures the polarity becomes important in achieving this.
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MexicanKeith
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(Original post by David Getling)
Take a look at the very first sentence of my first reply. Yes my university text (not some noddy school one) says that generally the order is HI > HBr > HCl > HF. This is clearly due to the bond strengths. However IF the question given in the title isn't wrong (maybe it is) then the only explanation I can think of for HBr reacting faster than HI is down to polarity. Yes, the colliding molecules need at least the activation energy, but they also need to be lined up suitably, and maybe, at cold (whatever is meant by that) temperatures the polarity becomes important in achieving this.
I think we can conclude that the original question is slightly incomplete, and should say HBr reacts in the cold whilst HCl and HF do not.

In terms of orientation based arguments, if we were to consider the orbitals actually involved in the reaction we have the alkene HOMO (pi bonding orbital) donating into the HX LUMO (sigma antibonding orbital). Thinking of the LUMO molecular orbital in terms of the linear combination of atomic orbitals we know that it's largest component must always be at the opposite end of the molecule from the largest component of the bonding orbital.

clearly as you move from HF to HI the bonding orbital coefficient on the halogen atom decreases and so the LUMO's coefficient on the hydrogen atom must also decrease.

From these arguments I quite agree with you that the chance of good spatial overlap between the HOMO and LUMO is highest for HF and smallest for HI (where the LUMO is more evenly shared between the hydrogen and the iodine) although given the small size of the diatomic and the fact the pi cloud protrudes out of the molecular plane on both sides of the alkene I suspect the steric factor would infact vary rather little between the various HX molecules.

This argument would indeed reverse the trend, but note in no way does it involve the polarity achieving successful alignment for reactive collision, instead it simply means that (for HF compared to HI) there are more angles of approach during collisions for which there is enough orbital overlap during the collision for the reaction to proceed. Clearly this argument is beyond the scope of an A-level course and perhaps its essence is what you were getting at with your original answer.
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