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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: shubhamrawal on June 30, 2015, 03:36:37 PM
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why does benzoin condensation take place with alcoholic KOH and not with aqueous KOH?
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Welcome. It is a forum policy that you must show an attempt before we can help you.
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i know that why alcoholic KOH shows elimination with haloalkanes and aqueous KOH shows substitution with them. in aqueous medium OH- has very less charge density to act as a base , so it can only act as a base.
now in alcoholic KOH , very small quantity of RO- is present . as it is a better base than OH- so it act as a base for elimination and the equilibrium of the reaction generating RO- shifts forward.
now for benzoin condensation i have read its full mechanism. i have understood it . but i don't find any step there indicating substitution or elimination. so i don't know anything in this . i am a 12th standard student . so please explain in simple words
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In general, what is the practical role of a solvent?
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Polar solvent supports sn1 because of the hydration or solvation energy released by the formation of carbocation while non polar solvent support sn2 .
Solvent also determine the charge density of a nucleophile whether it can act as a base or simply nucleophile. Like in polar aprotic solvent fluoride has a very large hydration sphere so its charge density decreases considerably. While in the case of iodide hydration sphere is not so big . So in polar protoic solvent iodide is a better nucleophile than fluoride. While in polar aprotic solvent they cannot hold anions . They can hold cations only . So in that fluoride is a better nucleophile than iodide
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Apart the solvent polarity, let me ask two additional questions.
According your opinion:
1). Does the given reaction mechanism demands a strong base or not and why?
2). Is ethanolic KOH a stronger base than aqueous KOH or not and why?
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In general, what is the practical role of a solvent?
Polar solvent supports sn1 because of the hydration or solvation energy released by the formation of carbocation while non polar solvent support sn2 .
Solvent also determine the charge density of a nucleophile whether it can act as a base or simply nucleophile. Like in polar aprotic solvent fluoride has a very large hydration sphere so its charge density decreases considerably. While in the case of iodide hydration sphere is not so big . So in polar protoic solvent iodide is a better nucleophile than fluoride. While in polar aprotic solvent they cannot hold anions . They can hold cations only . So in that fluoride is a better nucleophile than iodide
I'm getting an even simpler point: A solvent should dissolve the reagents. Benzaldehyde is poorly soluble in water, which might be why alcohols are preferred.
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In general, what is the practical role of a solvent?
Polar solvent supports sn1 because of the hydration or solvation energy released by the formation of carbocation while non polar solvent support sn2 .
Solvent also determine the charge density of a nucleophile whether it can act as a base or simply nucleophile. Like in polar aprotic solvent fluoride has a very large hydration sphere so its charge density decreases considerably. While in the case of iodide hydration sphere is not so big . So in polar protoic solvent iodide is a better nucleophile than fluoride. While in polar aprotic solvent they cannot hold anions . They can hold cations only . So in that fluoride is a better nucleophile than iodide
I'm getting an even simpler point: A solvent should dissolve the reagents. Benzaldehyde is poorly soluble in water, which might be why alcohols are preferred.
But benzaldehyde reacts with aqueous solvent also. There simply cyanohydrin is formed by nucleophilic addition.
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Apart the solvent polarity, let me ask two additional questions.
According your opinion:
1). Does the given reaction mechanism demands a strong base or not and why?
2). Is ethanolic KOH a stronger base than aqueous KOH or not and why?
Weaker the base, better it is the leaving group. As CN- is the leaving group in the final step so it must be a weak base.
As i earlier mentioned, In case of alcoholic KOH , RO- act as the base being a better base . But in aqueous KOH OH- is so much hydrated that it just act as a nucleophile
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Let’s try to put everything in an order.
1). Organic reactions are not only nucleophilic substitution ones but also condensations, percyclic, redox reactions, etc. Thus, principles being valuable for nucleophilic substitution, might not always be valuable for other kinds of organic reactions. So, forget the nucleophilic substitution for the moment. The said reaction is condensation catalyzed by nitrile anion and accompanied by basic co-catalysis.
2). “Weaker the base, better it is the leaving group” is not true. The truth is, the stronger the nucleophile, the easier leaving of the nucleophyge group, is. The confusion starts by the aphorism that “the stronger the base, the stronger the nucleophile is” and its many exceptions, e.g. tertiary amines are weak to medium bases (of course, not all and thus, another exception of the exception) but they are strong nucleophiles and therefore, they easily form quaternary ammonium salts (except fluorides).
3). There is no formation of ROK in alcoholic KOH solution and the base is still KOH. But the lower the ionization of the dissolving medium, the higher the strength of acids and bases, is. Therefore, alcoholic KOH is a stronger base than aqueous KOH.
4). In the given reaction mechanism there will be a proton exchange of the intermediary carbanions and alkoxydes, if not strong basic conditions are and thus, the reaction would stop. Alcoholic KOH comes in equilibrium (indeed, localized to the left) with the corresponding intermediary carbanions and alkoxydes.
5). As you mentioned before: Simply cyanohydrin is formed by nucleophilic addition (to carbonyl, in aqueous medium and thus, the reaction would stop). Besides, formation of cyanohydrin or of cyanohydrin ether is very difficult in alcoholic medium.
6). For your one education, the mechanism of benzoin condensation is still not very well clarified, yet. Apart the given mechanism, two other mechanism models have also been proposed. The one assumes a more active role of the base rather than a co-catalyst and the other one assumes the formation of a dianion in the first step and notably, the formation of both the carbanion and the alkoxyde after the nitrile attack to benzaldehyde. But for the moment, forget them until graduation.
7). Is it more clear, now?
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There is no formation of ROK in alcoholic KOH solution
This cannot be true.
-OH + ROH ::equil:: H2O + RO-
Water and alcohols have similar pKas, and with the alcohol in vast excess compared to hydroxide, there is certainly some alkoxide formation.
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It is true that water and alcohols have similar pKas (15.7 and 16, respectively), and with the alcohol in vast excess compared to hydroxide, there is certainly some alkoxide formation but still, the alkali hydroxide represents the biggest part of the base, in an alcoholic medium. The difference = 0.3 between pKa = 16 (alcohol) and pKa = 15.7 (water) may seem very low but pKa has logarithm values. In practice, water has the double acidity than alcohols. Besides, all above refer to completely anhydrous alcohols, otherwise alkoxide concentration is negligible. Please, remember, that benzoin condensation can also occur in aqueous-ethanolic medium, as well as in absence of base (but, with lower yields).
PS: Sodium ethoxide is prepared by reacting metallic sodium with super dry ethanol and not by a simple mixture of NaOH in azeotropic ethanol (95.5 % w/w) or absolute ethanol.
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There is no formation of ROK in alcoholic KOH solution
there is certainly some alkoxide formation
These statements are directly contradictory.
The difference = 0.3 between pKa = 16 (alcohol) and pKa = 15.7 (water) may seem very low but pKa has logarithm values. In practice, water has the double acidity than alcohols.
I understand that, but the concentration of the alcohol will be well over an order of magnitude higher than the concentration of water. Le Chatelier in action.
We can do a back-of-an-envelope calculation. I may be making some assumptions and approximations that are too wild here (hopefully Borek can comment), but imagine you (carefully) add 1 mol K to 1 L of 10:1 alcohol:water.
We will assume water has approximately double the acidity of an alcohol (as you state)
We will also assume that the ratio of water:alcohol is approximately 10:1 after the reaction has occurred.
Ka1 = [HO-][H+]/[H2O]
Ka2 = [RO-][H+]/[ROH]
We assume Ka1 = 2*Ka2
We approximate that [ROH] = 10[H2O]
So:
Ka1/Ka2 = 2 = [HO-][ROH]/[RO-][H2O]
2[H2O]/[ROH] = [HO-]/[RO-]
2/10 = [HO-]/[RO-]
[RO-] = 5[HO-]
i.e. there is 5 times more alkoxide than hydroxide at equilibrium.
If you have a system with even less water (i.e. KOH in dry ethanol) you will have even less hydroxide at equilibrium.
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1). “Practically, there is no significant formation of ROK in alcoholic KOH solution.” Is it OK, now?
2). 10exp(-15.7)/10exp(-16) = 1.995
3). But it may also be 1:10 water:alcohol or 1:100 water:alcohol . The results are completely different.
4). In dry ethanol:
K = [RO-][H2O]/ [HO-][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
that does not lead to less hydroxide at equilibrium.
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4). In dry ethanol:
K = [RO-][H2O]/ [HO-][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
that does not lead to less hydroxide at equilibrium.
I agree that:
K = [RO-][H2O]/ [HO-][ROH]
K = Ka(ROH)/Ka(water)
= 0.5
That's fine, so we have K = 0.5
Now we can do a calculation:
ROH + -OH ::equil:: RO- + H2O
Assuming the molarity of pure ROH is 16 M, and we will add 0.1 mol of KOH to a litre of ROH. We can then use a standard ICE table where the change in concentration is x:
[RO-] = x
[ROH] = 16-x
[HO-] = 0.1-x
[H2O] = x
K = [RO-][H2O]/[HO-][ROH]
0.5 = (x)(x)/(0.1-x)(16-x)
since x<<16
0.5 = x2/16(0.1-x)
x2 = 8(0.1-x)
x2 + 8x - 0.8 = 0
x = 0.1
i.e. there is negligible hydroxide present. It's all alkoxide for a dilute alcoholic KOH solution.
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The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.
ROH + KOH → ROK + H2O
KOH → K(+) + OH(-)
ROK → RO(-) + K(+)
ROH → RO(-) + H(+)
H2O → HO(-) + H(+)
In addition, positive and negative charges must be equilibrated.
[K(+)] + [H(+)] = [OH(-)] + [RO(-)]
So, start re-calculating by tacking all above equilibria constants, into account and by the assumption that the total volume does not change, when mixing ROH with water (if needed) and additionally, that no thermal expasion occurs during exothermic solvolysis.
Good luck.
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The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.
ROH + KOH → ROK + H2O
KOH → K(+) + OH(-)
ROK → RO(-) + K(+)
ROH → RO(-) + H(+)
H2O → HO(-) + H(+)
In addition, positive and negative charges must be equilibrated.
[K(+)] + [H(+)] = [OH(-)] + [RO(-)]
So, start re-calculating by tacking all above equilibria constants, into account and by the assumption that the total volume does not change, when mixing ROH with water (if needed) and additionally, that no thermal expasion occurs during exothermic solvolysis.
Good luck.
Burden of proof you are right lies on you. We know the general ideas about how to deal with the equilibrium calculations, so your post adds nothing new to the discussion. Either show results of your calculations, or stop posting at all, as you are just trolling now.
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Dear Sir,
Please accept the following comments:
1). Burden of proof me are right lies on me but according the forum policy rules, I have to show only the insights but not the full answer.
Are there different policy rules for the forum stuff than the forum members?
2). These are not general ideas but the basic principles of chemical equilibria.
3). Considering basic principles as nothing new to the educational forum, consists a violation of the first paragraph of the forum’s registration agreement.
Are there different registration terms for the forum stuff than the forum members?
4). The results of calculations are already shown above:
K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2) → [RO(-)]/[OH(-)] = 1/100
Please, read my posts more carefully before being the strict lecturer.
5). Please, feel free to restrict my posts that you do not want to judge.
6).”Trolling”? – Please, go back to the comments No1 and No3.
Sincerely yours
pgk
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K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2) → [RO(-)]/[OH(-)] = 1/100
Please, read my posts more carefully before being the strict lecturer.
Your equation is incorrect:
K = Ka(ROH)/Ka(water)
Ka(water) ≠ Kw
Ok, if 0.1 M is too dilute, let's try 1 M.
The same calculation as my previous gives x = 0.9, i.e. 90:10 alkoxide:hydroxide
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I have presented several calculations now that demonstrate that the alkoxide is the major anion present in alcoholic hydroxide solutions, despite the (generally) weaker acidity of alcohols compared to water (i.e. despite the fact that K < 1)
The reason for this is the vast excess of alcohol present. It's a quantitative demonstration of Le Chaterlier's principle.
You appear to dismiss these calculations as invalid on the basis of the assumptions made (i.e. because autodissociation was not considered). While the assumptions I have made in my calculations will affect the accuracy of the result, I think it is safe to say that they have qualitative value in predicting the major anions present. If the result of my calculations ([RO-] >> [HO-] in dilute alcoholic hydroxide solution) is reversed by a more sophisticated mathematical model, then you have to demonstrate it. That's not a forum rule, it's just that unless you can demonstrate it, I'm not going to take your word for it.
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Edit: due to circumstances it took me a long time to write this post. Dan was faster, and his post addresses basically the same problems.
1). Burden of proof me are right lies on me but according the forum policy rules, I have to show only the insights but not the full answer.
Are there different policy rules for the forum stuff than the forum members?
These rules are designed for helping those solving homework questions, we are long past this point.
2). These are not general ideas but the basic principles of chemical equilibria.
Which is why they don't add anything new. At the moment it is not a discussion on the HS level, so we can safely assume we all know basic principles. No need to list them as if they were eye-openers.
4). The results of calculations are already shown above:
K = Ka/Kw → K = 10exp(-16)/10exp(-14) → K= 10exp(-2) → [RO(-)]/[OH(-)] = 1/100
And they are wrong. K is not [RO-]/[OH-]. You have ignored concentrations of ROH and H2O, and their ratio can substantially change the result:
[tex]\frac{[RO^-]}{[OH^-]} = K \frac {[ROH]}{[H_2O]}[/tex]
If you start with anhydrous ethanol, ratio [itex]\frac {[ROH]}{[H_2O]}[/itex] is quite large. Dan calculations took it into account and he have shown to you how to find the result he got. If you think he is wrong, show precisely where Dan is wrong, or show your full calculations. So far your arguments are handwavy and don't look reasonable:
The KOH 0.1M is a quite low concentration, when compared to ROH 16M (1/160). Therefore, autoionization of both ROH and water must also be taken into account.
Autoionization of water and alcohol yields concentrations of products several orders of magnitude lower than the concentrations of RO- and OH- from the equilibria taken into account so far. Experience tells me they will not change anything in the full picture. Feel free to solve the full system and prove me wrong.
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Indeed, Kw is the ionization expression of per number of moles, contrary to Ka that is the ionization expression per molarity (even better, normality) of water and therefore, pkw = 14 and pKa = 15.7.
In other words, Kw measures the water ionization degree, contrary to pKa that measures the acidity of water and helps the comparison with other compounds. But when being in mixtures and solutions, the ratio of ionized/noninized molecules of water, remains 10exp(-14) and therefore, Kw is the predominant and valuable constant, for further mathematical applications.
So, let’s do it again:
ROH + OH(-) → RO(-) + H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
ATTENTION: The ratio of concentrations [RO(-)]/[HO(-)] per volume unit does not change, regardless the changes of concentrations of ROH and water that occured during the alkali exchange reaction.
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The experimental verification of all above seems quite easy because alkoxides have different α- and β- chemical shifts in 1H-NMR and different ipso-C shift in 13C-NMR than the corresponding alcohols.
I will really appreciate any literature and experimental data, regarding the issue.
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So, let’s do it again:
ROH + OH(-) → RO(-) + H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
No matter how many times you will repeat it, K will not become a ratio of concentrations of RO- and OH-. Some things do cancel, [ROH] and [H2O] do not, they are left in the equation. Yes, you have shown how to calculate K for the reaction we are interested in from known Ka and Kw, no, it doesn't show what you claim it shows.
ATTENTION: the ratio of concentrations [RO(-)]/[HO(-)] per volume unit does not change, regardless the changes of concentrations of ROH and water
This is exactly the problem with your reasoning. Why do you think the above is true? The correct expression describing the ratio of concentrations, derived from the ROH/OH- equilibrium, is
[tex]\frac{[RO^-]}{[OH^-]} = K \frac {[ROH]}{[H_2O]}[/tex]
and the ratio of concentrations of water and alcohol plays an important role. You can't ignore it, unless you have a good reason to do so. So far you have failed to show what the reason is.
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Because the total volume does not change (or practically, it does not significantly change).
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So, let’s do it again:
ROH + OH(-) → RO(-) + H2O
Kreaction = [RO(-)][H2O]/ [HO(-)][ROH]
If you multiply and divide that function by [H+], you finally get:
K = Ka(ROH diss.)/Kw
I don't think you do.
Kreaction = [RO-][H2O]/ [HO-)][ROH]
= ([RO-)][H+]/[ROH])*([H2O]/[HO-][H+])
= Ka(ROH)*Ka(H2O)
The term in red is Ka(H2O), not Kw. Kw = [HO-][H+] in dilute aqueous solution - I don't think it's applicable here. You are incorrectly ignoring [H2O].
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Or and in other words, the function Kw = [HO-][H+] = 10exp(-14) is not valuable and applicable for concentrated aqueous solutions, as well as for pure water. Of course, it is!
Please, note that the dissolution medium is not water, hereby and thus, [H2O] is not the basis for calculations of [H+] and [OH-].
Besides and for similar reasons:
KaKb = kw
and:
KaKb ≠ Ka(water)
Once again, the terms Ka(water) and pKa(water) refer to pure water (free of electrolytes and solutes) and they are the expression of water acidity, in comparison with Arrhenius acids.
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Or and in other words, the function Kw = [HO-][H+] = 10exp(-14) is not valuable and applicable for concentrated aqueous solutions, as well as for pure water. Of course, it is!
Kw is based on the assumption that [H2O] is constant. That is not necessarily true for concentrated solutions.
Kw = [H2O]*Ka(H2O]) = 10-14 if [H2O] = 55.5 M (i.e. pure water)
But that is really beside the point. We are not discussing aqueous solutions, we are discussing alcoholic hydroxide solutions. Kw has no place in these calculations. [H2O] is certainly not constant in this scenario, and it is certainly not 55.5 M.
I think you have misused Kw to incorrectly calculate the equilibrium constant. I calculate K = 0.5, but you get K = 0.01 because you have erroneously divided by the concentration of pure water.
If you can show a valid calculation that supports your assertion that alkoxide formation in alcoholic KOH is practically negligible, please do so. I don't mind being proved wrong, it wouldn't be the first time, but otherwise I think it's time to stop.
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Actually Kw is sometimes expressed not as 10-14, but as 1.8×10-16 (10-14/55.5) (and in this case that would be the more correct value).
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Because the total volume does not change (or practically, it does not significantly change).
How is it related to problem in question? Are you saying that because the volume doesn't change significantly, concentrations don't change as well? If you start with an anhydrous alcohol containing several ppm of water, and you add 1 mL of water to 1 L of alcohol, the concentration ratio change is hundredfold, while the volume change is for most practical purposes insignificant.
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Assuming that the total volume does not significantly change, when mixing ROH with water:
Mixing aqueous KOH 1N and alcoholic KOH 1N, we get an initial mixture of KOH 1N per total volume unit that will immediately be transformed (ionic reaction) to (say) KOH 0.995 N and ROK 0.005 N per total volume and (say) KOH 1.98 N and ROK 0.02 N per ROH volume or a ratio 0.99/0.01
The relative concentrations of KOH and ROK may change by further dilution with either water of ROH or their mixture but the concentration ratio 0.99/0.01 will remain constant, when calculated per ROH partition.
As a reminder, the latter discussion refers to anhydrous ethanolic KOH.
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Assuming that the total volume does not significantly change, when mixing ROH with water:
Mixing aqueous KOH 1N and alcoholic KOH 1N,
In what ratio?
we get an initial mixture of KOH 1N per total volume unit that will immediately be transformed (ionic reaction) to (say) KOH 0.995 N and ROK 0.005 N per total volume and (say) KOH 1.98 N and ROK 0.02 N per ROH volume or a ratio 0.99/0.01
Please show your working. Writing things doesn't make them true.
The relative concentrations of KOH and ROK may change by further dilution with either water of ROH or their mixture but the concentration ratio 0.99/0.01 will remain constant, when calculated per ROH partition.
Relative concentration and concentration ratio are the same thing.
As a reminder, the latter discussion refers to anhydrous ethanolic KOH.
So your example is irrelevant?
Can you show a valid calculation that supports your assertion that alkoxide formation in alcoholic KOH is negligible or not?
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Sorry but I do not have any more time to spend for this discussion. Besides, the principal question of the discussion, has already been answered.