kilomentor

A friend is having a problem. He can’t tell me much about it; so often the case. It is proprietary. It involves the reduction of a nitro group (presumably to an amine) with the reagent sodium sulfide. The problem is the yield. The question for you readers: How much help can one be when one doesn’t know the details of a particular situation?

Perhaps you can’t be much help at all; but the situation does provide the opportunity to be mindful of two rules of thumb, that, if they don’t work in this case, will upon repeated recollection be useful and solve problems. This reduction of a nitro functionality to an amine using sodium sulfide gives me the chance to commend both of them to you, again, at the same time.

Rule 1

More problems result from impure starting materials than from any other cause

Sodium sulfide is easy to get in an impure form. In Fieser and Fieser Vol. 1 (1967) pg. 1104 it is reported that the reagent sodium sulfide nonahydrate decomposes on contact with air; a freshly opened bottle should be used. In Organic Syntheses, Coll. Vol. IV (1963) pg.32 sodium sulfide nonahydrate is used also for the reduction of an aromatic nitro. Note 1 reads, “Merck’s reagent grade of sodium sulfide nonahydrate was used. Since sodium sulfide decomposes on contact with air, a freshly opened bottle should be employed. “Sodium sulfhydrate” (Hooker Electrochemical Company hydrated sodium hydrosulfide) is also satisfactory; the amount should be based upon the formula NaHS.2H2O and an equivalent amount of sodium hydroxide ……is require.” The same advice about instability but here an additional problem is noted. There are other reagent combinations which can be used to replace the reagent but the degree of hydration of these is sufficiently obscure that Organic Syntheses feels it needs to tell you what constitution to assume.

In Fieser & Fieser Vol. 3 (1972) pg. 269 under the heading, Sulfides, the authors list four types

(a) sodium sulfide nonahydrate crystalline, Na2S.9 H2O(

b) technical fused chip sodium sulfide, 60% Na2S (mol. wt. 78.06)

(c ) Sodium polysulfide. Sulfur (at. wt. 32.07) is dissolved by heating and stirring in aqueous or alcoholic sodium sulfide.

(d) Ammonium sulfide. aqueous or alcoholic ammonia is saturated with hydogen sulfide

My first questions are do we know exactly what we are using? Do we have a valid analysis and have we taken the required precautions to keep the reagent intact before it encounters the substrate in the reactor?

I can imagine that on a kilo-scale ,weighing out and storing the reagent in the absence of air could be a problem. I don’t know that it is a problem, but this reagent is not inert and given that problems with starting materials are frequent, it is a question worth asking.

Rule 2

Organic chemists have a bad habit of not writing balanced equations and in oxidation and reduction reactions this has an annoying habit of causing them to get the stoichiometry wrong. So do you have the stoichiometry correct?

R-NO2 + 3Na2S + 6NH4Cl gives R-NH2 + 3S +2H2O + 6NaCl + 6NH3

Note that all the sulfur from the sodium sulfide ends up as zero valent elemental sulfur which is a co-product. All the six equivalents of acid (here NH4Cl) are used to protonate the amine being generated (2 protons) and the two oxygens being converted into water (4 protons). The sulfide is just providing a supply of electrons.

What this equation may also be telling us is that if we get the amount of acid too low we may also stop the reaction. Six equivalents of acid are required to meet the stoichiometry.

The essential nature of these hydrogens are also suggested at by a paragraph in Fieser & Fieser Vol. 9 pg. 434 treating the reagent sodium sulfide-thiophenol:

‘Nitroalkenes substituted with at least one aryl group are reduced to alkenes rapidly at room temperature by reaction with sodium sulfide nonahydrate and thiophenol in DMF. Thiophenol is essential as a proton source for this reduction; it is converted into diphenyl disulfide during the reaction.” [my italics]

So again I repeat stoichiometry may well not be the problem in this particular real life case, but knowing nothing else except that it is a redox reaction- take heed check the stoichiometry. The error is so frequent it ought to be laughable but I can tell you from experience- it isn't.

The above are repetitions of two CAS registry numbers. The reason they are entered here is explained below. This is not one of the kilomentor typical chemical eduction blogs.

I see that several Chinese companies have put up blogs on this site that are actually advertisements for their chemical products. These are probably an improper use of the site but I am curious how effective they are. This posting is not a fraud or an abuse to legitimate searchers because I am actually the owner of technology for the kilo-scale production of the two compounds listed above, bis-(3-bromophenyl) methanone and bis-(3-methoxyphenyl)methanone respectively. I am interested to know if these substances are important to anyone. I can be contacted at kilomentor@sympatico.ca.

For my faithful chemical education process development readers, my apologies for this experiment.

General schemes have been devised for examining unknown mixtures such as those one learns as an undergraduate chemist for a laboratory examination. Such schemes as those in the classic text by Shriner, Fuson and Curtin. [The Systematic Identification of Organic compounds: A Laboratory Manual. Fifth Edition John Wiley & Son 1964] The more complex of these schemes, no matter what their wisdom, have never to my knowledge been adopted by real chemists to work up real reaction mixtures. The reason is simple. These schemes are designed to handle complete unknowns. The bench chemist always knows something about his mixture, even if it is no more than the fervent hope that a particular product will be present; therefore, the working chemist or chemical engineer is in a more knowledgeable position and so the protocol for a complete unknown is going to be inefficient.

At least the functional groups and molecular weight of the desired product are know and some educated guesses can be made about the likely physical properties and so a theoretical proposal can be made for a rational separation. For example, if the hoped for product is a neutral lipophilic aldehyde, a mild aqu. acid extraction and a mild aqu. base extraction can be applied to an ether solution. Then some aldehyde specific reagent such as bisulfite or Girard’s P or T can be contemplated.

Another reason that these classical schemes are set aside is that they apply simple solubility tests in carbon tetrachloride and benzene, which to-day are unacceptably toxic solvents.

What would be more interesting and have more likelihood of application would be a scheme, which can be adapted to predicted estimated properties of a particular reaction mixture and has an inherent logic with respect to the most helpful procedures for separations at scale.

To explore this, let us assume first that the reactor content, at the end of the reaction period, is homogeneous and the TLC or other in-process check is encouraging. Let us also assume that the trial reaction has been conducted on a scale of at least several grams. I make this assumption first because improving the throughput in a series of reactions in a process scheme should focus on the early steps and these are the ones using the less expensive materials.

If the content is not homogeneous, the phases should be separated and treated separately. This practice is based on the wise rule that the chemist should never refuse a phase separation offered by nature in the course of working out a separation. Specifically, if there is a solid in a liquid, filter the solid and retain the filtrate. If there are two liquids, cut the phases and examine each. Such natural separations are not likely to be quantitative recoveries of any component, but the constitution of the phases may provide a guide to a modification that delivers a quantitative separation.

Supposing then that the phase is a single one or that we are examining multiply phases separately. The chemist has a very good idea that the main constituent of this phase is solvent and the solvent is known. The object will be to remove that solvent completely without exposing the other constituents to conditions so severe that they may resume reacting with themselves. The chemist is the best judge of how to achieve this and to make sure that it happens that way.

Let us suppose now that the solvent has been removed under mild conditions and the residue, either as an oil, a mixture of solids and liquids, or a glass is ready to be examined.

It is not possible to separate effectively a viscous oil from a solid. In this situation oil and solid probably must be examined together although again this is a situation where the chemist’s powers of observation and judgment are more useful than any rule that can be offered by me out of the blue.

The mixture will be examined with respect to its solubility properties much as in the classic approach in Shriner and Curtin but here the solvents are chosen based on a different principle. The solvents, which will be examined, are hexane, acetonitrile, methanol and ethyl ether.

The first three solvents are chosen because methanol and hexane in the presence of a few percent of water gives two immiscible phases as does the combination of acetonitrile and hexane. As a consequence differential solubility information can not only direct us to a trituration step but possibly to a partitioning between two immiscible solvents.

The tests in diethyl ether are more standard. Although diethyl ether is not a solvent acceptable in a general purpose chemical plant, its remarkable ability not to form emulsions makes it irreplaceable for acid and basic aqueous extraction tests.

So in practice, one gram of the mixture is placed in a small r.b. flask and treated with 7 ml of methanol and swirled. If any solid remains undissolved the slurry can be cooled in ice to maximize the quantity of solid and filtered cold and washed with a little cold methanol. Such solid is examined.

If there is no solid in the solution we could add 7 ml/gm of hexane(s) and mix the phases together. Again we look for any solid, which might separate and treat it appropriately. If two immiscible liquid layers are not present a drop or two of water is added in to make the methanol layer separate. The two phases are separated and each is evaporated to dryness, pumped under vacuum and weighted. Each phase should be examined, if it is convenient, by the analysis that was used for the reaction’s in-process check. The combination of the weights obtained and the analyses of the separate layers are useful properties of the mixture. They may provide the first hints of the most efficient methods of isolation. If one phase or the other contains essentially all the contents of the mixture all one can say is that the mixture is substantially polar or apolar, depending to which solvent it has migrated.

If the mixture is substantially apolar take a new sample of the mixture in a small r.b. flask and tread it with 7 ml of acetonitrile and swirl, repeating the procedure that was used with methanol. In the case of acetonitrile, water will very rarely be needed to get two liquid phases upon adding hexane. In fact try to avoid using water here. The two phases are separated and each is evaporated to dryness, pumped under vacuum and weighted. Each phase should be examined, if it is convenient, by the analysis that was used for the reaction’s in-process check. The combination of the weights obtained and the analyses of the separate layers again are useful properties of the mixture. They may provide the first hints at the most efficient methods of isolation. If one phase or the other contains essentially all the contents of the mixture all one can say is that the mixture is substantially polar or apolar.

Water is not used in these tests. Nevertheless there is a good likelihood that if an inorganic salt is present it will be insoluble in one of methanol or acetonitrile and will have been filtered off.

A frequent result of this work will be that the substantial majority of the reaction mixture will remain in the hexane layer. This is to be expected since the vast majority of organic compounds are substantially lipophilic and non - crystalline when present as mixtures; nevertheless, when a useful separation is made it at this stage may be particularly valuable.

Take a new portion of the mixture and try to dissolve it in 7 ml/g of diethyl ether. Again if there is a solid separate it. Now in the classical way extract the diethyl ether with an equal volume of 1N aq. HCl and separate the phases. Adjust the pH of the aqueous phase back to neutrality observing any cloudiness or solid separation and then back extract the murky neutralized water with ether, dry, evaporate to dryness, and weight.

In the same classical fashion extract the ether, which has been acid extracted with aq. base of pH about 9.0 and recovery the acid fraction.

Recover the neutral constituents from the residual ether.

Each phase should be examined if it is convenient by the analysis, which was used for the reaction’s in-process check. The combination of the weights obtained and the analyses of the separate layers are useful properties of the mixture.

Quite often very little more will have been accomplished than would have been achieved following the tried and true rules of thumb, but a useful number of times something really exciting and simplifying will have been drawn to your attention.

If the material, which you are seeking is either in the acidic or the basic fractions, even if it is still a serious mixture, your problems are well on their way to resolution because the means for rugged separations of such mixtures on scale are plentiful and these ways I explore elsewhere. See for example Kilomentor’s blog on extractive crystallization.

If the substance you are seeking still seems only to be found in the hexane or neutral diethyl ether phases more sophisticate means need to be applied. If 30% or more of a target substance has ended up in the methanol or acetonitrile phases there is reason to hope that more intensive extractions may give you what you need.

If TLC of the methanol, acetonitrile or hexane solutions showed a substantial amount of material remaining at the origin, the presence of high molecular weight or even polymeric materials is likely. If the mixture is strongly colored and the product sought is not expected to show color, polymer and tars are likely and the mixture should be cleaned up before looking for the desired species. Filtering through a plug of adsorbant, which retains the origin material is usually successful. Charcoaling a portion in an alcohol solvent often works. Sometimes steam distillation, regular distillation, or codistillation with a high boiling hydrocarbon can be useful. In codistillation with kerosene be mindful that you will need to get the mixture back from the high boiling solvent!

Because cyclohexane combined with nitromethane or nitroethane or any mixture of the two, also form two immiscible phases; the same methods illustrated above can be applied in this system. The same goes for combinations of nitromethane /nitroethane with cyclohexane and apparently cyclohexane and mixtures of dimethylformamide/dimethylacetamide. Wth these combinations the temperature needs to be kept not to far above ambient to preserve the two phase behaviour.

What does pKa mean? The pKa of a neutral molecule or ion is the negative logarithm of dissociation constant of a particular hydrogen atom under defined solvent conditions. Thus for a molecule A with a particular attached hydrogen H that we designate as A-H then Ka=[H⁺] [A^- ] / [H-A] this is the hydrogen ion concentration in the solution multiplied by the concentration of the anion of A all divided by the concentration of the undeprotonated H-A. pKa is –log Ka = -log [H⁺] –log [A^- ] + log [H-A].

pKa = pH –log [A^- ] + log [H-A]

pKa is best understood as the pH at which equal amounts of H-A and A^- exist in the solution. Put another way the pK is the pH at which H-A is one-half deprotonated in the reference solution.

The concept is not just applied to neutral substances or substances that are commonly recognized as Bronsted acids. Basic molecules are also characterized by pKa values but these are the pKas of the corresponding fully protonated base.

then Ka=[H⁺] [B] / [H-B^+; ] pKa is –log Ka = -log [H⁺] –log [B] + log [H-B⁺]

Thus pKa= pH –log [B] + log [H-B⁺]

The solvent systems commonly used are aqueous sulphuric acid for measuring the pKa of strong acids. The pKas of strong acids are usually negative numbers. The more negative the number, the stronger the acid. The strongest commonly known acid is hydrogen iodide.

Water is the solvent used for measuring the pKa of moderate acids and DMSO is common for measuring the pKas of weak acids such as the important class of carbon acids (hydrogens bonded directly to a carbon).

As an example of proper interpretation of the pKa acidity of a particular proton, we can note the two pKas important for the common solvent methanol, The pKa of Me-OH₂ is -2.5 and the pKa of the hydrogen bonded to oxygen in neutral methanol, Me-OH, is 15.5. What the first number says is that at a pH of -2.5 (something like molar sulphuric acid), methanol in the solution is one-half protonated. The second number tells me that even in the strongest aqueous base (pH 14) methanol is not yet half deprotonated.

In the same vein one can look at the pKas of common protic solvents.

The table below shows the pKas of different common carboxylic acids.

The pka of the conjugate acids of solvents are a guide to how active a Bronsted acid will be in different solvents, The more negative the pKa the more reactive the proton will be.

Probably the most valuable table is the one showing carbon acids alongside other reference acids.

This helps when deciding how strong a base is required for a particular deprotonation.