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Offline shubhamrawal

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benzoin condensation
« on: June 30, 2015, 03:36:37 PM »
why does benzoin condensation take place with alcoholic KOH and not with aqueous KOH?

Online Babcock_Hall

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Re: benzoin condensation
« Reply #1 on: June 30, 2015, 04:04:45 PM »
Welcome.  It is a forum policy that you must show an attempt before we can help you.

Offline shubhamrawal

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Re: benzoin condensation
« Reply #2 on: June 30, 2015, 05:22:26 PM »
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

Offline Dan

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Re: benzoin condensation
« Reply #3 on: July 01, 2015, 04:39:16 AM »
In general, what is the practical role of a solvent?
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Offline shubhamrawal

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Re: benzoin condensation
« Reply #4 on: July 01, 2015, 10:54:16 AM »
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

Offline pgk

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Re: benzoin condensation
« Reply #5 on: July 01, 2015, 11:16:25 AM »
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?

Offline Dan

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Re: benzoin condensation
« Reply #6 on: July 01, 2015, 12:32:19 PM »
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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Offline shubhamrawal

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Re: benzoin condensation
« Reply #7 on: July 01, 2015, 04:10:29 PM »
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.

Offline shubhamrawal

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Re: benzoin condensation
« Reply #8 on: July 01, 2015, 04:21:14 PM »
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

Offline pgk

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Re: benzoin condensation
« Reply #9 on: July 02, 2015, 01:32:58 PM »
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?

Offline Dan

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Re: benzoin condensation
« Reply #10 on: July 03, 2015, 02:25:10 AM »
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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Offline pgk

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Re: benzoin condensation
« Reply #11 on: July 03, 2015, 11:41:09 AM »
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.
« Last Edit: July 03, 2015, 12:41:45 PM by pgk »

Offline Dan

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Re: benzoin condensation
« Reply #12 on: July 03, 2015, 01:29:12 PM »
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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Offline pgk

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Re: benzoin condensation
« Reply #13 on: July 03, 2015, 02:08:12 PM »
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.

Offline Dan

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Re: benzoin condensation
« Reply #14 on: July 04, 2015, 03:21:31 PM »
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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