The Difference Between XtalFluor-M and XtalFluor-E

I was questioning about the difference between these two reagents: XtalFluor-M and XtalFluor-E. I know from the structure, clearly, we can see the difference, but what else?

Then, I found out the history of how these two reagents were invented. Couturier and co-workers in 2009, were capturing the necessity of more stable and safe deoxyfluorinating reagents to cover the weakness of DAST and Deoxo-Fluor. To make it brief, the weakness of DAST and Deoxo-Fluor are liquid, thermally unstable, and unsafe as they generate highly toxic HF. 

Then, if we went back for over three decades, Markovskii et al was the one reported that DAST, dimethylamino, piperidino, and morpholino analogues all react with BF3.Et2O to form dialkylaminodiflurosulfinium tetrafluoroborates 3-6 (see below figure) which is in a salt form (not a liquid). Other researchers were also invented other analogues 7-11 of this reagents. However, the reactivity of this reagent derivatives have not been studied further toward alcohol, even aldehyde, ketone, or carboxylic acid, as well as the termal stability. Therefore, Couturier studied this. 

From their studies, when comparing using derivatives with different fluoride ion acceptors such as BF3, PF5, SeF4, SbF5, and AsF5, they found out that the derivatives with BF3, dialkylaminodifluorosulfinium tetrafluoroborate salts precipitated directly out of solution so from the manufacturing overview, it will make the isolation procedure at ease. Furthermore, they got derivatives 4 and 6 as the reagents with good thermal stability. 

How about the reactivity towards alcohol compound? At first, it was not a success, the reaction with geraniol, ended up with a mixture. They presumed that the problem was from the solvent used. After they studied the mechanistic pathways, and as we knew already that the mechanistic pathway is similar with DAST, they both involve a common dialkylaminodifluorosulfane intermediate, when DAST generated HF, Xtal-Fluor E and M generated tetrafluoroboric acid which the reactions then are a fluoride starved, that's why side reactions occur. 

In order to avoid the side reaction, exogenous fluoride addition is required as a promoter. They found out that the use of Et3N.3HF successfully improved the yield and selectivity. Then, the order of the reaction was a key parameter in this reaction as I already mentioned in this post.

Talking about the difference, here is the difference between these two reagents: thermal stability. Xtal-Fluor-M was found out to have greater thermal stability than XtalFluor-E (regardless of both are having greater stability compared to DAST and Deoxo-Fluor). Thermal stability was measured using DSC (Differential Scanning Calorimetry).

DAST and Deoxo-Fluor showed a decomposition at 155 (released 1641 J/g) and 158 (released 1031 J/g) degree Celcius respectively. While for the case of XtalFluor-E, it showed at 205-degree Celcius (released 1260 J/g). Moreover, XtalFluor-M showed greater stability as it showed at a higher temperature and a lower exothermic heat (243-degree Celcius and releasing only 773 J/g).

Another difference maybe about the price, but I found out, the price in Sigma Aldrich is comparable, not so different in price. 

The last, I think the difference is the scope of the reaction. Couturier and co-workers in 2010 have already studied the scope, but there is not so much difference, both reagent did work well for some substrate, but I believe, in another scope of substrate, a certain substrate may only work with just XtalFluor-E or XtalFluor-M.

In conclusion, about the difference between XtalFluor-M and XtalFluor-E, the most vivid is the thermal stability that XtalFluor-M is having greater stability than XtalFluor-E. Regardless of this difference, both XtalFluor-E and M are a great alternative for a stable, safe, and selective deoxyfluorination.

References:

Beaulieu, F., Beauregard, L., Courchesne, G., Couturier, M., & Heureux, A. L. (2009). Tetrafluoroborate Salts as Stable and Crystalline Deoxofluorinating Reagents, 57(3), 3–6.

Heureux, A. L., Beaulieu, F., Bennett, C., Bill, D. R., Clayton, S., Mirmehrabi, M., … Couturier, M. (2010). Aminodifluorosulfinium Salts : Selective Fluorination Reagents with Enhanced Thermal Stability and Ease of Handling †,‡, 3401–3411. https://doi.org/10.1021/jo100504x

Bruice Organic Chemistry Note #5: Acids and Bases

Once I dove in chemical synthesis, I realize how important it is to understand the concept of acids and bases. I realize how acids and bases are central of organic chemistry. So here in this post, I am going to discuss Chapter 2 in Bruice Organic Chemistry Book: Acids and Bases: Central to Understanding Organic Chemistry. 

Based on Bronsted and Lowry, acid is a species that loses a proton, while base is a species that receive the proton. In acids and bases reaction, there are 2 terms that we have to be familiar with, they are pKa and pH.

pKa is the degree of acidity possessed by a compound, a tendency of a compound to lose its proton. In a reaction of acids and bases, the degree of acid to dissociates is called the acid dissociation constant (Ka). The pKa value depends on Ka value. The larger the acid dissociation constant, the stronger the acids. Below is the equation:

The degree of the acidity of a compound based on its pKa value can be seen from the below table:

How about pH, pH is not a value possed by a compound, the value is obtained from the solution. So, pH is defined as the concentration of proton in a solution and the equation is shown below.

There are 4 compounds in organic chemistry that as an organic chemist we have to be familiar with because these 4 compounds are always popping up in an organic reaction mostly: (1) carboxylic acids, (2) alcohols, (3) amines, and (4) protonated compounds.

Carboxylic acids, for example, acetic acids and formic acid. 

Alcohols, for example, methyl alcohol and ethyl alcohol.

Amines, for example, methylamine and ammonia.

Protonated compounds, in order to understand this, there is an important state you have to know:

"The stronger the acid, the weaker its conjugate base". Look at the pKa value of these protonated compounds.

Protonated methylamine has pKa 10.7 compared with protonated ethylamine that pKa is 11.0, protonated methylamine is a stronger acid, means that from the states above, protonated methylamine has a weaker conjugate base than ethylamine.

Another example, protonated methyl alcohol, pKa -2.5, protonated ethyl alcohol, pKa -2.4, while protonated acetic acid, pKa -6.1 means that protonated acetic acid has a weaker conjugate base than ethyl alcohol, ethyl alcohol has a weaker conjugate base than methyl alcohol. 

By the way, as those compounds are very important to be remembered. Below is the table of the approximate pKa value of those compounds, so from the table, it can be easily recited. 

Some of the compounds can act as acid or base, depend on the situation of the reaction. So, how to predict it? Based on the pKa of the compounds involved in the reaction. 

For example, a reaction involving HCl (pKa -7) and H2O (pKa 15.7), we can see that HCl is stronger as an acid compared to H2O, so in this reaction HCl acts as the acid, while H2O as the base. 

Meanwhile, in different reaction involving NH3 (pKa 36) and H2O (pKa 15.7). H2O is a stronger acid than NH3, so H2O acts as the acid, while NH3 acts as the base. 

Now, after knowing which compound acts as the acid or the base, it's time to know, which part of the reaction is favored (reactant or product formation). As we know already from several reaction in equilibrium, the arrow is shown in 2 direction. However, there are some cases when one of the arrow is drawn longer than the other to show the part which is favored more. 

For example is the reaction below:

So, to know which one is favored, we can see from the pKa, once again, is all about the pKa value. The reaction is favored after comparing the pKa of the acid in both part of the reaction. The longer arrow will direct to the weaker acid formation. So from the reaction above, it's clear that the reaction favored to the product formation. Compared to the reaction below, the reaction is favored to the reactant side where the weaker acid is formed. 

I know it's always hard to recite numbers. It's not impossible, but it's too troublesome to recite all the pKa value of  all compounds. So, no need to recite all. Nevertheless, at least, we have to know the approximate pKa value of the compound. There are at least, 3 factors that can affect the pKa value: (1) electronegativity, (2) hybridization, and (3) size. 

Talking about electronegativity, it can be applied when the compounds being compared have the same size, I mean in the same bar of Periodic Table, such as C, N, O, and F. From that case, we can see the difference of the electronegativity that F is more electronegative than O > N > C. So, based on the electronegativity, the more electronegative the compound is, the stronger the acid. 

Hybridization is about sp3, sp2, or sp. Those differ in single, double, or triple bond. We also knew already that sp is more electronegative than sp2, and sp2 is more electronegative than sp3. So sp is a stronger acid than sp2, sp2 > sp3. 

Back to the statement I mentioned earlier, the stronger the acid, the weaker the conjugate base. A weaker conjugate base also means the more stable it is as a base. The stability of a base also depends on the size. The larger the size, the more stable it is. So in periodic table, in a column going down, between HF, HCl, HBr, and HI, the most stable base is HI, means the strongest acid. Even though it is the lowest in electronegativity, but the effect of electronegativity is put aside by the size. Size does matter more than the electronegativity. That's why, when we are talking about the effect of electronegativity, the size of the compounds compared must be in the same size. 

These there factors affect the pKa value directly. There is another factor that can indirectly affect the pKa value which is the substituent. See the example below.

When hydrogen is substituted with Br, Cl, and F, the pKa value is changing, and the compound becomes a stronger acid when substituted with the more electronegative atom. That's why the pKa becomes smaller when substitute with Fluorine atom. But how come substituent can affect the pKa? This is due to the inductive electron withdrawal. The electron in oxygen atom is pulled toward the halogen, so the electron density is reduced and stabilize it as it is being stabilized, the stronger the acid is. That's why the acidity increases. 

How about the position of the substituent, will it also affect the pKa value? The answer is yes. The closer the substituent, the stronger the acid. 

Now, let's flashback to a table that showed the pKa value of carboxylic acid and methanol. pKa value of carboxylic acid approximately around 5, while methanol around 15. How come carboxylic acid is stronger than methanol? This is also due to inductive electron withdrawal from the oxygen. In addition, delocalization also occurs there between both oxygen. As it is delocalized, it becomes more stable. While in methanol, there is only localization that happens. As the carboxylic acid is more stable, the stronger the acid than methanol. 

Here, we come to the end of acids and bases discussion. The effect of pH to determine the structure of the organic compound. When there is a compound in a solution with a certain pH, the structure is determined by the pH of the solution. A compound can be in acidic or basic form.

a. If pH of the solution is the same with the pKa of the compound, the compound will be in 50% acidic form and 50% basic form.

b. If pH is less than pKa, the compound will be in its acidic form.

c, If pH is more than pKa, the compound will exist in its basic form.

That's all from me. I apologize if there are some mistakes regarding my explanation. Thank you for visiting. See you in my next post!

Reference:
Bruice, P. Y. 2017. Organic Chemistry Eighth Edition. England: Pearson Education Limited.

Bruice Organic Chemistry Notes #4

This post is the continuation of this post. Here, I will explain from number 7 to 16. 

7. HOW SINGLE BONDS ARE FORMED IN ORGANIC COMPOUNDS

To visualize how the single bonds are formed, let we study that based on methane case. Methane (CH4) has 4 covalent C-H bonds. There are so many ways to represent methane such as based on the perspective formula, ball-stick model, space-filling model, and electronic potential map.

The model from the perspective formula lets the reader knows the 3-dimensional structure of a molecule on a paper easily by sketching it using wedges and dashes which the central atom is always assumed to be in the plane of the paper. A bond with wedges is drawn to visualize the bond that is projected out of the plane of the paper toward the viewer, while the bond with the dashes, projected back from the paper away from the viewer.

Ball-stick model visualizes the structure of the molecule that the ball shows the atoms as spheres and the bond as sticks. While the space-filling model is a model that can visualize the relative size of the atoms or molecules to understand an important point if needed. For the electronic potential map, we have discussed this before in the previous post.

From those model as the representation of a molecule, we can see the bonds that connect atom to atom. In the case of methane, the atom Carbon only has 2 unpaired valence electrons, so how this atom can make up to 4 bonds? So the answer of how the single bonds are formed is through HYBRIDIZATION.

Hybridization is a concept proposed by Linus Pauling in 1931. Basically, the theory is created to simplify the understanding of how bonds formed. So, The hybridization of the hybrid orbital theory comes to make things fit to give us a very good picture of the bonding in organic compounds.

Simply, hybridization is a concept of mixing orbitals or combining orbitals. So to make the bond, one s and 3 p orbitals are all combined into 4 equal orbitals. This hybridization for a single bond is called sp3 orbital, so it contains one part s and 3 parts p orbitals.

An important thing about single bond in methane is the angle formed. As it is sp3 orbitals, the orbitals move as far from each other that the most stable angle they can fit is 109.5 degree which then form a regular tetrahedron.

While, below is the bond in ethane.

Another important thing that will differentiate between single bond and other bonds is the length of carbon-carbon bond, in single bond, the length is 1.54 Armstrong.

Look also at the electrostatic potential map, the color shows there is no specific electronegativity difference.

8. HOW A DOUBLE BOND IS FORMED: THE BONDS ETHENE
The example of a compound with double bond is ethene. As it has double bond connected 2 carbons, it has only 3 atoms bonded. Then, the hybridization needed is sp2, as only 1 s orbital with 2 p orbitals are mixed to form 3 bonds, so 1 p orbital is not needed and becoming the unhybridized p orbital.

Next, I will discuss about these parts below. See you!

9. HOW A TRIPLE BOND IS FORMED: THE BONDS IN ETHYNE

10. THE BONDS IN THE METHYL CATION, THE METHYL RADICAL, AND THE METHYL ANION

11. THE BONDS IN AMMONIA AND IN THE AMMONIUM ION

12. THE BONDS IN WATER

13. THE BOND IN A HYDROGEN HALIDE

14. HYBRIDIZATION AND MOLECULAR GEOMETRY

15. SUMMARY HYBRIDIZATION, BOND LENGTHS, BOND STRENGTH, AND BOND ANGLES

16. DIPOLE MOMENTS OF MOLECULES


Reference
Bruice, P. Y. 2017. Organic Chemistry Eighth Edition. England: Pearson Education Limited.

Neighboring Group Participation

Neighboring group participation (NGP) is also known as anchimeric assistance (Greek: anchi means neighbor), a lone pair of electrons in an atom or the electrons present in a sigma bond or pi bond contained in the parent molecule but not conjugated with the reaction center (so it never be in alpha position) that influences or interacts with a reaction center. Because of the interaction from the neighboring group, (1) the reaction rate can be enhanced, (2) the stereochemical outcomes can be abnormal, and (3) the isomeric products can be isolated (Proust, Gallucci, & Paquette, 2009).

The group is not in alpha position, but it can be positioned either in beta, gamma, or delta position.  Then, by the interaction, they will create 3, 4, 5, or 6-membered rings. Besides that, because of this interaction, the stereochemical outcome is also always retention. It will never proceed with inversion. Therefore, in substitution nucleophilic reaction, it always goes with SN1 or usually called as SN1 anchimeric assistance. 

Below is the mechanism of how NGP interacts with the reaction center. 

Another example:

A paper written by Proust, Galluci, and Paquette (2009) found that substituent para to the sulfonyl protecting groups which act as NGP in a reaction have an influence on nucleophilicity of the sulfonamide nitrogens and the reactivity of the double bond of the diazocines. See the reaction mechanism below:

So, when the substrate has different substituents, there will be 2 possible pathways mean 2 possible products of dibromination. The substituents are various, some are electron withdrawing group such as -NO2, and some others are electron donating group such as -OMe. 

Substrates with different substituents

Two possible products of dibromination

From the table above, electron-withdrawing groups tended to decrease the amount of rearranged dibromides, while electron-donating groups tended to enhance the amount of rearranged dibromide. This is because the nucleophilicity of sulfonamido nitrogen was enhanced by the electron-donating group. Meanwhile, the existence of electron-withdrawing group reduced the nucleophilicity of the nitrogen. It was clearly shown from substrate 8g from entry 7 compared to entry 5 which in entry 5 when both are introduced with electron-donating group, -OMe, the amount of 13 is the highest, but then when one of them is electron-withdrawing group, -NO2, the amount was reduced as the nitrogen nucleophilicity was reduced. 

So, that's how neighboring groups interact with the reaction center and how a substituent in a neighboring group also can influence the reactivity.

That's all from me. I hope it can be useful for you. I apologize if there are any mistakes. Thanks for reading my blog! 

Reference:

Proust, N., Gallucci, J. C., & Paquette, L. A. (2009). Effect of Sulfonyl Protecting Groups on the Neighboring Group Participation Ability of Sulfonamido Nitrogen, (4), 2897–2900.

SNi (Substitution Nucleophilic Internal)

I only familiar with SN1 and SN2, I never know there is another version of substitution nucleophilic. It is substitution nucleophilic internal that I am going to talk more here. 

Cowdrey was introducing the name in 1937 which occur with retention of configuration. Do you remember that what makes SN1 differ with SN2 is that SN1 occurs with retention, while SN2 occurs with inversion? So, what is the difference between SNi and SN1? Here is the difference, in SNi the ion pair is not completely dissociated so the carbocation is not really formed, while as we know the rate-limiting step of SN1 is the formation of carbocation. The example of the reaction is a reaction between alcohol with thionyl chloride. 

Now, imagine you have a reaction, and you want to confirm that it occurs with SNi. So, how do you do it?

This paper, written by Tanaka et al in 2018, explained how they confirmed their reaction that occurs with SNi by using KIE and DFT. KIE stands for Kinetic Isotope Effect, the change in the reaction rate of a chemical reaction when one of the atom in the reactions is replaced by one of its isotopes. Let me explain about DFT later, focus on KIE first.

There are two type of KIE, primary and secondary KIE. Primary KIE occurs when the bond where the isotope connected is being broken or formed. Meanwhile, secondary KIE occurs when the bond is being broken between a carbon and an atom, not between the carbon and the isotope, the isotope just positioned in the same carbon where the bond is broken.

In primary KIE for deuterium, as it is heavier than H, when the bond is being broken (dissociation), the ratio of reaction rate between C-H and C-D became more than 1, as deuterium makes the reaction rate slower. However, if it is an association, the condition becomes different, so it is called as inverse, the reaction rate of the deuterium becomes faster, so the ratio of the reaction rate less than 1. 

So, how to determine the mechanism reaction using KIE? Tanaka et al used a comparison from another paper that stated 13C KIE for SN1 reaction are 0.995-1.01, while 13C KIE for SN2 is >1.07. Their experiments obtained 13C KIE at the anomeric carbon to be 0.9986 and 0.9999 which is in the range of SN1 reaction. While a paper they referred for the 2H KIE, it stated that 2H KIE for SN 1.08 and 1.05, while for SN2 1.02. From their experiment, they got 1.055 which also in the range of SN1. From 13C and 2H KIE, we could see, both corresponded to SN1. However, if the case is SN1, there is also a possibility to be SNi as the stereochemical preference is the same, which is retention. In order to confirm it, they did further calculation with DFT.

DFT is the abbreviation for Density Functional Theory which has become the most prevalent and efficient tool in studying organic reaction mechanisms (Zhang et al., 2019). From the DFT calculation, they calculated the 13C and 2H KIE by QUIVER, then comparing the results with the experiment data. They found out that the 13C KIE (1.009) to be close with data experimental (0.9986 and 0.9999). While for the 2H KIE (1.072), after being optimized, it was found to be close with data experimental (1.055). The most important part is after the IRC scan (Intrinsic Reaction Coordinate: the path of a chemical reaction after being traced from the transition state to the products and/or reactants using IRC method), the transition structure indicated that the glycosylation mechanism of the 1,2-anhydro donor and glycosyl-acceptor-derived boronic ester was a concerted SNi mechanism. 

[Picture Source: Tanaka et al, 2018]

[Picture Source, Tanaka et al, 2018)

The detail of the mechanism was then investigated by Mayer natural bond order or the reaction coordinate, from the shadowed part, it showed that a glycosidic linkage began to form before the bond breakage between the anomeric carbon and epoxy oxygen, indicating that this reaction mechanism is a concerted SNi mechanism. 

[Picture Source: Tanaka et al, 2018]

If it is SN1, absolutely, the bond must be broken first before the formation of the glycosidic linkage. Hence, it was proved from KIE experimental data and DFT calculated data, both agreed to the SNi reaction mechanism, not SN1. 

That's all from me. If there are some mistakes during the explanation. I apologize. I hope this is useful to you. Thank you for visiting.


Reference:

Tanaka, M., Nakagawa, A., Nishi, N., Iijima, K., Sawa, R., Takahashi, D., & Toshima, K. (2018). Boronic-Acid-Catalyzed Regioselective and 1,2- cis -Stereoselective Glycosylation of Unprotected Sugar Acceptors via S. https://doi.org/10.1021/jacs.7b12108

Zhang, H., Bai, H., Guo, Y., Wei, D., & Zhu, Y. (2019). A density functional theory study on mechanism and substituent effects of a base-free and catalyst-free synthesis of functionalized dihydrobenzoxazoles, 1(September 2018), 1–8. https://doi.org/10.1002/qua.25836