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Table 3.4: Procedural summary for single solvent crystallization.
Heat some solvent on a heat source to a boil (include boiling stones).
To the impure solid in an Erlenmeyer flask, add a small portion of hot solvent.
If a hot filtration step is expected, use boiling stones with the impure solid, or a boiling stick or stir bar if a filtration is not expected.
Put both flasks on the heat source and bring to a boil.
Add the minimum amount of boiling solvent needed to dissolve the impure solid, with swirling.
For $100 \: \text{mg}$-$1 \: \text{g}$ quantities, add $0.5$-$2 \: \text{mL}$ solvent each time. For smaller scales, add solvent dropwise.
Wait for each portion of solvent to come to a boil before adding more.
Possibly use charcoal and hot filtration at this point if colored or insoluble impurities are present.
When dissolved, remove the solution from the heat source and remove the boiling stick or stir bar if used. Allow the solution to slowly cool atop some paper towels, and with a small watch glass over the mouth of the Erlenmeyer.
If a hint of cloudiness is seen in the solution, or if the solution has cooled a good deal without crystallizing, scratch with a glass stirring rod to initiate crystallization.
A proper crystallization takes between 5-20 minutes to complete.
When the solution is at room temperature, place it in an ice bath for at least 10 minutes to maximize crystal formation.
Also chill a rinse solvent in the ice bath.
Collect the crystals by suction filtration.
3.6C: Using Solvents Other Than Water
This section describes a few key differences between a crystallization using water and one using volatile organic solvents. It is expected that readers have previously read or performed a crystallization using water as the solvent.
Ethanol, Methanol, Ethyl Acetate, and Hexanes
Ethanol, methanol, ethyl acetate, and hexanes are flammable and have moderate volatility, thus these solvents necessitate some different approaches than when using water as the crystallization solvent.
1. Although a steam bath is the preferred heat source for these solvents, if a hotplate is chosen to be used carefully it's essential that you:
1. Keep the hotplate on low (Figure 3.54a) and monitor the temperature of the solvent with a thermometer, being sure to patiently wait for the solvent to come to a boil instead of "cranking" up the heat.
2. Use the hotplate in a fume hood to prevent a "blanket" of solvent vapors from forming around the hotplate, which have the potential to ignite.
3. Monitor the hotplate the entire time while the solvent is heating.
2. It may be more controllable to use a pipette to transfer portions of hot solvent to the solid instead of pouring (Figure 3.54b+c). Pouring has a greater possibility of spilling, and if solvent drips onto the hotplate surface, it has the potential to ignite. Since solvent tends to cool in a pipette, care must be taken to be sure the solution is returned to a boil before adding more solvent.
3. All organic solvents have lower heat capacities than water, so tend to cool more quickly. Be sure to set the cooling sample atop several paper towels to encourage a slow crystallization (Figure 3.54d).
Diethyl Ether, Acetone, and Petroleum Ether (Low-Boiling)
Diethyl ether, acetone, and low-boiling petroleum ether are flammable and highly volatile, and thus necessitate further considerations than when using water or other organic solvents as the crystallization solvent.
• As these solvents have very low boiling points, they MUST be heated with a steam bath (Figure 3.55a), not a hotplate, or ignition may occur on the hotplate's surface.
• Steam is much hotter than is necessary to bring these solvents to a boil, and they must therefore be closely monitored during heating. To prevent too vigorous boiling, flasks may need to be lifted periodically off the steam bath, or simply hovered above the steam bath (Figure 3.55a) instead of resting directly atop the bath (Figure 3.55c). To control boiling, the steam rate should also be turned down.
• Using a Pasteur pipette with warm solvent is impractical with very volatile solvents, as liquid will undoubtedly drip out the end of the pipette before it is able to be delivered. Instead of heating the solvent beforehand, simply add cold solvent to the solid by pipette each time (Figure 3.55b), and then be sure to allow each portion to come to a boil before adding more.
Liquid may drip out the tip of a pipette even when dispensing cold solvent, which happens as solvent evaporates into the pipette's headspace, and the additional vapor causes the headspace pressure to exceed the atmospheric pressure. To prevent a pipette from dripping, withdraw and expunge solvent into the pipette several times. Once the headspace is saturated with solvent vapors, the pipette will no longer drip.
• These solvents boil so readily that it's possible they will vaporize as quickly as a new portion of solvent is added (Figure 3.55c shows vigorous boiling). It may feel as if you keep adding solvent and are "getting nowhere" with the additions. It is important to be watchful of the solvent volume in the flask, and if the additions appear to be quickly "disappearing", raise the flask above the steam bath or turn down the steam to moderate the rate of heating.
• As these solvents are not much warmer than room temperature when boiling, they will cool very quickly. Thus, it may be helpful to cover the flask with an inverted beaker to create an insulating atmosphere in addition to setting the cooling flask atop several paper towels (Figure 3.55d).
3.6D: Mixed Solvent Crystallization
The crystallization pictured in this section shows purification of a roughly $1 \: \text{g}$ sample of trans-cinnamic acid. Trans-cinnamic acid is soluble in methanol and insoluble in water, and this crystallization uses a mixed solvent of methanol and water to give a $74\%$ recovery.
It is assumed that students performing this technique have previously performed or read about a single-solvent crystallization.
1. Determine two miscible solvents that can be used for the crystallization (Figure 3.57a): the desired compound should be soluble in one solvent (called the "soluble solvent") and insoluble in the other solvent (called the "insoluble solvent").
2. Transfer the impure solid to be crystallized into an appropriately sized Erlenmeyer flask (Figure 3.57b).
3. Place some of the "soluble solvent" into the flask (Figure 3.57c), add a boiling stick (or boiling stones if preferred), then heat atop a steam bath (Figure 3.57d). A hotplate can be used cautiously if using the mixed solvents methanol/water or ethanol/water.
1. Add more of the "soluble solvent" in portions until the solid just dissolves (Figure 3.58a). Be sure to allow time in between additions, and allow each addition to come completely to a boil before adding more.
2. Add the "insoluble solvent" in portions with heating until the solution becomes just cloudy (Figure 3.58c).
3. Add the "soluble solvent" dropwise with heating until the solution again clarifies (Figure 3.58d).
1. Remove the flask from the heat source, remove the boiling stick and set the flask atop a paper towel folded several times. Cover the mouth of the Erlenmeyer flask with a watch glass, and allow the solution to slowly cool to room temperature (Figure 3.59a).
2. As the solution cools, eventually solid crystals should form (Figure 3.59b). Use a glass stirring rod to scratch the flask and initiate crystallization if necessary. Place the crystals in an ice-water bath for 10-20 minutes and collect the solid by suction filtration.
3.6E: Mixed Solvent Summary
Table 3.5: Procedural summary for mixed solvent crystallization.
Find a miscible pair of solvents: one in which the desired compound is soluble (called the "soluble solvent") and one in which the compound is insoluble (called the "insoluble solvent").
Place the impure solid in an Erlemeyer flask along with a boiling stick (or boiling stones if preferred).
Add the soluble solvent in portions while heating until the solvent just dissolves.
Allow time in between additions, and allow each addition to come completely to a boil before adding more.
Add the insoluble solvent in portions with heating until the solution becomes faintly cloudy. Add the "soluble solvent" dropwise with heating until the solution is clarified (transparent).
Remove the boiling stick (if used) and allow the system to come to room temperature while sitting atop some paper towels and with the flask's mouth covered by a watch glass.
Scratch with a glass stirring rod to initiate crystallization if necessary.
Submerge in an ice bath for 10-20 minutes.
Collect the crystals by suction filtration. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.06%3A_Step-by-Step_Procedures/3.6B%3A_Crystallization_Summary.txt |
Crystallization is Too Quick
Rapid crystallization is discouraged because impurities tend to become incorporated into the crystal, defeating the purpose of this purification technique. It may be acceptable for crystallization to start immediately after removing the flask from the heat source, but if a large amount of solid is formed then the compound is crystallizing too fast.
An ideal crystallization has some crystals forming in approximately 5 minutes, and growth continuing over a period of 20 minutes. Below are methods that can be used to slow the growth of crystals:
1. Place the solid back on the heat source and add extra solvent (perhaps $1$-$2 \: \text{mL}$ for $100 \: \text{mg}$ of solid), so that you have exceeded the minimum amount of hot solvent needed to dissolve the solid. Although more compound will dissolve in the mother liquor, the compound will stay soluble longer once set aside to cool.
For example, in the crystallization of trans-cinnamic acid with a mixed solvent of methanol and water, use of the minimum amount of hot solvent to dissolve the solid (Figure 3.60a) resulted in the solid immediately crashing out of solution when the solution was taken off the heat source (Figure 3.60b). To remedy this problem, the solid was placed back on the steam bath, additional methanol (soluble solvent) was added, brought to a boil to dissolve the solid, and then cooled. The solid then began to crystallize much more slowly (Figure 3.60c), taking a period of 15 minutes to fully crystallize.
1. If the minimal amount of hot solvent needed to dissolve the solid reached a height of less than $1 \: \text{cm}$ in the flask, the flask may be too big for the crystallization. A shallow solvent pool has a high surface area, which leads to fast cooling. Transfer the solution to a smaller flask (using some solvent to rinse the flask, and then boil away the same amount of solvent used for rinsing) and repeat the crystallization.
2. Be sure to use a watch glass over the top of the Erlenmeyer flask to trap heat, and set the flask atop some material to insulate the bottom (several paper towels, a wood block, or cork ring). An inverted beaker could also be tried to create an insulating atmosphere around the cooling flask.
Crystallization Doesn't Happen
It can be quite frustrating to set aside the dissolved solution to cool and have no crystals form at all. Methods to initiate crystallization were discussed in great detail previously. To summarize, here are the methods that can be tried (in hierarchical order) to form crystals depending on the appearance of the solution:
1. If the solution is cloudy, scratch the flask with a glass stirring rod.
2. If the solution is clear,
1. First try scratching the flask with a glass stirring rod.
2. Add a seed crystal (a small speck of crude solid saved from before the crystallization was begun, or a bit of pure solid from the reagent jar).
3. Dip a glass stirring rod into the solution, remove it, and allow the solvent to evaporate to produce a thin residue of crystals on the rod (Figure 3.61). Then touch the rod to the solution's surface, or stir the solution with the rod to dislodge small seed crystals.
4. Return the solution to the heat source an boil off a portion of solvent (perhaps half), then cool again.
5. Lower the temperature of the cooling bath.
3. If very few crystals are seen, there is likely too much solvent. Return the solution to the heat source and boil off a portion of solvent, then cool again.
4. If all else fails, the solvent can always be removed by rotary evaporation to recover the crude solid. Another crystallization can be attempted, perhaps with a different solvent system.
The Yield is Poor
A crystallization may result with a really poor yield (e.g. less than $20\%$). There are several reasons why this might happen:
1. Too much solvent may have been used during the crystallization, and therefore large quantities of compound were lost to the mother liquor. If the mother liquor (the filtrate after suction filtration) has not been disposed of, this can be tested by dipping a glass stirring rod into the mother liquor and letting it dry. If the solvent evaporates to leave a large residue on the rod, there is a lot of compound left in solution. Additional compound may be recovered by boiling away some of the solvent and repeating the crystallization (this is called "second crop crystallization"), or by removing all of the solvent by rotary evaporation and repeating the crystallization with a different solvent.
2. Too much solvent may have been used while attempting to dissolve semi-insoluble impurities. If this may have been the case, a hot filtration could have been attempted to remove the impurities. (The solid would have to be recovered from the mother liquor first through rotary evaporation in order to attempt the crystallization again.)
3. Too much charcoal may have been used to decolorize the solution (a pitch black solution has too much charcoal). Too much charcoal decreases the yield as charcoal can adsorb the desired compound along with impurities. There is no way to recover the product once it is adsorbed by charcoal.
4. If a hot filtration step was used, compound may have been lost in the filter paper and/or on the stem of the funnel. Additionally, too much solvent may have been used when adding a portion to get the system hot before filtration. The extra solvent before filtration adds to the "minimum amount of hot solvent" and if in substantial excess, can cause a loss of compound to the mother liquor.
5. If the impure solid was the product of a chemical reaction, the reaction may not have worked well. This could be assessed if a crude mass had been obtained: if the crude mass was very low to begin with, then the low crystallized yield was due to problems with the reaction, not the crystallization.
If a crude mass was not obtained, assessment of the rough volumes of solid before and after crystallization can be used. If the volumes were similar, the loss of yield may not be due to problems with the crystallization. Note that crystallized solids tend to be a lot "fluffier" than crude solids (Figure 3.62), so "eyeballing" solid volumes is only a very rough assessment of quantity.
Liquid Droplets Form (The Solid "Oils Out")
When cooling, a compound may come out of solution as a liquid rather than a solid (Figure 3.63). This process is called "oiling out" and happens when the melting point of the solid is lower than the solution's temperature. This is a problem in crystallization because when compounds liquefy first, they rarely form pure crystals. This is due to the fact that impurities often dissolve better in the liquid droplets than they do in the solvent. Figure 3.63c shows a sample of crude acetanilide that has oiled out (the droplets are impure liquid acetanilide), and the sample is contaminated with a methyl red impurity (which appears red in the low pH of the solution, an artifact of how the crude solid was synthesized). The oily acetanilide droplets appear more colored than the solution, indicating a higher quantity of dissolved methyl red impurity. If an oiled out liquid eventually solidifies, it often forms an impure glass-like non-crystalline solid.
The reasons for oiling out are several, and it can happen while dissolving the solid and during crystallization. It may be that the melting point of the solid is naturally low. It may also be that a solid is so impure that its melting point is dramatically lowered (as impurities lower the melting point).
There are several ways to attempt to fix an oiled out solution:
1. Return the sample to the heat source and add a bit more solvent, then cool the solution again. (If using a mixed solvent system, add more of the "soluble solvent"). The solid may have been coming out of solution too quickly (and thus at a temperature above its melting point), so it may stay soluble longer if there is more solvent.
2. Add a charcoal step if it was not already a part of the crystallization. The solid may be melting because there are large quantities of impurities, which charcoal can remove. This especially might work if a colored tint is noticed in the hot solution.
If either of these methods fail, recover the crude solid by rotary evaporation and attempt another crystallization. If the failed attempt used a mixed solvent, try a single solvent if possible. Or choose another solvent with similar solubility properties, but with a lower boiling point. For example, if ethanol were used as the solvent the first time, repeat the crystallization using methanol. Methanol has similar solubility properties as ethanol, but its lower boiling point may allow for the solid to come out of solution above its melting point. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/03%3A_Crystallization/3.06%3A_Step-by-Step_Procedures/3.6F%3A_Troubleshooting.txt |
"Extraction" refers to transference of compound(s) from a solid or liquid into a different solvent or phase. In the chemistry lab, it is most common to use liquid-liquid extraction, a process that occurs in a separatory funnel. A solution containing dissolved components is placed in the funnel and an immiscible solvent is added, resulting in two layers that are shaken together. It is most common for one layer to be aqueous and the other an organic solvent. Components are "extracted" when they move from one layer to the other.
• 4.1: Prelude to Extraction
Organic solvents and colored aqueous solutions in separatory funnels
• 4.2: Overview of Extraction
"Extraction" refers to transference of compound(s) from a solid or liquid into a different solvent or phase. When a tea bag is added to hot water, the compounds responsible for the flavor and color of tea are extracted from the grounds into the water. In the chemistry lab, it is most common to use liquid-liquid extraction, a process that occurs in a separatory funnel.
• 4.3: Uses of Extraction
There are several reasons to use extraction in the chemistry lab. It is a principal method for isolating compounds from plant materials. Extraction moves compounds from one liquid to another, so that they can be more easily manipulated or concentrated. It also enables the selective removal of components in a mixture.
• 4.4: Which Layer is Which?
It is essential that you know whether the aqueous layer is above or below the organic layer in the separatory funnel, as it dictates which layer is kept and which is eventually discarded. Two immiscible solvents will stack atop one another based on differences in density. The solution with the lower density will rest on top, and the denser solution will rest on the bottom.
• 4.5: Extraction Theory
When a solution is placed in a separatory funnel and shaken with an immiscible solvent, solutes often dissolve in part into both layers. The components are said to "partition" between the two layers, or "distribute themselves" between the two layers. When equilibrium has established, the ratio of concentration of solute in each layer is constant for each system, and this can be represented by a value K (called the partition coefficient or distribution coefficient).
• 4.6: Step-by-Step Procedures For Extractions
Steps are given for a single and multiple step extraction protocol.
• 4.7: Reaction Work-Ups
A key step in conducting a reaction and isolate the product comes immediately after the reaction is complete, and is called the reaction "work-up" . The work-up refers to methods aimed at purifying the material, and most commonly occur in a separatory funnel. Solutions are added to the funnel to either extract or wash the mixture, with the goal of isolating the product from excess reagents, catalysts, side products, solvents, or compounds formed from side reactions.
• 4.8: Acid-Base Extraction
A modification of the extractions previously discussed in this chapter is to perform a chemical reaction in the separatory funnel in order to change the polarity and therefore partitioning of a compound in the aqueous and organic layers. A common method is to perform an acid-base reaction, which can convert some compounds from neutral to ionic forms (or vice versa).
04: Extraction
Image: Organic solvents (clear top layer) and colored aqueous solutions (bottom layer) in separatory funnels
Image: Food dyes dissolving in aqueous solution.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
4.02: Overview of Extraction
"Extraction" refers to transference of compound(s) from a solid or liquid into a different solvent or phase. When a tea bag is added to hot water, the compounds responsible for the flavor and color of tea are extracted from the grounds into the water (Figure 4.1a). Decaffeinated coffee is made by using solvents or supercritical carbon dioxide to extract the caffeine out of coffee beans. Bakers use the extract of vanilla, almond, orange, lemon, and peppermint in their dishes, essences that have been extracted from plant materials using alcohol (Figure 4.1b).
Figure 4.1: Examples of extraction: a) Tea, b) Baking extracts, c) Plant pigments extracted into water droplets after sprinklers hit a fallen leaf on the sidewalk.
In the chemistry lab, it is most common to use liquid-liquid extraction, a process that occurs in a separatory funnel (Figure 4.2). A solution containing dissolved components is placed in the funnel and an immiscible solvent is added, resulting in two layers that are shaken together. It is most common for one layer to be aqueous and the other an organic solvent. Components are "extracted" when they move from one layer to the other. The shape of the separatory funnel allows for efficient drainage and separation of the two layers.
Compounds move from one liquid to another depending on their relative solubility in each liquid. A quick guide to solubility is the "like dissolves like" principle, meaning that nonpolar compounds should be readily extracted into nonpolar solvents (and vice versa). The compounds responsible for the taste and color of tea must be polar if they are readily extracted into hot water. When allowed to equilibrate between two liquids in a separatory funnel, the majority of a compound often ends up in the layer that it is more soluble. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.01%3A_Prelude_to_Extraction.txt |
There are several reasons to use extraction in the chemistry lab. It is a principal method for isolating compounds from plant materials. Extraction moves compounds from one liquid to another, so that they can be more easily manipulated or concentrated. It also enables the selective removal of components in a mixture.
Extracting Natural Compounds
Fruit and plant leaves are primarily composed of cellulose and water, but also contain "essential oils." a greasy mixture of compounds that capture the "essence" of the plant material's smell and taste. Orange oil is roughly \(95\%\) limonene (Figure 4.3b), and due to its nonpolar structure, can be extracted from its rind into an organic solvent like hexanes or dichloromethane (Figure 4.3a). The oil can then be concentrated and used to flavor or scent foods, cleaning supplies, and candles.
Figure 4.3: a) Orange rind extracted into dichloromethane, b) GC spectrum of orange oil.
In the chemistry lab, essential oils are often extracted from their source using solvents, and analyzed using gas chromatography or spectroscopy.
Transferring Compounds From Layers
Another method for extracting essential oils from fragrant plant materials is through steam distillation (Figure 4.4b). This process often results in the lovely smelling compounds suspended in the aqueous distillate (Figure 4.4c). In order to concentrate the oil, the aqueous suspension is often extracted with a low-boiling organic solvent (Figure 4.4d), which can then be easily removed from the oil.
Figure 4.4: a) Whole cloves, b) Steam distillation of cloves, c) Milky distillate composed of oil and water, d) Using extraction to separate the oil from the water.
Selective Removal of Components
When conducting an experiment that synthesizes a chemical product, a reaction is often complete whenever stirring or heating is ceased. And yet, there are always more steps in the procedure! What commonly happens directly afterwards is to "work-up" the reaction in some way. A work-up refers to methods aimed at isolating the product from the reaction mixture, and often begins by using a separatory funnel and extractions.
For example, imagine that acetic acid and isopentanol have been heated in the presence of an acid catalyst for one hour (Figure 4.5) in order to make isopentyl acetate, an ester that smells of bananas (see reaction scheme in Figure 4.6). After the one-hour time period, there is unfortunately not just the banana-smelling ester in the flask. The flask will also contain byproducts (the water in this case), leftover starting materials if the reaction is incomplete, as well as any catalysts used (\(\ce{H_2SO_4}\) in this case). In this example, there could be five compounds in the reaction flask after heating is ceased (Figure 4.7)!
When "working up" this reaction, the resulting mixture is often poured into a separatory funnel along with some water and organic solvent. This produces two layers in the separatory funnel: an aqueous layer and an organic layer.
After shaking this heterogeneous mixture, the compounds distribute themselves based on their solubility. Compounds that have high water solubility favor the aqueous layer while less polar compounds favor the organic layer. In this example, the acid catalyst and residual carboxylic acid or alcohol would likely be drawn into the water layer. The ester would have a greater affinity for the organic layer than the aqueous layer, causing it to be isolated from the other components in the reaction mixture (Figure 4.7).
In this example, it is possible that small amounts of alcohol are also drawn into the organic layer, but they could likely be removed with a water "wash." In a wash, the desired compound (e.g. the isopentyl acetate), remains in its current layer of the separatory funnel (in this example the organic layer), and unwanted compounds are removed, or "washed away" into another layer (e.g. the aqueous layer). A wash is different than an extraction, because in an extraction the desired compound moves from its current location (i.e. moves from an aqueous layer to an organic layer), while in washing the desired compound stays in its current layer. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.03%3A_Uses_of_Extraction.txt |
Density
It is essential that you know whether the aqueous layer is above or below the organic layer in the separatory funnel, as it dictates which layer is kept and which is eventually discarded. Two immiscible solvents will stack atop one another based on differences in density. The solution with the lower density will rest on top, and the denser solution will rest on the bottom.
Most non-halogenated organic solvents have densities less than 1 g/mL, so will float atop an aqueous solution (if they are immiscible). A notable exception is that halogenated solvents are denser than water (have densities greater than 1 g/mL), and so will instead sink below aqueous solutions (Table 4.1 and Figure 4.8).
Table 4.1: Density of common solvents at room temperature.
Solvent Density (g/mL)
Pentane 0.626
Petroleum Ether (mixture of C5 - C6 hydrocarbons) 0.653
Hexanes (mixture of 6 hydrocarbons) 0.655
Diethyl ether 0.713
Ethyl acetate 0.902
Water 0.998
Dicholoromethane (CH2Cl2) 1.33
Chloroform (CHCl3) 1.49
Many solutions used in separatory funnels are fairly dilute, so the density of the solution is approximately the same as the density of the solvent. For example, if mixing diethyl ether and a $10\% \: \ce{NaOH} \left( aq \right)$ solution in a separatory funnel, knowledge of the exact density of the $10\% \: \ce{NaOH}$ solution is not necessary. A $10\% \: \ce{NaOH} \left( aq \right)$ solution is $90\%$ water (by mass), meaning the density should be fairly close to the density of water (approximately $1 \: \text{g/mL}$). The actual density of a $10\% \: \ce{NaOH} \left( aq \right)$ solution is $1.1089 \: \text{g/mL}$, a value only slightly greater than the density of water. The diethyl ether will be the top layer in this situation.
There are times, however, when so may solute particles are dissolved that a solution's density is much greater than the solvent density. For example, a saturated $\ce{NaCl} \left( aq \right)$ solution has a density around $1.2 \: \text{g/mL}$ (significantly greater than the density of water), and can cause separation problems with solvents of similar densities like dichloromethane.
How to Determine the Aqueous Layer
Solvent densities may be used to predict which layer is organic and which is aqueous in a separatory funnel, but there are other methods that can be useful in this determination. If unsure which layer is aqueous and which layer is organic, do one of the following things:
1. Add a bit of water from a squirt bottle to the separatory funnel (Figure 4.9a) and watch where the water droplets go.
If the top layer is aqueous, the water droplets should mix with the top layer, and they will look as if they disappear. If the bottom layer is aqueous, the water droplets will fall through the top layer to mix with the bottom layer (as indicated by an arrow in Figure 4.9b+c). If it is difficult to track where the water droplets go, also keep track of the volume of the layers: whichever layer increases with the addition of water is the aqueous layer.
1. Consider relative volumes of aqueous and organic solvents, based on quantities used in the experiment.
Figure 4.10a shows a $125 \: \text{mL}$ separatory funnel containing $10 \: \text{mL}$ hexane and $100 \: \text{mL}$ water (tinted with blue dye). If these were the quantities used in an experiment, the aqueous layer would have to be the lower layer as it is so much larger. Although unequivocal in this case, it is important to know that the odd shape of the separatory funnel may cause you to misjudge volumes. A separatory funnel with equal volumes of aqueous and organic layers is shown in Figure 4.10b, although the layers rise to different heights in the funnel.
$^1$The solvents listed in Table 4.1 are pure compounds except for petroleum ether and hexanes. "Petroleum Ether" contains pentane, 2-methylbutane, 2,2-dimethylpropane, n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. "Hexanes" contains 2-methylpentane, 3-methylpentane, n-hexane, and methylcyclopentane. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.04%3A_Which_Layer_is_Which.txt |
Partition/Distribution Coefficient $\left( K \right)$
When a solution is placed in a separatory funnel and shaken with an immiscible solvent, solutes often dissolve in part into both layers. The components are said to "partition" between the two layers, or "distribute themselves" between the two layers. When equilibrium has established, the ratio of concentration of solute in each layer is constant for each system, and this can be represented by a value $K$ (called the partition coefficient or distribution coefficient).
$K = \dfrac{\text{Molarity in organic phase}}{\text{Molarity in aqueous phase}}$
For example, morphine has a partition coefficient of roughly 6 in ethyl acetate and water.$^2$ If dark circles represent morphine molecules, $1.00 \: \text{g}$ of morphine would distribute itself as shown in Figure 4.11.
Note that with equal volumes of organic and aqueous phases, the partition coefficient represents the ratio of particles in each layer (Figure 4.11a). When using equal volumes, a $K$ of $\sim 6$ means there will be six times as many morphine molecules in the organic layer as there are in the water layer. The particulate ratio is not as simple when the layer volumes are different, but the ratio of concentrations always equals the $K$ (Figure 4.11b).
The partition coefficients reflect the solubility of a compound in the organic and aqueous layers, and so is dependent on the solvent system used. For example, morphine has a $K$ of roughly 2 in petroleum ether and water, and a $K$ of roughly 0.33 in diethyl ether and water.$^2$ When the $K$ is less than one, it means the compound partitions into the aqueous layer more than the organic layer.
Choosing a Solvent with Solubility Data
The partition coefficient $K$ is the ratio of the compound's concentration in the organic layer compared to the aqueous layer. Actual partition coefficients are experimental, but can be estimated by using solubility data.
\begin{align} K &= \dfrac{\text{Molarity in organic phase}}{\text{Molarity in aqueous phase}} \[4pt] & \approx \dfrac{\text{Solubility in organic phase}}{\text{Solubility in aqueous phase}} \end{align}
The $K$'s calculated using molarity and solubility values are not identical since different equilibria are involved. The true $K$ represents the equilibrium between aqueous and organic solutions, while solubility data represent the equilibrium between a saturated solution and the solid phase. The two systems are related however, and $K$'s derived from solubility data should be similar to actual $K$'s.
Table 4.2: Solubility of caffeine in different solvents.$^3$
Solvent 1 g caffene dissolves in solvent (at 25 °C)
water 45 mL
Diethyl Ether 530 mL
Benzene 100 mL
Chloroform 5.5 mL
Solubility data can therefore be used to choose an appropriate solvent for an extraction. For example, imagine that caffeine (Figure 4.12) is intended to be extracted from tea grounds into boiling water, then later extracted into an organic solvent. Solubility data for caffeine is shown in Table 4.2.
Both diethyl ether and benzene at first glance appear to be poor choices for extraction because caffeine is more soluble in water than in either solvent (if a gram of caffeine dissolves in $46 \: \text{mL}$ water, but $100 \: \text{mL}$ of benzene, caffeine is more soluble in water). When extracting with either of these solvents, the $K$ would be less than one (see calculation below) and it would be an "uphill battle" to draw out the caffeine from the water. However, caffeine is more soluble in chloroform than water, so chloroform would be the best choice of the solvents shown in terms of the maximum extraction of caffeine.
\begin{align} K_\text{benzene} &\sim \dfrac{\left( \dfrac{1 \: \text{g caffeine}}{100 \: \text{mL benzene}} \right)}{\left( \dfrac{1 \: \text{g caffeine}}{46 \: \text{mL water}} \right)} \sim 0.46 \[4pt] K_\text{chloroform} &\sim \dfrac{\left( \dfrac{1 \: \text{g caffeine}}{5.5 \: \text{mL chloroform}} \right)}{\left( \dfrac{1 \: \text{g caffeine}}{46 \: \text{mL water}} \right)} \sim 8.4 \end{align}
Another consideration when choosing a solvent for extraction is toxicity: chloroform is carcinogenic and therefore is probably not the best option despite its excellent solvation ability. A further consideration is the solubility of other components present in a mixture. If the goal is to extract caffeine preferentially and leave behind other components in the tea, one solvent may be more selective in this regard.
Quantitating Single Extraction
Hyoscyamine is an alkaloid from a plant in the nightshade family (Figure 4.13a), and is used medicinally to provide relief for a variety of gastrointestinal disorders. Its solubility data is shown in Figure 4.13b.
Imagine that a nearly saturated solution of $0.50 \: \text{g}$ hyoscyamine in $150 \: \text{mL}$ water is to be extracted into $150 \: \text{mL}$ diethyl ether. How much hyoscyamine would be extracted into the diethyl ether layer in this process?
This quantity can be approximated using the solubility data. Taking the ratio of the compound's solubility in diethyl ether compared to water gives an approximate $K$ of 4.
\begin{align} K &\sim \dfrac{\text{organic solubility}}{\text{water solubility}} \[4pt] &\sim \dfrac{\left( 1.44 \: \text{g hyoscyamine}/100 \: \text{mL diethyl ether} \right)}{\left( 0.354 \: \text{g hyoscyamine}/100 \: \text{mL water} \right)} \[4pt] &\sim \textbf{4.07} \: \text{(approximate} K \text{)} \end{align}
If "$x$" is the gram quantity of hyoscyamine extracted into the diethyl ether layer, then "$0.50 \: \text{g} - x$" would remain in the aqueous layer after equilibrium is established. Knowing the value of $K$, the value of $x$ can be solved for using the equation below.
$4.07 = \dfrac{\left( \dfrac{x}{150 \: \text{mL ether}} \right)}{\left( \dfrac{0.50 \: \text{g} - x}{150 \: \text{mL water}} \right)}$
After solving the algebra, $x = \textbf{0.40 g}$. This result means that $0.40 \: \text{g}$ of the original $0.50 \: \text{g}$ of hyoscyamine is extracted into the diethyl ether using a single extraction. This process is summarized in Figure 4.14.
In this example, a single extraction resulted in extraction of $80\%$ of the hyoscyamine $\left( 100\% \times 0.40 \: \text{g}/0.50 \: \text{g} \right)$ from the aqueous layer into the organic layer. The partitioning of the compound between the two layers caused the sample to be incompletely extracted.
Multiple Extractions
Overview of Multiple Extractions
Depending on the partition coefficient for a compound in a solvent, a single extraction may be all that is needed to effectively extract a compound. However, more often than not a procedure calls for a solution to be extracted multiple times in order to isolate a desired compound, as this method is more efficient than a single extraction (see journal article in Figure 4.15b for an example of where this process is used).
In a multiple extraction procedure, a quantity of solvent is used to extract one layer (often the aqueous layer) multiple times in succession. The extraction is repeated two to three times, or perhaps more times if the compound has a low partition coefficient in the organic solvent. Figure 4.16 shows a diagram of an aqueous solution being extracted twice with diethyl ether. Diethyl ether has a density less than $1 \: \text{g/mL}$, so is the top organic layer in the funnel.
In a multiple extraction of an aqueous layer, the first extraction is procedurally identical to a single extraction. In the second extraction, the aqueous layer from the first extraction is returned to the separatory funnel (Figure 4.16b), with the goal of extracting additional compound. Since the organic layer from the first extraction had already reached equilibrium with the aqueous layer, it would do little good to return it to the separatory funnel and expose it to the aqueous layer again. Instead, fresh diethyl ether is added to the aqueous layer, since it has the potential to extract more compound.
The process is often repeated with a third extraction (not shown in Figure 4.16), with the aqueous layer from the second extraction being returned to the separatory funnel, followed by another portion of fresh organic solvent. In multiple extractions, the organic layers are combined together,as the goal is to extract the compound into the organic solvent.
When an aqueous solution is extracted with an organic solvent that is denser than water (for example dichloromethane, $\ce{CH_2Cl_2}$), the only procedural difference is that there is no need to ever drain the aqueous layer from the separatory funnel. After draining the organic layer from the first extraction, fresh solvent can be added to the aqueous layer remaining in the funnel to begin the second extraction (Figure 4.17b).
Quantitating Multiple Extraction
To demonstrate the effectiveness of a multiple extraction, let's return to the problem from the single extraction section, where a solution of $0.50 \: \text{g}$ hyoscyamine in $150 \: \text{mL}$ water is to be extracted into diethyl ether. Instead of using one $150 \: \text{mL}$ portion, let's instead split the solvent into three $50 \: \text{mL}$ portions of diethyl ether. How much hyoscyamine would be extracted with this method?
In the previous section, solubility data was used to estimate the partition coefficient $K$, and it was found to be 4.07. As before, we can assign the quantity of hyoscyamine extracted into the diethyl ether the value "$x$", which would leave "$0.50 \: \text{g} - x$" remaining in the aqueous layer of the first extraction. Using $K$, the calculation is identical to the previous discussion, differing only in the smaller volume of the organic layer ($50 \: \text{mL}$ instead of $150 \: \text{mL}$).
$4.07 = \dfrac{\left( \dfrac{x}{50 \: \text{mL ether}} \right)}{\left( \dfrac{0.50 \: \text{g} - x}{150 \: \text{mL water}} \right)}$
After solving the algebra, $x = 0.29 \: \text{g}$. This result means that $0.29 \: \text{g}$ is extracted into the diethyl ether in the first extraction and $0.21 \: \text{g}$ remains in the aqueous layer $\left( 0.50 \: \text{g} - 0.29 \: \text{g} \right)$. As the aqueous layer is returned to the separatory funnel, the residual $0.21 \: \text{g}$ is the quantity to be further extracted, which alters the calculation for the second extraction by replacing the $0.50 \: \text{g}$ value.
$4.07 = \dfrac{\left( \dfrac{x}{50 \: \text{mL ether}} \right)}{\left( \dfrac{0.21 \: \text{g} - x}{150 \: \text{mL water}} \right)}$
After solving the algebra, $x = 0.12 \: \text{g}$. This result means that $0.12 \: \text{g}$ is extracted into the diethyl ether in the second extraction and $0.09 \: \text{g}$ remains in the aqueous layer $\left( 0.21 \: \text{g} - 0.12 \: \text{g} \right)$. The calculation for the third extraction is as follows:
$4.07 = \dfrac{\left( \dfrac{x}{50 \: \text{mL ether}} \right)}{\left( \dfrac{0.09 \: \text{g} - x}{150 \: \text{mL water}} \right)}$
After solving the algebra, $x = 0.05 \: \text{g}$. This result means $0.04 \: \text{g}$ remains in the aqueous layer $\left( 0.09 \: \text{g} - 0.05 \: \text{g} \right)$ after the third extraction. The results of the calculations in this section are summarized in Figure 4.18.
If the $50 \: \text{mL}$ diethyl ether extracts are combined in this example (Figure 4.19), there would be a total of $0.46 \: \text{g}$ of hyoscyamine in the combined organic extracts. Of the $0.50 \: \text{g}$ of hyoscyamine in the original aqueous layer, $92\%$ of the material is extracted into the organic layer $\left( 100\% \times 0.46 \: \text{g}/0.50 \: \text{g} \right)$. This is a greater quantity than was obtained using a single extraction of $150 \: \text{mL}$ diethyl ether, which resulted in only $0.40 \: \text{g}$ of hyoscyamine extracted ($80\%$).
These calculations demonstrate that using multiple portions of a solvent maximizes the extractive power of the solvent. In general, three extractions are the optimal compromise between expended effort and maximizing the recovery of material.
$^2$The partition coefficients were approximated using solubility data found in: A. Seidell, Solubilities of Inorganic and Organic Substances, D. Van. Nostrand Company, 1907.
$^3$From: The Merck Index, 12$^\text{th}$ edition, Merck Research Laboratories, 1996. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.05%3A_Extraction_Theory.txt |
Single Extraction
The pictures in this section show a single extraction of methyl red (colored compound, Figure 4.21) from an aqueous solution (bottom layer) into $25 \: \text{mL}$ of ethyl acetate (top layer). The aqueous solution originally has a pink color, as the methyl red appears red in acidic solution (the aqueous solution was made from $50 \: \text{mL}$ water, 5 drops of $0.1 \: \text{M} \: \ce{HCl}$ and 5 drops of $1\%$ methyl red indicator solution). The methyl red has a large partition coefficient and is extracted from the aqueous layer into the ethyl acetate in this process.
Prepare the Setup (for single extraction)
1. Obtain a separatory funnel (Figure 4.23a).
1. If the separatory funnel has a Teflon stopcock, reassemble the stopcock if it was taken apart to dry, placing the parts in the appropriate order (Figure 4.23b). Be sure that the Teflon stopcock is moderately tight so that it can still easily turn, but is not so loose that liquid can seep around the joint.
2. If using a glass stopcock (Figure 4.23c), it likely needs no further preparation. There should be a very thin layer of grease used to seal the stopcock and prevent freezing. If both glass and Teflon stopcocks are available, Teflon is a better choice as there is always a possibility that solvent can dissolve the grease used with glass stopcocks and contaminate the sample.
3. Also obtain a stopper (Teflon or ground glass) that fits well in the top joint of the funnel (Figure 4.23a).
2. Place the separatory funnel in a ring clamp attached to a ring stand or latticework. The funnels are easy to break, so cushion the funnel in the metal clamp using pieces of slit rubber or plastic tubing (Figure 4.23d).
Add the Solutions (for single extraction)
1. Before pouring anything into a separatory funnel, be sure that the stopcock is in the "closed" position, where the stopcock is horizontal (Figure 4.24a).
As a fail-safe, always position an Erlenmeyer flask beneath the separatory funnel before pouring (Figure 4.24b). This can catch liquid in case the stopcock is accidentally left open, or if the stopcock is loose and liquid leaks through unintentionally.
2. Using a funnel, pour the liquid to be extracted into the separatory funnel (Figures 4.24b + 4.25). A separatory funnel should never be used with a hot or warm liquid.
The ground glass joint atop a separatory funnel is more prone to stick to the stopper if there was liquid in the joint at some point. Pouring liquid into the separatory funnel using a short-stemmed funnel avoids getting the joint wet, so that it will be less likely to freeze during mixing.
1. Pour a quantity of the extractive solvent into the separatory funnel, as indicated by the procedure (Figure 4.24c).
It is unnecessary to use precise quantities of solvent for extractions, and the volumes can be measured in a graduated cylinder. If a procedure calls for $20 \: \text{mL}$ of solvent, it is acceptable if between $20$-$25 \: \text{mL}$ is used each time.
Mix the Solutions (for single extraction)
1. Place the stopper on the funnel, and hold the funnel such that the fingers of one hand securely cover the stopper, while the other hand grips the bottom of the funnel (Figure 4.26a).
2. Gently invert the funnel (Figure 4.26b), and swirl the mixture a little.
Although it is not uncommon for some liquid to creep into the ground glass joint when inverted, it should be minimal. If liquid drips onto your fingers or gloves when you invert the funnel, the stopper is probably the wrong size.
3. Pressure may build up inside the separatory funnel when solutions are mixed, so immediately after swirling, and with the funnel still inverted, "vent" the funnel by briefly opening the stopcock to allow for a release of pressure (Figure 4.26c).
Pressure builds in the funnel as solvent evaporates into the headspace and contributes additional vapor to the initial $\sim$1 atmosphere of air pressure in the funnel. With highly volatile solvents (like diethyl ether), a definite "swoosh" can be heard upon venting, and small amounts of liquid may even sputter out the stopcock. If liquid spits out the stopcock, try to allow it to drain back into the funnel. The noise associated with venting normally ceases after the second or third inversions, as the headspace becomes saturated with solvent vapors and the pressures inside and outside the funnel are equalized.
Safety note: Never point the stopcock toward someone as you vent, as it's possible some liquid may splatter onto him or her.
4. Close the stopcock and mix the solutions a bit more vigorously, periodically stopping to vent the system.
There are differences of opinion on how vigorously solutions should be mixed in separatory funnels, and for how long. As a general guide, a mild mixing for 10-20 seconds should be enough. With some solutions (e.g. dichloromethane), care should be taken to not shake too vigorously, as these solutions often form emulsions (where the interface between the solutions doesn't clarify). With solutions prone to emulsions, a funnel should be gently rocked for one minute.
5. Place the separatory funnel upright in the ring clamp to allow the layers to fully separate. The interface between the layers should settle rather quickly, often within 10 seconds or so. If the interface is clouded or not well defined (an emulsion has formed), see the troubleshooting section for tips.
Separate the Layers (for single extraction)
1. Liquid will not drain well from a separatory funnel if the stopper remains on, as air cannot enter the funnel to replace the displaced liquid. If liquid did drain from the funnel without replacement by an equal volume of air, a negative pressure would form in the funnel. Thus, before draining liquid from a separatory funnel, remove the stopper (Figure 4.27a).
2. Drain the majority of the bottom layer into a clean Erlenmeyer flask, positioning the ring clamp so that the tip of the separatory funnel is nestled in the Erlenmeyer flask to prevent splashing (Figure 4.27b). Stop draining when the interface is within $1 \: \text{cm}$ of the bottom of the stopcock.
3. Gently swirl the funnel to dislodge any droplets clinging to the glass (Figure 4.27c). A glass stirring rod can be used to knock down stubborn clinging droplets.
4. Further drain the bottom layer, stopping when the interface just enters the stopcock chamber (Figure 4.27d). Label the Erlenmeyer flask (e.g. "bottom layer").
When labeling flasks, it's often best to use terminology that is without question correct, such as "top layer" or "bottom layer". If the layers are labeled with statements like "organic layer" or "aqueous layer", it's possible that the layers have been incorrectly identified. It may be best to use combined statements like "top organic layer" as it's quite useful to track the aqueous and organic layers. If the layers have been incorrectly identified, at least the "top" part of the label will always be correct.
Flasks can be labeled using labeling tape or by writing directly on the glass with a permanent marker (e.g. Sharpie). Marker ink can be removed from glass by rubbing it with a KimWipe moistened with a bit of acetone.
1. Pour out the top layer from the top of the separatory funnel into another clean Erlenmeyer flask (Figure 4.28a), making sure to again label this flask (Figure 4.28b).
It is proper technique to drain the bottom layer through the stopcock, and to pour out the top layer from the top of the funnel. This method minimizes re-mixing the solutions, as only the lower layer touches the stem of the funnel.
2. Never throw away any liquids from an extraction until you are absolutely sure that you have the desired compound. Undesired layers can be properly disposed of when the desired compound is in your hands (e.g. after the rotary evaporator has removed the solvent).
Mistakes made during extractions (e.g. carrying on with the wrong layer), can be solved as long as the solutions have not been placed in the waste container! The layers should also be saved until after evaporation because the desired compound may not be very soluble in the solvent used. If the compound failed to extract in one solvent, a different solvent could be tried later, again only if the layers had not yet been thrown away.
Clean Up (for single extraction)
1. To clean a separatory funnel, first rinse it with acetone into a waste container. Then wash the funnel with soap and water at your benchtop. Disassemble the Teflon stopcock (if used). After rinsing with distilled water, allow the parts to dry separated in your locker (Figure 4.28c).
Single Extraction Summary
Table 4.3: Procedural summary for single extraction.
Use slit tubing to cushion the separatory funnel in the ring clamp.
Close the stopcock on the separatory funnel and position an Erlenmeyer flask beneath the setup, in case it drips.
Into the separatory funnel pour the liquid to be extracted using a funnel: this prevents liquid from getting on the ground glass joint which can cause it to stick.
Pour the extractive solvent into the funnel.
Hold the separatory funnel so that your fingers firmly cover the stopper.
Invert the funnel and shake gently for 10-20 seconds.
Periodically "vent" the funnel (open the stopcock while inverted to release pressure). Never point the tip at someone while venting.
Return the separatory funnel to the ring clamp, and allow the layers to separate.
Remove the stopper (it won't drain otherwise).
Drain the majority of the bottom layer into an Erlenmeyer flask.
Stop when roughly $1 \: \text{cm}$ of the bottom layer is in the funnel, and swirl to dislodge clinging droplets.
Drain the rest of the bottom layer, stopping when the interface is inside the stopcock.
Label the flask (e.g. "bottom aqueous layer").
Pour out the top layer into another Erlenmeyer flask (and label it).
Don't throw away either layer until you are sure you've accomplished the goal of the extraction.
Multiple Extractions
Organic Layer is on the Top
In this section are stepwise instructions on how to extract an aqueous solution with an organic solvent that is less dense than water (the organic layer will be on the top). As an example, the instructions are written to extract an aqueous solution three times using $25 \: \text{mL}$ diethyl ether each time ($3 \times 25 \: \text{mL}$ diethyl ether). A procedural summary of the first two extractions is in Figure 4.29.
Extraction #1
1. Perform a single extraction using approximately $25 \: \text{mL}$ of diethyl ether (an exact amount is not necessary), as described previously, making sure to appropriately label each layer (e.g. "top organic layer" and "bottom aqueous layer").
Extraction #2
1. Return the aqueous layer to the separatory funnel. There is no need to wash the funnel in between extractions.
2. Add a fresh $25 \: \text{mL}$ portion of diethyl ether to the separatory funnel. Stopper the funnel, invert and shake with venting, then allow the layers to separate.
At this step, there should be two layers in the separatory funnel. If two layers aren't present, it's likely that the wrong layer was added to the funnel in step 2 (a common mistake). One way to test if this was the mistake is to add a bit of water from a squirt bottle. If the layer returned to the separatory funnel is the organic layer (incorrect), the squirt bottle water will not mix with the solution, and will instead fall as droplets to the bottom.
If the organic layer (incorrect) was accidentally returned to the separatory funnel, there is no harm done, as the organic layer was simply diluted. Pour the liquid back into the flask designed for the organic layer, and instead add the aqueous solution to the funnel.
3. Drain the bottom aqueous layer into an Erlenmeyer flask: it is acceptable to use the same flask that was used for the aqueous layer in the first extraction (that may have been labeled "bottom aqueous layer").
4. Since it is most common to combine the organic layers in multiple extractions, the top organic layer can be poured out of the separatory funnel into the same flask that was used for the organic layer in the first extraction (that may have been labeled "top organic layer"). In this flask, there should be roughly $50 \: \text{mL}$ of diethyl ether from the two extractions.
Extraction #3
1. Repeat the extraction a third time by adding the aqueous layer from the second extraction into the separatory funnel, followed by another fresh $25 \: \text{mL}$ portion of diethyl ether. Stopper the funnel, invert and shake with venting, then allow the layers to separate.
2. Drain the aqueous layer into the appropriate flask, and again pour the top layer into the organic layer flask, where there should be roughly $75 \: \text{mL}$ of diethyl ether from the three extractions.
Organic Layer is on the Bottom
In this section are stepwise instructions on how to extract an aqueous solution with an organic solvent that is denser than water (the organic layer will be on the bottom). As an example, the instructions are written to extract an aqueous solution three times using $25 \: \text{mL} \: \ce{CH_2Cl_2}$ each time ($3 \times 25 \: \text{mL} \: \ce{CH_2Cl_2}$, Figure 4.30).
Extraction #1
1. Perform a single extraction using approximately $25 \: \text{mL}$ of dichloromethane ($\ce{CH_2Cl_2}$, an exact amount is not necessary), as described previously, with the following differences:
1. As $\ce{CH_2Cl_2}$ is prone to emulsions, invert the funnel and shake gently for one minute with venting.
2. After allowing the layers to separate in the funnel, drain the bottom organic layer into a clean Erlenmeyer flask (and label the flask, e.g. "bottom organic layer"). Do not drain the top aqueous layer from the funnel.
Extraction #2
1. To the aqueous layer remaining in the funnel, add a fresh $25 \: \text{mL}$ portion of dichloromethane. Stopper the funnel, invert and shake gently for 1 minute with venting, then allow the layers to separate.
2. Since it is most common to combine the organic layers in multiple extractions, the bottom organic layer can be drained from the separatory funnel into the same flask that was used for the organic layer in the first extraction (that may have been labeled "bottom organic layer"). In this flask, there should be roughly $50 \: \text{mL}$ of dichloromethane from the two extractions.
Extraction #3
1. Repeat the extraction by adding another fresh $25 \: \text{mL}$ portion of dichloromethane to the aqueous layer in the separatory funnel. Stopper the funnel, invert and shake gently with venting, then allow the layers to separate.
2. Drain the bottom organic layer into the flask used previously, where there should be roughly $75 \: \text{mL}$ of dichloromethane from the three extractions.
Troubleshooting
This section descries common problems and solutions in extractions.
There is Only One Layer
The most common reason for having only one layer in a separatory funnel when there should be two (as in when the procedure tells you to "separate the layers"), is to have made a mistake. What likely happened is that the wrong layer was added to the separatory funnel - for example the organic layer was unknowingly added instead of the aqueous layer. When organic solvent is added to an organic layer in the separatory funnel, the result is only one layer. The mistake can be remedied as long as the layers have not yet been thrown away! If the correct layer is added to the funnel, everything will work out as planned.
To prevent making this mistake in the future, be sure to label the Erlenmeyer flasks. Also, be sure to never throw away a layer until you are absolutely sure that you've done everything correctly.
An occasional reason that only one layer forms in a separatory funnel is if there are large quantities of compounds present that dissolve in both solvents, for example if large amounts of ethanol are present, which dissolve well in both aqueous and organic solvents. In this situation, the best approach is to remove the troublesome compound (i.e. the ethanol) on a rotary evaporator before extraction.
There are Three Layers
The most common reason for three layers in a separatory funnel is inadequate mixing (Figure 4.31a). If the funnel is shaken with more vigor it will likely settle into two layers (Figure 4.31b).
It is also possible that a middle third layer is an emulsion, where the two layers are not fully separated.
There is Insoluble Material at the Interface
A small amount of insoluble film between two layers is not uncommon during an extraction. Polymeric materials tend to rest between layers as solvent interactions are minimized at the interface. A minor film is not something to worry about because if a small amount does make it into the organic layer, a subsequent drying and filtration step will often remove it.
The Interface Cannot be Seen
On occasion the compounds in a separatory funnel are so dark that they obscure the interface between the two layers. If this happens, there are several methods that might help you see the interface. One is to hold the separatory funnel up to the light, or to shine a flashlight onto the glass (Figure 4.32b). Additional light sometimes allows you to see the interface. A second method is to carefully observe the layers while tilting the funnel back and forth to the side (Figure 4.32c). Your eye can sometimes pick up on subtle differences in the way the liquids flow. A third method is to add a bit more solvent to the funnel to somewhat dilute one of the layers, or to add a different solvent to alter the index of refraction.
The Layers Don't Separate Well (An Emulsion Formed)
Emulsions are when tiny droplets of one layer are suspended in the other layer, resulting in no distinct interface between the two layers (Figure 4.33). Often an emulsion looks like a bubbly mess near the interface, and can even appear to be an odd-looking third layer.
Emulsions can happen for several reasons:
1. The density of each layer may be so similar that there is weak motivation for the liquids to separate.
2. There may be soap-like compounds or other emulsifying agents present that dissolve some of the components in one another.
Emulsions can be very difficult to rectify, and it's best if they are avoided in the first place by shaking solutions that are prone to emulsions (e.g. dichloromethane with highly basic or dense solutions) gently in the separatory funnel. Nonetheless, if an emulsion does form, there are some ways to attempt to clarify them:
1. For mild emulsions, gently swirl the layers and try to knock down suspended droplets with a glass stirring rod.
2. Allow the solution to sit for a period of time (even until the next lab period) if possible. With enough time, some solutions do settle out on their own. This of course may not be practical.
3. For small volumes, use a centrifuge if one is available. A centrifuge hastens the process of letting an emulsion settle on its own. Remember that a centrifuge needs to be balanced or it may wobble off the benchtop. Divide the solutions equally, putting tubes of equal volume opposite one another inside the centrifuge.
4. If an emulsion is formed because the two layers have similar densities, try to alter the density of each layer to make them more different. To help clarify an emulsion, try to decrease the density of the top layer or increase the density of the bottom layer.
For example, if an emulsion occurs with ethyl acetate (top layer) and an aqueous solution (bottom layer), add some $\ce{NaCl}$. $\ce{NaCl}$ will dissolve in the aqueous layer and increase the density of the aqueous solution. Alternatively add additional ethyl acetate, which will dilute the organic layer and lower its density. As a last resort add some pentane, which will mix with the top organic layer and decrease its density (pentane is one of the least dense organic solvents). The addition of pentane is used as a final effort as it will negatively affect the ability of the organic layer to extract somewhat polar compounds.
If an emulsion occurs with an aqueous solution (top layer) and dichloromethane (bottom layer), add some water from a squirt bottle to dilute the top layer and decrease its density. This method worked well to clarify the emulsion in Figure 4.32c, as evidenced by Figure 4.32d.
5. Try decreasing the solubility of one component in the other. One method is to add $\ce{NaCl}$ or $\ce{NH_4Cl}$ to the separatory funnel, which dissolves in the aqueous layer and decreases the ability of organic compounds to dissolve in water ("salting out").
Microscale Extractions
Microscale work involves the manipulation of less than $300 \: \text{mg}$ of compound, and usually involves solvent volumes of $5 \: \text{mL}$ or less. A separatory funnel would be impractical when working with such small quantities, and conical vials (Figure 4.35) or centrifuge tubes are typically used instead.
The pictures in this section show the extraction of $2 \: \text{mL}$ of a mildly acidic aqueous solution containing a single drop of methyl red solution into $2 \: \text{mL}$ of ethyl acetate. The color (methyl red), is extracted from the aqueous layer (bottom) into the ethyl acetate layer (top).
Mix the Solutions (for microscale extraction)
1. Pour the contents to be extracted into a conical vial, or a glass tube with a tapered end (e.g. centrifuge tube). As these containers are prone to tip, use a beaker (Figure 4.36a) or inverted cork ring (Figure 4.36b) for support.
2. Add the extractive solvent by pipette (Figure 4.36a). If using a conical vial, the volume markings on the glass may be helpful.
3. Gently mix the two solutions using one of the following methods:
1. Secure a cap firmly on the vial (Figure 4.36c+d) then invert and shake the tube for 10-20 seconds (Figure 4.35). Conical vials and centrifuge tubes tend to be less airtight than separatory funnels, so there should be no need to vent the system during shaking unless $\ce{NaHCO_3}$ or $\ce{Na_2CO_3}$ solutions are used.
2. Alternatively, manually mix the layers using a pipette. Withdraw a pipette-full of the bottom layer from the vial, and then vigorously expunge the solution through the top layer (Figure 4.36e). Do this repeatedly for at least one minute. Manual mixing is not recommended when using low-boiling solvents (e.g. diethyl ether), as the volume often decreases dramatically after mixing. Instead use the first mixing method described.
Separate the Layers (for microscale extraction)
1. Separate the layers with a Pasteur pipette. The design of conical vials and centrifuge tubes allows for efficient separation of the layers through withdrawal of the bottom layer by pipette. This means that even if the top layer is to be reserved, the bottom layer still needs to be removed first.
1. Hold the conical vial or tapered tube in the same hand as a container for the bottom layer (label it). Withdraw the majority of the bottom layer by Pasteur pipette, and dispense into the container (Figure 4.37a).
2. When withdrawing, always place the pipette tip to the point of the conical vial or tapered tube (Figure 4.37b).
3. It may be difficult to remove the very last drop of bottom layer from the point of the vial. To do so, withdraw the entirety of the bottom layer and a small amount of the top layer into the pipette. Allow the layers to separate inside the pipette (Figure 4.37c), then delicately expel the bottom layer from the pipette into the container. Return the rest of the top layer to the conical vial.
2. If the bottom layer is the desired layer, and another extraction is to be done, add fresh organic solvent to the top layer still in the conical vial and repeat the extraction and separation.
3. If the top layer is the desired layer, remove it from the conical vial using a fresh pipette into a clean container. If another extraction is to be done, return the bottom layer to the conical vial, add fresh solvent and repeat the extraction and separation. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.06%3A_Step-by-Step_Procedures_For_Extractions.txt |
Purpose of a Work-Up
When the goal of an experiment is to conduct a reaction and isolate the product, the general sequence of events is shown in Table 4.4.
Table 4.4: Typical reaction sequence of events.
a) Conduct the reaction. b) Perform multiple extractions and/or washes to partially purify the desired product. c) Remove trace water with a drying agent.
d) Filter or decant the drying agent.
e) Remove the solvent with a rotary evaporator
f) Further purify if necessary.
A key step in this sequence comes immediately after the reaction is complete, and is called the reaction "work-up" (step b) in Table 4.4). The work-up refers to methods aimed at purifying the material, and most commonly occur in a separatory funnel. Solutions are added to the funnel to either extract or wash the mixture, with the goal of isolating the product from excess reagents, catalysts, side products, solvents, or compounds formed from side reactions.
Common Washes
Water
The most common wash in separatory funnels is probably water. Water is cheap, non-hazardous, and works well to remove may impurities found alongside a desired product.
Water can potentially remove water-soluble impurities from an organic layer, as long as they are present in quantities that do not exceed their water solubility. The following are common materials that can be removed with a water wash: unconsumed acid or base, many ionic salts, and compounds that can hydrogen bond with water (have an oxygen or nitrogen atom) and are relatively small (e.g. $\ce{CH_3CH_2OH}$ or $\ce{CH_3COCH_3}$).
To demonstrate the effectiveness of a water wash, a Fischer esterification reaction was conducted to produce isoamyl acetate (Figure 4.38). In this reaction, an excess of acetic acid is used to drive the reaction through Le Chatelier's principle, and the acetic acid had to be removed from the product during the purification process.
The $\ce{^1H}$ NMR spectrum in Figure 4.39a was taken of the reaction mixture immediately after ceasing heating and before the work-up. As expected, a significant signal for acetic acid is seen at $2.097 \: \text{ppm}$.
The reaction was then "worked up" by pouring the reaction mixture into a separatory funnel and washing the organic layer with water, sodium bicarbonate, and brine in succession. The main purpose of the water wash was to remove the majority of the catalytic sulfuric acid and the excess acetic acid, while the sodium bicarbonate wash neutralized the rest. The $\ce{^1H}$ NMR spectrum of the final product (Figure 4.39b) showed the washes were effective as the acetic acid signal at $2.097 \: \text{ppm}$ is absent.
The sodium bicarbonate wash in this example was necessary (and discussed in the next section) because a water wash alone may not fully remove the acetic acid. It's important to know that when a compound is "water soluble" it does not necessarily mean it is "organic insoluble", a common misconception that arises from the "like dissolves like" principle. For example, acetic acid has a $K$ of 0.5 when partitioning between diethyl ether and water, meaning acetic acid favors the aqueous layer only twice as much as the organic layer.$^4$ The ability of acetic acid and other polar compounds to dissolve in the organic layer of a separatory funnel should not be ignored.
Sodium Bicarbonate and Sodium Carbonate
How They Work
A normal part of many work-ups includes neutralization. It is important to neutralize any organic solvent that was exposed to an acidic or basic solution as trace acid or base may cause undesired reactions to occur when the solutions are concentrated. Also, samples intended for GC analysis must be neutral as acidic solutions degrade the polymeric coating of the GC column. In addition, it is preferable to manipulate neutral materials rather than acidic or basic ones, as spills are then less hazardous.
Aqueous solutions of saturated sodium bicarbonate $\left( \ce{NaHCO_3} \right)$ and sodium carbonate $\left( \ce{Na_2CO_3} \right)$ are basic, and the purpose of these washes is to neutralize an organic layer that may contain trace acidic components. Even if an organic layer should not in theory dissolve very polar components such as acid, acid sometimes "hitches a ride" on polar components that may dissolve in an organic layer, such as small amounts of alcohols or water.
The following reactions occur between bicarbonate ion (1), carbonate ion (2) and acid $\left( \ce{H^+} \right)$ during a wash:
$\ce{HCO_3^-} \left( aq \right) + \ce{H^+} \left( aq \right) \rightarrow \ce{H_2CO_3} \left( aq \right) \rightleftharpoons \ce{H_2O} \left( l \right) + \ce{CO_2} \left( g \right) \tag{1}$
$\ce{CO_3^{2-}} \left( aq \right) + \ce{H^+} \left( aq \right) \rightarrow \ce{HCO_3^-} \left( aq \right) \tag{2}$
The initial product of reaction (1) is carbonic acid $\left( \ce{H_2CO_3} \right)$, which is in equilibrium with water and carbon dioxide gas. This means that solutions of bicarbonate often bubble during a neutralization wash in a separatory funnel. The product of reaction (2) is the bicarbonate ion, which can subsequently undergo reaction (1). This means that solutions of carbonate ion also often bubble during neutralizations.
Safety note: To prevent excess pressure form being generated by the release of carbon dioxide gas into a separatory funnel during neutralization, the layers should be gently swirled together before placement of the stopper. They should be vented directly after inversion, and more frequently than usual. Figure 4.41 shows a strongly acidic organic layer (top) in contact with an aqueous solution of $10\%$ sodium bicarbonate (bottom). A vigorous stream of bubbles is seen originating from a small portion of organic layer trapped on the bottom of the funnel. The bubbling was even more vigorous when the layers were mixed together.
Testing the pH After a Wash
To test whether a base wash with $\ce{NaHCO_3}$ or $\ce{Na_2CO_3}$ was effective at removing all the acid from an organic layer, it is helpful to test the pH. It is not possible to test the pH of an organic solution directly, however it is possible to test the pH of an aqueous solution that the organic solution has been in contact with. If the aqueous layer is on the top of a separatory funnel, insert a glass stirring rod into the top layer and touch the wet rod to blue litmus paper. An acidic solution turns blue litmus paper pink (or red), while a neutral or basic solution gives blue litmus paper only a darkened "wet" appearance (Figure 4.42d). If the litmus paper turns pink at all$^5$, the base wash has not fully neutralized the organic layer, and subsequent base washes are needed.
If the aqueous layer is on the bottom of the separatory funnel, test an "aliquot" of the aqueous layer (or tiny sample) on litmus paper through the following method:
1. With a finger placed atop a glass pipette, insert the pipette into the separatory funnel so the tip is positioned in the bottom aqueous layer (Figure 4.42a).
2. Remove the finger on the pipette to allow a sample of the aqueous layer to enter the pipette through capillary action (Figure 4.42b).
3. With a finger placed atop the glass pipette again, remove the pipette from the separatory funnel. A bit of liquid should remain in the pipette tip, an aliquot of the bottom layer (Figure 4.42c).
4. Touch the aliquot to blue litmus paper and observe the color (Figure 4.42d).
5. If the litmus paper turns pink at all, the base wash has not fully neutralized the organic layer, and subsequent base washes are needed.
Brine (Saturated $\ce{NaCl}$)
In some experiments, an organic layer may be washed with brine, which is a saturated solution of $\ce{NaCl} \left( aq \right)$. The purpose of this wash is to remove large amounts of water than may be dissolved in the organic layer. Although the organic layer should always be later exposed to a drying agent (e.g. anhydrous sodium sulfate, magnesium sulfate, or calcium chloride), these reagents at best remove only small amounts of water.
The organic solvents that require a brine wash before exposure to a solid drying agent are diethyl ether and ethyl acetate. These solvents dissolve large quantities of water in comparison to other solvents (Table 4.5).
Table 4.5: Quantity of water dissolved in various solvents.$^6$
Solvent Grams water dissolved in 100 mL solvent
Diethyl ether 1.24 g
Ethyl acetate 2.92 g
Dichloromethane ($CH_2Cl_2$) 0.32 g
Hexanes 0.007 g
Brine works to remove water from an organic layer because it is highly concentrated (since $\ce{NaCl}$ is so highly water soluble). A saturated $\ce{NaCl} \left( aq \right)$ solution is highly ordered, causing a large motivation for water to draw into the solution from the organic layer to increase the entropy of the salt solution (to dilute the solution).
Figure 4.44 shows a qualitative difference in the amount of water present in an organic layer with and without the use of a brine wash. Ethyl acetate was shaken with water (Figure 4.44a), then dried with a portion of anhydrous $\ce{MgSO_4}$. The large clumps of drying agent in Figure 4.44b indicate that this ethyl acetate layer is still noticeably wet. Ethyl acetate was then shaken with brine (Figure 4.44c), and dried with the same quantity of anhydrous $\ce{MgSO_4}$. There is little clumping of the drying agent in this ethyl acetate layer, and fine particles are seen (Figure 4.44d), signifying this layer contained very little water.
If drying agents are used to remove water, you might wonder "Why bother with brine; why not use lots of drying agent when the time comes?" The main reason to limit the amount of water present in an organic solution before the drying agent step is that the drying agent will often adsorb compound along with water. Using as little as possible will maximize the yield.
To demonstrate, Figure 4.45 shows an ethyl acetate solution that has a faint pink tint because it contains some dissolved red food dye. the solution was swirled with white anhydrous $\ce{MgSO_4}$, and the drying agent turned pink as it adsorbed the red food dye compound (Figure 4.45a). Addition of more anhydrous $\ce{MgSO_4}$ made the drying agent pinker (Figure 4.45b), as more dye was removed from solution. In this example, even after filter and rinsing the drying agent with additional solvent, the drying agent remained pink (Figure 4.45c). Thus, the more drying agent that is used, the more compound that may be irrecoverably lost.
Decreasing Water Solubility of Organic Compounds ("Salting Out")
Saturated ionic solutions may be used to decrease the solubility of organic compounds in the aqueous layer, allowing more of a compound to dissolve in the organic layer. If a desired product can hydrogen bond with water and is relatively small, it may be difficult to keep it in the organic layer when partitioning with an aqueous phase ($K$ will be <1). However, the equilibrium can favor the organic layer if all aqueous washes contain high concentrations of ions (e.g. saturated $\ce{NaHCO_3}$, $\ce{NaCl}$, or $\ce{NH_4Cl}$). With water being so tightly "occupied" in dissolving the ions in these solutions, they are less capable of dissolving organic compounds. Additionally, ionic solutions have high dielectric constants, making them less compatible with organic compounds.
Figure 4.47 shows how brine affects the partitioning of red food dye in ethyl acetate and aqueous solutions. Figure 4.47a shows addition of one drop of red food dye to a layer of water in a separatory funnel, and the dye dissolves easily even without swirling. Figure 4.47b shows the water layer containing the dye after shaking with a portion of ethyl acetate. The organic layer has only a very faint pink color, signifying that little dye has dissolved. The dye has obviously partitioned toward the aqueous layer, which is consistent with its very polar structure (Figure 4.46).
Figure 4.47c shows addition of one drop of red food dye to a brine solution, and the dye does not appear to mix with the brine at all. Figure 4.47d shows the brine layer containing the dye after shaking with a portion of ethyl acetate. The organic layer is pinker, signifying that more dye has now partitioned toward the organic layer. the polar dye molecules are much less soluble in the brine solution than in pure water (they have been "salted out"). In fact, some of the dye precipitated in the funnel (Figure 4.47d) as it had such low solubility in both brine and ethyl acetate.
Drying Agents
Why They are Used
An organic layer is always treated with a drying agent after having been exposed to water in a separatory funnel (step c) in Table 4.4). Drying agents are anhydrous inorganic materials that favorably form "hydrates", which incorporate water molecules into their solid lattice structure (for example, $\ce{Na_2SO_4} \cdot 7 \ce{H_2O}$). A drying agent is swirled with an organic solution to remove trace amounts of water.
Many organic solvents dissolve a significant portion of water (Table 4.6) that must be removed before rotary evaporation, or else water will be found in the concentrated product. After solvent removal using a rotary evaporator, it occasionally happens that so much water is present that droplets or a second layer is seen amongst the oily liquid in a round-bottomed flask. The presence of water with the product makes the yield inaccurate, and water also must be removed before GC-MS analysis, as water is incompatible with mass-spectrometer detectors.
Table 4.6: Quantity of water dissolved in various solvents.$^7$
Solvent Grams water dissolved in 100 mL solvent
Diethyl ether 1.24 g
Ethyl acetate 2.92 g
Dichloromethane ($CH_2Cl_2$) 0.32 g
Hexanes 0.007 g
Drying agents must be used with even relatively nonpolar organic solvents that do not theoretically dissolve much water, as water may cling to the sides of the separatory funnel and inadvertently travel with the organic layer while draining. Additionally, solutes dissolved in an organic layer with polar functional groups (e.g. alcohols, carboxylic acids) can hydrogen-bond with water and increase the likelihood of water dissolving in the organic layer.
Types of Drying Agents
Drying agents (Figure 4.48) remove trace amounts of water from organic solutions by forming hydrates. The most useful drying agents indicate when they have completely absorbed all of the water from the solution. Anhydrous magnesium sulfate $\left( \ce{MgSO_4} \right)$ is a fine, loose powder (Figure 4.49a), but its hydrate is clumpy and often clings to the glass (Figure 4.49b). A typical drying procedure is to add anhydrous $\ce{MgSO_4}$ to an organic solution until it stops clumping and fine particles are seen, which indicate that there is no longer water available to form the clumpy hydrates.
Anhydrous calcium sulfate $\left( \ce{CaSO_4} \right)$, can be purchased containing a cobalt compound that is blue when dry and pink when wet (this is then sold under the name Drierite, Figure 4.49c+d). In this way, blue Drierite can be used as a visual indicator for the presence of water.$^8$
The most common drying agents used to remove water from organic solutions are anhydrous sodium sulfate $\left( \ce{Na_2SO_4} \right)$ and anhydrous magnesium sulfate $\left( \ce{MgSO_4} \right)$. Many chemists consider $\ce{MgSO_4}$ the "go-to" drying agent as it works quickly, holds a lot of water for its mass, and the hydrates are noticeably chunkier compared to the anhydrous form, making it easy to see when you've added enough. A drawback to using $\ce{MgSO_4}$ is that it is a fine powder, and so the solutions must be subsequently filtered to remove the drying agent. Another drawback to $\ce{MgSO_4}$ is that all fine powders heavily adsorb product on their surface (which is why they must be rinsed with solvent after filtration), and sometimes more granular drying agents are used to minimize the loss of product by adsorption.
In some procedures $\ce{Na_2SO_4}$ or $\ce{CaCl_2}$ are used if they seem to work just as well as $\ce{MgSO_4}$, or if the solution is incompatible with $\ce{MgSO_4}$ (see Table 4.8). A procedural advantage to these drying agents is that their granules are not easily dispersed, allowing for the solutions to be easily decanted (poured). In many situations drying agents are interchangeable (see Table 4.8 for a survey of drying agents). However, it is most common for desiccators and drying tubes to use $\ce{CaSO_4}$ or $\ce{CaCl_2}$ (Figure 4.50), as they can be easily manipulated in their pellet or rock forms.
Table 4.8: Survey of drying agents.
Drying Agent Hydrate formula(s) Practical Comments Other Comments
Magnesium sulfate $\ce{MgSO_4} \cdot 7 \ce{H_2O}$ Quickly removes most water, and can hold a lot for its mass ($0.15$-$0.75 \: \text{g}$ water per $\text{g}$ desiccant).$^9$ Is a fine powder, so must be gravity filtered. Its high surface area means it will somewhat adsorb compound: be sure to rinse after filtering. $\ce{Mg(H_2O)_4^{2+}}$ is somewhat acidic, so is incompatible with highly acid-sensitive groups.
Sodium Sulfate
$\ce{Na_2SO_4} \cdot 7 \ce{H_2O}$
$\ce{Na_2SO_4} \cdot 10 \ce{H_2O}$
Removes water at a moderate rate, so the solution should be allowed to sit with the drying agent for some time. Can hold a lot of water for its mass ($1.25 \: \text{g}$ water per $\text{g}$ desiccant), but may leave small amounts of water remaining. Solutions with $\ce{Na_2SO_4}$ can usually be decanted. Cannot dry diethyl ether well unless a brine wash was used.
Calcium chloride
$\ce{CaCl_2} \cdot 2 \ce{H_2O}$
$\ce{CaCl_2} \cdot 6 \ce{H_2O}$
Quickly removes water well, although larger quantities are needed than other drying agents (holds $0.30 \: \text{g}$ water per $\text{g}$ desiccant). If using a fine powder, the solution must be gravity filtered and drying agent rinsed. If using pellets, the solution should be allowed to sit for a few minutes, then decanted. Absorbs water as well as methanol and ethanol.
Calcium sulfate (Drierite)
$\ce{CaSO_4} \cdot \frac{1}{2} \ce{H_2O}$
$\ce{CaSO_4} \cdot 2 \ce{H_2O}$
Quickly removes water, but needs large quantities as it holds little water per gram. Are most often used in desiccators and drying tubes, not with solutions.
Drying Agents Procedure
1. The organic solution to be dried must be in an Erlenmeyer flask, as solutions can easily splash out of beakers when swirled.
2. First inspect the solution to see if it's homogenous, or if there is a second layer of liquid (typically a puddle on the bottom). If a second layer is noticed, this is probably water and the majority of it should be pipetted out before continuing on (Figure 4.51a). It can be difficult to completely remove a water layer by pipette, so leaving a tiny bit is acceptable.
3. Add a small portion of drying agent to the flask,the size of one pea for macroscale work (Figure 4.51b), and swirl the solution (Figure 4.51c). Be sure to close the jar of drying agent when not in use, as the reagents are hygroscopic. After a short period of time, inspect the mixture closely.
1. If the entire drying agent clumps into pieces that are much larger than the original size (Figure 4.52b+c), there is still water remaining in the flask. Add another portion of drying agent and swirl.
2. A solution is nearing dryness when fine particles are noticed that don't cling to other particles (Figure 4.52a+c) or to the glass when swirled (Figure 4.53a). A wet organic solution can be cloudy, and a dry one is always clear.
3. If using anhydrous $\ce{Na_2SO_4}$, allow the solution to sit for at least 5 minutes before declaring the solution dry, as this reagent takes time to work.
1. When the solution is dry, separate the drying agent from the solution:
1. If using $\ce{Na_2SO_4}$, $\ce{CaCl_2}$ pellets, or $\ce{CaSO_4}$ rocks, carefully decant the solution into an appropriately sized round-bottomed flask (Figure 4.53b), being sure to fill the flask no more than halfway. Reminder: a mass of the empty flask should be obtained if the solvent will be evaporated on the rotary evaporator.
2. If using $\ce{MgSO_4}$, gravity filter the solution into an appropriately sized round-bottomed flask (Figure 4.53c). When pouring, leave the solid behind as long as possible (essentially decant the solution, but into the funnel lined with filter paper). Solid can slow drainage in the filter paper.
3. With all drying agents, rinse the drying agent (in the flask and in the filter funnel) with a few $\text{mL}$ of fresh organic solvent, and add the rinsing to the round-bottomed flask (Figure 4.53d). Remove the solvent using a rotary evaporator.
$^4$A. Seidell, Solubilities of Inorganic and Organic Substances, D. Van Nostrand Company, 1907.
$^5$When assessing the result of a litmus paper test, look at the center of the drop. The center is the most concentrated spot, and it's possible a color change may not be seen on the outside where the solution has spread and diluted. Any pink seen on blue litmus paper means the solution is acidic.
$^6$From: Fessenden, Fessenden, Feist, Organic Laboratory Techniques, 3$^\text{th}$ ed., Brooks-Cole, 2001.
$^7$From: Fessenden, Fessenden, Feist, Organic Laboratory Techniques, 3$^\text{th}$ ed., Brooks-Cole, 2001.
$^8$Blue Drierite is expensive, so is commonly used by mixing it together with white Drierite ($\ce{CaSO_4}$ without the cobalt indicator). Pink (wet) Drierite can be dried by spreading it on a watch glass and drying in a $110^\text{o} \text{C}$ oven overnight.
$^9$Grams water per gram of desiccant values are from: J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 1999, Sect. 11.2. $\ce{CaCl_2}$ value is quoted for the formation of $\ce{CaCl_2} \cdot 2 \ce{H_2O}$. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.07%3A_Reaction_Work-Ups.txt |
How They Work
A modification of the extractions previously discussed in this chapter is to perform a chemical reaction in the separatory funnel in order to change the polarity and therefore partitioning of a compound in the aqueous and organic layers. A common method is to perform an acid-base reaction, which can convert some compounds from neutral to ionic forms (or vice versa).
For example, imagine that a mixture of benzoic acid and cyclohexane is dissolved in an organic solvent like ethyl acetate in a separatory funnel. To separate the components, a water wash may be attempted to remove benzoic acid, but benzoic acid is not particularly water-soluble due to its nonpolar aromatic ring, and only small amounts would be extracted into the aqueous layer (Figure 4.54a).
Separation of a mixture of benzoic acid and cyclohexane is however possible using a wash with a base such as $\ce{NaOH}$. Due to its acidic nature, benzoic acid can undergo a reaction with $\ce{NaOH}$ as follows, resulting in the carboxylate salt sodium benzoate.
$\begin{array}{ccccccccc} \ce{PhCO_2H} \left( aq \right) & + & \ce{NaOH} \left( aq \right) & \rightarrow & \ce{H_2O} \left( l \right) & + & \ce{PhCO_2Na} \left( aq \right) & & \left( \text{or } \ce{PhCO_2^-} \ce{Na^+} \right) \ \text{Benzoic acid} & & & & & & \text{Sodium benzoate} & & \end{array}$
The solubility properties of carboxylic acids are substantially different than their corresponding carboxylate salts. Sodium salicylate is roughly 350 times more soluble in water than salicylic acid due to its ionic character (Figure 4.55), and it is rather insoluble in organic solvents such as diethyl ether.
Therefore, a wash with $\ce{NaOH}$ would convert benzoic acid into its ionic carboxylate form, which would then be more soluble in the aqueous layer, allowing for the sodium benzoate to be extracted into the aqueous layer. Cyclohexane would remain in the organic layer as it has no affinity for the aqueous phase, nor can react with $\ce{NaOH}$ in any way. In this manner, a mixture of benzoic acid and cyclohexane can be separated (Figure 4.54b). The aqueous layer may be later acidified with $\ce{HCl} \left( aq \right)$ if desired to convert the benzoic acid back to its neutral form.
Sodium Bicarbonate Washes
An acid-base extraction can be used to extract carboxylic acids from the organic layer into the aqueous layer. As was discussed in the previous section, $\ce{NaOH}$ can be used to convert a carboxylic acid into its more water-soluble ionic carboxylate form. However, if the mixture contains a desired compound that can react with $\ce{NaOH}$, a milder base such as sodium bicarbonate should be used. A similar reaction occurs:
$\begin{array}{ccccccccccc} \ce{PhCO_2H} \left( aq \right) & + & \ce{NaHCO_3} \left( aq \right) & \rightarrow & \ce{PhCO_2Na} \left( aq \right) & + & \ce{H_2CO_3} \left( aq \right) & \rightleftharpoons & \ce{H_2O} \left( l \right) & + & \ce{CO_2} \left( g \right) \ \text{Benzoic acid} & & & & \text{Sodium benzoate} & & & & & & \end{array}$
One difference in using the base $\ce{NaHCO_3}$ instead of $\ce{NaOH}$ is that the byproduct carbonic acid $\left( \ce{H_2CO_3} \right)$ can decompose to water and carbon dioxide gas. When shaking an acidic solution with sodium bicarbonate in a separatory funnel, care should be taken to swirl gently and vent more frequently to release pressure from the gas.
An example of a reaction that often uses sodium bicarbonate wash in the work-up is a Fischer Esterification reaction. To demonstrate, benzoic acid was refluxed in ethanol along with concentrated sulfuric acid in order to form ethyl benzoate (Figure 4.56a+b). A TLC plate of the reaction mixture at 1 hour of reflux showed residual unreacted carboxylic acid (Figure 4.56c), which is not uncommon due to the energetics of the reaction.
The residual carboxylic acid can be removed from the desired ester product using an acid-base extraction in a separatory funnel. A wash with sodium bicarbonate converts benzoic acid into its more water-soluble sodium benzoate form, extracting it into the aqueous layer (Figure 4.57). Additionally, the sodium bicarbonate neutralizes the catalytic acid in this reaction.
Sodium bicarbonate is preferable to $\ce{NaOH}$ in this process, as it is a much weaker base; washing with $\ce{NaOH}$ could cause hydrolysis of the ester product.
Mixtures of Acids and Bases
As has been discussed previously, the acid-base properties of compounds can be utilized to selectively extract certain compounds from mixtures. This strategy can be extended to other examples
Extracting Bases
Basic compounds such as amines can be extracted from organic solutions by shaking them with acidic solutions to convert them into more water-soluble salts. In this way, they can be extracted from an organic layer into an aqueous layer.
$\begin{array}{ccccccc} \ce{PhNH_2} \left( aq \right) & + & \ce{HCl} \left( aq \right) & \rightarrow & \ce{PhNH_3Cl} \left( aq \right) & & \left( \text{or } \ce{PhNH_3^+} \ce{Cl^-} \right) \ \text{Basic amine} & & & & \text{Ammonium salt} & & \end{array}$
Extracting Carboxylic Acids vs. Phenols
As previously discussed, carboxylic acids can be extracted from an organic layer into an aqueous layer by shaking them with basic solutions, which converts them into their more water-soluble salts.
$\begin{array}{ccccccccc} \ce{PhCO_2H} \left( aq \right) & + & \ce{NaOH} \left( aq \right) & \rightarrow & \ce{H_2O} \left( l \right) & + & \ce{PhCO_2Na} \left( aq \right) & & \left( \text{or } \ce{PhCO_2^-} \ce{Na^+} \right) \ \text{Carboxylic acid} & & & & & & \text{Carboxylate salt} & & \end{array}$
A similar reaction occurs with phenols $\left( \ce{PhOH} \right)$, and they too can be extracted into an aqueous $\ce{NaOH}$ layer (Figure 4.58a).
However, phenols are considerably less acidic than carboxylic acids, and are not acidic enough to react completely with $\ce{NaHCO_3}$, a weaker base. Therefore, a solution of bicarbonate can be used to separate mixtures of phenols and carboxylic acids (Figure 4.58b).
Extracting Acid, Base, and Neutral Compounds
The acid-base properties previously discussed allow for a mixture containing acidic (e.g. $\ce{RCO_2H}$), basic (e.g. $\ce{RNH_2}$), and neutral components to be purified through a series of extractions, as summarized in Figure 4.59 (which uses an organic solvent less dense than water).
It is assumed that readers conducting this type of experiment are familiar with performing single and multiple extractions. In this section are described differences between general extraction procedures and the process as summarized in Figure 4.59.
1. Isolating the Acidic component:
1. When the acidic component is in the aqueous layer in an Erlenmeyer flask, it can be converted back to the neutral component through addition of $2 \: \text{M} \: \ce{HCl} \left( aq \right)$ until the solution gives a pH of 3-4 (as determined by pH paper). If large quantities of acid are present such that acidification would require too great a volume of $2 \: \text{M} \: \ce{HCl} \left( aq \right)$, concentrated $\ce{HCl} \left( aq \right)$ may be instead added dropwise. Lower concentrations of $\ce{HCl} \left( aq \right)$ are less hazardous, but increasing the volume of the aqueous layer by a large amount would affect the efficiency of subsequent extractions and filtering steps.
2. After acidification, two routs may be taken, depending on if the acidic component is solid or liquid.
1. If a solid forms upon acidification of the ionic salt, it can be collected through suction filtration. This method should only be used if large quantities of large-sized crystals are seen. If fine crystals form (which are quite common), they will clog the filter paper and interfere with adequate drainage. If only a small amount of solid is seen compared to the theoretical quantity, it is likely the compound is quite water-soluble, and filtration would lead to low recovery.
2. If no solid forms upon acidification (or if fine crystals or low quantity of solid forms), extract the acidic component back into an organic solvent ($\times 3$). As a general rule of thumb, use one-third as much solvent for the extractions as the original layer (e.g. if using $100 \: \text{mL}$ aqueous solution, extract with $33 \: \text{mL}$ organic solvent each time). Be sure to first cool the aqueous solution in an ice bath before extraction if the acidification created noticeable heat. Follow up with a brine wash ($\times 1$) if using diethyl ether or ethyl acetate, dry with a drying agent, and remove the solvent via rotary evaporator to leave the pure acidic component.
2. Isolating the Basic component:
1. Use a similar process as the isolation of the acidic component, except basify the solution using $2 \: \text{M} \: \ce{NaOH} \left( aq \right)$ until it gives a pH of 9-10 as determined by pH paper.
3. Isolating the Neutral component:
1. The neutral component will be the "leftover" compound in the organic layer. To isolate, wash with brine ($\times 1$) if using diethyl ether or ethyl acetate, dry with a drying agent, and remove the solvent via rotary evaporator to leave the pure neutral component. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/04%3A_Extraction/4.08%3A_Acid-Base_Extraction.txt |
Distillation is a purification method for liquids, and can separate components of a mixture if they have significantly different boiling points. In a distillation, a liquid is boiled in the "distilling flask," then the vapors travel to another section of the apparatus where they come into contact with a cool surface. The vapors condense on this cool surface, and the condensed liquid (called the "distillate") drips into a reservoir separated from the original liquid. In the simplest terms, a distillation involves boiling a liquid, then condensing the gas and collecting the liquid elsewhere
• 5.1: Overview of Distillation
Several distillation variations are used in the organic laboratory depending on the properties of the mixture to be purified.
• 5.2: Simple Distillation
A simple distillation is used if the components have widely different boiling points (greater than a 100 °C difference in boiling points). However, if a simple distillation is attempted on a mixture where the components have more similar boiling points (less than a 100 °C difference in boiling points), it will fail to purify the mixture completely.
• 5.3: Fractional Distillation
A simple distillation is incapable of significant purification if the boiling points of the components are too close. When the difference in boiling points is less than 100 ˚C, a modification is necessary, namely insertion of a fractionating column between the distilling flask and three-way adapter.
• 5.4: Vacuum Distillation
Boiling commences when the vapor pressure of a liquid or solution equals the external or applied pressure (often the atmospheric pressure). Thus, if the applied pressure is reduced, the boiling point of the liquid decreases. This behavior occurs because a lower vapor pressure is necessary for boiling, which can be achieved at a lower temperature.
• 5.5: Steam Distillation
Steam distillation is analogous to simple distillation, the main difference being that steam (or water) is used in the distilling flask along with the material to be distilled. Experimentally the setups are arranged more or less the same, with small differences being how the steam is added to the flask: either indirectly if a steam line is available in the building, or directly by boiling water in the flask.
• 5.6: Rotary Evaporation
The preferred method for solvent removal in the laboratory is by use of a rotary evaporator (also known as a "rotovap"), A rotary evaporator is essentially a reduced pressure distillation: a solution in a round bottomed flask is placed in the water bath of the apparatus, and rotated while the system is partially evacuated (by a water aspirator or vacuum pump). The reduced pressure in the apparatus causes the solvent to boil at a lower temperature than normal.
05: Distillation
Several distillation variations are used in the organic laboratory depending on the properties of the mixture to be purified. The apparatus in Figure 5.1 is used to perform a simple distillation and is used if the components have widely different boiling points (greater than a 100 °C difference in boiling points). If a simple distillation is attempted on a mixture where the components have more similar boiling points (less than a 100 °C difference in boiling points), it will fail to purify the mixture completely. Instead, fractional distillation can be used to improve the chances of purification. Vacuum distillation may be used when the boiling points of the mixture's components are very high (>150 °C), or steam distillation if the components are very water insoluble. The distillation variations are summarized in Table 5.1, and are discussed in detail in this chapter.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.01%3A_Overview_of_Distillation.txt |
• 5.2A: Uses of Simple Distillation
Distillation is the method used to concentrate fermented solutions and produce hard liquors. Distillation is also an excellent purification tool for many liquids, and can be used to purify products from a chemical reaction.
• 5.2B: Separation Theory
Distillation of mixtures may or may not produce relatively pure samples. As distillation involves boiling a solution and condensing its vapors, the composition of the distillate is identical to the composition of the vapors. Several equations can be used to describe the composition of vapor produced from a solution.
• 5.2C: Step-by-Step Procedures
• 5.2D: Microscale Distillation
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
5.02: Simple Distillation
Concentration of Alcohol
Fermented grains can produce a maximum alcohol content of about $16\%$, the upper limit of most beers and wines, as the yeast organism used in fermentation cannot survive in more concentrated alcoholic solutions. However, spirits such as gin, vodka, and rum usually have an alcohol content of roughly $40\%$ by volume. Distillation is the method used to concentrate fermented solutions and produce hard liquors. A distillation of fermented grape juice is commonly done in academic laboratories (Figure 5.2).
Figure 5.3 shows the distillation of red wine. Notice how the wine is noticeably purified in the process as the distillate is clear (Figure 5.3d). All concentrated alcoholic products at first distill clear, but some become colored during the aging process (from components leaching out of the barrels they are stored in, or from oxidation).
The distillation of red wine can be proven to have concentrated the alcohol through a flame test, which is the origin of the term "proof". In the 16$^\text{th}$ century when rum was a bartering item, it was regularly tested to ensure that it had not been watered down. An alcoholic sample that is $57\%$ alcohol by volume is capable of catching on fire, while more dilute solutions cannot, enabling ignition to serve as "proof" of alcohol content. The red wine used in Figure 5.3 did not ignite, which is consistent with its label that states it to be only $13\%$ alcohol (Figure 5.4a). The distillate easily caught on fire (Figure 5.4c+d), and density analysis showed it to be $67\%$ alcohol.
Distilled Water
Access to clean, fresh water is a major problem facing the world today. In countries neighboring the ocean, seawater desalination is sometimes used to provide the country with drinkable water. Distillation is one of the main methods$^1$ used to purify ocean water and works well since salt, microorganisms, and other components of seawater are non-volatile. The main disadvantage of distilling water is that the process requires a lot of energy, and unless engineered creatively, the economics can be a major deterrent to using the method. Heat released from a power plant is often used to provide the energy for the distillation of seawater, and Saudi Arabia and Israel mainly use coupled power plants and distilleries to obtain roughly half the fresh water needed for their countries.$^2$
Along with drinking water, distilled water is necessary for scientific lab work as dissolved salts in tap water may interfere with some experiments. Many science buildings have their own distilled water generators (Figure 5.5a), enabling wash bottles and carboys to be filled with a turn of the spigot (Figure 5.5b).
Purification of Reagents and Products
Distillation is an excellent purification tool for many liquids, and can be used to purify products from a chemical reaction. Figure 5.6 shows the distillation of a crude sample of isoamyl acetate, formed through a Fischer esterification reaction. The crude sample was originally yellow (Figure 5.6a), but the distillate was colorless (Figure 5.6d), making obvious the removal of some materials through the process.
Distillation can also be used to purify reagents that have degraded over time. Figure 5.7a shows a bottle of benzaldehyde that has partially oxidized, as evidenced by the crystals of benzoic acid seen adhering to the inside of the glass (indicated with an arrow). A simple distillation purified the benzaldehyde sample, and its effectiveness was demonstrated through comparison of the infrared (IR) spectra of the original material (Figure 5.8a) and the distillate (Figure 5.8b). The broad region indicated by an arrow in Figure 5.8a represent the $\ce{O-H}$ stretch of a carboxylic acid (benzoic acid), and the distillate's IR spectrum lacks this feature.
$^1$More often, vacuum distillation is used for desalination, as the lowered pressure allows for boiling to occur at a lower temperature (which requires less energy). Vacuum distillation of seawater is still energy intensive. Reverse osmosis is also used for seawater desalination.
$^2$Pyper, Julia, "Israel is creating a water surplus using desalination," E+E News, February 7, 2014. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2A%3A_Uses_of_Simple_Distillation.txt |
Raoult's and Dalton's Laws
Distillation of mixtures may or may not produce relatively pure samples. As distillation involves boiling a solution and condensing its vapors, the composition of the distillate is identical to the composition of the vapors. Several equations can be used to describe the composition of vapor produced from a solution.
Raoult's law states that a compound's vapor pressure is lessened when it is part of a solution, and is proportional to its molar composition. Raoult's law is shown in Equation \ref{1}. The equation means that a solution containing $80 \: \text{mol}\%$ compound "A" and $20 \: \text{mol}\%$ of another miscible component would at equilibrium produce $80\%$ as many particles of compound A in the gas phase than if compound A were in pure form.
$P_A = P_A^o \chi_A \label{1}$
where $P_A^o$ is the vapor pressure3 of a sample of pure A, and $\chi_A$ is the mole fraction of $A$ in the mixture.
Dalton's law of partial pressures states that the total pressure in a closed system can be found by addition of the partial pressures of each gaseous component. Dalton's law is shown in Equation \ref{2}.
$P_\text{total} = P_A + P_B \label{2}$
A combination of Raoult's and Dalton's laws is shown in Equation \ref{3}, which describes the vapor produced by a miscible two-component system (compounds A + B). This combined law shows that the vapors produced by distillation are dependent on each component's vapor pressure and quantity (mole fraction).
$P_\text{solution} = P_A^o \chi_A + P_B^o \chi_B \label{3}$
A compound's vapor pressure reflects the temperature of the solution as well as the compound's boiling point. As temperature increases, a greater percentage of molecules have sufficient energy to overcome the intermolecular forces (IMF's) holding them in the liquid phase. Therefore, a compound's vapor pressure always increases with temperature (see Table 5.2), although not in a linear fashion.
Table 5.2: Vapor pressure data for selected compounds.$^4$
Compound Boiling Point (ºC) Vapor Pressure at 0 ºC Vapor Pressure at 20 ºC
Diethyl Ether 34.6 183 mmHg 439 mmHg
Methanol 64.7 30 mmHg 94 mmHg
Benzene 80.1 24.5 mmHg 75 mmHg
Toluene 110.6 6.8 mmHg 22 mmHg
When comparing two compounds at the same temperature, the compound with weaker IMF's (the one with the lower boiling point) should more easily enter the gas phase. Therefore, at any certain temperature, a compound with a lower boiling point always has a greater vapor pressure than a compound with a higher boiling point (see Table 5.2). This last concept is the cornerstone of distillation theory.
A compound with a lower boiling point always has a greater vapor pressure than a compound with a higher boiling point
Purification Potential
A simple distillation works well to purify certain mixtures, specifically to separate a liquid from non-volatile impurities (e.g. solids or salts), or from small amounts of significantly higher or lower boiling impurities. A general guideline is that a simple distillation is capable of separating components if the difference in boiling point of the components is greater than $100^\text{o} \text{C}$. A simple distillation does not work well to purify a mixture that contains components with similar boiling points (when the difference in b.p. is $< 100^\text{o} \text{C}$).
To demonstrate, a mixture containing $75 \: \text{mol}\%$ 1-chlorobutane (normal b.p. $78^\text{o} \text{C}$) and $25 \: \text{mol}\%$ p-cymene (normal b.p. $177^\text{o} \text{C}$, Figure 5.10b) was purified using a simple distillation to collect the first two $\text{mL}$ of distillate. The difference in boiling point of these two components is $99^\text{o} \text{C}$.
Gas chromatograph (GC) analysis of the initial mixture and distillate allowed for quantitation, although the reported percentages were not proportional to the $\text{mol}\%$ values because of the detector used (often higher molecular weight compounds have abnormally high integration values with mass spectrometers). A calibration curve was necessary to translate the reported percentages into accurate values (Figure 5.9). To generate the calibration curve, the GC integration values were recorded for several known mixtures of 1-chlorobutane and p-cymene. For example, a sample with $95 \: \text{mol}\%$ 1-chlorobutane (x-axis) and $5 \: \text{mol}\%$ p-cymene was reported by the GC instrument to be $42 \: \text{mol}\%$ 1-chlorobutane (y-axis). This was recorded as the data point (95,42).
Table 5.3: Recorded GC$%$ for the simple distillation of a $75 \: \text{mol}\%$ 1-chlorobutane (CB) and $25 \: \text{mol}\%$ p-cymene (PC) mixture. The actual $\%$ were estimated using the calibration curve in Figure 5.9.
Sample Component Recorded GC % Actual %
Initial Solution 1-chlorobutane 22% 75%
Initial Solution p-cymene 78% 25%
Distillate 1-chlorobutane 86% 99.5%
Distillate p-cymene 14% 0.5%
GC analysis (Figure 5.10) and a calibration curve (Figure 5.9 showed the distillate to be nearly pure 1-chlorobutane ($\sim 99.5 \: \text{mol}\%$, Table 5.3). This is an example of a mixture that is purified well through simple distillation.
A second distillation was performed in a similar manner using a mixture containing $75 \: \text{mol}\%$ ethylbenzene (normal b.p. $136^\text{o} \text{C}$) and $25 \: \text{mol}\%$ p-cymene (normal b.p. $177^\text{o} \text{C}$, Figure 5.12b). The difference in boiling point of these two components is $41^\text{o} \text{C}$.
Table 5.4: Recorded GC$\%$ for the simple distillation of a $75 \: \text{mol}\%$ ethylbenzene (EB) and $25 \: \text{mol}\%$ p-cymene (PC) mixture. The actual $\%$ were estimated using the calibration curve in Figure 5.11.
Sample Component Recorded GC % Actual %
Initial Solution ethylbenzene 56% 75%
Initial Solution p-cymene 44% 25%
Distillate ethylbenzene 68% 90%
Distillate p-cymene 32% 10%
GC analysis (Figure 5.12) and a calibration curve (Figure 5.11) showed the simple distillation did increase the quantity of ethylbenzene in the mixture (from $75 \: \text{mol}\%$ to $90 \: \text{mol}\%$, see Table 5.4), but a significant portion of p-cymene remained in the distillate. This is an example of a mixture that is not completely purified by simple distillation.
The two distillations in this section both started with $75/25 \: \text{mol}\%$ mixtures, and yet the two distillations resulted in distillates with different purity. This result can be explained by analysis of Raoult's and Dalton's laws (Equations \ref{1} and \ref{2}). The equation describing the composition of vapor produced from the $75 \: \text{mol}\%$ 1-chlorobutane/$25 \: \text{mol}\%$ p-cymene mixture is shown in Equation \ref{4}:
\begin{align} P_\text{solution} &= P^o_\text{1-chlorobutane} \chi_A + P^o_\text{p-cymene} \chi_B \[4pt] &= P^o_\text{1-chlorobutane} \left( 0.75 \right) + P^o_\text{p-cymene} \left( 0.25 \right) \label{4} \end{align}
In this distillation, the distillate was "enriched" in 1-chlorobutane, as it changed from $75 \: \text{mol}\%$ 1-chlorobutane in the distilling flask to $99.5 \: \text{mol}\%$ 1-chlorobutane in the distillate. The distillate contained nearly pure 1-chlorobutane, as it had the greater contribution to the solution's vapor. This is because:
• 1-chlorobutane was present in a greater quantity in the distilling flask ($75\%$, or had a mole fraction of 0.75)
• 1-chlorobutane had the lower boiling point of the components ($78^\text{o} \text{C}$ compared to $177^\text{o} \text{C}$) and so had a significantly higher vapor pressure than p-cymene $\left( P^o_\text{1-chlorobutane} \gg P^o_\text{p-cymene} \right)$.
In this distillation, the difference in boiling point of the components was so large ($\Delta$ b.p. of $99^\text{o} \text{C}$), that the vapor pressure of p-cymene (and thus its contribution to the vapor phase) was essentially insignificant during the distillation. The vastly different vapor pressures (along with the differences in mole fraction), were the reason the simple distillation resulted in a nearly pure distillate.
In the distillation of the $75 \: \text{mol}\%$ ethylbenzene/$25 \: \text{mol}\%$ p-cymene mixture, the distillate was also enriched in the lower boiling component (ethylbenzene) as it had the greater vapor pressure. However, the distillate still contained $10 \: \text{mol}\%$ p-cymene, because there was only a $41^\text{o} \text{C}$ difference in boiling point between the two components. This meant that the higher boiling component (p-cymene) had a smaller but not negligible vapor pressure, leading to a significant quantity of p-cymene in the distillate.
Distillation Curves
Equation (5) describes the vapor produced by a miscible two-component system (compounds A + B). This mathematical equation can be used to generate phase diagrams, or distillation curves, which graphically correlate temperature to the molar composition of the liquid and vapor phases.
$P_\text{solution} = P^o_A \chi_A + P^o_B \chi_B \label{5}$
Figure 5.13 shows a generic distillation curve for a two-component system. Molar composition is on the x-axis, with the left side of the plot corresponding to a sample that is pure compound A and the right side of the plot corresponding to a sample of pure compound B. In between the two sides represent mixtures of A and B. Temperature is on the y-axis, and the boiling point of each compound are marked ("bp A" and "bp B").
You may have been previously exposed to phase diagrams for pure compounds (e.g. of water or carbon dioxide), where only a single line would be used to denote a liquid-gas equilibrium. In the distillation of mixtures, there is a difference in composition between the liquid and gas phases, which is the reason for the tear-shaped appearance of the phase diagram. The tear-shaped region represents conditions where both liquid and gas coexist, during boiling of the liquid.
Imagine a $25 \: \text{mol}\%$ A/$75 \: \text{mol}\%$ B mixture is to be distilled, and this mixture is described by the distillation curve in Figure 5.13a. When the temperature reaches the lower line of the tear drop (temperature x and point a in Figure 5.13b), the solution begins to boil. Since liquid and gas have the same temperature during boiling, the vaporization process can be thought of as following the horizontal line from position a to b in Figure 5.13b. After vaporization, the gas condenses into the distillate, which can be thought of as following the vertical line from position b to c in Figure 5.13b. Under perfect equilibrating conditions, the distillate in this example would be $74 \: \text{mol}\%$ A/$26 \: \text{mol}\%$ B, and is "enriched" in A as it has the lower boiling point and thus the higher vapor pressure.
In a distillation curve, the leftward horizontal "movement" followed by downward vertical "movement" represents one vaporization-condensation event (see Figure 5.13b). This is often referred to as a "theoretical plate", historical terminology related to the collection of distillate onto plates or trays, and represents the purification potential of a simple distillation.
Every mixture has its own distillation curve, reflective of the boiling points of the components. A mixture containing two components whose boiling points are vastly different ($\Delta$ b.p. $> 100^\text{o} \text{C}$) is shown in Figure 5.14.
Imagine a $25 \: \text{mol}\%$ A/$75 \: \text{mol}\%$ B mixture is to be distilled, and this mixture is described by the distillation curve in Figure 5.14a, where the components have significantly different boiling points. One vaporization-condensation cycle (from a to b to c in Figure 5.14b) will produce a distillate that is nearly $100\%$ A, the compound with the much lower boiling point. When distilling a mixture with vastly different boiling points, the two components act almost independently and can be easily separated by simple distillation.
Distilling Temperatures
A pure compound distills at a constant temperature, its boiling point. When distilling mixtures, however, the temperature does not often remain steady. This section describes why mixtures distill over a range of temperatures, and the approximate temperatures at which solutions can be expected to boil. These concepts can be understood by examination of the equation that describes a solution's vapor pressure (Equation (6)) and distillation curves.
$P_\text{solution} = P_A + P_B = P^o_A \chi_A + P^o_B \chi_B \label{6}$
Imagine an ideal solution that has equimolar quantities of compound "A" (b.p. = $50^\text{o} \text{C}$) and compound "B" (b.p. = $100^\text{o} \text{C}$). A common misconception is that this solution will boil at $50^\text{o} \text{C}$,at the boiling point of the lower-boiling compound. The solution does not boil at A's boiling point because A has a reduced partial pressure when part of a solution. In fact, the partial pressure of A at $50^\text{o} \text{C}$ would be half its normal boiling pressure, as shown in Equation \ref{7}:
$\text{At } 50^\text{o} \text{C}: \: \: \: \: \: P_A = \left( 760 \: \text{torr} \right) \left( 0.50 \right) = 380 \: \text{torr} \label{7}$
Since the partial pressures of A and B are additive, the vapor contribution of B at $50^\text{o} \text{C}$ is also important. The partial pressure of B at $50^\text{o} \text{C}$ would equal its vapor pressure at this temperature multiplied by its mole fraction (0.50). Since B is below its boiling point $\left( 100^\text{o} \text{C} \right)$, its vapor pressure would be smaller than $760 \: \text{torr}$, and the exact value would need to be looked up in a reference book. Let's say that the vapor pressure of B at $50^\text{o} \text{C}$ is 160 torr.
At 50ºC:
$P_B = \left( 160 \: \text{torr} \right) \left( 0.50 \right) = 80 \: \text{torr} \label{8}$
The solution boils when the combined pressure of the components equals the atmospheric pressure (let's assume $760 \: \text{torr}$ for this calculation). This equimolar solution would therefore not boil at $50^\text{o} \text{C}$, as its combined pressure is less than the external pressure of $760 \: \text{torr}$ (Equation \ref{9}).
At 50 ºC:
\begin{align} P_\text{solution} &= P_A + P_B \[4pt] &= \left( 380 \: \text{torr} \right) + \left( 80 \: \text{torr} \right) \[4pt] &= 460 \: \text{torr} \label{9} \end{align}
The initial boiling point of this solution is $66^\text{o} \text{C}$, which is the temperature where the combined pressure matches the atmospheric pressure (Equation \ref{10}, note: all vapor pressures would have to be found in a reference book).
At 66 ºC:
\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \[4pt] &= \left( 1238 \: \text{torr} \right) \left( 0.5 \right) + \left( 282 \: \text{torr} \right) \left( 0.5 \right) \[4pt] &= \left( 619 \: \text{torr} \right) + \left( 141 \: \text{torr} \right) = 760 \: \text{torr} \label{10} \end{align}
These calculations demonstrate that a solution's boiling point occurs at a temperature between the component's boiling points.$^5$.
This can also be seen by examination of the distillation curve for this system, where the solution boils when the temperature reaches position a in Figure 5.15a, a temperature between the boiling point of the components.
It is important to realize that mixtures of organic compounds containing similar boiling points ($\Delta$ b.p. $< 100^\text{o} \text{C}$) boil together. They do not act as separate entities (not "A boils at $50^\text{o} \text{C}$, then B boils at $100^\text{o} \text{C}$").
The mixture in this example begins boiling at $66^\text{o} \text{C}$, but after a period of time boiling would cease if maintained at this temperature. This happens because the composition of the distilling pot changes over time. Since distillation removes more of the lower boiling A, the higher boiling B will increase percentagewise in the distilling pot. Imagine that after some material has distilled, the distilling pot is now $42.2\%$ A and $57.8\%$ B. This solution no longer boils at $66^\text{o} \text{C}$, as shown in Equation \ref{11}.
At 66 ºC:
\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \[4pt] &= \left( 1238 \: \text{torr} \right) \left( 0.422 \right) + \left( 282 \: \text{torr} \right) \left( 0.578 \right) \[4pt] &= \left( 522 \: \text{torr} \right) + \left( 163 \: \text{torr} \right) \[4pt] &= 685 \: \text{torr} \label{11} \end{align}
At the new composition in this example, the temperature must be raised to 70 ºC to maintain boiling, as shown in Equation \ref{12}. This can also be shown by examination of the distillation curve in Figure 5.15b, where boiling commences at temperature c, but must be raised to temperature e as the distilling pot becomes more enriched in the higher boiling component (shifts to the right on the curve).
At 70 ºC:
\begin{align} P_\text{solution} &= P^o_A \chi_A + P^o_B \chi_B \[4pt] &= \left( 1370 \: \text{torr} \right) \left( 0.422 \right) + \left( 315 \: \text{torr} \right) \left( 0.578 \right) \[4pt] &= \left( 578 \: \text{torr} \right) + \left( 182 \: \text{torr} \right) \[4pt] &= 760 \: \text{torr} \label{12} \end{align}
These calculations and analysis of distillation curves demonstrate why a mixture distills over a range of temperatures:$^6$ as the composition in the distilling pot changes with time (which affects the mole fraction or the x-axis on the distillation curve), the temperature must be adjusted to compensate. These calculations also imply why a pure compound distills at a constant temperature: the mole fraction is one for a pure liquid, and the distilling pot composition remains unchanged during the distillation.
The calculations and distillation curves in this section enable discussion of another aspect of distillation. The partial pressures of each component at the boiling temperatures can be directly correlated to the composition of the distillate. Therefore, the distillate at $66^\text{o} \text{C}$ is $81 \: \text{mol}\%$ A ($100\% \times 619 \: \text{torr}/760 \: \text{torr}$, see Equation \ref{10}) while the distillate at $70^\text{o} \text{C}$ is $76 \: \text{mol}\%$ A ($100\% \times 578 \: \text{torr}/760 \: \text{torr}$, see Equation \ref{12}). The initial distillate contains the greatest quantity of the lowest boiling component, and the distillate degrades in purity as the distillation progresses. This can also be seen in the distillation curve in Figure 5.15b, where the initial distillate composition corresponds to position d (purer), while the distillate at $70^\text{o} \text{C}$ corresponds to position f (less pure).
Azeotropes
Pure compounds distill at a constant temperature, while most solutions containing more than one volatile component distill over a range of temperatures. There is an exception to this statement, as some mixtures of certain composition also distill at a constant temperature. These mixtures are called "azeotropes", which display non-ideal behavior and do not follow Raoult's law (Equation \ref{13}).
One of the best-known azeotropes is a mixture containing $95.6\%$ ethanol and $4.4\%$ water. When distilling a mixture containing ethanol and water (for example when concentrating fermented grains to produce hard liquor), $95.6\%$ is the highest percentage of ethanol possible in the distillate. It is impossible to distill $100\%$ pure ethanol when water is in the distilling pot, as the ethanol and water co-distill acting as a pure substance. For this reason, most ethanol used by chemists is the $95\%$ version as it can be purified through distillation and is the least expensive. "Absolute ethanol" can also be purchased (which is $> 99\%$ ethanol), but is more expensive as further methods (e.g. drying agents) are required to remove residual water before or after distillation.
Many compounds form azeotropes, which are specific ratios of compounds that co-distill at a constant composition and temperature. Lange's Handbook of Chemistry$^7$ lists roughly 860 known azeotropes composed of two or three components each. A selection of these azeotropic mixtures are in Table 5.5.
Table 5.5: Select azeotropes from Lange's Handbook of Chemistry.$^7$
Azeotropic Composition Boiling Point (ºC)
91% Benzene, 9% Water 69
92% Cyclohexane, 8% Water 70
96% Ethanol, 4% Water 78
80% Toluene, 20% Water 84
59% Acetone, 41% Hexane 50
74% Benzene, 19% Ethanol, 7% Water 65
69% Cyclohexane, 31% Ethanol 65
68% Benzene, 32% Ethanol 68
32% Toluene, 68% Ethanol 77
61% Benzene, 39% Methanol 58
29% Toluene, 71% Methanol 64
83% Ethyl acetate, 9% Water, 8% Ethanol 70
Azeotropes form when a solution deviates from Raoult's law (Equation (13)), meaning that the vapor pressure produced from an azeotropic solution does not directly correlate to its mole fraction. Deviations occur when the components are either attracted to or repelled by one another, namely if they have significantly different strengths of intermolecular forces (IMF's) to themselves (e.g. A-A or B-B) as they do with the other components (e.g. A-B). In other words, a solution will deviate from Raoult's law if the enthalpy of mixing is not zero.
$P_A = P^o_A \chi_A \label{13}$
Azeotropic mixtures come in two forms: one whose boiling point is lower than any of its constituents (called a "minimum boiling azeotrope") or ones whose boiling point is higher than any of its constituents (called a "maximum boiling azeotrope"). For example, the $61\%$ benzene/$39\%$ methanol azeotrope is a minimum boiling azeotrope as its boiling point is $58^\text{o} \text{C}$, which is lower than the boiling point of benzene $\left( 80^\text{o} \text{C} \right)$ and methanol $\left( 65^\text{o} \text{C} \right)$. Minimum boiling azeotropes are much more common than maximum boiling azeotropes.
Minimum boiling azeotropes occur when the component's IMF's are stronger to themselves (A-A/B-B) than to each other (A-B). This generally occurs with components that lack an affinity for one another, for example when one component can hydrogen bond but the other cannot. In these situations, the components aggregate somewhat in solution, which decreases the solution's entropy and makes it more favorable to become a gas. These solutions therefore have a higher vapor pressure than predicted by Raoult's law, leading to a lower boiling temperature. Maximum boiling azeotropes have the opposite effect, resulting from attractions in the solution that lead to lower vapor pressures than predicted, and thus higher boiling temperatures.
A distillation curve for an ethanol-water mixture is shown in Figure 5.16 (not drawn to scale in order to show detail). The minimum boiling azeotrope is represented by the intersection of the two tear-drop shapes, as indicated in Figure 5.16a. this curve demonstrates why an ethanol-water mixture cannot be distilled to produce a distillate with greater than $95.6\%$ ethanol. Imagine distilling an ethanol/water mixture of the composition at position a in Figure 5.16b). After each vaporization-condensation event (a to b to c, then c to d to e in Figure 5.16b), the ethanol concentration increases. However, material is always funneled toward the lowest point on the curve, the azeotropic composition. Concentrations higher than $95.6\%$ ethanol can only be distilled when there is no longer any water in the system (at which point the phase diagram in Figure 5.16 is no longer applicable). Since water forms minimum-boiling azeotropes with many organic compounds, it should be clear why water must always be carefully removed with drying agents before distillation: failure to remove trace water will result in a wet distillate.
$^3$Recall that vapor pressure is the partial pressure of a compound formed by evaporation of a pure liquid into the headspace of a closed container.
$^4$Data from Handbook of Chemistry and Physics, 50$^\text{th}$ ed., CRC Press, Cleveland, 1970, p. 148.
$^5$Azeotropic solutions do not follow this same pattern, as they do not obey Raoult's law.
$^6$Azeotropic solutions again do not follow this generalization, and instead boil at a constant temperature.
$^7$J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 1999, Sect. 5.78 and 5.79. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2B%3A_Separation_Theory.txt |
Condenser Hoses
The condenser is an intricate piece of glassware, and allows for cold water to circulate through the distillation apparatus. The circulating water does not mix with the sample to be purified, but instead passes through another jacket surrounding the hollow tube where the gaseous sample travels. It is important that the water jacket be full of cold water, to maximize the efficiency of condensing the gaseous sample. It is for this reason that the water hoses must be attached to the condenser in a certain way.
A hose should connect from the water spigot to the lower arm of the condenser, forcing water to travel against gravity through the condenser (this is shown correctly in Figure 5.17b). The hose connecting the upper arm of the condenser should then drain to the sink. By forcing the water uphill, it will completely fill the condenser. The hoses are connected incorrectly in Figure 5.17a, with water flowing into the upper arm of the condenser. With a downhill flow of water, only the lower portion of the condenser fills with the circulating water, leading to inefficient cooling of gas traveling through the inner tube.
Whether the condenser hoses point up or down seems to be a personal choice, as there are pros and cons to both orientations. When the hoses point downward, there is a small portion of the condenser that does not fill with cooling water (Figure 5.18a), but the effect is likely minimal. When the hoses point upwards, there is sometimes a tendency for the hoses to bend and pop off, spraying water all over the lab (this can sometimes be avoided by securing wire around the hose joints). Personal judgment may be applied to whether the condenser hoses point up or down.
Simple Distillation Procedure
An assembled simple distillation apparatus is shown in Figure 5.19. Assembly of this complicated apparatus is shown in this section piece by piece. The glassware used for this apparatus is quite expensive, and undoubtedly your instructor would appreciate care being taken when using this experiment. A single condenser costs $72, and a complete kit containing all the glassware needed for distillation costs$550!$^8$
Assemble the Apparatus:
1. To visualize the assembly of the apparatus, it may be helpful to first lay out the glassware on the benchtop before assembling the parts (Figure 5.21a).
2. Pour the liquid to be distilled into a round bottomed flask, trying to avoid pouring liquid on the ground glass joint. If liquid drops onto the joint, wipe it off with a KimWipe. Alternatively use a funnel to be sure no liquid ends up on the joint, which can sometimes cause the joint to freeze.
The flask should ideally be between one-third to one-half full of the liquid to be distilled. If the flask is more than half full, it will be difficult to control the boil. If less than a third full, the recovery may be compromised, as there is a quantity of vapor required to fill the flask that will not distill over (this is called the "holdup volume", and later condenses when the flask is cooled).
3. Add a few boiling stones or magnetic stir bar to the solution to prevent bumping during heating.
4. Use a metal extension clamp to secure the round bottomed flask containing the sample to the ring stand or latticework at least 4 inches above the benchtop (to leave room for the heat source). The clamp should securely hold the joint below the glass protrusion on the flask (Figure 5.21b, not FIgure 5.21c).
1. Attach a three-way adapter (or "distilling adapter") to the round bottomed flask (Figure 5.22a). [If using grease as recommended by your instructor, lightly grease all joints.]
2. Attach a rubber fitting to the top of a thermometer adapter (a cylindrical tube, Figure 5.22b) by stretching it over the glass. Then delicately insert a thermometer into the hole of the rubber fitting.
Safety note: While inserting the thermometer, position your hands near the joint (Figure 5.22c), not far from the joint or it may snap (Figure 5.22d). A prepared thermometer adapter with inserted thermometer is shown in Figure 5.22e.
1. Connect the thermometer adapter to the three-way adapter, securing the joint with a plastic clip (sometimes called a "Keck clip", the yellow clip in Figure 5.23a). The clip is directional, and if it doesn't easily snap on it is probably upside down. Check that the clip is not broken, and if it is, replace it.
A plastic clip should not be used to connect the round bottomed flask to the three-way adapter (Figure 5.23b) for two reasons. First, this is one of the hottest parts of the apparatus, and could cause the plastic to melt (especially if the compound has a high boiling point or a strong heat source is used). Secondly, a plastic clip in this location interferes with a secure grip by the metal extension clamp. A secure clamp on the flask is necessary in order to maintain the integrity of the system while lowering the heat source at the end of the distillation.
2. Adjust the thermometer so the bulb is just below the arm of the adapter (Figure 5.23c). If the bulb is positioned too high, it will not register the correct temperature of the vapors as they make their turn toward the condenser.
3. Prepare the condenser: inspect two rubber hoses and cut off any cracked sections in the rubber with scissors. Wet the ends of two hoses using the faucet or by dipping into a beaker of water, then twist the wet ends of the two arms on the condenser. The hoses should fit onto the condenser arms higher than one centimeter or else water pressure may cause them to pop off.
4. Use another clamp to secure the condenser to the ring stand or latticework (not so tight or it may crack), and position the condenser at roughly the location it will eventually be, with a slight downward angle (Figure 5.23d).
1. Connect the condenser to the rest of the apparatus (Figure 5.24a); this is probably the most difficult part of the whole distillation assembly. While keeping the clamp's arm connected to the condenser, slightly loosen the clamp's attachment to the ring stand or latticework so that it can rotate in all directions. Then wiggle the condenser toward the three-way adapter, holding onto the clamp near the ring stand. When the two joints connect in a perfect match, secure them with a plastic clip (must not be broken!), then tighten the clamps attached to the ring stand.
2. Adjust the clamps and height of the condenser so that the thermometer is perfectly vertical in the apparatus. A tilted distillation apparatus sometimes refluxes instead of distilling (where gas condenses and drips back into the distilling flask).
Safety note: Be sure that there is a secure connection between the distilling flask and the three-way adapter (as pointed to in Figure 5.24a). This connection can be enforced by a strong hold on the condenser. If there is an opening between the distilling flask and adapter, vapors can escape. Being so close to the heat source, the vapors have the greatest potential to ignite at this location.
3. Connect the rubber hose attached to the lower arm of the condenser to the water spigot and allow the hose attached to the upper arm of the condenser to drain to the sink. The hoses may point either up or down. If the drainage hose just barely reaches the sink, secure it with lab tape (Figure 5.24b) so that it does not spray all over the benchtop.
4. Connect a vacuum adapter to the end of the condenser with a plastic clip. Then use another clip to connect a round bottomed flask (receiving flask) that has enough volume to just hold the quantity of expected distillate (Figure 5.24c). Variations in receiving flasks are shown in a later section.
5. A completed distillation apparatus is shown in Figure 5.25. No parts should be able to jiggle or they are not adequately clamped. Although it may appear to be a closed system (which would be dangerous to heat), the system is actually open to the atmosphere at the arm in the vacuum adapter.
1. Prepare the heat source beneath the flask:
1. Heating mantles, oil baths, and heat guns are recommended heat sources for distillation. Bunsen burners are not recommended for use with volatile organic compounds, but may be used in some situations at the discretion of the instructor.
2. Position the heat source beneath the flask, using an adjustable platform that allows for some mechanism by which the heat can be lowered and removed at the end of the distillation. A ring clamp and wire mesh platform is a good way to hold a heating mantle or oil bath beneath the flask.
Begin Distilling
1. When ready to begin the distillation, start circulating water in the condenser. The flow of water should be more than a trickle, but should not be so strong that the hose flops around from the high water pressure.
2. Since it is so important, check once again that there is a secure connection between the distilling flask and the three-way adapter (as indicated in Figure 5.26). If there is a crack between the distilling flask and adapter, vapors can escape the apparatus and possibly ignite!
1. Begin heating the flask. Depending on the size of the flask and the necessary temperature, the liquid may start boiling between 1-20 minutes. Condensation will eventually be seen on the sides of the flask, but the reading on the thermometer will not increase until the vapors immerse the thermometer bulb. It is not unusual that the liquid boils while the thermometer remains at room temperature.
2. Eventually condensation should be seen in the three-way adapter, and the thermometer temperature will rise. Not all compounds produce obvious droplets (as in Figure 5.26), but condensation may be seen by close inspection and subtle movement near the thermometer.
Safety note: If the compound's boiling point is $> 75^\text{o} \text{C}$, the three-way adapter will be hot to the touch!
3. Eventually a droplet may be seen traveling down the length of the condenser and liquid will begin to collect in the receiving flask. An appropriate distillation collects distillate at a rate of 1 drop per second. Distilling at too fast a rate prevents proper equilibration between liquid and gas phases, and results in poor separation.
Although a slow distillation is necessary, heating the solution such that nothing is dripping into the receiving flask simply wastes time. There is no reason to heat so gradually that nothing distills.
Record the Temperature and Pressure
1. Pay attention to the temperature on the thermometer throughout the entire time that liquid is distilling. Record the temperature when it has plateaued, or the highest temperature seen. If you know a mixture is distilling, record the temperature range over which liquid is actively dripping into the receiving flask.
The thermometer bulb must be fully immersed in vapors to register an accurate temperature (Figure 5.27d). Therefore, the temperature near the very beginning and end of the distillation tends to be inaccurate. Only record temperatures that correspond to active distillation and full immersion of the thermometer with vapor.
2. As boiling points are so dependent on temperature, also record the barometric pressure.
Stop Distilling
1. Cease the distillation when one of the following takes place:
1. If the liquid is nearly gone in the distilling flask.
Safety note: It is unsafe to distill a flask to dryness as side reactions can occur when components are concentrated. This is especially dangerous with compounds that can form peroxides,$^9$ as these can become explosive when reacting with concentrated solutions. Additionally, when the entire sample is in the gas phase, the system is not longer restricted to the boiling point of the liquid and may reach dangerous temperatures.
2. If the thermometer temperature was higher during the distillation but has dropped significantly. This normally corresponds to a lull between distilling two components, and means that one component has essentially completed distilling.
3. If the thermometer temperature has a dramatic spike. This normally corresponds to the beginning of distillation of a higher boiling component, and would contaminate the distillate if allowed to continue.
4. If anything surprising or unusual happens, such as thick smoke, darkening/thickening in the distilling flask, or uncontrollable bumping.
2. To stop the distillation, lower and remove the heat source from the round bottomed flask (Figure 5.28a).
1. Keep water circulating in the condenser until the liquid in the distilling flask is just warm to the touch, then turn off the condenser water and disassemble the apparatus. After a few minutes of air cooling, cooling of the flask can be hastened by immersion with a container of tap water (Figure 5.28b).
Simple Distillation Summary
Table 5.6: Procedural summary of simple distillation.
Assembly tips
Fill the distilling flask with sample 1/3-1/2 full. Always use an extension clamp on the distilling flask.
Add a few boiling stones or stir bar to flask.
Position the thermometer bulb just below the arm of the three-way adapter, where vapors turn toward the condenser.
Wet condenser hoses with water before attaching.
Connect the condenser hoses such that water flow uphill: bring water from the faucet into the lower arm, and drain out the upper arm.
Be sure all of the connections are secure (especially between the distilling flask and 3-way adapter: potential of fire!)
Begin distillation
Turn on the condenser water.
Apply the heat source to the distilling flask.
Collect distillate at a rate of 1 drop per second.
Record the temperature where liquid is actively distilling and thermometer bulb is immersed.
Record the pressure.
Cease distillation
Stop the distillation when the temperature changes dramatically or if the distilling flask is nearly empty (never distill to dryness!)
Lower and remove the heat source, but keep water circulating until the flask is just warm to the touch.
Variations
A few variations of the simple distillation setup are commonly encountered:
1. When a certain volume of distillate is desired, the distillate may be collected directly into a graduated cylinder (Figure 5.30a). This method allows for cylinders to be easily exchanged, perhaps for analysis of different fractions of distillate.
2. The receiving flask is sometimes submerged in an ice bath (Figure 5.30b), particularly when the distillate is volatile (has a boiling point $< 100^\text{o} \text{C}$). Any container can be used for the ice bath. Be sure to use a combination of both ice and water in the bath, as ice alone does not have as good contact with the flask as a slurry of ice-water.
1. A Claisen adapter (Figure 5.30c) may be inserted between the round bottomed flask and three-way adapter in anticipation of foaming or bumping. The Claisen adapter acts as a buffer so that uncontrolled activity in the distilling flask does not immediately carry over to the receiving flask. If the solution foams, the Claisen adapter provides time to adjust the heat without ruining the distillate. A Claisen adapter also potentially allows for more vaporization-condensation events (or theoretical plates).
2. For distillation of water-sensitive materials, glassware can be oven or flame dried, and a drying tube can be attached to the vacuum adapter (Figure 5.30d).
Troubleshooting
Boiling, but Temperature isn't Rising/How to Insulate
It is not uncommon for new organic chemistry students to expect the temperature to rise on the thermometer the very moment boiling is seen in the distilling flask. Students should remember that the thermometer measures the temperature at the location of the bulb, which is a distance away from the boiling liquid. Therefore, when a solution begins to boil, it may still take a while for the thermometer to register anything other than room temperature.
It is important to record a temperature only when the thermometer bulb is fully engulfed by vapors. Figure 5.31a shows a solution that is boiling but not yet distilling - the temperature at this point would not represent the boiling point of the liquid. Figure 5.31b shows an actively distilling solution, and notice that the thermometer bulb is completely enveloped in vapors and condensation; the temperature at this point would correspond to the boiling point of the solution.
For high-boiling liquids, it may be difficult for vapors to reach the condenser as they too quickly are cooled by the glassware which is in contact with the air in the room. It may be helpful to insulate the distilling flask and three-way adapter to better retain heat and allow the sample to remain in the gas phase longer.
To insulate a portion of the distillation, wrap the parts prior to the condenser with glass wool (Figure 5.32c), then secure with an outer wrapping of aluminum foil (Figure 5.32d). A small gap can be left in the insulation in order to "peek in" on activity inside the apparatus. Glass wool has an appearance similar to cotton, but unlike cotton is not flammable so is useful as an insulating material when an apparatus is to be heated. Glass wool comes in two forms: a fibrous form and a cottony form. The fibrous form (Figure 5.32a) has tiny fibers that can become embedded in skin in the most painful way, similar to several simultaneous paper cuts. This type of glass wool should not be manipulated with bare hands, but only when wearing thick gloves. The more cotton-like glass wool (Figure 5.32b), however, does not hurt when handling with bare hands.
If glass wool is unavailable, aluminum foil can be used alone to insulate a portion of the distillation. It will not insulate well if the foil is wrapped too tightly to the glass, but works well if a small pocket of air is allowed between the foil and glass.
If the expected boiling point of the compound to be distilled is quite high $\left( > 150^\text{o} \text{C} \right)$ and distillation continues to be difficult, vacuum distillation might be attempted.
Vapor is Escaping from Vacuum Adapter
If vapor is noticed escaping out of the vacuum adapter like a tea kettle, the condenser is not doing a good enough job of trapping the gas (Figure 5.33). Reasons for this may be that you have forgotten to turn on the water in the condenser, the water stream is too weak, or the heating is too vigorous.
$^8$Prices were found in the Chemglass catalog in March 2016.
$^9$Certain compounds (e.g. ethers, dioxanes, tetrahydrofuran) are well known to form peroxides when left in contact with air over a period of time. However, many compounds are somewhat susceptible to peroxide formation, including compounds with allylic hydrogen atoms (e.g. cyclohexene). A more complete listing of compounds that form peroxides can be found in: Handbook of Chemistry and Physics, CRC Press, 84$^\text{th}$ edition, 2003-2004, 16-6.
5.2D: Microscale Distillation
There are a variety of methods used to distill small amounts of material $\left( < 10 \: \text{mL} \right)$. The goal of all of these variations is to minimize the loss of material. A certain amount of sample is always necessary to fill the apparatus before distillation occurs, and this quantity (called the "holdup volume") normally condenses after cooling, often back into the original distilling flask. Therefore, minimizing the path length between distilling and receiving flasks can increase the recovery with distillation. Another approach to increase recovery on the microscale is to minimize the number of joints in the apparatus, which may not be perfectly airtight, and can contribute to leaking of material.
Most microscale versions are simple distillations, as use of a fractional column adsorbs too much material. Therefore, these techniques are generally used to remove non-volatile or very high-boiling or low-boiling impurities.
Semi-Microscale
A semi-microscale apparatus is essentially a macroscale apparatus but lacks the condenser (Figure 5.34). This shortens the path length between distilling and receiving flask. It can be used for $5$-$10 \: \text{mL}$ of material.
The distillation needs to be done with care, as too high of temperatures may cause vapor to be lost out of the vacuum adapter.
Hickman Head
A Hickman head is provided in many microscale kits, and can be used to distill $1$-$3 \: \text{mL}$ of material. It is often connected to a conical vial, using an O-ring and screw-capped connector to secure the joint (Figure 5.35a). The solution boils, condenses on the sides of the flask and collects in the lip of the Hickman head (indicated with an arrow in Figure 5.35b). The bulb of the thermometer should be positioned below the lip (Figure 5.35a), although it may be difficult to accurately measure the vapor temperature as so little material is used. After cooling, the flask must be tilted to retrieve the distillate by pipette (Figure 5.35c), and the lip holds at the most $1 \: \text{mL}$ of distillate so may need to be emptied several times. Some Hickman heads have a side port so distillate can be removed without stopping the distillation (Figure 5.35d).
Short-Path Distillation
An all-in-one distillation apparatus, called a "short-path distilling head", is sometimes used to distill small quantities of material (Figure 5.36). These pieces of glassware are quite expensive (\$200 each),$^{10}$ and so are normally used in research settings, not teaching labs.
When assembling the glassware, it is important that the joint connecting the distilling flask to the apparatus is held securely: verify it is not loose by grasping the pieces with your fingers (Figure 5.36b). The apparatus requires a thermometer that has a ground-glass fitting (Figure 5.36d).
$^{10}$Prices were found in the Chemglass catalog in March 2016. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2C%3A_Step-by-Step_Procedures.txt |
A simple distillation is incapable of significant purification if the boiling points of the components are too close. When the difference in boiling points is less than 100 ˚C, a modification is necessary, namely insertion of a fractionating column between the distilling flask and three-way adapter.
• 5.3A: Theory of Fractional Distillation
A simple distillation is incapable of significant purification if the boiling points of the components are too close. The distillate of a simple distillation is always enriched in the lower boiling compound.Every vaporization-condensation event (called a "theoretical plate") is similar to a simple distillation, and each event enriches the distillate in the lower boiling component.
• 5.3B: Fractionating Columns
The choice of what fractionating column to use for which application depends in part on availability and the task at hand. Different columns have different surface areas and numbers of theoretical plates, and thus differ in their ability to separate close-boiling components. They also differ somewhat in the quantity of compound that will be sequestered through wetting the column.
• 5.3C: Uses of Fractional Distillation
Fractional Distillation is used for both oil refining and purification of reagents and products. Fractional distillation is used in oil refineries (Figure 5.41) to separate the complex mixture into fractions that contain similar boiling points and therefore similar molecular weights and properties. Gasoline, diesel fuel, kerosene, and jet fuel are some of the different fractions produced by an oil refinery.
• 5.3D: Step-by-Step Procedures for Fractional Distillation
Other columns may be substituted. It is assumed that readers have previously performed a simple distillation, so in this section are described differences between simple and fractional distillation.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
5.03: Fractional Distillation
A simple distillation is incapable of significant purification if the boiling points of the components are too close. When the difference in boiling points is less than $100^\text{o} \text{C}$, a modification is necessary, namely insertion of a fractionating column between the distilling flask and three-way adapter (Figure 5.37a).
The distillate of a simple distillation is always enriched in the lower boiling compound. As previously discussed, the simple distillation of a $75 \: \text{mol}\%$ ethylbenzene/$25 \: \text{mol}\%$ p-cymene mixture resulted in a distillate that was $90 \: \text{mol}\%$ ethylbenzene (ethylbenzene had the lower boiling point). Imagine if a subsequent distillation were performed on the $90 \: \text{mol}\%$ solution: the distillate of that process would contain an even higher percentage of ethylbenzene.
A fractionating column essentially allows for many successive distillations to take place at once, without dismantling the apparatus. A fractionating column contains indentations (a Vigreux column, Figure 5.37) or a packing material with lots of surface area. The vapors temporarily condense on these surfaces (see Figure 5.37b) and the heat of the distillation allows those pools of liquid to vaporize again. Every vaporization-condensation event (called a "theoretical plate") is similar to a simple distillation, and each event enriches the distillate in the lower boiling component.
The concepts of a fractional distillation can be shown through a distillation curve. In the distillation of an equimolar mixture of compounds A + B (which is described by the distillation curve in Figure 5.38), one vaporization-condensation event is represented in Figure 5.38 by following points a to b to c. This process represents one theoretical plate, and would produce a distillate that is $81\%$ A and $19\%$ B. A second vaporization-condensation event is represented in Figure 5.38 by following points c to d to e, and would produce a distillate that is $96\%$ A and $4\%$ B.
A $50\%$/$50\%$ mixture of two components whose boiling points differ by only $20$-$30^\text{o} \text{C}$ would require at least three theoretical plates to obtain a distillate with $> 95\%$ purity. In practice, fractional distillations still often produce mixtures. The best chance of obtaining purity via fractional distillation is when there is very little impurity to begin with. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.03%3A_Fractional_Distillation/5.3A%3A_Theory_of_Fractional_Distillation.txt |
The choice of what fractionating column to use for which application depends in part on availability and the task at hand. Several columns are shown in Figure 5.39: a) Vigreux column with glass indentations, b) Steel wool column made simply be loosely inserting steel wool into the cavity of a fractionating column (similar to a West condenser, but wider), c) Glass beads filled in the cavity of a fractionating column.
These columns have different surface areas and numbers of theoretical plates, and thus differ in their ability to separate close-boiling components. They also differ somewhat in the quantity of compound that will be sequestered through wetting the column. A Vigreux column has the least surface area, making it the least capable of separating close-boiling components. And yet with the lowest surface area, it can have the highest recovery, making it the optimal choice if a separation is not particularly difficult. Glass beads have a high surface area, so are a good choice for separation of close-boiling components. And yet, beads will suffer the greatest loss of material. Steel wool columns have intermediate surface areas and their effectiveness can depend on how tightly the wool is packed in the column. Steel wool columns also cannot be used with corrosive vapors like those containing acid.
To demonstrate the effectiveness of different fractionating columns, a mixture containing $75 \: \text{mol}\%$ ethylbenzene (normal b.p. $136^\text{o} \text{C}$) and $25 \: \text{mol}\%$ p-cymene (normal b.p. $177^\text{o} \text{C}$) was distilled using several methods. The first two $\text{mL}$ of distillate was collected in each process (Figure 5.40).
Gas chromatograph analysis of the initial mixture and distillates allowed for quantitation of the mixtures, although a calibration curve was necessary to translate the reported percentages into meaningful values. As the two components had only a $41^\text{o} \text{C}$ difference in boiling points, a fractional distillation was more effective than a simple distillation, although in some cases only slightly (see Table 5.7). The distillate was never $> 98\%$ pure with any method as there was a large quantity of the minor component in the distilling pot.
Table 5.7: GC results for various distillations of a $75 \: \text{mol}\%$ ethylbenzene (EB) and $25 \: \text{mol}\%$ p-cymene (PC) mixture. The actual percentage values were estimated from the calibration curve in Figure 5.11.
Method Components GC% Actual %
Initial Solution ethylbenzene 55.6% 75%
p-cymene 44.4% 25%
Simple Distillation ethylbenzene 67.6% 90%
p-cymene 32.4% 10%
Fractional (Beads) ethylbenzene 87.7% 97%
p-cymene 12.3% 3%
Fractional (Steel Wool) ethylbenzene 71.4% 92%
p-cymene 28.6% 8%
Fractional (Vigreux) ethylbenzene 70.4% 92%
p-cymene 29.6% 8%
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5.3C: Uses of Fractional Distillation
Oil Refining
Crude oil (petroleum) is composed of mostly hydrocarbons (alkanes and aromatics), and is a tremendous mixture of compounds consisting of between 5 and 40 carbon atoms.$^{11}$ The components in oil are incredibly useful as fuels and lubricants, but not when they are mixed together. Fractional distillation is used in oil refineries (Figure 5.41) to separate the complex mixture into fractions that contain similar boiling points and therefore similar molecular weights and properties. Gasoline, diesel fuel, kerosene, and jet fuel are some of the different fractions produced by an oil refinery.
Purification of Reagents and Products
Cyclopentadiene is used in many chemical reactions, including Diels-Alder reactions and polymerizations. The reagent is so reactive, however, that it undergoes a Diels-Alder reaction with itself in the reagent bottle to form dicyclopentadiene (Figure 5.42a). Therefore, chemical companies do not sell cyclopentadiene, and chemists are instead required to distill commercial dicyclopentadiene (Figure 5.42b) to reverse the dimerization reaction and obtain cyclopentadiene (Figure 5.42c).
At temperatures above $150^\text{o} \text{C}$ the dimer reverts to the monomer through a retro Diels-Alder reaction (driven by the favorable change in entropy, Figure 5.42c). Distillation can be used to remove the monomer as it forms. Although the two components (dimer and monomer) have dramatically different boiling points, the temperature required for the reverse reaction is too similar to the boiling point of the dicyclopentadiene that its vapor pressure cannot be ignored. Therefore, a fractional distillation is required for this process.
$^{11}$About $6\%$ of crude oil contains hydrocarbons with greater than 40 carbon atoms, a fraction that eventually becomes used for asphalt. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.03%3A_Fractional_Distillation/5.3B%3A_Fractionating_Columns.txt |
Fractional Distillation Procedure
An assembled fractional distillation apparatus is shown in Figure 5.43, using glass beads in the fractionating column. Other columns may be substituted. It is assumed that readers have previously performed a simple distillation, so in this section are described differences between simple and fractional distillation.
1. If a beaded fractionating column is used, sometimes a wad of glass wool is inserted into the top so that the beads do not spill out. Before using the column, remove this wad as it may interfere with the passage of vapors (Figure 5.44a). If using a Vigreux column, check for broken glass indentations (which would case a leak in the column).
2. The distilling pot will need to be heated much more vigorously than with a simple distillation, as there is a greater distance for the vapors to travel before reaching the condenser. The vapors will tend to reflux in the column (condense and drip back into the distilling pot) unless stronger heating is applied.
A rule of thumb is that the distilling pot needs to be $30^\text{o} \text{C}$ hotter than the top of the column in order for material to ascend the column. If it is difficult to achieve more than a reflux, the column can be insulated by wrapping it with glass wool then aluminum foil (Figure 5.44b). This allows the column to maintain heat and the sample to remain in the gas phase longer. A small gap can be left in the foil or glass wool if desired to "peek in" on the activity in the column.
3. Ideally both liquid and gas should be seen in the fractionating column, as the sample needs to undergo many vaporization-condensation events (Figures 5.44 c+d). Droplets of liquid should be seen on the surfaces of the packing material, but there should never by a large pool of liquid. A "river" of liquid traveling up the column is called flooding (Figure 5.45). If a column floods, remove the heat until the liquid drains back into the distilling flask, then resume heating at a gentler rate.
1. Cleaning of a fractionating column:
1. Vigreux column: rinse with acetone. Don't use a scrub brush or the glass indentations may break.
2. Steel wool column. rinse with large amounts of acetone. Don't rinse with water as wet steel will rust over time.
3. Glass bead column: rinse with acetone, then replace the glass wool wad to prevent the beads from pouring out when horizontal. Alternatively, pour out the glass beads to be cleaned separately. Be delicate when using a scrub brush on the fractionating column as there are fragile indentations near the bottom joint which can break.
Fractional Distillation Summary
Table 5.8: Procedural summary of fractional distillation.
Most comments for a simple distillation apply to fractional as well.
The distilling pot will need to be heated more vigorously than with a simple distillation, as there is a greater distance for the vapors to travel before reaching the condenser.
Commonly the column will need to be insulated to maintain heat: wrap the column (and three-way adapter if desired) in glass wool followed by an outer layer of aluminum foil. Droplets of liquid should be seen in the fractional column, but there should never be a large pool of liquid (flooding). If the column floods, allow the liquid to drain back into the distilling flask and heat at a gentler rate. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.03%3A_Fractional_Distillation/5.3D%3A_Step-by-Step_Procedures_for_Fractional_Distillation.txt |
Boiling commences when the vapor pressure of a liquid or solution equals the external or applied pressure (often the atmospheric pressure). Thus, if the applied pressure is reduced, the boiling point of the liquid decreases. This behavior occurs because a lower vapor pressure is necessary for boiling, which can be achieved at a lower temperature.
• 5.4A: Overview of Vacuum Distillation
Boiling commences when the vapor pressure of a liquid or solution equals the external or applied pressure (often atmospheric pressure). Thus, if the applied pressure is reduced, the boiling point of the liquid decreases. This behavior occurs because a lower vapor pressure is necessary for boiling, which can be achieved at a lower temperature.
• 5.4B: Predicting the Boiling Temperature
The boiling point of a liquid or solution drops when the pressure is reduced in a distillation apparatus. It is helpful to be able to predict the altered boiling point depending on the pressure inside the apparatus.
• 5.4C: Step-by-Step Procedures for Vacuum Distillation
A vacuum distillation apparatus can be constructed from a simple distillation setup although a fraction distillation can also be used. It is assumed that readers have previously performed a simple distillation under atmospheric pressure, so in this section are described differences between atmospheric and reduced pressure distillations.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
5.04: Vacuum Distillation
Boiling commences when the vapor pressure of a liquid or solution equals the external or applied pressure (often atmospheric pressure). Thus, if the applied pressure is reduced, the boiling point of the liquid decreases (see the graph for cinnamyl alcohol in Figure 5.47a). This behavior occurs because a lower vapor pressure is necessary for boiling, which can be achieved at a lower temperature. For example, water produces a vapor pressure of $760 \: \text{mm} \: \ce{Hg}$ at $100^\text{o} \text{C}$ (see Figure 5.47b), and so water boils at sea level at $100^\text{o} \text{C}$. However, if water was subjected to a reduced pressure of only $100 \: \text{mm} \: \ce{Hg}$, it would boil at roughly $50^\text{o} \text{C}$ as this is the temperature that it produces a vapor pressure of $100 \: \text{mm} \: \ce{Hg}$ (see Figure 5.47b). This trend follow the Clausius-Clapyron equation.
The dependence of boiling point on applied pressure can be exploited in the distillation of very high boiling compounds (normal b.p. $> 150^\text{o} \text{C}$), which may decompose if heated to their normal boiling point. A vacuum distillation is performed by applying a vacuum source to the vacuum adapter of either a simple or fractional distillation (Figure 5.48). When the pressure is lowered inside the apparatus, solutions boil at a lower temperature.
$^{12}$Data from The Merck Index, 12$^\text{th}$ edition, Merck Research Laboratories, 1996.
$^{13}$Data from J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 1999, Sect. 5.28.
5.4B: Predicting the Boiling Temperature
The boiling point of a liquid or solution drops when the pressure is reduced in a distillation apparatus. It is helpful to be able to predict the altered boiling point depending on the pressure inside the apparatus.
The lowest pressure attainable inside the apparatus depends largely on the vacuum source and the integrity of the seal on the joints. Lower pressures are attainable when using a portable vacuum pump$^{14}$ than when using a water aspirator or the building's house vacuum (Figure 5.49). Due to the very low pressures possible with oil pumps in portable vacuums, these vacuum distillations should be conducted in the fume hood behind a blast shield.
Water aspirators are the most common vacuum source in teaching labs because they are inexpensive. When a water aspirator is used, the vacuum pressure is always limited by the intrinsic vapor pressure of water, which is often between $17.5 \: \text{mm} \: \ce{Hg}$ $\left( 20^\text{o} \text{C} \right)$ and $23.8 \: \text{mm} \: \ce{Hg}$ $\left( 25^\text{o} \text{C} \right)$.$^{15}$ the vacuum pressure is also very dependent on water flow, which can vary greatly. If an entire lab section uses the water lines at the same time, the water flow can be significantly compromised, leading to a much higher pressure than $25 \: \text{mm} \: \ce{Hg}$ inside an apparatus. The number of students using aspirators at one time should be limited as much as possible.
If a manometer is available, the distillation apparatus should be set up and evacuated without heating to measure the pressure. The expected boiling point of a compound can then be roughly estimated using a nomograph (found in a CRC or online) or through the general guidelines in Table 5.9. If a manometer is not available and a water aspirator is to be used, the expected boiling point can be estimated using an approximate pressure of $20 \: \text{mm} \: \ce{Hg}$, although the pressure will likely be higher than this.
Figure 5.9: Approximate boiling point of compounds at reduced pressure (all in ºC).$^{16}$
Boiling point at 760 mmHG 150 170 200 220 250 270 300
Boiling point at 20 mmHG 62 78 101 117 141 157 181
Boiling point at 18 mmHG 60 76 99 115 139 154 178
Boiling point at 16 mmHG 58 73 97 112 136 151 174
Boiling point at 14 mmHG 56 71 94 108 133 148 171
Boiling point at 12 mmHG 52 68 90 104 129 144 167
$^{14}$A Kugelrohr apparatus can obtain pressures as low as $0.05 \: \text{mm} \: \ce{Hg}$, as reported by the Sigma-Aldrich operating instructions.
$^{15}$J. A. Dean, Lange's Handbook of Chemistry, 15$^\text{th}$ ed., McGraw-Hill, 199, Sect 5.28.
$^{16}$Selected values from: A. J. Gordon and R. J. Ford, The Chemist's Companion. A Handbook of Practical Data, Techniques and References, Wiley & Sons, 1972, p 32-33. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.04%3A_Vacuum_Distillation/5.4A%3A_Overview_of_Vacuum_Distillation.txt |
Vacuum Distillation Procedure
A vacuum distillation apparatus is shown in Figure 5.50, using a simple distillation setup. A fraction distillation can also be used. It is assumed that readers have previously performed a simple distillation under atmospheric pressure, so in this section are described differences between atmospheric and reduced pressure distillations.
Prepare the Apparatus
1. Safety note: Inspect every piece of glassware to be used with the vacuum distillation, checking for stars, cracks, or other weaknesses in the glass, as these may allow for implosion when the pressure is reduced.
2. A stir bar needs to be used for bump prevention. Boiling stones cannot be used with vacuum distillation as air trapped in the stone's pores is rapidly removed under vacuum, causing the stones to fail to produce bubbles.
3. Although greasing is somewhat of a personal choice with simple and fractional distillations, all joints must be greased in vacuum distillations or the system will leak and fail to achieve a low pressure (Figures 5.51a+b).
4. Begin assembly of the apparatus near the vacuum source. If using a water aspirator, test to be sure that the aspirator works well as some are more functional than others. To test an aspirator, apply thick vacuum hosing to the nub on the aspirator, turn on the water and feel for suction at the end of the hose with your finger (Figure 5.51c).
5. A Claisen adapter should be included in the apparatus as solutions under vacuum tend to bump violently (a Claisen adapter is labeled in Figure 5.50a).
6. Attach thick-walled tubing to the vacuum adapter on the distillation apparatus (Figure 5.51d) and connect to a vacuum trap. A trap suitable for a water aspirator is shown in Figure 5.50b, but a more substantial trap cooled with dry ice and acetone should be used with a portable vacuum to prevent solvent vapors from degrading the oil pump.
Connect the trap to the vacuum source (aspirator or vacuum pump). It is best to not bend or strain the tubing as much as is practical, as this may create a leak in the system.
1. Insert a wood block (Figure 5.52a) or lab jack beneath the stirring plate to allow for lowering of the heat source when the distillation is complete.
Begin the Distillation
1. Before heating, turn on the vacuum source to begin reducing pressure inside the apparatus. There should not be a hissing sound or else there is a leak in the system.
The purpose of reducing the pressure before heating is for removal of very low-boiling liquids (e.g. residual solvent). If the system were heated at the same time, the low-boiling liquids might boil violently in the flask.
If a manometer is available, take note of the pressure inside the apparatus. This may be used to predict the boiling point of the sample.
2. When confident that the apparatus is adequately evacuated and any low-boiling compounds have been removed, begin heating the sample (Figure 5.52b).
3. If it is difficult to achieve more than a reflux, the Claisen and three-way adapter can be insulated by wrapping them tightly with glass wool then aluminum foil (Figure 5.52c). Insulation allows the column to maintain heat and the sample to remain in the gas phase longer. A small gap should be left in the insulation near the distilling flask to "peek in" and make sure the stirring mechanism continues to work properly.
4. Record the temperature over which material is collected, making sure the value corresponds to a temperature when the thermometer bulb is fully immersed in vapors. If a manometer is used, also record the pressure. If no manometer is used, record the vacuum source (e.g. aspirator).
Pure liquids do not always distill at a constant temperature when under vacuum, as variations in pressure so easily occur and affect the boiling temperature. A range of $5^\text{o} \text{C}$ is not uncommon for pure liquids. This is especially true when the vacuum source is a water aspirator, where variations in water flow alter the pressure.
5. If more than one fraction of distillate is desired, the distillation must be stopped before changing the receiving flask (see the next section for how). If available, a "cow" receiving flask can be used to collect different fractions without ceasing the vacuum (Figure 5.53a).
Stop the Distillation
1. To stop the distillation, first remove the heat source, cool the flask to room temperature then further cool in a tap water bath (Figure 5.53a).
2. Slowly reinstate the atmospheric pressure into the flask by opening the pinch clamp at the vacuum trap (Figure 5.53c), or by removing the rubber tubing at the vacuum adapter or aspirator (Figure 5.53d). You will know the system is open to the atmosphere when there is an increase in water flow at the aspirator, or if a hissing sound is heard. Then turn off the vacuum source.
It is important to first cool the system before allowing air back in as the superheated residue in the flask may react unexpectedly with oxygen in the air.
It is also important to first allow air back into the system before turning off the vacuum source. If the vacuum is turned off first, sometimes changes in pressure inside the apparatus (as it cools) cause back-suction. If a water aspirator is used, this may cause water from the sink to be pulled into the vacuum line. The vacuum trap prevents this back suction from ruining the distillate.
3. Disassemble and clean up the distillation apparatus as quickly as is practical, as the joints can sometimes freeze if left connected for prolonged periods.
Vacuum Distillation Summary
Table 5.10: Procedural summary of vacuum distillation.
Always use a stir bar, not boiling stones.
Grease all joints.
Use a Claisen adapter, as solutions tend to bump under vacuum.
Connect thick-walled hosing at the vacuum adapter to a trap, then to the vacuum source (water aspirator or vacuum pump).
Turn on the vacuum first, before heating, to remove very volatile components.
When confident the apparatus is maintaining a reduced pressure, then heat the sample.
Use glass wool or foil insulation if the sample is stubbornly refluxing instead of distilling.
Record temperature and pressure if using a manometer during active distillation.
Pure compounds may not distill over a constant temperature due to changes in pressure.
If multiple fractions will be collected, the system needs to be vented and cooled in between (or a cow receiving flask used).
To stop the distillation, remove the heat and cool the flask in a tap water bath.
Then open the apparatus to the atmosphere by opening the pinch clamp on the trap, or removing the tubing on the vacuum adapter.
Lastly turn off the vacuum.
Correct order: cool, open to atmosphere, then turn off vacuum. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.04%3A_Vacuum_Distillation/5.4C%3A_Step-by-Step_Procedures_for_Vacuum_Distillation.txt |
Steam distillation is analogous to simple distillation, the main difference being that steam (or water) is used in the distilling flask along with the material to be distilled. Experimentally the setups are arranged more or less the same, with small differences being how the steam is added to the flask: either indirectly if a steam line is available in the building, or directly by boiling water in the flask.
• 5.5A: Overview of Steam Distillation
Steam distillation is analogous to simple distillation, the main difference being that steam (or water) is used in the distilling flask along with the material to be distilled. Experimentally the setups are arranged more or less the same with small differences being how the steam is added to the flask: either indirectly if a steam line is available in the building, or directly by boiling water in the flask.
• 5.5B: Uses of Steam Distillation
The most common use of steam distillation is the extraction of natural products from plant materials. This is the main industrial method for obtaining plant essential oils, used in fragrances and personal hygiene products. As so many products can be isolated in this way, this technique is regularly performed in teaching labs.
• 5.5C: Separation Theory
egular" distillation, separation is attempted on a mixture of components that dissolve in one another. When the components in the distilling flask do not dissolve in one another, such as when water and nonpolar organic compounds are present, the vapor produced from these mixtures is different. The components act independently from one another, and the partial pressure from each component is no longer determined by its mole fraction. The partial pressure of each component is its vapor pressure.
• 5.5D: Step-by-Step Procedures for Steam Distillation
A steam distillation apparatuses are shown that uses boiling water in the distilling flask or a steam line. In this discussion of step-by-step procedure for steam distillation , it is assumed that readers have previously performed a simple distillation, so in this section are described differences between simple and steam distillations.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
5.05: Steam Distillation
Steam distillation is analogous to simple distillation, the main difference being that steam (or water) is used in the distilling flask along with the material to be distilled. Experimentally the setups are arranged more or less the same (Figure 5.56), with small differences being how the steam is added to the flask: either indirectly if a steam line is available in the building, or directly by boiling water in the flask.
5.5B: Uses of Steam Distillation
The most common use of steam distillation is the extraction of natural products from plant materials. This is the main industrial method for obtaining plant essential oils, used in fragrances and personal hygiene products. As so many products can be isolated in this way, this technique is regularly performed in teaching labs. An extraction and GC analysis of the essential oil in bay leaves is shown in Figure 5.57.
5.5C: Separation Theory
Although steam distillation appears almost identical to "regular" distillation, the principles behind the separation of components are quite different. In a "regular" distillation, separation is attempted on a mixture of components that dissolve in one another. The vapor produced from these mixtures can be described by a combination of Raoult's law and Dalton's law, as shown in Equation \ref{14} for a two-component mixture.
$\text{Miscible components:} \: \: \: \: \: P_\text{solution} =P_A + P_B = P_A^o \chi_A + P_B^o \chi_B \label{14}$
When the components in the distilling flask do not dissolve in one another, such as when water and nonpolar organic compounds are present, the vapor produced from these mixtures is different. The components act independently from one another (which makes sense considering they do not mix), and the partial pressure from each component is no longer determined by its mole fraction. The partial pressure of each component is simply its vapor pressure, and the vapor composition for a two-component mixture is described by Equation \ref{15}. Although the components do not mix in the liquid phase, they do in the gas phase, which allows for co-distillation of "incompatible components".
$\text{Immiscible components:} \: \: \: \: \: P_\text{solution} = P_A^o + P_B^o \label{15}$
The implications of Equation \ref{15} are several. First, since mole fraction is not a factor, it is possible that a minor component in the distilling flask can be a major component in the distillate if it has an appreciable vapor pressure. In the steam distillation of volatile plant materials, this means that the distillate composition is independent of the quantity of water or steam used in the distilling flask.
Secondly, since the vapor pressures of each component add, the mixture will always boil at a lower temperature than the boiling point of the lowest boiling component. For example, at $100^\text{o} \text{C}$ water has a vapor pressure of $760 \: \text{torr}$ since it is at its normal boiling point. If another volatile, non-water-soluble component was present with the water in a distilling flask at $100^\text{o} \text{C}$ (for example toluene, which has a boiling point of $111^\text{o} \text{C}$), it too would produce vapors that contributed to the total pressure. As boiling occurs when the combined pressure matches the atmospheric pressure (let's say $760 \: \text{torr}$ for the sake of this calculation), boiling would occur below $100^\text{o} \text{C}$. For example, a mixture of toluene and water boils at $85^\text{o} \text{C}$, as shown in Equation (16).
\text{Water/toluene mix at } 85^\text{o} \text{C}: \: \: \: \: \: \begin{align} P_\text{solution} &= P_\text{water}^o + P_\text{toluene}^o \ &= \left( 434 \: \text{torr} \right) + \left( 326 \: \text{torr} \right) = 760 \: \text{torr} \end{align} \label{16}
The distilling temperature in steam distillation is always below $100^\text{o} \text{C}$ (the boiling point of water), although in many cases the distilling temperature is very near or just under $100^\text{o} \text{C}$. This feature allows for plant essential oils (complex mixtures that often include components with very high boiling points, $> 250^\text{o} \text{C}$), to be extracted at lower temperatures than their normal boiling points, and thus without decomposition. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.05%3A_Steam_Distillation/5.5A%3A_Overview_of_Steam_Distillation.txt |
Steam Distillation Procedure
A steam distillation apparatus is shown in Figure 5.58 that uses boiling water in the distilling flask. An apparatus using a steam line is shown in Figure 5.59. It is assumed that readers have previously performed a simple distillation, so in this section are described differences between simple and steam distillations.
Prepare the Setup
1. Place the plant material to be extracted into a large round bottomed flask with a 24/40 joint (wide mouth), no more than half full. A large amount of material is often necessary as the bulk of plants is cellulose, not oil. Use of a 14/20 flask (more narrow mouth) would make it difficult to remove the plant materials after distillation.
2. Use an extension clamp to secure the large flask to the ring stand or latticework. If desired, further secure the setup with a platform (e.g. ring clamp and wire mesh as in Figure 5.58).
3. Add water to the distilling flask to just cover the plant material.
4. Always use a Claisen adapter in the setup as plant materials commonly foam with heating and there is often a great amount of turbulence in the distilling flask (see splashing in the flask in Figure 5.59).
5. Two variations of steam distillation are regularly used: one where steam is directly generated by heating water in the distilling flask with a Bunsen burner (Figure 5.58), and another where steam is indirectly generated by a steam line in the building (Figure 5.59).
Although burners are not commonly used with the distillation of organic compounds, they are relatively safe in steam distillations because: a) the majority of the vapor produced by the solution is typically water rather than organic compounds, so the vapor is not particularly flammable, b) The distillate may be somewhat flammable, but is a distance away from the heat source.
6. To the top of the Claisen adapter, connect one of the following, depending on the variation of distillation used:
1. If water will be heated in the distilling flask along with a burner:
• Attach a stopper to close the system (indicated by the arrow in Figure 5.60) if the quantity of water will be sufficient for the distillation.
• Attach a $125 \: \text{mL}$ separatory funnel (which must have a ground glass joint below the stopcock) if it is anticipated that water will need to be replenished during the distillation. Clamp and partially fill this separatory funnel with water (Figure 5.58).
2. If a steam line will be used:
• Insert a steam inlet tube, and position it so the outlet is just raised above the bottom of the flask. Connect this tube to a separatory funnel trap attached to the steam line (Figure 5.59).
1. The receiving flask can be a graduated cylinder, Erlenmeyer flask, or round bottomed flask.
Begin Distilling
1. Begin heating the flask:
1. If using a Bunsen burner, initially heat the flask at a rapid rate (it takes a lot of energy to boil water), but once the solution is boiling or foaming, turn down the burner and/or wave the burner in order to moderate the distillation rate (Figure 5.61a).
2. If using a steam line, open the stopcock in the separatory funnel steam trap. Turn on the steam, and allow the water in the steam line to drain through the funnel and into a container (Figure 5.61b). Close the stopcock when it appears that most of what is coming out of the steam line is steam and not water. Adjust the steam rate so that the steam bubbles vigorously through the liquid in the distilling flask (Figure 5.61c).
1. Collect distillate at a rate of 1 drop every second. The distillate will contain both water and oil, and may look one of two ways.
1. A second layer may form, or the distillate may appear to have oily droplets clinging to the sides of the flask. This happens when the oil is very nonpolar (e.g. Figure 5.62a+b show the distillate of orange oil, which is almost entirely composed of hydrocarbons).
2. The distillate may also appear cloudy or "milky" (Figure 5.62c shows the cloudy distillate of clove oil). A cloudy mixture forms when tiny insoluble particles are dispersed throughout the suspension. This tends to happen when the oil is water-insoluble but contains some polar functional groups (e.g. large compounds with a few alcohol or ether functional groups).
2. If using a steam line, the steam rate will need to be reduced once the distillation begins.
3. If not use a steam line, be sure to replenish the water in the distilling flask if the level gets too low (if the liquid in the distilling flask is getting thick). Heating a solution which is too concentrated can sometimes cause sugars in the plant material to caramelize, making cleanup difficult. To replenish water either:
1. Temporarily open the stopper on the Claisen adapter and add water via wash bottle.
2. If using a separatory funnel on the Claisen adapter, turn the stopcock to gradually pour in water. If water is drained too quickly so that it pools in the Claisen adapter, it will be difficult to drain into the distilling flask for the same reason that a separatory funnel doesn't drain with its stopper in place. Draining water from the separatory funnel slowly prevents this from happening.
Stop the Distillation
1. If following a procedure, collect the recommended quantity of distillate. If the distillate is milky, the distillation can be ceased when the distillate dripping from the vacuum adapter clarifies, indicating that only water is condensing.
2. If using a steam line, fully drain the liquid in the separatory funnel steam trap (and leave open) before turning off the steam. If water is in the funnel when the steam line is turned off, suction formed by the trap cooling faster than the distilling flask may pull liquid from the distilling flask into the steam trap (sometimes violently).
Isolate the Oil
1. If the oil separates from the distillate as in Figure 5.62a, it can be removed via pipette, dried with a drying agent (such as $\ce{Na_2SO_4}$) and analyzed directly. If milky as in Figure 5.62c, the oil must be extracted into an organic solvent, dried with a drying agent (such as $\ce{MgSO_4}$), and solvent removed with the rotary evaporator. Three extractions are typically necessary to fully extract the oil from a milky distillate into an organic solvent. In the extraction of the milky clove oil distillate from Figure 5.62c, notice how the aqueous layer (bottom) remains milky after the first extraction (Figure 5.63a). This milky appearance shows that the oil was not fully removed with one extraction. The aqueous layer clarifies with the second extraction (Figure 5.63b) demonstrating the importance of multiple extractions.
Steam Distillation Summary
Table 5.11: Procedural summary of steam distillation.
Place large amount of plant material in a large round bottomed flask (no more than half full), and just cover with water.
Always use a Claisen adapter, as there is often turbulence in the flask.
Two variations are common:
• Heat the water with a Bunsen burner to create steam directly.
• Add steam indirectly through a steam line from the building.
Collect the distillate at a rate of 1 drop per second.
The distillate may appear cloudy, or a second layer may form on the top.
If milky, the distillation can be ceased when the distillate is clear.
If a steam line was used, be sure to drain the liquid from the steam trap before turning off the steam, to prevent back suction.
Separated oil can be pipetted (then dried with $\ce{Na_2SO_4}$), while milky distillates need to be extracted. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.05%3A_Steam_Distillation/5.5D%3A_Step-by-Step_Procedures_for_Steam_Distillation.txt |
The preferred method for solvent removal in the laboratory is by use of a rotary evaporator (also known as a "rotovap"). A rotary evaporator is essentially a reduced pressure distillation: a solution in a round bottomed flask is placed in the water bath of the apparatus, and rotated while the system is partially evacuated (by a water aspirator or vacuum pump). The reduced pressure in the apparatus causes the solvent to boil at a lower temperature than normal, and rotating the flask increases the liquid's surface area and thus the rate of evaporation.
• 5.6A: Overview of Rotary Evaporation
It is very common for a desired compound to be dissolved in a solvent during regular manipulations in the laboratory. The preferred method for solvent removal in the laboratory is by use of a rotary evaporator, also known as a "rotovap". A rotary evaporator is essentially a reduced pressure distillation: a solution in a round bottomed flask is placed in the water bath of the apparatus , and rotated while the system is partially evacuated (by a water aspirator or vacuum pump).
• 5.6B: Step-by-Step Procedures for Rotary Evaporation
• 5.6C: Troubleshooting Rotary Evaporation
If the solvent fails to boil on the rotary evaporator even after one minute, consider whether a mistake may have been made. For example, it may be that you are trying to evaporate the aqueous layer from a separatory funnel step instead of the organic layer. Alternatively, the water bath may need to be heated to a higher temperature depending on the boiling point of the solvent.
Contributors and Attributions
• Lisa Nichols (Butte Community College). Organic Chemistry Laboratory Techniques is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Complete text is available online.
5.06: Rotary Evaporation
It is very common for a desired compound to be dissolved in a solvent during regular manipulations in the laboratory. Solvents are used in separatory funnel extractions and column chromatography, and the solvent must be removed in order to isolate te desired compound. Solvents are regularly chosen that have lower boiling points than the compound of interest, so that there is some mechanism for their removal. In theory, a solution could simply be placed on a heat source to boil away the lower-boiling solvent, but this approach is not often used.
The preferred method for solvent removal in the laboratory is by use of a rotary evaporator (Figure 5.65), also known as a "rotovap". A rotary evaporator is essentially a reduced pressure distillation: a solution in a round bottomed flask is placed in the water bath of the apparatus (Figure 5.66a), and rotated while the system is partially evacuated (by a water aspirator or vacuum pump). The reduced pressure in the apparatus causes the solvent to boil at a lower temperature than normal (see vacuum distillation), and rotating the flask increases the liquid's surface area and thus the rate of evaporation. The solvent vapor condenses when it comes into contact with a water condenser (Figure 5.65) and drips into a receiving flask (Figure 5.66b). When the solvent is removed, the concentrated compound is left in the flask. One difference between distillation and rotary evaporation is that the distillate is most often retained in distillation while the residue is retained in rotary evaporation.
Removal of solvent by a rotary evaporator is superior to evaporation under atmospheric pressure for many reasons. The process is much quicker (often takes less than 5 minutes), uses lower temperatures (so decomposition is unlikely), and uses less energy than boiling with a heat source. Since low pressure is used, a rotary evaporator is also quite efficient at removing the last traces of residual solvent from a solution. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.06%3A_Rotary_Evaporation/5.6A%3A_Overview_of_Rotary_Evaporation.txt |
Rotary Evaporation Procedure
1. If your institution uses a water circulator connected to the condenser to minimize water usage (open cabinet of Figure 5.65), be sure that there is ice in the water reservoir.
2. Pre-weigh a round bottomed flask, and fill it no more than half-full with the solution to be evaporated (Figure 5.67a).
3. Connect the flask to the evaporator's "bump trap" using a plastic clip ("Keck clip", Figure 5.67b).
The bump trap prevents foaming or splashing solutions ("bumping") from dirtying the condenser or collecting in the receiving flask where components cannot be recovered. A bump trap should be kept clean so that if solutions do collect there, they can be rinsed back into the original flask without worry about contamination from previously evaporated samples.
4. Use the joystick knob on the apparatus to lower the flask into the water (Figure 5.67c) so that the flask is partially submerged. Be sure the flask is not positioned so low that the joint with the plastic clip is in the water.
1. Turn on the vacuum source (Figure 5.69a), which may be a water aspirator or vacuum pump. A hissing sound should be heard from air being pulled through the stopcock. The partial vacuum will help hold the flask securely onto the bump trap.
2. Begin rotating the flask at a medium rate by adjusting the rotation notch (to roughly $110 \: \text{rpm}$, or to one-third of the maximum rotation value, Figure 5.69b).
3. Close the stopcock on the evaporator by turning it perpendicular to the bleed valve (Figure 5.69c). The hissing sound should stop, and the pressure inside the apparatus will decrease further.
1. Allow the solution to evaporate. Solvent should collect in the large round bottomed flask reservoir (indicated in Figure 5.70a). There is no set amount of time required for complete evaporation.
1. If the expected compound is a solid, keep evaporating until a solid or thin film appears (Figure 5.70b). A film may form if the temperature of the bath is above the solid's melting point, if crystallization is slow, or if the warmth of the bath prevents crystallization (Figure 5.71b shows a film that crystallized into Figure 5.71c after a day).
2. If the expected compound is a liquid, evaporate until the approximate expected volume is seen and it appears that the liquid level is no longer changing (Figure 5.71a). Evaporation can also be tested for by lifting the flask out of the water bath, drying the outside with paper towels, and feeling the flask with your hand as it rotates. If solvent continues to evaporate, the flask will feel cool.
3. If solvent bumps from the flask into the bump trap (Figure 5.67b), raise the flask slightly out of the water bath, and/or allow small amounts of air into the apparatus by partially turning the stopcock (Figure 5.69c). If large amounts of liquid have bumped, stop the evaporation and rinse the trap with solvent such that the rinsing is added back to the flask. Evaporate the solvent.
4. It is acceptable if solvent pools in the bump trap during normal evaporation (not bumping). It will sometimes later vaporize, and if it doesn't, it's still been removed from the flask. Pooling often happens if the solvent has a relatively high boiling point.
2. When it appears that the sample has completed evaporating, allow it to remain in the reduced pressure system an extra few minutes to remove any final solvent residue.
1. To stop the evaporation, reverse all the previous steps: open the stopcock, stop the rotation, turn off the vacuum source, lift the flask from the water bath, and remove the flask. Turn the evaporator off completely if you are the final user of the day.
2. The residue in the round bottomed flask should contain the desired compound, but may still contain trace amounts of solvent. If the compound is not particularly volatile (or if it's a non-volatile solid), the flask can be left open in a locker until a future lab session. After fully dry, the mass can be obtained and the compound analyzed.
Rotary Evaporation Summary
Table 5.12: Procedural summary of rotary evaporation.
Be sure there is ice in the water circulator (if used).
Fill a round bottomed flask no greater than half-full.
Connect to the bump trap with a plastic clip.
Lower the flask into the water bath to submerge the liquid (don't submerge the plastic clip).
Turn on the vacuum source (vacuum will hiss).
Rotate the flask at a moderate rate (one-third the maximum value).
Close the stopcock in the apparatus (hissing will stop).
Evaporate until solid forms or liquid level doesn't appear to change anymore, then evaporate an extra few minutes for good measure.
To stop evaporation, reverse all steps:
• Open the stopcock
• Stop the rotation
• Turn off the vacuum
• Lift the flask from the water bath
• Remove the flask
5.6C: Troubleshooting Rotary Evaporation
1. If the solvent fails to boil on the rotary evaporator even after one minute, consider whether a mistake may have been made. For example, it may be that you are trying to evaporate the aqueous layer from a separatory funnel step instead of the organic layer. Alternatively, the water bath may need to be heated to a higher temperature depending on the boiling point of the solvent.
2. If droplets are seen among the liquid residue in the flask after evaporation (Figure 5.73), most likely water was not effectively removed from the sample prior to evaporation. To remedy this problem, dissolve the sample is a portion of the previously used solvent, use a drying agent (e.g. anhydrous \(\ce{MgSO_4}\)), filter if necessary, and again evaporate the solvent. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.06%3A_Rotary_Evaporation/5.6B%3A_Step-by-Step_Procedures_for_Rotary_Evaporation.txt |
• 6.1: Melting Point
Measurement of a solid compound's melting point is a standard practice in the organic chemistry laboratory. The melting point is the temperature where the solid-liquid phase change occurs
• 6.2: Boiling Point
A compound's "normal boiling point" refers to its boiling point at a pressure of 760 mmHg. A compound's boiling point is a physical constant just like melting point, and so can be used to support the identification of a compound.
• 6.3: Sublimation
Some compounds are capable of sublimation, which is the direct phase change from solid to gas. Sublimation is an analogous process to boiling, as it occurs when a compound's vapor pressure equals its applied pressure (often the atmospheric pressure). Most solids do not have an appreciable vapor pressure at easily accessible temperatures, and for this reason the ability to sublime is uncommon. Compounds that are capable of sublimation tend to be those with weak intermolecular forces.
• 6.4: Chemical Tests
Before spectroscopic analysis (IR, NMR) became commonplace in the organic chemistry lab, chemical tests were heavily relied upon to support compound identification. A chemical test is typically a fast reaction performed in a test tube that gives a dramatic visual clue (a color change, precipitate, or gas formation) as evidence for a chemical reaction.
06: Miscellaneous Techniques
Measurement of a solid compound's melting point is a standard practice in the organic chemistry laboratory. The melting point is the temperature where the solid-liquid phase change occurs
• 6.1A: Overview of Melting Point
Measurement of a solid compound's melting point is a standard practice in the organic chemistry laboratory. The melting point is the temperature where the solid-liquid phase change occurs. In some reference books it is listed as a single value (e.g. 98˚C), but in chemical catalogs it is more often listed as a range of values (e.g. 96-98˚C). The melting "point" is therefore more of a melting "range," and in part, reflects how melting points are experimentally determined.
• 6.1B: Uses of Melting Points
There are several reasons to determine a compound's melting point: it is useful in supporting the identification of a compound, as well as serving as a rough guide to the relative purity of the sample.
• 6.1C: Melting Point Theory
• 6.1D: Step-by-Step Procedures for Melting Point Determination
There are a variety of methods by which a sample's melting point can be measured, with the newest being electrical probes (e.g. Vernier MeltStation). Presented in this section are traditional methods that use an electrical melting point apparatus and Thiele tube. Both methods use capillary samples that are prepared in the same manner.
• 6.1E: Mixed Melting Points
As previously discussed, there are a large number of compounds that have coincidentally identical melting points. Therefore, caution should be used in identifying a compound based solely on matching the literature melting point. However, mixed melting points offer an ability to almost certainly identify an unknown compound.
6.01: Melting Point
Measurement of a solid compound's melting point is a standard practice in the organic chemistry laboratory. The melting point is the temperature where the solid-liquid phase change occurs. In some reference books it is listed as a single value (e.g. $98^\text{o} \text{C}$), but in chemical catalogs it is more often listed as a range of values (e.g. $96$-$98^\text{o} \text{C}$). The melting "point" is therefore more of a melting "range," and in part, reflects how melting points are experimentally determined.
A melting point is determined by loading a small amount of sample into a capillary tube (Figure 6.1), and then slowly heating the sample. Figure 6.2 shows a close-up view of a sample inside a melting point apparatus, where the sample is slowly heated through contact with hot vertical metal blocks on either side of the capillary tube. The sample is kept small in this technique to ensure adequate heat transfer between the metal and sample.
The first value recorded for the melting range is with the very first appearance of liquid. As this temperature is approached, the solid may begin to glisten (Figure 6.2b), and the temperature is recorded with the first hint of liquid movement (a droplet) inside the tube (Figure 6.2c). The second value recorded for the melting range is with the melting of the entire sample, which occurs when all areas of opaque solid have turned into a transparent liquid (Figure 6.2h). | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.01%3A_Melting_Point/6.1A%3A_Overview_of_Melting_Point.txt |
There are several reasons to determine a compound's melting point: it is useful in supporting the identification of a compound, as well as serving as a rough guide to the relative purity of the sample.
Identification
As a compound's melting point is a physical constant, it can be used to support the identity of an unknown solid. The melting point can be looked up in a reference book (this value would then be called the "literature melting point"), and compared to the experimental melting point. For example, the literature melting point of ferrocene, is $172$-$174^\text{o} \text{C}$.$^1$ The author found the melting point of a ferrocene sample (Figure 6.3) to be $176$-$178^\text{o} \text{C}$,$^2$ and there is good agreement between these two values.
Care must be taken to refrain from jumping to conclusions about the identity of a compound based solely on a melting point. Millions of solid organic compounds exist, and most have melting points below $250^\text{o} \text{C}$. It is not uncommon for two different compounds to have coincidentally similar or identical melting points. Therefore, a melting point should be used as simply one piece of data to support the identification of an unknown.
Although coincidentally similar melting points are not unheard of, when used in the context of assessing the product of a chemical reaction, melting points can be a powerful identification tool. For example, three possible products of the nitration of benzaldehyde are 2, 3, or 4-nitrobenzaldehyde (Figure 6.4). Since these products have very different melting points, the melting point of the resulting solid (if pure) could be used to strongly suggest which product was formed.
Assessing Purity
A second reason to determine a compound's melting point is for a rough measure of purity. In general, impurities lower and broaden the melting range.
For example, the melting points of samples of benzoic acid contaminated with known quantities of acetanilide are summarized in Table 6.1. As the quantity of impurity increased, melting began at a lower temperature, and the breadth of the melting range increased.
Table 6.1: Melting points of benzoic acid/acetanilide mixtures (taken with a MelTemp apparatus).
Mol % Benzoic Acid Mol % Acetanilide Melting Point (ºC)
100% 0% 120 - 122
95% 5% 114 - 121
90% 10% 109 - 120
85% 15% 105 - 117
80% 20% 94 - 116
Figure 6.5 shows the time-lapse melting of three samples side by side in a melting point apparatus: pure benzoic acid (left), benzoic acid with $10 \: \text{mol} \%$ acetanilide impurity (middle), and benzoic acid with $20 \: \text{mol} \%$ acetanilide impurity (right). As the samples are heated, the sample with the greatest impurity (on the right) melts first. Interestingly, both of the impure samples complete melting before the pure sample (on the left) begins to melt.
A solid's melting point may be so reduced by impurity that it becomes a liquid at room temperature. For example, when piperonal (melting point of $35$-$39^\text{o} \text{C}$) is mixed with resorcinol (melting point of $109$-$112^\text{o} \text{C}$) in a 4:1 ratio by mass,$^3$ the mixture become slushy and eventually melts at room temperature (Figure 6.6). This sort of behavior is not uncommon for solids whose melting points are only marginally higher than room temperature.
$^1$Melting points are from the Aldrich Chemical Catalog.
$^2$As determined using a MelTemp melting point apparatus. The temperature values are uncorrected.
$^3$As published in Di Pippo, A. G., J. Chem. Ed, 1965, 42(5), p A413. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.01%3A_Melting_Point/6.1B%3A_Uses_of_Melting_Points.txt |
Melting Point Diagrams
The typical behavior of an impure solid containing two components is summarized by the general phase diagram in Figure 6.7a. The furthest left side of the graph represents a sample that is pure compound "A," while the furthest right side of the graph represents a sample of pure compound "B." The lines mark the solid-liquid transition temperature (melting points). The melting point decreases the further the composition is from purity, toward the middle of the graph. In many mixtures, the minimum melting temperature for a mixture occurs at a certain composition of components, and is called the eutectic point (Figure 6.7a). Some systems do not have any eutectic points and some have multiple eutectic points.
An impure solid is typically heterogeneous on the microscopic level, with pure regions of each component distributed through the bulk solid much like granite. When an impure solid is warmed, microscopic melting first occurs in a pure region by the component with the lower melting point (compound A in Figure 6.7a). This microscopic melting is not visible to the eye.
The preliminary melting of compound A in Figure 6.7a forms tiny pools of liquid that begin to dissolve compound B from the bulk solid. As compound B is dissolved into the melt (causing it to become more impure), the freezing point of this mixture is depressed. Compound B will continue to dissolve in the melt, until it reaches the eutectic composition (point a in Figure 6.7b), and the system will continue to melt at this composition until the entirety of the minor component (the impurity) is dissolved. Once the minor component is completely dissolved, further melting continues of the bulk component. This increases the purity of the melt, so the melting temperature increases somewhat. The system follows the melting line in Figure 6.7b either to the left or right of the eutectic temperature (depending on which side of the eutectic point is started), adjusting its melting temperature as the bulk component increases its concentration in the melt. This continues until the entire sample is melted.
Although microscopic melting begins at the eutectic temperature, the first value of the melting range (when a droplet of liquid is seen with the eye) is not necessarily recorded at this temperature. A droplet of liquid is not seen until approximately $10$-$20\%$ of the sample has melted. Depending on the quantity of impurity, the system may have progressed far from the eutectic temperature (perhaps to point b in Figure 6.7b) before liquid becomes visible to the eye. The final value of the melting range is at the highest the melting point of the pure solid, but is often lower, reflecting the depressed melting point of the bulk solid. For example, a solid that is $20\%$ compound A and $80\%$ compound B would have a final melting temperature of point c in Figure 6.7b. The recorded melting range for this system would be at the maximum between temperatures a and c, but if the first droplet is seen at point b, the recorded melting range would be between temperatures b and c.
Impurities Effect on the Melting Point
A melting point is a useful indicator of purity as there is a general lowering and broadening of the melting range as impurities increase. In this section is described the theory behind the phenomenon of melting point depression (which is identical to freezing point depression since freezing and melting are the same processes in reverse) and why an impure sample has a broad melting range.
Melting Point Depression (Lowering the M. P.)
Melting of a pure solid occurs at a higher temperature than melting of an impure solid, a concept called melting point depression (or freezing point depression). The melting point is the temperature where the solid and liquid phases are in equilibrium with each other, and the change in free energy $\left( \Delta G^\text{o} \right)$ for the process (solid $\rightleftharpoons$ liquid) is zero. $\Delta G^\text{o}$ is dependent on both the changes in enthalpy $\left( \Delta H^\text{o} \right)$ and entropy $\left( \Delta S^\text{o} \right)$ during the process (see versions of the Gibbs free energy equation in Figure 6.8b), but the changes in enthalpy are similar when melting a pure and impure solid as similar intermolecular forces are broken. Melting point depression is the result of different changes in entropy when melting a pure and impure solid.
As solids are restricted in atomic motion, there is little difference in entropy between a pure and impure solid. However, there is a more significant difference in entropy between a pure and impure liquid, and an impure liquid has greater disorder and greater entropy. Melting of an impure solid into an impure liquid therefore has a larger change in entropy than melting a pure solid into a pure liquid (Figure 6.8a). A larger change in entropy corresponds to a lower melting temperature. This can be rationalized either mathematically or conceptually. A mathematical description is in Figure 6.8b: as $\Delta S^\text{o}$ is the denominator in the final equation, a larger $\Delta S^\text{o}$ corresponds to a smaller $T_\text{melting}$. A conceptual approach is to consider that melting occurs when the enthalpy $\left( \Delta H^\text{o} \right)$ and entropy components $\left( T \Delta S^\text{o} \right)$ are equal in magnitude (when $\Delta G^\text{o} = 0$). A larger $\Delta S^\text{o}$ means that a smaller temperature will be required to "match" the enthalpy component.
Broadening of the Melting Point
The breadth of an experimentally determined melting point can often be correlated to the purity of the solid. Although all samples start melting at the eutectic temperature, the first droplet of liquid is not seen until approximately $10$-$20\%$ of the sample has microscopically melted. As the melting temperature does not rise above the eutectic temperature until the entirety of the impurity has melted, the quantity of impurity will determine how far the system will have progressed along the melting point line in the phase diagram before reaching the visible minimum of $10$-$20\%$ of solid.
For example, if a solid has a minor amount of impurity, the impurity will quickly melt at the eutectic temperature (point a in Figure 6.9a), and the melting temperature will increase, following the melting point line in the phase diagram. When $10$-$20\%$ of solid has melted and a droplet is visible, the system may have progressed far from the eutectic composition (perhaps to begin visibly melting at point b in Figure 6.9a). The solid will continue melting until perhaps point c in Figure 6.9a, to give a relatively narrow melting range (between points b and c). If instead the solid has a significant amount of impurity, it may take melting of nearly $10\%$ of the solid to fully dissolve the impurity, which means the melting temperature may not have progressed far from the eutectic temperature when a droplet becomes visible. A more impure solid may first visibly melt at perhaps point d in Figure 6.9b, to give a broader melting range (between points d and e).
It is for these reasons that a low melting range $\left( < 2^\text{o} \text{C} \right)$ is associated with purity, although it is also possible that the solid's composition could be coincidentally near a eutectic point. If the eutectic composition is, for example, $40\%$ A/$60\%$ B, and the solid's composition is $45\%$ A/$55\%$ B, nearly all of the impure solid will melt before the melting temperature will change from the eutectic temperature in the phase diagram. Therefore, mixtures with compositions near the eutectic composition also give a sharp melting range, even though they may be far from pure.
Whether a system is in fact pure, or sharply melting because it is at the eutectic composition, can be proven by performing a mixed melting point. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.01%3A_Melting_Point/6.1C%3A__Melting_Point_Theory.txt |
There are a variety of methods by which a sample's melting point can be measured, with the newest being electrical probes (e.g. Vernier MeltStation). Presented in this section are traditional methods that use an electrical melting point apparatus and Thiele tube. Both methods use capillary samples that are prepared in the same manner.
Sample Preparation
1. Obtain a glass capillary melting point tube, which has one end sealed and the other end open. Jab the open end of the tube into a pile of the solid to be analyzed (Figure 6.10a). The solid must be dry or the results will be affected as solvent can act as an impurity and affect the melting range. If the solid is granular, pulverize the solid somewhat before packing.
2. Invert the capillary tube and gently tap the tube on the benchtop to cause the solid to fall to the closed end (Figure 6.10b). Then, drop the capillary tube closed side down several times through a long narrow tube (glass tube or cut PVC pipe, Figure 6.10 c). The capillary tube will bounce as it hits the benchtop, and pack the solid into the bottom of the tube. Failure to pack the solid well may cause it to shrink when heating, which can cause confusion as to the correct melting temperature.
3. If needed, repeat the previous steps to load sample until it is a height of $2$-$3 \: \text{mm}$ in the tube (when packed, Figure 6.10d). It is important that the sample be no higher than $3 \: \text{mm}$ or the melting range will be artificially broad.
Melting Point Apparatus
1. Insert the capillary tube containing the sample into a slot behind the viewfinder of a melting point apparatus (Figure 6.11a). There are usually three slots in each apparatus, and multiple melting points can be taken simultaneously after gaining experience with the technique.
2. Turn on the apparatus and adjust the setting to an appropriate heating rate (Figure 6.11b). The rate of heating is often experimental and should be adjusted by careful monitoring of the thermometer on the apparatus.
3. Look through the viewfinder (Figure 6.11c) to see a magnified view of the sample in the apparatus, which should be illuminated.
4. If the expected melting point of the compound is known, heat at a medium rate to $20^\text{o} \text{C}$ below the expected melting point, then slow the rate of heating such that the temperature increases no more than $1^\text{o} \text{C}$ every 30 seconds (i.e., very slowly).
The temperature must be incremental as the melting point is approached so the system can reach equilibrium, making the thermometer temperature an accurate gauge of the solid's true temperature.
5. If the expected melting point of the compound is NOT known, heat the sample at a medium rate the entire time and determine an approximate melting point. Repeat the process with a fresh sample after allowing the apparatus to cool and use the recommendations in prompt 4 to perform a more careful assessment of the melting point.
A fresh sample is necessary for a second melting point trial; even if the first sample solidifies after cooling it should not be used again. Differences in crystal structure between the original solid and the previously melted solid could lead to different melting ranges.
1. The solid may be approaching its melting point if the solid is seen pulling away from the walls of the tube to form a cone of solid (Figure 6.12b), which is called "sintering." Melting will normally occur within a few degrees of this point. The solid may also shrink or compact before melting.
1. Record the first temperature of the melting range with the appearance of the first visible drop of liquid. At first it may seem as if the sides of the solid glisten (Figure 6.13b), and the temperature should be recorded when a droplet is seen on the side or bottom of the tube (a hint of movement will be noticed in the tube, Figure 6.13c).
Record the temperature reading to the nearest degree. Although some thermometers may read to greater precision, the imperfect heat transfer between the metal block and sample leaves the error larger than $0.1^\text{o} \text{C}$.
2. Record the second temperature of the melting range when the entire sample has just melted (Figure 6.13d), which occurs when all portions of the opaque solid have turned to a transparent liquid.
3. The following unusual situations may occur in the process:
1. The sample may begin to darken, which indicates decomposition is occurring before the sample is melting. Take note of the decomposition temperature, as it is sometimes as reliable a reference point as a compound's melting point. Use the letter "d" after a melting point to indicate decomposition (e.g. $251^\text{o} \text{C}$ d).
2. The sample may sublime instead of melting. Sublimation may be noticed by a ring of solid above where the sample is heated. Take note of this behavior in your lab notebook.
4. If another melting point trial is to be performed directly after the first, the metal block should be rapidly cooled to at least $20^\text{o} \text{C}$ below the next melting point by touching it with wet paper towels (Figure 6.11d) or cooling it with a jet of air.
Thiele Tube Method
1. Obtain a Thiele tube and clamp it to a ring stand or latticework (Figure 6.15b). The tube is normally filled with clear mineral oil, but it may have darkened from oxidation or spilled compounds. If the oil is quite dark, it should be replaced. The oil should be filled to at least $1 \: \text{cm}$ higher than the top triangular arm (an appropriate oil level is pointed to in Figure 6.15c), and if too low the oil will not circulate as needed (Figure 6.15d).
2. Insert a thermometer into a one-holed rubber stopper with a slit down one side. Attach the capillary sample to the thermometer with a tiny rubber band (as indicated in Figure 6.15a). These tiny rubber bands are often made by cutting pieces of small rubber tubing.
3. Position the capillary tube so that the solid sample is lined up with the middle of the thermometer bulb (Figure 6.15a).
4. Place the rubber stopper and thermometer assembly into the Thiele tube, adjusting the height so that the sample is midway inside the tube (Figure 6.15c). The rubber band should be adjusted so it is not submerged in the mineral oil, keeping in mind that the oil may expand somewhat during heating. The thermometer should not touch the sides of the glass, and if it does it should be clamped in such a way that it no longer touches.
5. Heat the apparatus gently on the side arm of the Thiele tube with a microburner if available or Bunsen burner using a back and forth motion (Figure 6.15d). As the oil warms and becomes less dense, it will rise and travel up the triangular portion of the tube. The cooler, denser oil will sink, thereby creating an oil current as shown in Figure 6.15d. This method is an excellent way to indirectly and slowly heat the sample.
6. Although bubbles should not be seen in the Thiele tube as it warms, they commonly are seen if the tube is used for other purposes (bubbles are seen in Figure 6.15d). For example, Thiele tubes can be used for boiling point determinations, and on occasion a sample falls into the oil and contaminates it. If the oil is not subsequently changed, the sample may boil when heated in the tube. If bubbles are seen upon heating a Thiele tube, the entire setup should be conducted in the fume hood.
7. If the expected melting point of the compound is known, heat at a medium rate to $20^\text{o} \text{C}$ below the expected melting point, then slow the rate of heating such that the temperature increases no more than $1^\text{o} \text{C}$ every 30 seconds (i.e., very slowly).
The temperature must be incremental as the melting point is approached so the system can reach equilibrium, making the thermometer temperature an accurate gauge of the solid's true temperature.
8. If the expected melting point of the compound is NOT known, heat the sample at a medium rate the entire time and determine an approximate melting point. Repeat the process with a fresh sample after allowing the oil to cool to at least $20^\text{o} \text{C}$ below the previous melting point, and use the recommendations in prompt 7 to perform a more careful assessment of the melting point.
A fresh sample is necessary for a second melting point trial; even if the first sample solidifies after cooling it should not be used again. Differences in crystal structure between the original solid and the previously melted solid could lead to different melting ranges.
1. Record the first temperature of the melting range with the appearance of the first visible drop of liquid. At first it may seem as if the sides of the solid glisten (Figure 6.13b), and the temperature should be recorded when a droplet is seen on the side or bottom of the tube (a hint of movement will be noticed in the tube, Figure 6.16b).
Record the temperature reading to the nearest degree. Although some thermometers may read to greater precision, the imperfect heat transfer between the oil and sample leaves the error larger than $0.1^\text{o} \text{C}$.
2. Record the second temperature of the melting range when the entire sample has just melted (Figure 6.16d), which occurs when all portions of the opaque solid have turned to a transparent liquid.
3. Take note if darkening or sublimation occur.
4. If another melting point trial is to be performed directly after the first, be sure to allow the oil to cool to at least $20^\text{o} \text{C}$ below the next melting point beforehand.
5. Cleanup note: drops of mineral oil on the benchtop do not easily wipe away, but can be cleaned by a paper towel soaked with either window cleaner or hexanes.
Melting Point Summary
Table 6.2: Procedural summary for obtaining a melting point.
Load the sample by jabbing the open end of a capillary tube into a pile of the sample.
With closed end down, drop the tube down a long hollow tube so that it hits the benchtop and packs the sample into the closed end of the tube.
Load the sample to a height of $2$-$3 \: \text{mm}$.
Place the sample into a slot in the MelTemp.
Turn the dial to begin heating.
Heat at a medium rate to $20^\text{o} \text{C}$ below the expected melting point.
Then heat very slowly ($1^\text{o} \text{C}$ every 30 seconds).
Record the temperature where the first droplet of liquid is seen (there is movement in the tube).
Record the second temperature when the entire sample liquefies (the entire sample changes from opaque to transparent).
Record a melting range, e.g. $120$-$122^\text{o} \text{C}$.
If another melting point trial is to be performed, cool the metal block to at least $20^\text{o} \text{C}$ below the next melting point, by wiping it with a wet paper towel or cooling with a jet of air.
Thiele Tube Variation:
Attach the sample to a thermometer with a tiny rubber band, positioning the sample flush with the bottom of the thermometer.
Insert the sample into a Thiele tube, so that the sample is near the middle of the tube.
6.1E: Mixed Melting Points
As previously discussed, there are a large number of compounds that have coincidentally identical melting points. Therefore, caution should be used in identifying a compound based solely on matching the literature melting point. However, mixed melting points offer an ability to almost certainly identify an unknown compound.
Imagine that the nitration of benzaldehyde (Figure 6.17), produces a solid that is determined to have a melting point of $54$-$57^\text{o} \text{C}$. This solid would be assumed to be 3-nitrobenzaldehyde due to the proximity of the experimental melting point to the literature melting point.
Although the product likely is as identified, if a pure sample of 3-nitrobenzaldehyde is available, there is a possibility of more strongly identifying the product. A mixed melting point can be taken, by measuring the melting point of a sample composed of roughly equal volumes of the unknown product and of known 3-nitrobenzaldehyde (ground together well with a mortar and pestle, as in Figure 6.18a). If the product is indeed 3-nitrobenzaldehyde, then this "mixture" would not be a mixture at all. Its melting point would be sharp and around the literature range of $55$-$58^\text{o} \text{C}$. If this result occurs, the two samples are almost certainly the same compound. If the product however is not 3-nitrobenzaldehyde, then this "mixture" would truly be very impure ($50\%$ of each component), and the resulting melting point would have a much lowered and broadened range. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.01%3A_Melting_Point/6.1D%3A__Step-by-Step_Procedures_for_Melting_Point_Determination.txt |
The boiling point of a compound is the temperature where the liquid-gas phase change occurs. In more technical terms, it is when a liquid's vapor pressure equals its applied pressure (typically the atmospheric pressure). Boiling points are very sensitive to changes in applied pressure, so all boiling points should be reported along with the measured pressure. A compound's "normal boiling point" refers to its boiling point at a pressure of 760 mmHg. A compound's boiling point is a physical constant just like melting point, and so can be used to support the identification of a compound.
• 6.2A: Overview of Boiling Point
A compound's boiling point is a physical constant just like melting point, and so can be used to support the identification of a compound. Unlike melting points however, boiling points are not generally used as a gauge of purity. Impure liquids do boil over a range of temperatures (similar to how melting points have breadth), but the temperature span does not correlate well to purity. Thus, measurement of a compound's boiling point is mainly used to support its identification.
• 6.2B: Step-by-Step Procedures for Boiling Point Determination
There are a variety of methods by which a sample's boiling point can be determined, including distillation, reflux, and by using a Thiele tube. The most straightforward method uses a Thiele tube, and has the advantage of using less than
6.02: Boiling Point
The boiling point of a compound is the temperature where the liquid-gas phase change occurs. In more technical terms, it is when a liquid's vapor pressure equals its applied pressure (typically the atmospheric pressure). Boiling points are very sensitive to changes in applied pressure, so all boiling points should be reported with the measured pressure. A compound's "normal boiling point" refers to its boiling point at a pressure of $760 \: \text{mm} \: \ce{Hg}$.
A compound's boiling point is a physical constant just like melting point, and so can be used to support the identification of a compound. Unlike melting points however, boiling points are not generally used as a gauge of purity. Impure liquids do boil over a range of temperatures (similar to how melting points have breadth), but the temperature span does not correlate well to purity. Thus, measurement of a compound's boiling point is mainly used to support its identification.
An experimental boiling point is often compared to the literature boiling point, which are typically reported for 1 atmosphere of pressure. If a boiling point is determined at any pressure significantly different than 1 atmosphere, the pressure should be corrected. A general rule of thumb is that for pressures within $10\%$ of one atmosphere, a $10 \: \text{mm} \: \ce{Hg}$ drop in pressure will account for a $0.3$-$0.5^\text{o} \text{C}$ drop in boiling point.$^4$ Another rule of thumb is that for every halving of pressure, the boiling point drops by about $10^\text{o} \text{C}$.
$^4$This generality is based on tables in the Handbook of Chemistry and Physics, CRC Press, 84$^\text{th}$ edition, 2003-2004, 15-19.
6.2B: Step-by-Step Procedures for Boiling Point Determination
There are a variety of methods by which a sample's boiling point can be determined, including distillation, reflux, and by using a Thiele tube. The most straightforward method uses a Thiele tube, and has the advantage of using less than $0.5 \: \text{mL}$ of material.
Distillation Method
There are simpler methods than a distillation to measure a compound's boiling point, and it is recommended to explore other options (e.g. Thiele tube) if this is the only goal. However, if materials are limited, or if a purification is planned anyhow, a distillation can be used to determine a compound's boiling point. The distillation technique is discussed in great detail in Chapter 5.
A simple distillation should suffice for most situations (Figure 6.19), and at least $5 \: \text{mL}$ of sample should be used in the distilling flask along with a few boiling stones or stir bar. As the bulk of the material distills, the highest temperature noted on the thermometer corresponds to the boiling point. A major source of error with this method is recording too low a temperature, before hot vapors fully immerse the thermometer bulb.$^5$ Be sure to monitor the thermometer periodically, especially when the distillation is active. Record the barometric pressure along with the boiling point.
Reflux Method
A reflux setup can also be used to determine a compound's boiling point. Reflux is when a liquid is actively boiling and condensing, with the condensed liquid returning to the original flask. It is analogous to a distillation setup, with the main difference being the vertical placement of the condenser.
If materials are available, the best reflux setup for this application is shown in Figure 6.20b and uses a microscale condenser and digital thermometer. The setup uses $5 \: \text{mL}$ of liquid, and a few boiling stones or stir bar. The condenser is attached to the round bottomed flask, with the lower water hose connected to the water spigot and the upper water hose draining to the sink. It is important to check that the joint connecting the flask and condenser is securely fastened. The liquid is brought to a boil on a sand bath, and the thermometer is placed low into the apparatus (Figure 6.20c) such that the bottom inch is between the boiling liquid and the bottom of the condenser. In this position, the thermometer can accurately measure the hot vapors and the temperature will stabilize at the compound's boiling point.$^6$ Record the barometric pressure along with the boiling point.
Although it might seem prudent to plunge the thermometer directly into the boiling liquid, it is possible the liquid may be superheated, or hotter than its boiling point. After determining the boiling point, the flask should be raised from the sand bath (Figure 6.20d) to cool, and condenser kept running until the flask is only warm to the touch. At this point the setup can be dismantled.
If a microscale condenser is not available, an alternative reflux method can also be used as shown in Figure 6.21. Roughly $5 \: \text{mL}$ of sample is placed in a medium test tube ($18$ x $150 \: \text{mm}$) with thermometer clamped inside so it does not touch the sides of the glass. The apparatus is carefully heated on a sand bath such that reflux happens controllably and vapors do not escape from the tube. The temperature during reflux will eventually stabilize (this takes some time), and the highest temperature noted corresponds to the compound's boiling point.$^6$ The boiling points measured with this method may have significant error if the boiling point is very low or high ($< 70^\text{o} \text{C}$ or $> 150^\text{o} \text{C}$) as low boiling compounds boil away too easily and high boiling compounds tend to cool too easily.
Thiele Tube Method
Thiele Tube Theory
The Thiele tube method is one of the simplest methods to determine a compound's boiling point, and has the advantage of using small amounts of material (less than $0.5 \:\text{mL}$ of sample). The sample is placed in a small tube along with an inverted capillary tube. The setup is attached to a thermometer (Figure 6.23a) and heated inside a Thiele tube (Figure 6.22a) to slightly higher than the compound's boiling point (which is evidenced by a continuous stream of bubbles emerging from the capillary tube). The tube is then allowed to cool, and the moment liquid is drawn into the capillary tube, the temperature is the compound's boiling point.
This method utilizes the definition of boiling point: the temperature where the compound's vapor pressure equals the applied (atmospheric) pressure. The inverted capillary tube acts as a reservoir to trap the compound's vapors. As the apparatus is heated, the air initially trapped in the capillary tube expands and causes bubbles to emerge from the tube (Figure 6.23b). With further heating, the compound's vapors eventually displace all of the trapped air, which is why heat is applied until there is a continuous stream of bubbles.
When the apparatus is cooled, eventually the pressure inside the capillary tube (due solely to the compound's vapors) will match the atmospheric pressure, at which point the bubbles will slow and liquid will be drawn into the tube. The temperature where this begins is the compound's boiling point (Figure 6.23d).
Thiele Tube Procedure
1. Obtain a Thiele tube and clamp it to a ring stand in the fume hood (Figure 6.24a). The tube is normally filled with clear mineral oil, but it may have darkened from oxidation or spilled compounds. If the oil is quite dark, it should be replaced. The oil should be filled to at least $1 \: \text{cm}$ higher than the top triangular arm (an appropriate oil level is indicated in Figure 6.24a), and if too low the oil will not circulate as needed (Figure 6.25c).
2. Insert a thermometer into a one-holed rubber stopper with a slit down one side. Attach a small glass vial ("Durham tube", or $6$ x $50 \: \text{mm}$ culture tube) to the thermometer with a small rubber band (Figure 6.24b). The bottom of the vial should be flush with the bottom of the thermometer.
3. Fill the vial about half-full with sample, which will require between $0.25$-$0.5 \: \text{mL}$ of sample (Figure 6.24c).
4. Insert a capillary tube into the sample (the same type that is used for melting points), open end down and sealed end up (Figure 6.24d).
1. Place the rubber stopper and thermometer assembly into the Thiele tube, adjusting the height so that the sample is midway (if possible) inside the tube (Figure 6.25a). The rubber band should be higher than the top of the mineral oil (Figure 6.25b), keeping in mind that the oil may expand somewhat during heating. The thermometer should not touch the sides of the glass, and if it does it should be clamped in such a way that it no longer touches.
2. Heat the oil gently on the side arm of the Thiele tube with a microburner if available, or Bunsen burner using a back and forth motion (Figure 6.25c). As the oil warms and becomes less dense, it will rise and travel up the triangular portion of the tube. The cooler, denser oil will sink, thereby creating a current as shown in Figure 6.25c). This method is an excellent way to indirectly and slowly heat the sample.
3. Although bubbles should not be seen in the Thiele tube as it warms, they commonly are seen if the tube had been used previously for boiling point determinations. In this method, the rubber band occasionally breaks causing the sample to fall into the oil and contaminate it. If the oil is not subsequently changed, the sample may boil when heated in the tube. It is okay to continue heating a Thiele tube if bubbles are seen.
4. Studies of this method$^7$ have determined that it is best to heat the oil gently and in a continual manner, as stopping and starting have caused the results to suffer.
5. Continue heating until a vigorous stream of bubbles emerges from the tip of the capillary tube (Figure 6.25d), such that individual bubbles can barely be distinguished. The purpose of this step is to expunge the air originally present in the capillary tube and replace it with the sample's vapor. Do not heat so vigorously that the entire sample boils away. when bubbles are vigorously emerging from the capillary tube, the vapor pressure inside the tube is greater than the atmospheric pressure (the oil is at a higher temperature than the boiling point).
6. Turn off the burner and allow the apparatus to cool. The bubbles will slow and eventually stop. At some point the vapor pressure inside the capillary tube will equal the atmospheric pressure and liquid will be drawn into the tube. The boiling point should be recorded as the temperature when liquid just begins to enter the capillary tube (Figure 6.26b).
1. Record the atmospheric pressure along with the boiling point.
Thiele Tube Summary
Table 6.3: Procedural summary for obtaining a boiling point.
Fill a small tube about half-full with sample and insert a capillary tube, closed end up.
Attach the tube to a thermometer with a small rubber band.
Insert the sample into a Thiele tube, so that the sample is near the middle of the oil.
Heat the arm of the Thiele tube with a burner, gently and continuously.
Heat until a vigorous stream of bubbles emerges from the capillary tube, such that individual drops can be barely distinguished.
Remove the heat and allow the oil to cool.
The boiling point is the temperature when the oil just begins to enter the capillary tube.
$^5$The author found the boiling point of ethanol to be $76^\text{o} \text{C}$ $\left(765 \: \text{mm} \: \ce{Hg} \right)$ with distillation (literature boiling point is $78^\text{o} \text{C}$.
$^6$The author found the boiling point of ethanol (literature boiling point of $78^\text{o} \text{C}$) to be $77.2^\text{o} \text{C}$ with the microscale condenser setup and $76^\text{o} \text{C}$ with the test tube reflux setup $\left( 765 \: \text{mm} \: \ce{Hg} \right)$. Note: Different thermometers were used with each method.
$^7$Blank, E. W., Ind. Eng. Chem. Anal. Ed., 1933, 5(1), p 74-75. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.02%3A_Boiling_Point/6.2A%3A_Overview_of_Boiling_Point.txt |
Some compounds are capable of sublimation, which is the direct phase change from solid to gas. Sublimation is an analogous process to boiling, as it occurs when a compound's vapor pressure equals its applied pressure (often the atmospheric pressure). The difference is that sublimation involves a solid's vapor pressure instead of a liquid's. Most solids do not have an appreciable vapor pressure at easily accessible temperatures, and for this reason the ability to sublime is uncommon. Compounds that are capable of sublimation tend to be those with weak intermolecular forces in the solid state.
• 6.3A: Overview of Sublimation
Some compounds are capable of sublimation, which is the direct phase change from solid to gas. Solid carbon dioxide is an example of a substance that sublimes readily at atmospheric pressure, as a chunk of dry ice will not melt, but will seem to "disappear" as it turns directly into carbon dioxide gas. Sublimation is an analogous process to boiling, as it occurs when a compound's vapor pressure equals its applied pressure (often the atmospheric pressure).
• 6.3B: Step-by-Step Procedures for Sublimation in the Laboratory
6.03: Sublimation
Some compounds are capable of sublimation, which is the direct phase change from solid to gas. Solid carbon dioxide is an example of a substance that sublimes readily at atmospheric pressure, as a chunk of dry ice will not melt, but will seem to "disappear" as it turns directly into carbon dioxide gas. Sublimation is an analogous process to boiling, as it occurs when a compound's vapor pressure equals its applied pressure (often the atmospheric pressure). The difference is that sublimation involves a solid's vapor pressure instead of a liquid's. Most solids do not have an appreciable vapor pressure at easily accessible temperatures, and for this reason the ability to sublime is uncommon. Compounds that are capable of sublimation tend to be those with weak intermolecular forces in the solid state. These include compounds with symmetrical or spherical structures. Examples of compounds that can be sublimed are in Figure 6.28.
As relatively few solids are capable of sublimation, the process can be an excellent purification method when a volatile solid is contaminated with non-volatile impurities. The impure solid is heated in the bottom of a vessel in close proximity to a cold surface, called a "cold finger" (Figure 6.29). As the volatile solid sublimes, it is deposited on the surface of the cold finger (where it can later be recovered), and is thus separated from the non-volatile substance left in the vessel. Sublimation is an example of a "green chemistry" technique, as no solvents are used and no waste is generated. The process, however, is not particularly efficient at separating volatile solids from one another.
Of the solids with appreciable vapor pressures at room temperature, many still require rather high temperatures to actively sublime (when their vapor pressure equals the atmospheric pressure of nearly $760 \: \text{mm} \: \ce{Hg}$). If these solids are heated to their sublimation points under atmospheric pressure, some will char and decompose during the process. For this reason, it is very common to perform sublimation under a reduced pressure (vacuum sublimation). Analogous to vacuum distillation in which liquid boils when its vapor pressure equals the reduced pressure in the apparatus, in vacuum sublimation solid sublimes when its vapor pressure equals the reduced pressure in the apparatus. In vacuum distillation, reducing the pressure allows for liquids to boil at a lower temperature. Similarly, reducing the pressure in vacuum sublimation allows for solids to sublime at a lower temperature, one which avoids decomposition.
$^8$As reported in D. D. Perrin, W. L. F. Armarego, Purification of Organic Chemicals, Pergamon Press, 3$^\text{rd}$ edition, 1988. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.03%3A_Sublimation/6.3A%3A_Overview_of_Sublimation.txt |
Under Atmospheric Pressure
The sublimation pictured in this section shows purification of $0.29 \: \text{g}$ of ferrocene, which grew in long needles on the bottom and top of the petri dishes ($90\%$ recovery).
1. Spread the crude, dry solid to be sublimed in a thin layer on a "bottom" petri dish (Figure 6.30a). If chunky, first crush with a mortar and pestle. Determine the empty mass of the top petri dish.
It is important that the solid to be purified is dry: if the sample is wet with solvent, condensation may form on the top petri dish during the sublimation. In the beginning stages of the sublimation, small amounts of condensation can be wiped off the top petri dish with a paper towel. However, too much condensation may wash crystals off from the top dish.
2. Cover the bottom petri dish with the top dish and set atop a wire mesh on a hotplate in the fume hood (Figure 6.30b) set to the appropriate temperature (depending on the sublimation point of the compound of interest, perhaps medium low). The wire mesh helps dissipate the heat evenly to the dish and minimizes charring.
3. Place a large $600 \: \text{mL}$ beaker filled with ice water atop the petri dish (Figure 6.30c).
1. Over time the sample will sublime and collect on the upper petri dish (Figure 6.31). Monitor the sublimation as compounds may char during the process (if it starts to blacken, turn down the heat). Continue the sublimation until it appears as if little (or no) solid remains on the bottom petri dish. It is very common for crystals to also grow along the sides of the bottom dish.
2. Delicately remove the petri dishes from the hotplate using cotton gloves (Figure 6.32a) or a silicone hot hand protector. Jostling the dishes will cause sublimated crystals to fall from the top petri dish.
Safety note: Allow the two dishes to cool intact on a ceramic tile in the fume hood (Figure 6.32b). Do not remove the top petri dish right away or noxious fumes may escape.
3. The crystals on the top petri dish are purified and should be retained (and their mass determined). Sometimes material on the bottom dish may also be saved if it appears crystalline (signifying it underwent a sublimation process) and doesn't appear contaminated with char (Figure 6.32c).
Under Reduced Pressure (Vacuum Sublimation)
The sublimation in this section shows purification of camphor on two scales $2.28 \: \text{g}$ (large scale, $77\%$ recovery), and roughly $0.2 \: \text{g}$ (small scale).
1. If the solid to be sublimated is chunky, first crush with a mortar and pestle (Figure 6.33a). Then place the crude, dry solid in the bottom of the sublimation apparatus (Figure 6.33b).
It is important that the solid is dry: if the sample is wet with solvent, condensation may form on the cold finger during the sublimation. Too much condensation may wash crystals off the cold finger.
2. Secure the apparatus to a ring stand or latticework (Figure 6.33c+d). For small scales, support the apparatus with a platform (Figure 6.33d). A large scale sublimator is shown in Figure 6.33c.
3. Lightly grease the joint that connects the two pieces of sublimation glassware. Grease can be easily applied with a syringe full of grease. If using ground glass, lightly grease the joint near the end that will not be in contact with the sample (Figure 6.33d).
1. Insert the top piece of the sublimation apparatus (cold finger), and twist the two pieces of glassware together to spread the grease in the joint. When using ground glass, the grease should cause the bottom half of the joint to become transparent all the way around (Figure 6.34a). If the entire joint becomes transparent, too much grease has been used and some should be wiped off.
2. Use thick-walled rubber tubing (clear hose in Figure 6.34a) to connect the apparatus to a vacuum source (vacuum line or water aspirator). Apply the vacuum. The setup should not hiss or there is a leak in the system.
3. Prepare the cold finger:
1. If the cold finger has a condenser, connect water hoses such that the lower arm connects to the water spigot and the upper arm drains to the sink (tan hoses in Figure 6.34a). Begin circulating water through the condenser.
2. If the cold finger is an empty tube, fill the cold finger to the brim with ice, then pour in enough water to fill the finger about three-quarters of the way (Figure 6.34b). In some cases, the cold finger could be filled with dry ice and acetone.
3. It is proper technique to apply the vacuum before cooling the finger to prevent water condensation from forming, which could wash crystals off the cold finger.
4. Heat the solid with a heat gun (Figure 6.34c) or Bunsen burner, beginning slowly with a back and forth motion and low heat. It is not recommended to use a sand bath or heating mantle for sublimation, as heating is often too slow and can only direct heat to the bottom of the apparatus, not the sides. Increase the rate of heating if the sublimation does not begin within a few minutes.
5. Over a short amount of time, solid should begin to deposit on the cold finger. It will undoubtedly also deposit on the outsides of the glassware (Figure 6.34d). Solid can be coaxed away from the outside of the glassware and toward the cold finger by waving the heat gun or burner periodically up the sides of the glass.
1. Continue the sublimation until all of the volatile substance is transferred from the bottom piece of glassware to the cold finger (Figure 6.35). If the compound begins to darken, decrease the rate of heating to prevent decomposition.
2. Remove the coolant from the cold finger:
1. If a condenser was used, turn off the circulating water and remove the water hoses from the apparatus (carefully, without making a large mess).
2. If an ice water coolant was used, scoop out the ice if possible, and remove the water by pipette (or for large scales with a turkey baster, Figure 6.36a).
3. Allow the system to come to room temperature.
4. Delicately reinstate air pressure to the apparatus (Figure 6.36b), noting that an abrupt opening of the system will cause air to violently enter the apparatus and will likely cause crystals to dislodge from the cold finger.
5. Delicately remove the emptied cold finger from the apparatus, and scrape the sublimed crystals onto a watch glass (Figure 6.36c). Alternatively, rinse the crystals from the cold finger with solvent through a funnel and into a round bottomed flask (Figure 6.36d), to later remove the solvent using a rotary evaporator.
Vacuum Sublimation Summary
Table 6.4: Procedural summary for vacuum sublimation.
Place the sample to be sublimed in the bottom of the sublimation apparatus.
Lightly grease all joints.
Use thick walled tubing to attach to the vacuum arm, and apply the vacuum.
The setup should not hiss or there is a leak.
Fill the cold finger, or run water through the condenser.
Be sure to apply the vacuum first, then coolant. If cooled before the vacuum, condensation may occur on the cold finger.
Wave a heat gun or Bunsen burner on the apparatus to heat the sample.
Sublimation should begin within a few minutes.
Coax solid deposited on the side of the glassware toward the cold finger by waving the heat gun/burner on the sides of the glass.
When the sublimation is complete:
Remove the coolant.
Allow the apparatus to come to room temperature.
Delicately reinstate the air pressure, noting that an abrupt opening of the vessel will cause air to knock crystals off the cold finger.
Remove the cold finger. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.03%3A_Sublimation/6.3B%3A_Step-by-Step_Procedures_for_Sublimation_in_the_Laboratory.txt |
Before spectroscopic analysis (IR, NMR) became commonplace in the organic chemistry lab, chemical tests were heavily relied upon to support compound identification. A chemical test is typically a fast reaction performed in a test tube that gives a dramatic visual clue (a color change, precipitate, or gas formation) as evidence for a chemical reaction.
• 6.4A: Overview of Chemical Tests
A chemical test is typically a fast reaction performed in a test tube that gives a dramatic visual clue (a color change, precipitate, or gas formation) as evidence for a chemical reaction. For example, addition of an orange chromic acid reagent to some compounds causes the chromium reagent to change to a blue-green color. This is considered a "positive" test result, and in this case indicates the presence of a functional group that can be oxidized (alcohol or aldehyde).
• 6.4B: Flowcharts
In some teaching labs, a combination of spectroscopy and chemical tests are used in determination of an unknown. If available, an infrared spectrometer is very useful in determining possible functional groups present in an unknown. The following flowcharts summarize key signals present in an IR spectrum, and chemical tests that can be used to support or narrow down structural identification.
• 6.4C: Chemical Test Summary
To follow is a visual summary of the various chemical tests. Procedures and details are provided for each in the following section.
• 6.4D: Individual Tests
The Beilstein test confirms the presence of a halogen in solution, although it does not distinguish between chlorine, bromine, or iodine. A copper wire is dipped into the halogen-containing solution and thrust into a flame. The copper oxide on the wire reacts with the organic halide to produce a copper-halide compound that gives a blue-green color to the flame.
6.04: Chemical Tests
Before spectroscopic analysis (IR, NMR) became commonplace in the organic chemistry lab, chemical tests were heavily relied upon to support compound identification. A chemical test is typically a fast reaction performed in a test tube that gives a dramatic visual clue (a color change, precipitate, or gas formation) as evidence for a chemical reaction. For example, addition of an orange chromic acid reagent to some compounds causes the chromium reagent to change to a blue-green color (Figure 6.37a). This is considered a "positive" test result, and in this case indicates the presence of a functional group that can be oxidized (alcohol or aldehyde). A negative test result is retention of the original color of the reagent, in this case the orange color (Figure 6.37b).
Performing chemical tests is commonly done in the teaching lab. Although the tests work well in general, when using a chemical test to support identification of a structure, caution should be used in interpretation of the results. For example, aldehydes are stated to give a positive result in the bromine test, which is when the compound turns the orange bromine solution clear. Figure 6.38 shows the reaction of two aldehydes with the bromine test: one gives a positive result (the left tube), and one gives a negative result (the right tube). Variation in chemical structure can sometimes interfere with "typical" results, leading to both false positives and false negatives. It is for this reason that spectroscopic methods are often more reliable in structure determination than chemical tests. Nonetheless, the ease of administration makes chemical tests preferable in certain applications, for example in roadside drug testing by police officers, and in environmental and chemical laboratories. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.04%3A_Chemical_Tests/6.4A%3A_Overview_of_Chemical_Tests.txt |
In some teaching labs, a combination of spectroscopy and chemical tests are used in determination of an unknown. If available, an infrared spectrometer (Figure 6.39) is very useful in determining possible functional groups present in an unknown. The following flowcharts summarize key signals present in an IR spectrum, and chemical tests that can be used to support or narrow down structural identification.
1. $\ce{O-H}$ Bond
An $\ce{O-H}$ bond strongly absorbs infrared radiation over a broad range of wavenumbers ($2400$-$3400 \: \text{cm}^{-1}$), and has a characteristic shape in IR spectra. The specific range of absorption can be used to identify whether the $\ce{O-H}$ is part of an alcohol or carboxylic acid functional group: a broad absorption centering around $3300 \text{cm}^{-1}$ corresponds to an alcoholic $\ce{O-H}$ bond (Figure 6.40a), while a broader absorption centering around $3000 \: \text{cm}^{-1}$ corresponds to a carboxylic acid $\ce{O-H}$ bond (Figure 6.40b).
Chemical tests (pH and bicarbonate tests) can be done to support the identification of a carboxylic acid. Specific structural features of alcohols (primary, secondary, or tertiary) can be determined using a variety of chemical tests, as summarized by the flowchart in Figure 6.41.
2. Carbonyl, $\ce{C=O}$ Bond
The carbonyl bond absorbs infrared radiation very strongly and sharply in the $1700 \: \text{cm}^{-1}$ region (Figure 6.42). The specific wavenumber ($1715 \: \text{cm}^{-1}$ or $1735 \: \text{cm}^{-1}$ for example) often correlates well to a specific functional group (aldehyde, ketone, ester, amide, or carboxylic acid), but it is not uncommon for compounds to absorb outside of their "expected" range. Chemical tests can be useful in narrowing down the specific functional group once a carbonyl bond is identified in an IR spectrum, as summarized by the flowchart in Figure 6.43.
3. Unsaturated Hydrocarbons (Alkenes, Alkynes, Aromatics)
Unsaturated compounds have the following characteristic signals in their IR spectrum:
Alkenes/Aromatics (Figure 6.44a):
• $\ce{C-H}$ stretch involving $sp^2$ hybridized carbon atom: medium strength signal slightly above $3000 \: \text{cm}^{-1}$, normally appearing as a shoulder on the $\ce{C-H}$ block of signals around $3000 \: \text{cm}^{-1}$.
• $\ce{C=C}$ stretch: medium strength signals in the $1480$-$1680 \: \text{cm}^{-1}$ region.
Alkynes (Figure 6.44b):
• $\ce{C-H}$ stretch involving $sp$ hybridized carbon atom (terminal alkynes only): strong, sharp signal around $3300 \: \text{cm}^{-1}$. This signal is in the same region as an $\ce{O-H}$ signal, but is much sharper.
• $\ce{C \equiv C}$ triple bond stretch: around $2100 \: \text{cm}^{-1}$, where few other signals occur. The signal is often weak, and may be absent if the compound is mostly symmetric about the triple bond.
Verification of the presence of an alkene or alkyne can be accomplished with the permanganate and bromine tests. Aromatic groups give no reaction in these tests.
4. Alkyl Halide
Presence of an alkyl halide is not obvious from an IR spectrum, as the $\ce{C-X}$ signals occur in the "fingerprint region" $\left( < 1500 \: \text{cm}^{-1} \right)$ and near the lower limit of the IR spectrum ($\ce{C-Cl}$ stretches occur at $550$-$850 \: \text{cm}^{-1}$ while $\ce{C-Br}$ stretches occur at $515$-$690 \: \text{cm}^{-1}$). However, several chemical tests can be used to verify the presence of halogens, the most reliable being the Beilstein test. Two other tests can be used to infer structural features (primary, secondary, tertiary), as shown by the flowchart in Figure 6.45.
6.4C: Chemical Test Summary
To follow is a visual summary of the various chemical tests. Procedures and details are provided for each in the following section.
Table 6.5: Chemical tests summary. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.04%3A_Chemical_Tests/6.4B%3A_Flowcharts.txt |
Beilstein Test
The Beilstein test confirms the presence of a halogen in solution, although it does not distinguish between chlorine, bromine, or iodine. A copper wire is dipped into the halogen-containing solution and thrust into a flame. The copper oxide on the wire reacts with the organic halide to produce a copper-halide compound that gives a blue-green color to the flame.
Procedure: In the fume hood, clean a looped copper wire by thrusting it into the tip of the blue cone of a Bunsen burner flame until it glows (Figure 6.46a). Be sure to "burn off" any residual liquid on the wire (make sure any green flames from previous tests are gone before you begin).
Allow the copper to cool to room temperature, then dip it into a test tube containing 5-10 drops of your sample, coating it as much as possible (Figure 6.46b). If the sample is a solid, adhere some of the solid to the copper wire by first wetting the wire with distilled water then touching it to the solid.
Immediately plunge the wire with sample into the blue cone of the flame. A positive result is a green flame, although it might be short-lived and faint (it may be easier to see if the fume hood light is turned off). A negative result is the absence of this green color (Figure 6.46c+d).
Benedict's Test
The Benedict's test can verify the presence of reducing carbohydrates: compounds that have hemiacetals in their structures and are therefore in equilibrium with the free carbonyl form (aldehyde or $\alpha$-hydroxyketone). The carbonyl forms are oxidized by the $\ce{Cu^{2+}}$ in the Benedict's reagent (which complexes with citrate ions to prevent the precipitation of $\ce{Cu(OH)_2}$ and $\ce{CuCO_3}$). An insoluble $\ce{Cu_2O}$ is the inorganic product of this reaction, which usually has a red-brown color (Figure 6.47). Carbohydrates with only acetal linkages are non-reducing sugars and give a negative result with this test.
Procedure: Dissolve $10$-$30 \: \text{mg}$ of solid or 3 drops liquid sample in a minimal amount of water $\left( 0.5 \: \text{mL} \right)$ in a small test tube ($13$ x $100 \: \text{mm}$). Add $2 \: \text{mL}$ of Benedict's reagent.$^9$ Warm the blue solution in a boiling water bath for 2 minutes (Figure 6.48a). A positive result is the formation of a reddish-brown solution or precipitate after some time, while a negative result is retention of the blue color (Figure 6.48c+d).
Conjugated aldehydes are unreactive in the Benedict's test, and the author found many non-conjugated aldehydes to also be unreactive. Formation of colloids seem to prevent the formation of the red precipitate (Figure 6.49 shows the appearance of propionaldehyde in the hot water bath, forming a cloudy colloid).
The reaction may only work for compounds that are water soluble (like carbohydrates), as the reaction seems to initiate at the surface (Figure 6.50), and the author found aldehydes that formed an insoluble layer on the surface to be unreactive.
The Benedict's test is related to the Fehling's test, which uses different ligands on the copper oxidizing species. The Fehling's reagent uses a $\ce{Cu^{2+}}$ ion complexed with two tartrate ions.
Bicarbonate Test
Carboxylic acids and sulfonic acids can react with sodium bicarbonate $\left( \ce{NaHCO_3} \right)$ to produce carbon dioxide and water (Figure 6.51). Other mainstream functional groups (most phenols and alcohols) are not acidic enough to produce a gas with bicarbonate.
Procedure: Add $2 \: \text{mL}$ of $5\% \: \ce{NaHCO_3} \left( aq \right)$ into a test tube and add 5 drops or $50 \: \text{mg}$ of your sample. Mix the solution by agitating the test tube. A positive test for carboxylic acids is the formation of bubbles or frothing (Figure 6.52).
Bromine Test
A solution of bromine in $\ce{CH_2Cl_2}$ is a test for unsaturation (alkenes and alkynes) and in some cases the ability to be oxidized (aldehydes). The bromine solution is orange and upon reaction the solution turns colorless due to the consumption of bromine. Bromine reacts with alkenes and alkynes through addition reactions and with aldehydes through oxidation (Figure 6.53). It gives no reaction with aromatics, making this a good test to distinguish alkenes from aromatics.
Procedure: Dissolve 4 drops or $50 \: \text{mg}$ of sample in $1 \: \text{mL}$ of dichloromethane $\left( \ce{CH_2Cl_2} \right)$ or 1,2-dimethoxyethane. Add 2 drops of the orange $5\% \: \ce{Br_2}$ in $\ce{CH_2Cl_2}$ solution to the test tube and observe. A positive result is the immediate disappearance of the orange color to produce a clear or slightly yellow solution (Figure 6.54). A negative result is the retention of the orange color. An aldehyde may require a small amount of time to decolorize the solution and produce a positive result (approximately 1 min, Figure 6.55) and conjugated aldehydes are unreactive (Figure 6.55).
Chromic Acid (Jones) Test
A solution of $\ce{CrO_3}$ in $\ce{H_2SO_4}$ is a test for polar functional groups that can be oxidized, which includes aldehydes, primary alcohols, and secondary alcohols (Figure 6.57). Tertiary alcohols give a negative result with this test (Figure 6.56). The orange $\ce{Cr^{6+}}$ reagent converts to a blue-green $\ce{Cr^{3+}}$ species, which often precipitates in acetone.
Procedure: Place $1 \: \text{mL}$ of acetone in a small test tube ($13$ x $100 \: \text{mm}$) and add 2 drops or $20 \: \text{mg}$ of your sample. While wearing gloves, add 2 drops of the orange chromic acid reagent$^{10}$ (safety note: the reagent is highly toxic!) and mix by agitating. A positive result is a blue-green color or dark precipitate, while a negative result is a yellow-orange solution or precipitate with no dark-colored precipitate (Figure 6.58).
Water works better than acetone to rinse chromium reagents into the waste beaker, although some time needs to be allowed for dissolution of the $\ce{Cr^{3+}}$ species.
2,4-DNPH (Brady's) Test
A solution of 2,4-dinitrophenylhydrazine (2,4-DNPH) in ethanol is a test for aldehydes or ketones (Figure 6.59). Most aldehydes or ketones will react with the orange reagent to give a red, orange, or yellow precipitate. Esters and other carbonyl compounds are generally not reactive enough to give a positive result for this test.
The color of the precipitate may give evidence for the amount of conjugation present in the original carbonyl: an orange precipitate forms for non-conjugated carbonyls (Figure 6.60c shows the result for 2-butanone), and a red precipitate forms for conjugated carbonyls (Figure 6.60d shows the result for cinnamaldehyde).
Procedure: Add 3 drops of sample to a small test tube ($13$ x $100 \: \text{mm}$), or dissolve $10 \: \text{mg}$ of solid sample in a minimal amount of ethanol in the test tube. While wearing gloves, add about $1 \: \text{mL}$ of the orange 2,4-DNPH reagent$^{11}$ (safety note: the reagent is highly toxic!) and mix the test tube by agitating.
A positive result is the immediate formation of a large amount of brightly colored precipitate (red, orange, or yellow). A negative result is the absence of this precipitate and a transparent yellow-orange solution (Figure 6.60).
Ferric Hydroxamate Test
The ferric hydroxamate procedure is a probe for the ester functional group. Esters heated with hydroxylamine produce hydroxamic acids, which form intense, colored complexes (often dark maroon) with $\ce{Fe^{3+}}$. A possible structure of these complexes is shown in Figure 6.61. This test is related to the phenol test, and as in that test, compounds with high enolic character can give a colored complex with $\ce{Fe^{3+}}$. Therefore, a preliminary test is performed to see if the carbonyl compound being tested produces enough enol to form a colored complex with $\ce{Fe^{3+}}$, which would lead to a false positive result.
Procedure: Perform a preliminary test to be sure that this test will not give a false positive. Add the following to a small test tube ($13$ x $100 \: \text{mm}$): $1 \: \text{mL}$ ethanol, 2 drops or $20 \: \text{mg}$ of your sample, $1 \: \text{mL}$ of $1 \: \text{M} \: \ce{HCl} \left( aq \right)$, and 2 drops of $5\% \: \ce{FeCl_3} \left( aq \right)$ solution. If the solution is clear or yellow (the color of the $\ce{FeCl_3}$, Figure 6.62a), this test will work and not produce a false positive (continue on). If a definite color other than yellow appears, this test will not work for your sample, as it forms a colored complex with $\ce{Fe^{3+}}$ even without hydroxylamine.
Into a clean medium sized test tube ($18$ x $150 \: \text{mm}$), add $1 \: \text{mL}$ of $0.5 \: \text{M}$ aqueous hydroxylamine hydrochloride $\left( \ce{NH_2OH} \cdot \ce{HCl} \right)$, $0.5 \: \text{mL}$ of $6 \: \text{M} \: \ce{NaOH} \left( aq \right)$, and 5 drops or $50 \: \text{mg}$ of sample. Heat the mixture in a boiling water bath for about 3 minutes (the volume will reduce by about half, Figure 6.62b).
Quickly cool the solution by immersing it in a tap water bath, then add $2 \: \text{mL}$ of $1 \: \text{M} \: \ce{HCl} \left( aq \right)$. If the solution becomes cloudy, add enough ethanol to clarify it. Then add 6-10 drops of a yellow $5\% \: \ce{FeCl_3} \left( aq \right)$ solution. Vigorously mix the tube.
A positive result is a deep burgundy, umber, or magenta color (red/brown) while a negative result is any other color (Figure 6.62c+d). Note: use water to rinse out the test tubes,and if a red result won't easily clean up, add a few drops of $6 \: \text{M} \: \ce{HCl}$.
Iodoform Test
A solution of iodine $\left( \ce{I_2} \right)$ and iodide $\left( \ce{I^-} \right)$ in $\ce{NaOH}$ can be used to test for methyl ketones or secondary alcohols adjacent to a methyl group. This is a very specific test that will give a positive result (formation of a canary yellow precipitate) only for compounds with the structure $\ce{RCH(OH)CH_3}$ or $\ce{RC=OCH_3}$ (Figure 6.63). It does not work for all alcohols or ketones, and does not work well for water-insoluble compounds.
Procedure: Add 10 drops sample to a small test tube ($13$ x $100 \: \text{mm}$) or $0.10 \: \text{g}$ dissolved in the minimal amount of 1,2-dimethoxyethane followed by $1 \: \text{mL}$ of $10\% \: \ce{NaOH} \left( aq \right)$. Next add 10 drops of the dark brown iodoform reagent$^{12}$ ($\ce{I_2}/\ce{KI}$ solution) and vigorously mix the test tube by agitating.
A positive result is a cloudy yellow solution, or a yellow precipitate. A negative result is a clear, yellow, or orange solution with no precipitate (Figure 6.64).
If the sample is not water soluble, a small organic layer separate from the solution may be seen (it will likely be on top). This layer may become dark yellow or brown from dissolving the iodine. Vigorously mix the tube to encourage a reaction, but if the darkened organic layer remains and no precipitate forms, this is still a negative result (Figure 6.64d).
Note: a false positive result may occur if the test tube was cleaned with acetone before use, and residual acetone remained in the tube.
Lucas Test
The Lucas reagent (concentrated $\ce{HCl}$ and $\ce{ZnCl_2}$) is a test for some alcohols. Alcohols can react through an $S_\text{N}1$ mechanism to produce alkyl halides that are insoluble in the aqueous solution and appear as a white precipitate or cloudiness. The test cannot be used for water-insoluble alcohols (generally > 5 carbon atoms), as they may produce a cloudiness or second layer regardless if any reaction occurred or not.
$2^\text{o} \: \text{or} \: 3^\text{o} \: \ce{ROH} + \ce{HCl}/\ce{ZnCl_2} \rightarrow \ce{RCl} \left( s \right)$
As the mechanism is $S_\text{N}1$, a tertiary alcohol should react immediately, a secondary alcohol react more slowly (perhaps in 5 minutes if at all) and primary alcohols often don't react at all. Benzylic alcohols $\left( \ce{Ph-C-OH} \right)$, allylic alcohols $\left( \ce{C=C-C-OH} \right)$ and propargylic alcohols $\left( \ce{C \equiv C-C-OH} \right)$ often give immediate results just like tertiary alcohols.
Procedure: Place $2 \: \text{mL}$ of the Lucas reagent$^{13}$ (safety note: the reagent is highly acidic and corrosive!) into a small test tube ($13$ x $100 \: \text{mm}$). Add 10 drops of sample, and mix by agitating the test tube.
A positive result is a white cloudiness within 5 minutes or a new organic layer $\left( \ce{RCl} \right)$ formation on the top.$^{14}$ A negative result is the absence of any cloudiness or only one layer (Figure 6.65).
Permanganate (Baeyer) Test
A potassium permanganate $\left( \ce{KMnO_4} \right)$ solution is a test for unsaturation (alkenes and alkynes) or functional groups that can be oxidized (aldehydes and some alcohols, Figure 6.66). The permanganate ion $\left( \ce{MnO_4^-} \right)$ is a deep purple color, and upon reduction converts to a brown precipitate $\left( \ce{MnO_2} \right)$. Permanganate cannot react with aromatics, so is a good test to discern between alkenes and aromatics. A positive reaction with alcohols is not always dependable (a negative result is seen with benzyl alcohols in Figure 6.67).
Procedure: Dissolve 4 drops or $40 \: \text{mg}$ of sample in $1 \: \text{mL}$ of ethanol (or 1,2-dimethoxyethane) in a small test tube ($13$ x $100 \: \text{mm}$). While wearing gloves, add 3 drops of the deep purple $1\% \: \ce{KMnO_4} \left( aq \right)$ solution to the test tube (safety note: reagent is corrosive and will stain skin brown!). Mix the test tube with agitation, and allow it to sit for 1 minute. A positive result is the appearance of a brown color or precipitate. A negative result is a deep purple with no precipitate (unreacted $\ce{KMnO_4}$, Figure 6.67).
pH Test
Carboxylic acids and sulfonic acids produce acidic aqueous solutions (Figure 6.68a), which can be confirmed by turning blue litmus paper pink. The paper changes color (Figure 6.68c) as the indicator molecules react in the lowered pH and form a structure that has a different color.
Procedure: Dissolve 3 drops or $30 \: \text{mg}$ of sample in $1 \: \text{mL}$ of water. Dip a glass stirring rod into the solution and touch the rod to blue litmus paper. A positive result is a pink or red color on the litmus paper (Figure 6.68c). If the sample doesn't dissolve in water, instead dissolve the same amount of unknown in $1 \: \text{mL}$ of ethanol. Add enough water to make the solution barely cloudy. Then add a few drops of ethanol to turn the solution clear again, and test with the litmus paper.
Phenol Test
A ferric chloride solution is a test for phenols, as they form intensely colored complexes with $\ce{Fe^{3+}}$ (often dark blue). The actual structure of these complexes is debated,$^{15}$ but may be of the general form in Figure 6.69. Some carbonyl compounds with high enol content can give false positives with this test.
Procedure: Place $1 \: \text{mL}$ water in a small test tube ($13$ x $100 \: \text{mm}$) along with either 3 drops or $30 \: \text{mg}$ of sample. Add 3 drops of the yellow $5\% \: \ce{FeCl_3} \left( aq \right)$ solution, and mix by agitating.
A positive result is an intense blue, purple, red, or green color while a negative result is a yellow color (the original color of the $\ce{FeCl_3}$ solution, Figure 6.70).
Silver Nitrate Test
A dilute solution of silver nitrate in ethanol is a test for some alkyl halides. Silver has a high affinity for halogens (forms strong $\ce{AgX}$ ionic bonds), and so encourages an $S_\text{N}1$ mechanism. For this reason, tertiary alkyl halides react faster than secondary alkyl halides (which may or may not react, even with heating), and primary alkyl halides or aromatic halides give no reaction. Benzylic $\left( \ce{PhCH_2X} \right)$ and allylic $\left( \ce{CH_2=CHCH_2X} \right)$ alkyl halides will also give a fast reaction. A positive test result is the formation of the insoluble $\ce{AgX}$ (Figure 6.71). $\ce{AgCl}$ and $\ce{AgBr}$ are white solids, while $\ce{AgI}$ is a yellow solid.
Procedure: In a small test tube ($13$ x $100 \: \text{mm}$), add $2 \: \text{mL}$ of $1\% \: \ce{AgNO_3}$ in ethanol solution. Add 4 drops of liquid sample or $40 \: \text{mg}$ fo solid dissolved in the minimal amount of ethanol. Mix the test tube by agitating. Some compounds will have an initial insolubility when first mixed, but the solid often dissolves with swirling. A positive result is a sustaining white or yellow cloudiness. If cloudiness does not occur within 5 minutes, heat the tube in a $100^\text{o} \text{C}$ water bath for 1 minute (Figure 6.72b). Absence of cloudiness even at $100^\text{o} \text{C}$ is a negative result (Figures 6.72+6.73).
For reactions that produce an intense precipitate, the solution may also turn blue litmus paper pink (Figure 6.73c+d). An analysis of the reaction mechanism can explain the source of this acidity.
Sodium Iodide (Finkelstein) Test
A solution of sodium iodide in acetone is a test for some alkyl chlorides and bromides. The mechanism is largely $S_\text{N}2$, so primary alkyl halides react faster than secondary alkyl halides, and tertiary alkyl halides generally give no reaction. The reaction is driven by the precipitation of the $\ce{NaCl}$ or $\ce{NaBr}$ in the acetone solvent. Therefore, a positive test result is the appearance of a white cloudiness ($\ce{NaX}$ solid).
$\begin{array}{ccccccccc} \ce{CH_3CH_2X} & + & \ce{NaI} \: \text{(acetone)} & \rightarrow & \ce{CH_3CH_2I} & + & \ce{NaX} \left( s \right) & & \left( \ce{X} = \ce{Cl}, \ce{Br} \right) \ & & & & & & \text{white solid} & & \end{array}$
Procedure: In a small test tube ($13$ x $100 \: \text{mm}$), add $2 \: \text{mL}$ of $15\% \: \ce{NaI}$ in acetone solution.$^{16}$ Add 4 drops of liquid sample or $40 \: \text{mg}$ of solid dissolved in the minimal amount of ethanol. Mix the test tube by agitating.
A positive result is a sustaining white cloudiness. If cloudiness does not occur within 5 minutes, heat the tube in a $50^\text{o} \text{C}$ water bath for 1 minute. Absence of cloudiness even at $50^\text{o} \text{C}$ is a negative reaction (Figures 6.74+6.75).
Tollens Test
The Tollens reagent $\left( \ce{Ag(NH_3)_2^+} \right)$ is a mild oxidizing agent that can oxidize aldehydes, but not alcohols or other carbonyl compounds. A positive test result is the formation of elemental silver (Figure 6.76), which precipitates out as a "silver mirror" on the test tube, or as a black colloidal precipitate.
Procedure: While wearing gloves, mix $1 \: \text{mL}$ of $5\% \: \ce{AgNO_3} \left( aq \right)$ (safety note: toxic!) with $1 \: \text{mL}$ of $10\% \: \ce{NaOH} \left( aq \right)$ in a medium sized test tube ($18$ x $150 \: \text{mm}$). A dark precipitate of silver oxide will form (Figure 6.77b). Add dropwise enough $10\% \: \ce{NH_4OH} \left( aq \right)$ to just dissolve the precipitate (note some time should be allowed between additions). This solution is now the Tollens reagent $\ce{Ag(NH_3)_2^+}$ (Figure 6.77c).
Dissolve 3 drops or $30 \: \text{mg}$ of sample in a few drops of diethyl ether (omit solvent if compound is water soluble). Add this solution to the $2$-$3 \: \text{mL}$ of previously prepared Tollens reagent. Mix the test tubes by agitating. A positive result is a silver mirror on the edges of the test tube, or formation of a black precipitate. A negative result is a clear solution (Figures 6.77d+6.78).
Clean-up: The reagent may form a very explosive substance (silver fulminate) over time, so the test should be immediately cleaned up. Acidify the solution with $5\% \: \ce{HCl} \left( aq \right)$, then dispose in a waste beaker. A silver mirror can be removed from the glassware by adding a small amount of $6 \: \text{M} \: \ce{HNO_3} \left( aq \right)$.
$^9$The Benedict's reagent is prepared as follows, as published by the Flinn Scientific catalog: $173 \: \text{g}$ of hydrated sodium citrate and $100 \: \text{g}$ of anhydrous sodium carbonate is added to $800 \: \text{mL}$ of distilled water with heating. The mixture is filtered, then combined with a solution of $17.3 \: \text{g}$ copper(II) sulfate pentahydrate dissolved in $100 \: \text{mL}$ distilled water. The combined solutions are diluted to $1 \: \text{L}$.
$^{10}$The chromic acid reagent is prepared as follows: $25.0 \: \text{g}$ of chromium(VI) oxide is added to $25 \: \text{mL}$ concentrated sulfuric acid, which is then added in portions to $75 \: \text{mL}$ of water. The reagent has a very long shelf life (10+ years).
$^{11}$Preparation of the 2,4-DNPH reagent, as published in B. Ruekberg, J. Chem. Ed., 2005, 82(9), p. A1310, is as follows: To a dry $125 \: \text{mL}$ Erlenmeyer flask is added $3 \: \text{g}$ 2,4-dinitrophenylhydrazine, $20 \: \text{mL}$ water and $70 \: \text{mL}$ of $95\%$ ethanol. The solution is cooled in an ice bath with stirring, and when at $10^\text{o} \text{C}$, $15 \: \text{mL}$ of concentrated sulfuric acid is added slowly in portions. If the temperature exceeds $20^\text{o} \text{C}$ during the addition, the solution should be allowed to cool to $10^\text{o} \text{C}$ before continuing. The solution is then warmed to $60^\text{o} \text{C}$ with stirring, and if solids remain, they are filtered. Finally, the solution is cooled.
$^{12}$Preparation of the iodoform reagent is as follows: $10 \: \text{g} \: \ce{KI}$ and $5 \: \text{g} \: \ce{I_2}$ is dissolved in $100 \: \text{mL}$ water.
$^{13}$Preparation of the Lucas reagent is as follows: $160 \: \text{g}$ of fresh anhydrous $\ce{ZnCl_2}$ is dissolved in $100 \: \text{mL}$ of cold concentrated $\ce{HCl}$.
$^{14}$Although chlorinated organics are typically denser than water, the Lucas reagent has a high quantity of solute, and chlorinated compounds tend to be less dense than the reagent.
$^{15}$See Nature, 24 June 1950, 165, 1012.
$^{16}$This solution often has a yellow tin to it. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/06%3A_Miscellaneous_Techniques/6.04%3A_Chemical_Tests/6.4D%3A_Individual_Tests.txt |
The following summary sheets are not intended as a substitute for reading the sections on each technique. The purpose of these summary sheets is to review the main points for students who have already read the sections. They can also be useful for students who have previously performed these techniques (earlier in a semester or in recent years) and wish to remind themselves about minor details
07: Technique Summaries
Table 7.1: Procedural summary for flame drying glassware.
Clamp the flask to be dried (and stir bar if using) to a ring stand or latticework with an extension clamp.
Remove any vinyl sleeves on the clamps as they will catch on fire.
Heat the flask with a Bunsen burner or heat gun for several minutes until all the fog is removed and glassware is scorching hot. Handle hot glassware with thick gloves. Cool in a water-free environment (in a desiccator, under a stream of inert gas, or with a drying tube).
7.02: Using Calibrated Glass Pipettes
Table 7.2: Procedural summary for using calibrated glass pipettes.
Place pipette tip in reagent bottle, squeeze pipette bulb, and connect to the pipette.
Partially release your hand to create suction. Do not let to completely or liquid will withdraw forcibly and possibly into the bulb.
Apply suction until liquid is withdrawn to just past the desired mark.
Remove the pipette bulb and place your finger atop the pipette.
Allow tiny amounts of air to be let into the top of the pipette by wiggling your finger or a slight release of pressure.
Drain the liquid to the desired mark.
Tightly hold the pipette with your finger, bring it to the transfer flask and deliver the reagent to the desired mark.
Touch the pipette to the side of the container to dislodge the drip at the end of the pipette.
If a pipette is drained to the tip,
• To-deliver (T.D.) pipettes and volumetric pipettes should not be blown out.
• To-contain (T.C.) pipettes should be "blown out".
Note: If pipette is wet with a different solution before use, obtain a fresh one or "condition" the pipette with two rinses of the reagent.
7.03: Inert Atmospheric Methods
Table 7.3: Procedural summary for inert atmospheric methods.
Prepare the balloon
Fill a balloon with an inert gas (nitrogen or argon) to 7-8 inches in diameter.
Twist the balloon to prevent gas from escaping, then attach a needle and insert into a rubber stopper to plug.
Prepare the reaction flask
Flame or oven dry a reaction flask (with stir bar), and fold a rubber septum over the joint while wearing thick gloves.
Clamp the hot flask to the ring stand or latticework and insert the balloon of inert gas.
Insert an "exit needle" and allow the flask to flush for ~5 minutes (to displace the air).
Remove the exit needle and allow the flask to cool.
Prepare the syringe
Obtain a needle from the hot oven and screw it into the tip of a freshly opened plastic syringe.
Wrap the needle/syringe joint with Teflon tape or Parafilm.
Flush the syringe with inert gas by using an empty flask attached to a balloon of inert gas. Withdraw a volume of gas and expunge into the air.
Insert the needle tip into a rubber stopper to plug.
Withdraw the reagent
Use the prepared syringe to slowly withdraw some reagent (or air may enter the syringe).
Push out the gas bubble. Withdraw reagent to a slightly greater volume than needed, then expunge liquid to the exact volume.
Place the needle in the headspace of the bottle and withdraw an "inert gas buffer", $\sim 20\%$ of the volume of the syringe.
Insert the needle into a rubber stopper for transport.
Deliver the reagent
Into the reaction flask (with inert gas balloon), deliver first the inert gas buffer, then the air-sensitive reagent.
Don't push out the residual liquid.
Withdraw an inert gas buffer into the mostly empty syringe.
Clean by rinsing with two portions (few $\text{mL}$ each) of solvent (similar to reagent solvent), then one portion of water and two portions of acetone. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.01%3A_Flame-Drying_Glassware.txt |
Table 7.4: Procedural summary for reflux.
Pour liquid into the flask along with a stir bar or boiling stones.
Use an extension clamp on the round bottomed flask to connect to the ring stand or latticework.
Attach the condenser, and connect the hoses so that water travels against gravity (cooling water comes into the bottom and drains out the top).
Be sure there is a secure connection between the round bottomed flask and condenser, as vapors escaping this joint have the potential to catch on fire.
Circulate water through the condenser, then begin heating the flask (by using a heating mantle, sand, water, or oil bath).
Use an adjustable platform so the heat can be lowered and removed at the end of the reflux, or if something unexpected occurs.
Heat so that the "reflux ring" is seen in the lower third of the condenser.
Turn down the heat if the refluxing vapors reach higher than halfway up the condenser.
At the end of the reflux period, lower the heat source from the flask or raise the apparatus.
Keep circulating water in the condenser until the flask is just warm to the touch.
After air cooling somewhat, the flask can be quickly cooled by immersing in a container of tap water.
7.05: Suction Filtration
Table 7.5: Procedural summary for suction filtration.
Clamp a side-armed Erlenmeyer flask.
Connect thick-walled hosing from the side arm to a vacuum trap and the water aspirator.
Place a vacuum sleeve on the Buchner (or Hirsch) funnel, then filter paper on the funnel so it arches downward.
Turn on the aspirator.
Add a few $\text{mL}$ of the same solvent used in the flask to wet the filter. The solvent should drain with suction.
Swirl the mixture to be filtered to dislodge the solid from the sides of the flask.
With a quick motion, pour the slurry into the funnel portions.
In some applications (e.g. crystallization), rinse with solvent:
• Open the apparatus to the atmosphere, then turn off the aspirator.
• Add a few $\text{mL}$ of cold solvent to the filter paper.
• Delicately swirl the solid in the solvent with glass rod.
Apply suction again for a few minutes (repeat the rinse step if necessary).
Dry the solid on a watch glass along with the filter paper, overnight if possible. The solid will flake off the paper when dried.
7.06: Hot Filtration
Table 7.6: Procedural summary for hot filtration.
Prepare a fluted filter paper and stemless or short-stemmed funnel clamped above the filter flask. Pour a few $\text{mL}$ of solvent through the funnel, and allow the solvent to boil and get the funnel hot. When the mixture to be filtered is boiling, pour the mixture into the funnel in portions, returning it to the heat source in between additions. Rinse the filter paper with hot solvent if crystals are seen on the paper.
7.07: Thin Layer Chromatography
Table 7.7: Procedural summary for thin layer chromatography.
Place a small portion of solvent ($5$-$10 \: \text{mL}$ for this chamber) into a TLC chamber with lid, along with a cut piece of filter paper. Dissolve liquid or solid samples (1 drop per $\sim 1 \: \text{mL}$ solvent) using a low boiling solvent (e.g. acetone or dichloromethane).
Draw a pencil line on a TLC plate $\sim 1 \: \text{cm}$ from the bottom with a ruler, and mark the lanes.
Don't put lanes too close to the edge or to each other.
Spot a dilute sample on the pencil line of the correct lane, making very small spots ($2 \: \text{mm}$ in diameter).
Rinse the spotter with a solvent (e.g. acetone) if going to use it for another sample.
Place the sample in the TLC chamber with forceps, cap it, and leave it alone.
Remove the plate when the solvent line is $\sim 0.5 \: \text{cm}$ from the top.
Immediately mark the solvent line with a pencil.
Visualize if necessary.
7.08: TLC Visualization Methods
Table 7.8: Procedural summary for TLC visualization methods.
Ultraviolet Light:
Press the short-waved button on a UV lamp positioned on a tilt over the TLC plate.
Alternatively view the TLC plate inside a box designed to protect your eyes from UV damage.
The TLC plate will fluoresce green and spots will be dark.
Circle the spots with a pencil.
Further visualize with another method if desired.
Iodine:
In the fume hood, place the TLC plate in a jar containing a few crystals of iodine (or alternatively iodine with silica). Close the lid.
If using the iodine-silica powder, submerge the plate under the powder for a few minutes.
Spots will appear yellow-brown.
Remove the plate and circle the spots as they will fade rather quickly.
Chemical Stains:
Wear gloves and work in the fume hood as many stains are highly acidic or oxidative.
Hold a TLC plate with forceps and quickly dip the plate into and out of the stain jar to cover the area where solvent traveled. Let excess stain drip off, then wipe residue off the back with a paper towel.
Wave a heat gun set to "high" (at first) back and forth over the top of the TLC plate until spots appear.
Turn the heat gun down to "low" if charring begins on the plate.
Further visualization is not possible after a TLC plate has been stained. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.04%3A_Reflux.txt |
Table 7.9: Procedural summary for macroscale column chromatography.
Make sure there is a frit or cotton plug in the bottom of the column.
Fill the column with silica or alumina to 5-6 inches in the fume hood.
Pour the adsorbent into an Erlenmeyer flask and add eluent to make a pourable slurry.
The eluent should give the desired component an $R_f$ of 0.35 by TLC.
Pour the slurry and immediately rinse the sides of the column with eluent and a pipette.
Jostle the column to remove air bubbles.
Use air pressure to pack the column. Keep it perfectly vertical.
Add $0.5 \: \text{cm}$ of sand.
Rinse the sand off the sides carefully with a swirling motion.
Don't disrupt the top surface of the silica or alumina with rinsing.
Adjust the eluent level to the sand layer, and then add the sample (pure liquid, dissolved in $\ce{CH_2Cl_2}$, or solid adsorbed onto a portion of silica).
Rinse the sides and use air pressure to force the eluent down onto the silica/alumina layer.
Fill the reservoir with eluent (carefully to not disrupt the top surface).
Use steady air pressure to elute the column.
Collect fractions in test tubes in a rack (keep in order).
Rinse the column tip if a component has finished coming off the column.
Possibly increase the eluent polarity to make components elute faster.
Never allow the eluent to drop below the top of the adsorbent column.
Use TLC to determine the purity of the fractions, and combine appropriate fractions.
Remove the solvent with the rotary evaporator.
7.10: Pipette Column Chromatography
Table 7.10: Procedural summary for pipette column chromatography.
Wedge a bit of cotton into the bottom of a pipette.
Use a scooping method to fill silica or alumina to 2-2.5 inches high.
Add a $0.5 \: \text{cm}$ layer of sand. Add eluent to the column and apply pressure with a pipette bulb to force eluent through the column to completely wet it.
Remember to break the seal before letting go of the pipette bulb, or suction will ruin the column.
Refill the column with eluent as necessary.
Adjust the eluent level to the sand layer, and then delicately add the sample.
Use pressure to push the eluent down onto the silica/alumina layer.
Rinse with one portion of eluent and push the solvent onto the column.
Fill the pipette with eluent and apply pressure to elute the column.
Collect liquid into test tubes.
Always keep the white column section wet (refill whenever the eluent level nears the sand layer).
Switch test tubes periodically (perhaps when they are $1 \: \text{cm}$ high in small test tubes) to collect different fractions.
Keep fractions organized in a test tube rack in the order they are eluted.
Use TLC to determine the purity of the fractions, and combine appropriate fractions.
Remove the solvent with the rotary evaporator.
7.11: Testing Solvents for Crystallization
Table 7.11: Procedural summary for testing solvents for crystallization.
Place $100 \: \text{mg}$ of the solid to be crystallized in a test tube and add $3 \: \text{mL}$ solvent.
Flick the tube to mix the contents.
If the solid is soluble, the solvent will not work for crystallization. If the solid appears to be insoluble, the solvent may work.
Bring the solid suspension to a boil on a steam bath or boiling water bath.
If the solid dissolves when hot, the solvent may work for crystallization. If it remains insoluble when hot, the solvent will not work.
Allow the hot solution to cool to room temperature (scratch the sides of the test tube to initiate crystallization if necessary).
Place the solution in an ice bath for 10-20 minutes.
If large quantities of crystals return when cool, the solvent should work for crystallization. If crystals do not return, the solvent will not work.
7.12: Testing Mixed Solvents for Crystallization
Table 7.12: Procedural summary for testing mixed solvents for crystallization.
Use $100 \: \text{mg}$ solid and $3 \: \text{mL}$ solvent to test the solubility of the compound.
Find a miscible pair of solvents: one in which the compound is soluble (called the "soluble solvent") and one in which the compound is insoluble (called the "insoluble solvent").
Place a fresh $100 \: \text{mg}$ of the solid to be crystallized in a test tube and add the "soluble solvent" dropwise with heating in a steam bath or hot water bath, until the solid just dissolves.
Immerse the suspension in the heat after each addition, and allow time in between additions for the sometimes slow dissolving process.
Add the "insoluble solvent" dropwise with heating until the solution becomes faintly cloudy.
Add the "soluble solvent" dropwise with heating until the solution is clarified (clear).
Allow the system to come to room temperature, then submerge in an ice bath for 10-20 minutes.
If crystals return, the mixed solvent pair may work for the crystallization.
7.13: Single Solvent Crystallization
Table 7.13: Procedural summary for single solvent crystallization.
Heat some solvent on a heat source to a boil (include boiling stones).
To the impure solid in an Erlenmeyer flask, add a small portion of hot solvent.
If a hot filtration step is expected, use boiling stones with the impure solid, or a boiling stick or stir bar if a filtration is not expected.
Put both flasks on the heat source and bring to a boil.
Add the minimum amount of boiling solvent needed to dissolve the impure solid, with swirling:
For $100 \: \text{mg}$-$1 \: \text{g}$ quantities,add $0.5$-$2 \: \text{mL}$ solvent each time. For smaller scales, add solvent dropwise.
Wait for each portion of solvent to come to a boil before adding more.
Possibly use charcoal and hot filtration at this point if colored or insoluble impurities are present.
When dissolved, remove the solution from the heat source and remove the boiling stick or stir bar if used. Allow the solution to slowly cool atop some paper towels,and with a small watch glass over the mouth of the Erlenmeyer.
If a hint of cloudiness is seen in the solution, or if the solution has cooled a good deal without crystallizing, scratch with a glass stirring rod to initiate crystallization.
A proper crystallization takes between 5-20 minutes to complete.
When the solution is at room temperature, place it in an ice bath for at least 10 minutes to maximize crystal formation.
Also chill a rinse solvent in the ice bath.
Collect the crystals by suction filtration. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.09%3A_Macroscale_Column_Chromatography.txt |
Table 7.14: Procedural summary for mixed solvent crystallization.
Find a miscible pair of solvents: one in which the desired compound is soluble (called the "soluble solvent") and one in which the compound is insoluble (called the "insoluble solvent").
Place the impure solid in an Erlenmeyer flask along with a boiling stick (or boiling stones if preferred).
Add the soluble solvent in portions while heating until the solid just dissolves.
Allow time in between additions, and allow each addition to come completely to a boil before adding more.
Add the insoluble solvent in portions with heating until the solution becomes faintly cloudy. Add the "soluble" solvent dropwise with heating until the solution is clarified (transparent).
Remove the boiling stick (if used) and allow the system to come to room temperature while sitting atop some paper towels and with the flask's mouth covered by a watch glass.
Scratch with a glass stirring rod to initiate crystallization if necessary.
Submerge in an ice bath for 10-20 minutes.
Collect crystals by suction filtration.
7.15: Single Extraction
Table 7.15: Procedural summary for single extraction.
Use slit tubing to cushion the separatory funnel in the ring clamp.
Close the stopcock on the separatory funnel and position an Erlenmeyer flask beneath the setup, in case it drips.
Into the separatory funnel pour the liquid to be extracted using a funnel: this prevents liquid from getting on the ground glass joint which can cause it to stick.
Pour the extractive solvent into the funnel.
Hold the separatory funnel so that your fingers firmly cover the stopper.
Invert the funnel and shake gently for 10-20 seconds.
Periodically "vent" the funnel (open the stopcock while inverted to release pressure). Never point the tip at someone while venting.
Return the separatory funnel to the ring clamp, and allow the layers to separated.
Remove the stopper (it won't drain otherwise).
Drain the majority of the bottom layer into an Erlenmeyer flask.
Stop when roughly $1 \: \text{cm}$ of the bottom layer is in the funnel, and swirl to dislodge clinging droplets.
Drain the rest of the bottom layer, stopping when the interface is inside the stopcock.
Label the flask (e.g. "bottom aqueous layer").
Pour out the top layer into another Erlenmeyer flask (and label it).
Don't throw away either layer until you are sure you've accomplished the goal of the extraction.
7.17: Microscale Extractions
Table 7.16: Procedural summary for microscale extractions.
Pour contents and solvent into a conical vial or centrifuge tube with tapered end.
As these containers are prone to tipping, stabilize them in a beaker or inverted cork ring.
Mix the solutions by one of these methods:
• Cap and shake the vial for 10-20 seconds (no need to vent periodically).
• Withdraw and vigorously expunge the bottom layer through the top layer by pipette for roughly 1 minute.
Separate the layers by pipette, always withdrawing the bottom layer first (even if the bottom layer is not desired).
Place the pipette tip all the way to the bottom of the tapered end.
If difficulty arises in removing a residual drop, withdraw some of both layers into the pipette, allow the layers to separate in the pipette, then delicately expunge the bottom layer.
Remove the top layer with a fresh pipette if saving.
If the top layer is to be further extracted, leave it in the vial and add fresh solvent.
Label each layer.
7.18: Testing the pH After a Wash
Table 7.17: Procedural summary for testing the pH after a wash.
If a wash was used to neutralize an organic layer, test the aqueous layer it is in contact with. Most often a wash is used to neutralize trace amounts of acid.
If the aqueous layer is the top layer, insert a glass stirring rod, and touch the rod to blue litmus paper.
An acidic solution turns the blue paper pink (or red), while a neutral solution makes the paper look wet.
If the aqueous layer is the bottom layer,remove and test the pH of an aliquot:
Place your finger atop a pipette and insert it so the tip is in the bottom layer.
Let go of your finger and allow a small amount of the bottom layer to withdraw into the pipette.
Place your finger atop the pipette again, then remove the aliquot from the funnel.
Test the pH of the aliquot on blue litmus paper.
If still acidic, repeat the neutralization washes on the organic layer. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.14%3A_Mixed_Solvent_Crystallization.txt |
Table 7.18: Procedural summary for using drying agents.
Drying agents are used to remove trace amounts of water from an organic solution.
Always use an Erlenmeyer flask, not a beaker.
If a second layer (water) is seen in the flask, remove it by pipette before addition of the drying agent.
Start by adding a small portion of drying agent (size of a pea) to the flask.
Swirl the flask.
Water will cause the drying agent to clump into particles larger than the original particle size, and possibly stick to the glass.
A solution is dry when small particles the size of the original drying agent are notice with the flask is swirled.
Add additional portions of drying agent until small particles are seen.
Allow extra time when exposing a solution to anhydrous sodium sulfate $\left( \ce{Na_2SO_4} \right)$, as this drying agent takes some time.
To remove the drying agent from the solution, decant (pour) the solution if using $\ce{Na_2SO_4}$, pellet $\ce{CaCl_2}$, or $\ce{CaSO_4}$.
Remove the drying agent with gravity filtration if $\ce{MgSO_4}$ or granular $\ce{CaCl_2}$ are used.
Rinse the drying agent with a few portions of fresh solvent, and combine the rinsings with the dried solution.
7.21: Simple Distillation
Table 7.19: Procedural summary for simple distillation.
Assembly tips
Fill the distilling flask with sample 1/3-1/2 full. Always use an extension clamp on the distilling flask.
Add a few boiling stones or stir bar to the flask.
Position the thermometer bulb just below the arm of the three-way adapter, where vapors turn toward the condenser.
Wet condenser hoses with water before attaching.
Connect the condenser hoses such that water flows uphill: bring water from the faucet into the lower arm, and drain out the upper arm.
Be sure all of the connections are secure (especially between the distilling flask and 3-way adapter: potential of fire!).
Begin distillation
Turn on the condenser water.
Apply the heat source to the distilling flask.
Collect distillate at a rate of 1 drop per second.
Record the temperature where liquid is actively distilling and thermometer bulb is immersed.
Record the pressure.
Cease distillation
Stop the distillation when the temperature changes dramatically or if the distilling flask is nearly empty (never distill to dryness!).
Lower and remove the heat source, but keep water circulating until the flask is just warm to the touch.
7.22: Fractional Distillation
Table 7.20: Procedural summary for fractional distillation.
Most comments for a simple distillation apply to fractional distillation as well.
The distilling pot will need to be heated more vigorously than with a simple distillation, as there is a greater distance for the vapors to travel before reaching the condenser.
Commonly the column will need to be insulated to maintain heat: wrap the column (and three-way adapter if desired) in glass wool followed by an outer layer of aluminum foil. Droplets of liquid should be seen in the fractional column, but there should never be a large pool of liquid (flooding). If the column floods, allow the liquid to drain back into the distilling flask and heat at a gentler rate.
7.23: Vacuum Distillation
Table 7.21: Procedural summary for vacuum distillation.
Always use a stir bar, not boiling stones.
Grease all joints.
Use a Claisen adapter, as solutions tend to bump under vacuum.
Connect thick-walled hosing at the vacuum adapter to a trap, then to the vacuum source (water aspirator or vacuum pump).
Turn on the vacuum first, before heating, to remove very volatile components.
When confident the apparatus is maintaining a reduced pressure, then heat the sample.
Use glass wool or foil insulation if the sample is stubbornly refluxing instead of distilling.
Record the temperature and pressure if using a manometer during active distillation.
Pure compounds may not distill over a constant temperature due to changes in pressure.
If multiple fractions will be collected, the system needs to be vented and cooled in between (or a cow receiving flask used).
To stop the distillation, remove the heat and cool the flask in a tap water bath.
Then open the apparatus to the atmosphere by opening the pinch clamp on the trap, or removing the tubing on the vacuum adapter.
Lastly turn off the vacuum.
Correct order: cool, open to atmosphere, then turn off vacuum.
7.24: Steam Distillation
Table 7.22: Procedural summary for steam distillation.
Place large amount of plant materials in a large round bottomed flask (no more than half full), and just cover with water.
Always use a Claisen adapter, as there is often turbulence in the flask.
Two variations are common:
• Heat the water with a Bunsen burner to create steam directly.
• Add steam indirectly through a steam line from the building.
Collect the distillate at a rate of 1 drop per second.
The distillate may appear cloudy, or a second layer may form on the top.
If milky, the distillation can be ceased when the distillate is clear.
If a steam line was used, be sure to drain the liquid from the steam trap before turning off the steam, to prevent back suction.
Separated oil can be pipetted (then dried with \(\ce{Na_2SO_4}\)), while milky distillates need to be extracted. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.19%3A_Using_Drying_Agents.txt |
Table 7.23: Procedural summary for rotary evaporation.
Be sure there is ice in the water circulator (if used).
Fill a round bottomed flask no greater than half-full.
Connect to the bump trap with a plastic clip.
Lower the flask into the water bath to submerge the liquid (don't submerge the plastic clip).
Turn on the vacuum source (vacuum will hiss).
Rotate the flask at a moderate rate (one-third the maximum value).
Close the stopcock in the apparatus (hissing will stop).
Evaporate until solid forms or liquid level doesn't appear to change anymore, then evaporate an extra few minutes for good measure.
To stop evaporation, reverse all steps:
• Open the stopcock
• Stop the rotation
• Turn off the vacuum
• Lift the flask from the water bath
• Remove the flask
7.26: Melting Points
Table 7.24: Procedural summary for obtaining a melting point.
Load the sample by jabbing the open end of a capillary tube into a pile of the sample.
With closed end down, drop the tube down a long hollow tube so that it hits the benchtop and packs the sample into the closed end of the tube.
Load the sample to a height of $2$-$3 \: \text{mm}$.
Place a sample into a slot in the MelTemp.
Turn the dial to begin heating.
Heat at a medium rate to $20^\text{o} \text{C}$ below the expected melting point.
Then heat very slowly. ($1^\text{o} \text{C}$ every 30 seconds).
Record the temperature where the first droplet of liquid is seen (there is movement in the tube).
Record the second temperature when the entire sample liquefies (the entire sample changes from opaque to transparent).
Record a melting range, e.g. $120$-$122^\text{o} \text{C}$.
If another melting point trial is to be performed, cool the metal block to at least $20^\text{o} \text{C}$ below the next melting point, by wiping it with a wet paper towel or cooling with a jet of air.
Thiele Tube Variation:
Attach the sample to a thermometer with a tiny rubber band, positioning the sample flush with the bottom of the thermometer.
Insert the sample into a Thiele tube, so that the sample is near the middle of the tube.
7.27: Boiling Points (Thiele Tube)
Table 7.25: Procedural summary for obtaining a boiling point.
Fill a small tube about half-full with sample and insert a capillary tube, closed end up.
Attach the tube to a thermometer with a small rubber band.
Insert the sample into a Thiele tube, so that the sample is near the middle of the oil.
Heat the arm of the Thiele tube with a burner, gently and continuously.
Heat until a vigorous stream of bubbles emerges from the capillary tube, such that individual drops can be barely distinguished.
Remove the heat and allow the oil to cool.
The boiling point is the temperature when the oil just begins to enter the capillary tube.
7.28: Vacuum Sublimation
Table 7.26: Procedural summary for vacuum sublimation.
Place the sample to be sublimed in the bottom of the sublimation apparatus.
Lightly grease all joints.
Use thick-walled tubing to attach to the vacuum arm, and apply the vacuum.
The setup should not hiss or there is a leak.
Fill the cold finger, or run water through the condenser.
Be sure to apply the vacuum first, then coolant. If cooled before the vacuum, condensation may occur on the cold finger.
Wave a heat gun or Bunsen burner on the apparatus to heat the sample.
Sublimation should begin within a few minutes.
Coax solid deposited on the side of the glassware toward the cold finger by waving the heat gun/burner on the sides of the glass.
When the sublimation is complete:
Remove the coolant.
Allow the apparatus to come to room temperature.
Delicately reinstate the air pressure, noting that an abrupt opening of the vessel will cause air to knock crystals off the cold finger.
Remove the cold finger. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.25%3A_Rotary_Evaporation.txt |
Learning Objective
• Naming Unbranched Alkanes (Organic Chemistry)
An alkane is a type of hydrocarbon (a compound consisting of only carbon and hydrogen atoms). When the carbon-carbon backbone consists only of single bonds, the hydrocarbon contains as many hydrogen atoms as possible, and is therefore saturated. Alkanes are saturated hydrocarbons.
Below is a table of the names of unbranched, saturated alkanes containing up to ten carbons, with their condensed structural formulas, molecular formulas, and boiling points.
Alkane Condensed Structural Formula Molecular Formula Boiling Point (degrees C)
Methane CH4 CH4 -161
Ethane CH3CH3 C2H6 -89
Propane CH3CH2CH3 C3H8 -42
Butane CH3CH2CH2CH3 -0.5
Pentane CH3(CH2)3CH3 36
Hexane CH3(CH2)4CH3 69
Heptane 98
Octane 126
Nonane 151
Decane 174
Table of names, condensed structural formulas, molecular formulas, and boiling points of unbranched saturated alkanes containing up to ten carbon atoms.
Practice Questions
1. Fill in the blank spaces in the table.
2. Draw the structures of these ten unbranched alkanes, using Figure \(1\) (hexane) as a model.
3. What do the names of these molecules have in common?
4. What is the relationship between the number of carbon atoms and the number of hydrogen atoms in an unbranched alkane? Provide a general formula (use n for the number of carbon atoms).
5. Draw a graph of boiling points versus number of carbon atoms in unbranched alkanes.
6. How do the boiling points vary with the number of carbons? Propose an explanation for this observation.
1.02: Constitutional Isomers
Learning Objective
• Introduction to constitutional isomers using alkanes
Evaluate the two molecules below (Figure \(1\)).
• How are they the same?
• How are they different?
Two molecules which have the same molecular formula but different structural formulas, or bonding arrangements, are known as constitutional isomers.
Pentane is an alkane with five carbon atoms. It has three constitutional isomers, shown below.
Practice Questions
1. Draw three constitutional isomers of heptane.
2. Circle the constitutional isomers of hexane in the figure below. Why are the remaining molecules not considered to be constitutional isomers of hexane?
3. Draw the constitutional isomer of hexane that is missing.
4. What is the minimum number of carbons in the chain of an alkane to be able to have a constitutional isomer?
1.03: Alkyl Substituents
Learning Objective
• How to name organic molecules with alkyl substituents
An alkane can be appended onto an existing chain to create a branched molecule. This branched piece of the molecule is called a substituent. A substituent made from an alkane is called an alkyl group. Alkyl groups are named similarly to unbranched alkane chains.
Alkane Condensed Structure Alkyl Group Condensed Structure
Methane CH4 Methyl CH3-
Ethane CH3CH3 Ethyl CH3CH2-
Propane CH3CH2CH3 Propyl CH3CH2CH2-
Butane CH3CH2CH2CH3 Butyl CH3CH2CH2CH2-
Alkyl groups can also be branched. For example, there are three constitutional isomers of the butyl substituent. In these diagrams, the negative charge on the carbon indicates the site of the bond from the substituent to the rest of the molecule.
When naming molecules according to the IUPAC system of substitutive nomenclature, remember prefix-parent-suffix (like un-believe-able).
1. prefix: what are the substituents?
2. parent: how many carbons in the parent chain?
3. suffix: what is the family of compounds?
In the case of an alkane, the suffix is -ane.
Practice Questions
1. The name of this molecule is 2-methylhexane. Identify the alkyl group. Label the parent chain carbons from 1-6.
2. The name of this molecule is 3-methylheptane. Identify the alkyl group. Label the parent chain carbons from 1-7.
3. Identify the alkyl group in Molecule A. Label the parent chain carbons from 1-8. What is the name of this molecule?
4. Identify the alkyl group in Molecule B. Label the parent chain carbons. What is the name of this molecule?
5. The name of this molecule is 2,3-dimethylpentane.
What is the name of Molecule C?
6. The name of this molecule is 4-ethyl-2-methyloctane.
What is the name of Molecule D?
7. The name of this molecule is 4-isopropylnonane.
What is the name of Molecule E?
8. The name of this molecule is cyclobutane.
What is the name of Molecule F?
9. The name of this molecule is methylcyclobutane.
The name of this molecule is 1-ethyl-2-methylcyclohexane.
What is the name of Molecule G?
10. Write the steps that you use to name an alkane, in order, as instructions for a student who doesn't know how to do it.
11. Draw any alkane and go through the steps in naming your molecule.
1.04: Alkenes and Alkynes
Learning Objective
• How to name alkenes and alkynes.
An alkane is a saturated hydrocarbon, meaning that the molecule contains all the possible hydrogen atoms because all the carbon-carbon bonds are single bonds. If one of those carbon-carbon bonds is a double bond, the resulting hydrocarbon is unsaturated and called an alkene.
This alkene is named propene.
If one of the carbon-carbon bonds is a triple bond, the resulting hydrocarbon is called an alkyne.
Practice Question
1. This alkyne is named ethyne.
What is the name of Molecule A?
The double or triple bond is called a functional group, and is often the site where chemical reactions occur. Like a substituent, it is specified in the molecular name. When naming molecules according to the IUPAC system of nomenclature, remember prefix-parent-suffix (like un-believe-able).
prefix: what are the substituents?
parent: how many carbons? If there is a double or triple carbon-carbon bond in the molecule, both carbons in that bond must belong to the parent carbon chain, even if that chain does not have the greatest number of carbons.
suffix: what is the family of compounds?
Practice Questions
1. This molecule is named 2-pentene.
What is the name of Molecule B?
What is the name of Molecule C?
2. This molecule is named 4-methyl-2-pentene.
What is the name of Molecule D?
3. This molecule is named 3-isobutyl-1-octyne.
What is the name of Molecule E?
4. This molecule is named cyclohexene.
What is the name of Molecule F?
5. This molecule is named 4-methylcyclohexene. Number the carbons.
What is the name of Molecule G?
6. This molecule is named 1,3-pentadiene.
What is the name of Molecule H?
7. The location of substituents relative to the double bonds can lead to a type of constitutional isomer known as a positional isomer.
The name of this molecule is 5-methyl-1,3-cyclohexadiene.
What is the name of Molecule I?
Double or triple carbon-carbon bonds are rigid and planar. Since the carbons cannot rotate freely around the bond, cis/trans isomers are common, and the orientation may be important for chemical reactions.
Practice Questions
1. Write the steps that you use to name an alkene and an alkyne, in order, as instructions for a student who doesn't know how to do it.
2. Draw any alkene or alkyne and go through the steps in naming your molecule. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Nomenclature_Workbook_(O'Donnell)/1.01%3A_Unbranched_Alkanes.txt |
Learning Objective
• How to name halogenated hydrocarbons
The halogens are elements belonging to Group 7A. Fluorine, chlorine, bromine, and iodine can be added to hydrocarbons through reactions with their diatomic forms or when bound to hydrogen as hydrogen halides.
When a halogen atom is bound to an otherwise saturated carbon atom, the molecule is known as an alkyl halide.
Substituent Symbol Name
Fluorine Fl fluoro-
Chlorine Cl chloro-
Bromine Br bromo-
Iodine I iodo-
When naming molecules according to the IUPAC system of nomenclature, remember prefix-parent-suffix (like un-believe-able).
• prefix: what are the substituents?
• parent: how many carbons?
• suffix: what is the family of compounds?
Practice Questions
1. This molecule is named 2-chlorobutane.
What is the name of Molecule A?
2. This molecule is named 2-fluoro-3-methylpentane.
This molecule is named 4-iodo-2-methylhexane.
What is the name of Molecule B?
3. This molecule is named 1-chloro-2-butene.
This molecule is named 5-fluoro-2-hexene.
What is the name of Molecule C?
4. This molecule is named 5-fluoro-1,3-cyclohexadiene.
What is the name of Molecule D?
5. Write the steps that you use to name an alkyl halide, in order, as instructions for a student who doesn't know how to do it.
6. Draw any alkyl halide and go through the steps in naming your molecule.
1.06: Benzene and Conjugation
Learning Objective
• Understanding and naming benzene derivatives.
Aromatic hydrocarbons, like this benzene ring, have unexpected chemistry.
We would expect to name this molecule 1,3,5-cyclohexatriene and see its double bonds react like other double bonds. However, these double bonds do not react in the same way as double bonds in a standard alkene. Observe the result of the below experiment under the appropriate reaction conditions.
This surprising stability of the double bonds in the benzene ring is not due to the cyclic arrangement of the carbon chain. Under the appropriate reaction conditions, we see the following results for adding hydrogen across the double bonds of the molecules below.
Measuring the bond lengths gives the following result.
Bond Type Bond Length (Angstroms)
Single 1.54
Double 1.34
Benzene 1.4
Therefore, the three “double bonds” of the benzene ring are not true double bonds. The electrons are shared across all of the carbons in the ring, an arrangement called conjugation which is better represented by the below structure.
When depicting benzene rings using the double bond drawing, remember that the ring is a hybrid between two equally likely resonance structures.
Therefore both structures below would be named 1-chlorobenzene.
When a benzene ring has two substituents, they are named based upon their position to each other rather than by numbers.
Practice Questions
1. What is the name of Molecule A?
2. A benzene ring can be a substituent.
The name of this molecule is 2-phenyldecane.
What is the name of Molecule B?
Conjugation is not limited to cyclic structures. One of the two molecules below does not react with halogen halides under standard reaction conditions. Which molecule do you predict is the more stable one? Why? What prediction would you make about the lengths of its carbon-carbon bonds?
Practice Questions
1. Write the steps that you use to name a benzene derivative or a molecule containing a benzene ring as a substituent, in order, as instructions for a student who doesn't know how to do it.
2. Draw any benzene-containing molecule and go through the steps in naming your molecule.
1.07: Alcohols
Learning Objective
• How to name alcohols and phenols.
In organic chemistry, any alkyl group can be abbreviated as R. An alcohol, in which a hydroxy (-OH) group is attached to a carbon of the alkyl group, can be abbreviated as R-OH.
Practice Questions
1. The name of this molecule is ethanol.
What is the name of Molecule A?
2. This molecule is named 2-butanol.
This molecule is named 6-isopropyl-5-nonanol. Number the carbons.
Number the carbons. What is the name of Molecule B?
3. The name of this molecule is 3-propyl-2-octanol. Number the carbons.
4. Number the carbons. What is the name of molecule C?
5. This molecule is named 3-fluorocyclohexanol. Number the carbons.
6. Number the carbons. What is the name of Molecule D?
Molecule D
7. This molecule is named 4-hexen-2-ol. Number the carbons.
8. Number the carbons. What is the name of Molecule E?
9. This molecule is named 2,3-pentanediol.
What is the name of Molecule F?
If a hydroxy (-OH) group is attached to an aromatic ring system, it is called a phenol.
This molecule is named 2-chlorophenol or o-chlorophenol.
Practice Questions
1. Name Molecule G using 1) numbers and 2) the o/m/p nomenclature.
2. What additions do we make to our existing naming rules to name alcohols?
3. Write the steps that you use to name an alcohol in order, as instructions for a student who doesn't know how to do it.
4. Draw any alcohol and go through the steps in naming your molecule.
1.08: Ethers
Learning Objective
• How to name ethers.
In organic chemistry, any alkyl group can be abbreviated as R. An ether, in which the carbons of two alkyl groups are linked to the same oxygen, can be abbreviated as R-O-R.
This molecule is often referred to simply as “ether”. Its common name is diethyl ether, and its IUPAC name is ethoxyethane.
The common name of this molecule is ethyl methyl ether, and its IUPAC name is methoxyethane.
Practice Questions
1. What are the common and IUPAC names of Molecule A?
2. The name of this molecule is 2-methoxy-3-methylbutane. Number the carbons.
3. Number the carbons. What is the name of Molecule B?
4. The name of this molecule is 4-butoxy-2-butanol. Number the carbons.
5. Number the carbons. What is the name of Molecule C?
6. What additions do we make to our existing naming rules to name ethers?
7. Write the steps that you use to name an ether in order, as instructions for a student who doesn't know how to do it.
8. Draw any ether and go through the steps in naming your molecule. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Nomenclature_Workbook_(O'Donnell)/1.05%3A_Halogens.txt |
Learning Objective
• How to name aldehydes and ketones.
A carbonyl group consists of a carbon atom double-bonded to an oxygen atom, written as C=O. Aldehydes and ketones are two compounds which contain the carbonyl group.
Aldehydes and ketones are constitutional isomers. For example, the aldehyde and ketone below both have the molecular formula C3H6O.
The simplest aldehyde is methanal, commonly known as formaldehyde, and used as a preservative.
The name of this molecule is butanal.
Practice Questions
1. What is the name of Molecule A?
2. This molecule is named 5-methylhexanal. Number the carbons.
3. Number the carbons. What is the name of Molecule B?
4. This molecule is named 3-ethyl-4-methylhexanal. Number the carbons.
5. Number the carbons. What is the name of Molecule C?
6. This molecule is named 3-pentenal. Number the carbons.
7. Number the carbons. What is the name of Molecule D?
8. A well-known ketone is 2-propanone, commonly known as acetone, and used as a nail polish remover.
This molecule is named 2-pentanone.
What is the name of Molecule E?
9. The name of this molecule is 5-ethyl-2-heptanone. Number the carbons.
10. Number the carbons. What is the name of Molecule F?
11. This molecule is named 6-chloro-4-ethyl-3-heptanone. Number the carbons.
12. Number the carbons. What is the name of Molecule G?
13. The name of this molecule is 3-bromocyclohexanone. Number the carbons.
14. Number the carbons. What is the name of Molecule H?
15. The name of this molecule is 4-hydroxy-2-butanone. Number the carbons.
16. Number the carbons. What is the name of Molecule I?
17. This molecule is named 5-chloro-4-oxohexanal. Number the carbons.
18. Number the carbons. What is the name of Molecule J?
19. What additions do we make to our existing naming rules to name aldehydes and ketones?
20. Write the steps that you use to name an aldehyde or ketone in order, as instructions for a student who doesn't know how to do it.
21. Draw any aldehyde or ketone and go through the steps in naming your molecule.
1.10: Carboxylic acids and Esters
Learning Objective
• How to name carboxylic acids and esters.
A carbonyl group consists of a carbon atom double-bonded to an oxygen atom, written as C=O. When a hydroxy (-OH) group is also bound to the carbonyl carbon, the resulting group is known as a carboxy group. Carboxylic acids contain the carboxy group, and one type of carboxylic acid derivative is an ester.
Carboxylic acids are often abbreviated as R-COOH or R-CO2H. Some simple carboxylic acids, such as acetic acid and benzoic acid, are referred to primarily by their common names.
The common name of this molecule is valeric acid. Its formal name is pentanoic acid.
Practice Questions
1. The common name of Molecule A is caprylic acid. What is the formal name of Molecule A?
2. The name of this molecule is hexanedioic acid.
What is the name of Molecule B?
3. The name of this molecule is cyclohexanecarboxylic acid.
What is the name of Molecule C?
4. The name of this molecule is 1,2-benzenedicarboxylic acid.
What is the name of Molecule D?
5. The name of this molecule is 4-bromopentanoic acid. Number the carbons.
6. Number the carbons. What is the name of Molecule E?
Molecule E
7. This molecule is named 2-hydroxybutanoic acid (also known as lactic acid).
What is the name of Molecule F?
8. This molecule is named 3,6-dioxo-4-hydroxyhexanoic acid.
What is the name of Molecule G?
Esters are often manufactured to provide fragrance. One example is isobutyl methanoate, which smells like raspberries.
The process of esterification involves adding an alcohol to a carboxylic acid in the presence of hydrogen ions.
To name an ester, first we name the alcohol used, and then the carboxylic acid. Below is shown the esterification reaction of ethanol and propanoic acid to create ethyl propanoate.
Practice Questions
1. What is the name of Molecule H?
2. What is the name of Molecule I?
3. This molecule is named dimethylbutanedioate.
What is the name of Molecule J?
4. This molecule is named 2-chloroethyl propanoate.
What is the name of Molecule K?
5. This molecule is named 2-bromopropyl 4-chlorobutanoate. Number the carbons.
6. Number the carbons. What is the name of Molecule L?
7. What additions do we make to our existing naming rules to name carboxylic acids and esters?
8. Write the steps that you use to name a carboxylic acid in order, as instructions for a student who doesn't know how to do it.
9. Draw any carboxylic acid and go through the steps in naming your molecule.
10. Write the steps that you use to name an ester in order, as instructions for a student who doesn't know how to do it.
11. Draw any ester and go through the steps in naming your molecule.
1.11: Amines and Amides
Learning Objective
• How to name amines and amides.
An amino group consists of a nitrogen atom bonded to two hydrogen atoms, written as -NH2. If the hydrogens are replaced by R groups, the group is referred to as a substituted amino group. Amines and amides are two compounds which contain amino or substituted amino groups.
Amines are designated as primary, secondary, or tertiary amines based upon the degree of substitution of the amino group. For example, an amine in which all three of the potential nitrogen bonds are with R groups instead of hydrogens is called a tertiary amine (see the right-most molecule below).
Despite the complexity of the below molecule, because the nitrogen is directly bound to two hydrogens and only one R group, it is a primary amine.
There are multiple substitutive nomenclatures for naming amines, with the two most common being IUPAC and Chemical Abstract Service. Both will be described below.
This molecule is named 2-pentanamine (CAS) or 2-aminopentane (IUPAC).
Practice Questions
1. What are the names of Molecule A?
Molecule A
2. This molecule is named 1,6-hexanediamine (CAS) or 1,6-diaminohexane (IUPAC).
What are the names of Molecule B?
3. This molecule is named N,N-dimethylethanamine (CAS) or dimethylaminoethane (IUPAC).
N,N-dimethylethanamine (CAS) / dimethylaminoethane (IUPAC)
What are the names of Molecule C?
4. This molecule is named 4-amino-2-pentanone (CAS) or 4-aminopentan-2-one (IUPAC). Number the carbons.
5. Number the carbons. What are the names of Molecule D?
6. This molecule is named 2-(N-methylamino)ethanol (CAS) or 2-methylaminoethanol (IUPAC). Number the carbons.
7. Number the carbons. What are the names of Molecule E?
An amide is a carboxylic acid derivative in which the carboxyl -OH has been replaced with an amino or substituted amino group. Amides are also described as primary, secondary, or tertiary depending on the number of R groups bound directly to the nitrogen.
The naming of simple amides is based on the carboxylic acid nomenclature, and keeps the same name in the CAS and IUPAC systems.
The name of this molecule is 3-methylbutanamide.
Practice Questions
1. What is the name of Molecule F?
Molecule F
2. The name of this molecule is N-methylbenzamide.
What is the name of Molecule G?
3. What additions do we make to our existing naming rules to name amines and amides, in the substitutive naming system that you use?
4. Write the steps that you use to name an amine in order, as instructions for a student who doesn't know how to do it.
5. Draw any amine and go through the steps in naming your molecule.
6. Write the steps that you use to name an amide in order, as instructions for a student who doesn't know how to do it.
7. Draw any amide and go through the steps in naming your molecule. | textbooks/chem/Organic_Chemistry/Organic_Chemistry_Nomenclature_Workbook_(O'Donnell)/1.09%3A_Aldehydes_and_Ketones.txt |
• 1.1: Introduction to Pericyclic Reactions
This page provides an overview of the three types of pericyclic reactions: cycloadditions, electrocyclic reactions, and sigmatropic rearrangements.
• 1.2: Cycloaddition Reactions
This page covers common cycloaddition reactions, including the Diels-Alder reaction, ene reaction, photo [2+2], ketene [2+2], and 1,3-dipolar cycloadditions, and discusses why some are promoted by heat and others by light.
• 1.3: Electrocyclic Reactions
This page covers electrocyclic reactions promoted by heat and light including 4 pi, 6 pi, and 8 pi. It focuses on the molecular orbital analysis to explain conrotatory and disrotatory reaction pathways. It also covers ionic electrocyclic reactions including the Nazarov reaction.
• 1.4: Sigmatropic Rearrangements
This chapter highlights synthetically useful sigmatropic rearrangements including hydride shifts, the Cope rearrangement, the Claisen rearrangement, and the Wittig rearrangement.
01: Pericyclic Reactions
Learning Objectives
After completing this section, you should be able to:
1. identify a reaction as a cycloaddition, electrocyclic reaction, or sigmatropic rearrangement.
2. draw curved arrows to explain the electron movement in a pericyclic reaction.
Key Terms
Make certain that you can define, and use in context, the key terms below.
• pericyclic reaction
• cycloaddition
• electrocyclic reaction
• sigmatropic rearrangement
Study Notes
Pericyclic reactions proceed by a rearrangement of electrons through a cyclic transition state and are distinct from polar and radical reactions that students first learn about in introductory organic chemistry. This brief chapter introduces the topic of pericyclic reactions and provides examples of the three classes of reactions. The following chapters go into depth for cycloadditions, electrocyclic reactions, and sigmatropic rearrangements.
Content
Prior to 1965, pericyclic reactions were known as "no mechanism reactions" since no one could adequately explain why reaction outcomes changed depending on whether reactants were exposed to heat or light. In 1965 Robert Burns Woodward and Roald Hoffmann used Frontier Molecular Orbital Theory, initially proposed by Kenichi Fukui, to develop their Theory of Conservation of Orbital Symmetry where outcomes of pericyclic reactions are explained by examining the Highest Occupied Molecular Orbital (HOMO) or Lowest Unoccupied Molecular Orbital (LUMO) of the reacting system. Their analysis of cycloadditions, electrocyclic reactions, and sigmatropic rearrangements is commonly referred to as the Woodward-Hoffmann Rules. A brief overview of these three reaction types is shown below and a detailed analysis is provided in the subsequent chapters.
All pericyclic reactions are concerted, they occur in one step with no intermediates formed. They are highly stereoselective, thus providing excellent methods for the synthesis of stereocenters. Product formation depends on three things: 1) Structure of the reactant, 2) Number of electrons (orbitals) involved, and 3) Conditions (heat or light). Understanding the outcome of pericyclic reactions is only possible by looking at the molecular orbitals involved, and only in-phase orbitals can overlap to form bonds during pericyclic reactions.
Cycloadditions
Cycloaddition reactions can be inter- or intramolecular and involve two different pi systems combining to form two new sigma bonds. (Cycloadditions are the only pericyclic reactions that can involve intermolecular reactions. The reverse of a cycloaddition is a cycloreversion.) They are the most convergent and synthetically useful pericyclic reactions. Common examples of cycloadditions include the Diels-Alder reaction to form 6-membered rings, dipolar cycloadditions to form 5-membered rings, and photo [2+2] cycloadditions to form 4-membered rings. Examples are shown below with the newly formed sigma bonds in the products highlighted in magenta. It is important to note that the first two examples are possible with heat while the final example only happens with light.
To understand these results, as mentioned above, we must look carefully at the pi system HOMO and LUMO involved in each reaction. An example of this is provided below. In a standard Diels-Alder reaction, an electron rich diene reacts with an electron poor dienophile, meaning the diene reacts from its pi HOMO (psi 2) while the dienophile reacts from its pi LUMO (psi 2). (Note: The dashed black lines in the figure below represent nodes in the pi molecular orbitals of the diene and dienophile.) The two new sigma bonds, shown as dashed magenta lines below, are formed from constructive overlap of the dienophile orbitals with the terminal orbitals of the diene.
We will explore a more detailed analysis of these orbital interactions in the next chapter. For now, we can summarize the most common outcomes in the table below. We should note that there is one important cycloaddition, the [2+2] thermal reaction of ketenes that does not conform to this generalized analysis.
Generalized Statement of Woodward-Hoffmann Rules for Cycloadditions
Number of Electrons Thermal Photochemical
4n + 2 Allowed Forbidden
4n Forbidden Allowed
Electrocyclic Reactions
Electrocyclic reactions are intramolecular reactions that are ring closing (form a sigma bond) or ring opening (break a sigma bond). The key sigma bond, either formed or broken, must be at the terminus of a pi system so that either the product or reactant must be a fully conjugated diene, triene, etc. These reactions are often reversible and are classified by the number of pi electrons involved. Thus, 4 pi reactions involve the forming/breaking of 4 membered rings, 6 pi reactions involve the forming/breaking of 6 membered rings, and so on. The 4 pi and 6 pi variants are by far the most common and are illustrated below with the key sigma bond highlighted in magenta.
Electrocyclic reactions happen from the HOMO of the molecule and differ in their outcomes based on orbital rotation for the forming/breaking sigma bond. If the orbitals involved rotate in the same direction (both counterclockwise or both clockwise), the process is called conrotatory. If the orbitals involved rotate in opposite directions (one clockwise and one counterclockwise), the process is called disrotatory. These differences in rotation are critically important when stereocenters are formed or broken.
As illustrated below, treatment of the susbstituted hexatriene shown yields the cis product when heated but the trans product when treated with light. This can be explained by looking at the pi HOMO for each case. For the thermal reaction, the HOMO is psi 3 which rotates in a disrotatory fashion to yield the cis product while the trans product is formed under irradiation due to the conrotatory motion of the HOMO psi 4.
We will delve deeper into this analysis in a subsequent chapter. In general, electrocyclic reactions behave according the generalized rules outlined below.
Generalized Statement of Woodward-Hoffmann Rules for Electrocyclic Reactions
Number of Electrons Thermal Photochemical
4n + 2 Disrotatory Conrotatory
4n Conrotatory Disrotatory
Sigmatropic Rearrangements
Sigmatropic rearrangements involve an intramolecular rearrangement of a pi system where a sigma bond in the reactant is broken and reformed in a different position in the product. The reactant and product have the same number and type of bonds, just different bond locations. The most common examples include hydrogen shifts across a diene system (called a [1,5] H shift) and rearrangements of double allyl-type systems (called [3,3] rearrangements). Sigmatropic rearrangements are labeled based on the position of the sigma bond in the reactant that is broken compared to the position of the sigma bond in the product that is formed. As shown in the examples below, the atoms on the sigma bond in the reactant that is broken (magenta bond) are both labeled "1", and the numbering of atoms on each side of that sigma bond continue until the atoms connected by the new sigma bond in the product (magenta) are reached. Thus, the H shift is [1,5] because the key sigma bond in both the reactant and product is to the H while the H moves from C-1 to C-5. For the [3,3] rearrangement, the broken sigma bond migrates across two allyl-type systems and forms between atoms "3" and "3" in the product. This particular example is a Claisen rearrangement since an allyl vinyl ether is transformed into a 1,4-carbonyl alkene.
Sigmatropic rearrangements can occur on one face of the molecule (think top or bottom, like a syn addition to an alkene) which is called a suprafacial reaction or from one face to the other (think from top to bottom or vice versa, like an anti addition to an alkene) which is called antarafacial. Suprafacial reactions are much more common. For example, a [1,5] H shift like the one shown below illustrates a hydrogen atom (s orbital) moving across a 5 atom pi system in a suprafacial fashion. The bond breaking and the bond forming, magenta dashed lines, are both on the bottom face of the pi system. Since three curved arrows are involved in the mechanism, this is a 6 pi reaction.
We will explore these ideas further in a subsequent chapter. The general rules for sigmatropic rearrangements are shown in the table below.
Generalized Statement of Woodward-Hoffmann Rules for Sigmatropic Rearrangements
Number of Electrons Thermal Photochemical
4n + 2 Suprafacial Antarafacial
4n Antarafacial Suprafacial | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/01%3A_Pericyclic_Reactions/1.01%3A_Introduction_to_Pericyclic_Reactions.txt |
Objectives
After completing this section, you should be able to
1. Identify cycloaddition reactions including the Diels-Alder reaction, dipolar cycloaddition, photo [2+2] reaction, thermal ketene [2+2] reaction, and ene reaction.
2. Understand the orbital analysis of cycloaddition reactions
3. Draw curved arrows to illustrate electron flow for cycloaddition reactions
4. Given cycloaddition starting materials, accurately predict the product including stereochemistry
5. Use retrosynthetic analysis to determine starting materials given a cycloaddition target
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Diels-Alder reaction
• Diene
• Dienophile
• HOMO
• LUMO
• Dipolar cycloaddition
• Dipole
• Dipolarophile
• Photo [2+2] cycloaddition
• Ketene thermal [2+2] cycloaddition
• Ketene
• Ketenophile
• Ene reaction
• Ene
• Enophile
Study Notes
Cycloadditions are highly synthetically useful reactions that provide access to 4-, 5-, and 6-membered rings with excellent stereocontrol. Your goal is to understand the theoretical basis of these reactions using a molecular orbital analysis and to apply these reactions in synthesis. You should focus on accurately predicting products with a focus on stereochemistry and, given a complicated cyclic product, to accurately predict the starting materials used to generate it.
The previous section provided an overview of cycloaddition reactions. They are one class of pericyclic reactions that involve combination of two different pi systems to yield a cyclic product. They can be inter- or intramolecular and can produce all carbon and heterocyclic ring systems. They are powerful tools that synthetic chemists often employ to generate ring structures central to complex natural products. The following scheme highlights the four major types of cycloaddition reactions that we will study in this chapter: Diels-Alder reaction, dipolar cycloaddition, photo [2+2] reaction, and thermal ketene [2+2] reaction. We will also see an additional cycloaddition reaction, the ene reaction, that is related to the Diels-Alder reaction but doesn't form a ring.
Diels-Alder Reaction
The Diels-Alder reaction is a [4+2] cycloaddition (4 pi electrons from one reactant and 2 pi electrons from the other reactant) that yields a functionalized 6-membered ring product. It is the most useful cycloaddition reaction due to the ubiquity of 6-membered rings and its ability to reliably control stereochemistry in the product. Students generally learn about this reaction in introductory organic chemistry, so this section will start with an overview of key points before addressing more advanced concepts.
Regiochemistry
In standard Diels-Alder reactions, an electron rich diene (4 pi component) reacts with an electron poor dienophile (2 pi component). To determine how they combine, you must identify the most electron rich terminal atom of the diene and have that atom form a bond with the most electron poor atom of the dienophile. This is sometimes called the Ortho/Para Rule which can lead to confusion since we generally aren't making benzene products. We will avoid this memorization tool and instead focus on drawing resonance structures to find the key electron rich and electron poor atoms as demonstrated below. The examples below also demonstrate that the dieophile can be an alkyne which yields a cyclohexadiene product.
Stereochemistry - Stereospecific for Diene and Dienophile Atoms
Due to the concerted mechanism for cycloaddition reactions, the geometry of atoms on the dienophile or the diene maintain their orientation in the product. This is a critical point for the 4 atoms (both dienophile atoms and the terminal atoms of the diene) that become sp3 hybridized and thus are potential stereocenters in the product. (As a reminder, when chiral products are formed, we obtain a racemic mixture of enantiomers.) As highlighted below, cis dienophiles yield cis substituents in the product, while trans dienophiles yield trans product substituents. Substituents on the terminal atoms of the diene also can become stereocenters and this analysis is a little less straightforward than for dienophile substituents. The way to think about the diene substituents is whether the are pointing "outside" or "inside" the diene. These orientations are illustrated below. When groups are both pointing "outside" or "inside", we can consider them to be cis and they will end up cis in the product. When one group is pointing "outside" and one "inside", we can consider them as trans and they will be trans in the product. Note: CHO is a common abbreviation for an aldehyde.
Stereochemistry - Alder Endo Rule
The Alder Endo Rule enables us to predict the product stereochemistry when new stereocenters are created on atoms that originated in both the diene and dienophile. This rule is often described using the terms endo and exo in relation to the structure of bicyclic products. This is often confusing and difficult to apply in many circumstances. Instead we will define it in relation to substituents on the diene and dienophile. (We will explain the orbital origin of this rule shortly.) The Alder Endo Rule states the the electron withdrawing group on the dienophile (W group) ends up cis to groups pointing "outside" the diene in the cyclohexene product. The two reactions below illustrate the Alder Endo Rule and how to apply it. The second reaction highlights what we will see often in this section; cyclic dienes react with cyclic dienophiles to create complicated polycyclic products. We need to pay careful attention to stereochemistry in these cases and can use either option for drawing the result.
Transition State
Like all cycloaddition reactions, the Diels-Alder is concerted. Its mechanism has one transition state and no intermediates. The orientation of the reactants in the transition state determines the structure of the product. As previously mentioned, the way we can understand the mechanism is to look at the molecular orbitals involved. This will illustrate that there are primary orbital interactions (shown below in magenta), the orbitals that overlap to form new sigma bonds, and secondary orbital interactions (shown in blue), the orbitals that interact to give rise to the Endo Rule. The reactive conformation and transition state shown below illustrate that the stereochemistry is set when the reactants come together in the transition state with the W group pointing the opposite direction of the CH2 in the cyclopentadiene ring. Thus, the bridge in the bicyclic product (the CH2 from the reactant) is trans to the ketone.
Diels-Alder Drawing Tool
One way to keep track of stereochemistry in the Diels-Alder reaction is to use a drawing tool that mimics the reactive conformation. Note: Follow the rule for determining the regiochemistry and be sure to place the most electron rich terminal atom of the diene next to the most electron poor atom of the dienophile when using this tool. It is easy to get the incorrect product if you forget this step. First, draw the structure of the cyclohexene product ignoring stereochemistry. (The next steps will enable you to determine the stereoechemistry in the product.) Second, draw the diene with the opening pointing to the right. Third, draw the dienophile underneath the diene with the W group pointing left. Fourth, identify substituents that are positioned "outside" the diene and substituents that are positioned "inside" the diene. All groups "outside" will be cis in the product. All groups "inside" will be cis in the product. "Outside" groups will be trans to "inside" groups in the product. The figure below illustrates how to use the drawing tool. Fifth, redraw the product with all of stereochemistry.
Exercise \(1\)
Predict the product of this intermolecular Diels-Alder reaction. Clearly show both enantiomers of the product.
Answer
Exercise \(2\)
Predict the product of this intramolecular Diels-Alder reaction. Clearly show both enantiomers of the product.
Hints: For intramolecular reactions, regiochemistry considerations are often ignored because one bicyclic structure is much more stable than the other possibility. Be sure to number out the starting material to identify the cyclohexene that will form and the size of the other ring that is produced as a result of the Diels-Alder reaction. Try your best to redraw the molecule in a reactive conformation. The Drawing Tool also works well for intramolecular reactions to help determine the stereochemistry of the product.
Answer
Exercise \(3\)
Propose a synthesis of the following substituted cyclohexane.
Hints: This takes more than one step. Try to think retrosynthetically. Focus on forming the stereocenters using the Diels-Alder reaction. Think about the electronic requirements of the Diels-Alder reaction.
Answer
This target is a substituted cyclohexane. The Diels-Alder reaction yields a substituted cyclohexene. So, this product was likely made by alkene hydrogenation of a Diels-Alder product. When thinking backwards, this means we need to determine which position in the six-membered ring contained the double bond. We know that the Diels-Alder reaction is an excellent method for making stereocenters, so we should assume the carbons with the aldehyde and isopropyl groups aren't part of the alkene. We also know that standard Diels-Alder reactions have electron withdrawing groups on the dienophile, so we should place the alkene in a position where the aldehyde starts on the dienophile. This leads to two possibilities for cyclohexene targets.
In option A, the atoms that will become stereocenters originate on both the diene and dienophile. This means that the isopropyl group must point "inside" the diene to yield the trans product according to the Endo Rule. This type of diene conformation is sterically very hindered as can be seen just looking at the image, so the diene will prefer to be in an unreactive conformation (rotate the singe bond between the two alkenes) and this reaction will occur slowly. Option B has no drawbacks. We have a simple and reactive diene and a trans dienophile that will yield the desired trans product.
The synthesis follows in two steps from the retrosynthetic analysis.
Forming Heterocycles with the Diels-Alder Reaction
There are many fascinating applications of the Diels-Alder reaction. We have already seen in the problems above that we will encounter intramolecular Diels-Alder reactions to enable us to make complicated polycyclic targets. We will also encounter hetero Diels-Alder reactions where one or more atoms of the diene or dienophile contains atoms other than carbon. Oxygen and nitrogen are most popular and will enable us to make six-membered heterocyclic rings, as shown below. Additionally, it is important to note that Diels-Alder reactions can be promoted not only with heat but also with Lewis acids. These reagents bind to the dienophile electron withdrawing group, making the most reactive carbon more electron poor and, thus, more reactive.
With the right combination of diene, dienophile, and an interesting cascade mechanism, the Diels-Alder reaction can yield benzene products. The reaction shown below is one example. This is a two step process involving a cycloaddition and a cycloreversion. Try to generate a mechanism to explain this interesting reaction. Why is this a favorable process?
Exercise \(4\)
Propose a mechanism and explain why this is a favorable reaction.
Answer
The first step is the Diels-Alder reaction that we would expect to yield the bicyclic intermediate that then breaks down via a retro Diels-Alder reaction (cycloreversion). The intermediate isn't the product because the cycloreversion yields a stable aromatic product and generates a stable gaseous product that bubbles out of the reaction.
Diels-Alder Reaction and Retrosynthetic Analysis
In one of the problems above, we introduced the idea of thinking backwards using retrosynthetic analysis to help us determine the diene and dienophile that could react to yield a target six-membered ring. This is a skill that we will use often when analyzing complicated targets. The key structural fragment, or retron, that we look for in a molecule that alerts us that we can make it by a Diels-Alder reaction is a six-membered ring. More specifically, it's a six-membered ring containing one pi bond. For a traditional all-carbon Diels-Alder reaction, the retron is a cyclohexene. The following examples highlight how we can use retrosynthetic analysis for standard synthesis problems.
Exercise \(5\)
Propose a Diels-Alder reaction that will yield each of the target molecules.
Answer
Begin your retrosynthetic analysis by performing a retro Diels-Alder reaction to reveal the diene and dienophile that can combine to make your target. Draw your first curved arrow by starting at the alkene in the six-membered ring. Move the electrons clockwise or counterclockwise then draw the resulting diene and dienophile. The stereochemistry in the product will dictate how the subsitutuents are arranged on the starting materials. In this case, the dienophile contains a nitro electron withdrawing group and a trans methyl. The methoxy substituent on the diene must be pointing out since it ends up cis to the nitro group in the product. The methyl group on the diene must be pointing in because it is trans to the methoxy in the product.
It is often helpful to number all of the atoms in the product before beginning your retrosynthetic analysis. For complicated structures, like this bicyclic target, it is essential to number the atoms. We use the same strategy as the first problem by beginning our curved arrows for the retro Diels-Alder at the alkene. This reveals that we are breaking C1-C10 and C4-C9 and that we must perform an intramolecular Diels-Alder to generate the target. Be sure to pay attention to stereochemistry which reveals that the dienophile stereochemistry is trans (since C8 and C11 are trans in the product). Since C5 and the ketone withdrawing group are cis, C5 must be pointing out on the diene.
Synthetically Useful Diels-Alder Dienes
There are several other specific target structures that we need to become familiar with since they will show up in future synthetic applications. The first is a bicyclic 6,6 system with one benzene ring. This is a retron for a Diels-Alder reaction using an ortho-quinodimethane diene. Dienes of this type are highly reactive and are generally made by two different strategies. One is an electrocyclic ring opening, a reaction we will study in more detail in the next chapter, and the other is a fluoride promoted vinylogous elimination reaction.
Conjugated cyclohexenones are highly valuable synthetic building blocks. These types of structures are often made using enolate chemistry, specifically the Robinson annulation. They can also be synthesized via the Diels-Alder reaction using a diene developed by Sam Danishefsky which is known as Danishefsky's diene. The Diels-Alder retron for use of this diene is shown below along with the complete retrosynthetic analysis and the structure of Danishefsky's diene. The Diels-Alder product of this reaction yields the target cyclohexenone upon treatment with acid. Propose a mechanism for this process which is a good review of acid promoted carbonyl-type chemistry.
Exercise \(6\)
Propose a mechanism for the conversion of the Diels-Alder product from use of Danishefsky's diene into the target cyclohexenone product upon treatment with acid.
Answer
Diels-Alder Reaction Orbital Analysis
To understand why the Diels-Alder reaction occurs thermally and not photochemically, we must analyze the molecular orbitals involved in the reaction. This type of analysis, often called the Woodward-Hoffmann Rules, was introduced in the previous chapter. We will take a deeper look in this section. As mentioned previously, cycloadditions combine two independent pi systems to form two new sigma bonds. We must have constructive overlap between the Highest Occupied Molecular Orbital (HOMO) of one system (the nucleophile) and the Lowest Unoccupied Molecular Orbital (LUMO) of the other system (the electrophile). There are two possible modes or overlap: suprafacial, where both sigma bonds form on the same side of the pi system, and antarafacial, where the sigma bonds form on opposite sides of the pi system. Examples similar to these modes of overlap are also found in alkene reactions. Suprafacial reactions are similar to alkene syn additions (e.g., H2, cat. Pd/C) while antarafacial reactions are similar to alkene anti additions (e.g., Br2). The Diels-Alder reaction is suprafacial with respect to both diene and dienophile, as illustrated below. (The diene is on the top; the dienophile on the bottom. Dashed black lines are nodes; dashed magenta lines are constructive overlap where new sigma bonds can form.) Two examples of antarafacial reactions (that don't actually occur) are shown for comparison. Antarafacial cycloadditions are very unusual, and we will only discuss one example, the ketene thermal [2+2] reaction that will be introduced later in this chapter.
The molecular orbital analysis for cycloadditions involves several steps. First, determine how many atoms from each component are involved in the reaction and draw out their pi molecular orbitals. For the Diels-Alder reaction, there are 4 atoms with p orbitals in the diene and 2 atoms with p orbitals in the dienophile. The reactive orbitals for the Diels-Alder reaction are pi molecular orbitals, so we will focus on the pi MOs for both the diene and dienophile. The dienophile has one pi bond, so we will look at the pi MOs for a 2 atom system. The diene has two pi bonds, so we will look at the pi MOs for a 4 atom system. The details are shown below. We will use these MO Diagrams for all types of pericyclic reactions, they are not specific to the Diels-Alder reaction or cycloadditions.
Second, add electrons to the pi MOs to determine the HOMO and LUMO for each reactant. The dienophile has 1 pi bond, so it has 2 pi electrons. The diene has 2 pi bonds, so it has 4 pi electrons. For the dienophile, the HOMO is psi 1 and the LUMO is psi 2*. For the diene, the HOMO is psi 2 and the LUMO is psi 3*.
Third, determine if the reaction is possible by combining the HOMO of one component with the LUMO of the other. In a standard Diels-Alder reaction, an electron rich diene reacts with an electron poor dienophile. This means the diene is the nucleophile and reacts from the HOMO, while the dienophile is the electrophile and reacts from the LUMO. The opposite electronic arrangement, diene LUMO plus dienophile HOMO, is also possible. This is known as an Inverse Electron Demand Diels-Alder reaction. Both possibilities are illustrated below. These diagrams clearly demonstrate these reactions are suprafacial with respect to both the diene and dienophile, and, thus, they will happen. (Note: As always, dashed black lines are nodes and dashed magenta lines are positive orbital overlap that results in new sigma bond formation. Solid curved lines showing positive orbital overlap have been omitted for clarity.) To avoid potential confusion, we will delay analysis of photochemical cycloadditions until a little later in the chapter when we get to photo [2+2] reactions. If we did analyze the Diels-Alder under photochemical conditions, it would confirm that this is not possible.
Dipolar Cycloaddition Reactions
Another useful thermal reaction is the dipolar cycloaddition which provides ready access to a variety of 5-membered ring heterocycles. The retron for a dipolar cycloaddition is a five-membered ring heterocycle. The key reaction component is a 3-atom dipole generally consisting of a pi bond adjacent to a negatively charged atom with a lone pair. This provides 4 electrons for the reaction, 2 electrons in the pi bond and 2 electrons from the delocalized lone pair. Thus, this reaction is electronically equivalent to the Diels-Alder reaction with the dipole contributing 4 pi electrons and the dipolarophile contributing 2 pi electrons. Two common 3 atom dipoles that many students encounter in previous organic chemistry classes are diazomethane and ozone. In fact, the ozonolysis mechanism, pictured below, is a wonderful example of dipolar cycloadditions in action. The first step is a dipolar cycloaddition which is followed by a retro dipolar cycloaddition (a cycloreversion). (Note: Retro dipolar cycloadditions are very unusual.) After a molecular rotation of one or the other component, a second dipolar cycloaddition occurs to yield an intermediate that is stable until the reaction is worked up under either oxidative or reductive conditions. This is a very complicated transformation. Problems that we will see using dipolar cycloadditions will generally be much more straightforward. As we will see in the problems that follow, the dipole and dipolarophile can contain triple bonds and, like all cycloadditions, this reaction can occur intramolecularly.
Exercise \(7\)
What product is generated in each of the following reactions?
Answer
Part a) demonstrates the utility of a dipolar cycloaddition to make an aromatic heterocycle. Think about the regiochemistry of this one based on the electronic analysis we did for the Diels-Alder reaction. In this case, the dipole is the nucleophile with the negatively charged oxygen. Think about why the internal carbon on the alkyne is the most electron poor.
Part b) is complicated because of the challenging tricyclic structure that forms in the reaction. This is an example where numbering the atoms is critical and building a model is a huge help.
The orbital analysis of dipolar cycloadditions is very similar to the Diels-Alder reaction, though it requires us to understand how to generate the pi molecular orbital picture for a 3 atom system (illustrated below). If you have already studied pi molecular orbitals for an allyl system (radical, carbocation, or anion), the dipole in our reactions can be treated as an allyl anion. This is consistent with what we said previously about the dipole being a 3 atom system that has 4 pi electrons. In standard dipolar cycloadditions, the electron rich component is the dipole which reacts from the psi 2 HOMO while the electron poor dipolarophile reacts from the psi 2* LUMO. The orbital picture of the dipolarophile is identical to the Diels-Alder dienophile, both are two atom pi systems. Like the Diels-Alder reaction, the electronics can be reversed so that an electron poor dipole reacts with an electron rich dipolarophile, psi 3* LUMO plus psi 1 HOMO, respectively. Both options are shown below.
Photo [2+2] Cycloadditions
Photochemical [2+2] cycloadditions are excellent reactions for the synthesis of strained products containing 4-membered rings. They produce all carbon and heteroatom rings by both inter- and intramolecular reactions. One of the reaction partners must be conjugated so that it can absorb light and become an excited state molecule. These reactions produce strained 4-membered rings but are not reversible because the products lack conjugation and, thus, can't absorb light to facilitate a cycloreversion.
Stereochemistry in photo [2+2] reactions is more straightforward than the Diels-Alder reaction. The steochemistry present in each reactant is maintained in the product, so cis alkenes yield cis product substituents and vice versa. When comparing possible stereochemistry between the two reactants, the least hindered product is favored. For example, in the reaction below, the all cis product is not formed; the trans structure is generated.
Because this is a photochemical reaction, predicting regiochemistry is the opposite of what we would expect from our standard analysis of thermal reactions like in the Diels-Alder reaction. This result is illustrated below. Instead of combining the most electron rich atom in one reactant with the most electron poor atom of the other reactant, it looks like we are combining the most electron rich atom with the other most electron rich atom (or vice versa). Why does this happen? It is important to remember that, as we will see in more detail shortly, one of the reactants participates in the reaction in its excited state. Molecules in the excited state have the opposite electron configuration than in the ground state. So, the most electron rich atom becomes the most electron poor atom and vice versa. One way to accurately predict the correct regiochemistry is to determine the correct ground state regiochemistry then switch the substituents to get the correct photoreaction regiochemistry.
So, why won't this reaction happen thermally? How does our molecular orbital analysis help us understand the importance of this being a photochemical reaction? First, let's look at the orbital analysis if we tried to do a thermal [2+2] reaction. As shown below, we cannot get suprafacial overlap for both of the 2 pi reactants when trying to combine psi 1 HOMO with psi 2* LUMO. This means that it is not favorable to convert the two reactant pi bonds into two new product pi bonds.
What happens when we shine light on the reaction? Light creates an excited state molecule by promoting an electron in the HOMO to the LUMO, as shown below. This means the excited state HOMO is the ground state LUMO. We need to understand a few key points about photoreactions before doing our molecular orbital analysis. Excited state molecules are very short lived, relaxing back to the ground state very quickly. Therefore, it is practically impossible for two excited state molecules to find each other in a reaction. Instead, reactions occur between one excited state molecule and one ground state molecule. When only one molecule is conjugated, that is the molecule that will form the excited state. If both reactants are conjugated, either can form the excited state. The orbital analysis is shown below. First, we see the orbital picture when a ground state molecule absorbs light to form an excited state. Second, when we analyze the reaction, it is now psi 2* HOMO of the excited state molecule reacting with psi 2* LUMO of the ground state molecule. This gives suprafacial constructive overlap for both orbitals and the 2 reactant pi bonds can be converted into two product sigma bonds. The reaction occurs!
Exercise \(8\)
What are the products of the following two reactions? Hints: For a), this is an example of a hetero photo [2+2] cycloaddition. For b), your product is the result of two cycloadditions.
Answer
Part a is known as a Paterno-Bucchi reaction. It's a photo [2+2] reaction that combines a carbonyl with an alkene to form an oxetane. Also, like many organic reactions, it is very unusual for benzenes to participate in cycloadditions. Don't forget to think about stereochemistry and regiochemistry when drawing your product. Since the two benzenes start on each of the reactants, they will end up trans in the product because this is sterically favored. Performing our ground state analysis, it becomes clear that the molecule in the box is the product from the photoreaction.
Part b combines a Diels-Alder reaction with a photo [2+2] to make a very interesting and highly strained product. Using photochemistry to generate strain in a target molecule is a very useful synthetic strategy.
Ketene Thermal [2+2] Cycloaddition
Ketenes are unusual functional groups that react in unexpected ways. (Ketenes contain an sp hybridized carbon connecting a carbonyl directly to an alkene. The all carbon equivalent is an allene. The 2 pi bonds in ketenes and allenes are not conjugated; they are perpendicular.) The most common ketene reaction is a thermal [2+2] cycloaddition. We will explore the orbital explanation for this shortly. First, let's look at a common method for the synthesis of ketenes and an example of a reaction. Ketenes can be reliably synthesized upon reaction of an acid chloride with a non-nucleophilic amine base, in this case triethylamine, via an elimination reaction. Once formed, the ketene undergoes a [2+2] cycloaddition to yield a cyclobutane product. This reaction follows the same stereochemistry rules as photo [2+2] cycloadditions. Since this is a thermal reaction, the regiochemistry is straightforward. Find the most electron rich atom in the ketenophile and react it with the highly electron poor carbonyl carbon of the ketene.
The orbital analysis is exactly the same as our discussion of the photo [2+2] reaction above except that the thermal reaction that doesn't occur in that analysis is now happening. How is that possible? The critical factors are that the ketene carbonyl carbon is both electron poor and unhindered since no other substituents are attached. This means that the ketenophile reacts in a suprafacial orientation from HOMO psi 1 while the ketene reacts in an antarafacial fashion from LUMO psi 2*, as depicted below. This antarafacial reactivity is only possible because of the sp hybridization and resulting lack of steric hinderance for the approaching ketenophile. Thinking about this in three dimensions helps visualize how this is possible (see representation below).
It is normally impossible to see the result of this suprafacial plus anatarafacial reactivity, but the following intramolecular ketene [2+2] reaction clearly illustrates the reaction mechanism. Instead of the standard fused bicycylic product that would result from a suprafacial plus suprafacial reaction, the more complicated bridged bicyclic product is generated. This is only possible because of the suprafacial plus antarafacial reactivity.
Exercise \(9\)
Predict the product of the following reaction.
Answer
This problem demonstrates to power of the ketene thermal [2+2] reaction to make heterocycles. Combination of a ketene with an imine yields a 4-membered ring amide which is known as a beta lactam. This functional group is present in bioactive molecules like penicillin.
Ene Reaction
Our final cycloaddition reaction is another transformation discovered by Kurt Alder. The Diels-Alder reaction was originally know as the "diene reaction". This transformation involves an alkene, so it was named the "ene reaction". An example with the mechanism is shown below. The alkene is the 4 electron component reacting with its pi bond and an allylic C-H bond. The enophile is the 2 electron component making this another thermal [4+2] reaction. Intermolecular ene reactions happen only with very electrophilic alkenes or alkynes and often with a Lewis acid catalyst. Intramolecular reactions are much more common and occur with heating or Lewis acid catalysis.
So, is this really a cycloaddition reaction since we aren't using pi bonds in the mechanism? Yes, it is. As we will see in subsequent chapters, sigma bonds, especially C-H bonds, can participate in pericyclic reactions. The orbital picture is shown below where we treat the sigma plus pi bonds of the ene as a 4 atom system. Thus, the reaction occurs between HOMO psi 2 of the ene and LUMO psi 2* of the enophile. With 4 electrons from the ene and 2 electrons from the enophile, this is another 6 electron, thermally allowed cycloaddition, just like the Diels-Alder reaction and dipolar cycloaddition.
Exercise \(10\)
What products are formed in the following reactions? Hint: For b, the product contains a new 5-membered ring.
Answer
The most common version of the ene reaction is known as the "carbonyl ene reaction" where the enophile is a carbonyl, usually an aldehyde. This generates a product that is a homoallylic alcohol which is the retron for a carbonyl ene reaction. An example is shown below.
An application of the carbonyl ene reaction is the two-step conversion of citronellal to menthol, as shown below. The first step is an intramolecular carbonyl ene reaction followed by an alkene hydrogenation. An interesting question is how is the stereochemistry controlled in the ene reaction? Like many reactions that we will see in subsequent chapters, the key is a chair-like 6-membered ring transition state. With the stereogenic methyl positioned equatorial, this enables the developing stereocenters, the OH and the propene, to be equatorial in the transition state. Try to remember this strategy when thinking about stereochemistry for intramolecular reactions in the future, not just the ene reaction.
Summary Problems
Exercise \(1\)
In 2012, Tom Hoye from the University of Minnesota reported a novel Diels-Alder reaction called the Hexadehydro-Diels-Alder reaction with a diyne as the diene and another alkyne as the dienophile. Hint: Benzyne plays a critical role in this reaction mechanism.
a) Here’s an example of this novel type of Diels-Alder reaction. Propose a mechanism for this transformation. Hint: The first step is an intramolecular Diels-Alder reaction.
b) Using the mechanism you determined in part a, determine the product of this reaction.
Answer
DOI - 10.1038/nature11518 (Nature 2012)
a) The Diels-Alder reaction yields a cyclic allene that is one of the resonance structures for benzyne. So, this is a cycloaddition that yields a benzyne intermediate! Under these reaction conditions, the benzyne reacts as its polar resonance structure to first deprotonate acetic acid and then combine with acetate to yield the final product.
b) In this example, without the addition of acid, the benzyne intermediate deprotonates the sulfonamide proton. The resulting negatively charged nitrogen forms an intramolecular C-N bond resulting in the synthesis of a new 5-membered ring.
Exercise \(2\)
In 1964, Prof. Phil Eaton at the University of Chicago achieved the first synthesis of cubane. A key molecule in his synthesis is shown below. This compound can be prepared in two steps via dimerization (one molecule reacting with itself) of a monocyclic compound followed by an intramolecular reaction. Determine the structure of the starting material and the two reactions needed to prepare the key synthetic intermediate. Hint: Both reactions are cycloadditions. (Don’t worry about how to convert the dibromide into cubane.) As a bonus question, how many peaks do you predict will be in the 1H and 13C NMR spectra for cubane?
Answer
DOI - 10.1021/ja01069a041 (Journal of the American Chemical Society 1964)
There are several possible photo [2+2] disconnections, but this one reveals a tricyclic compound that can be made by a Diels-Alder reaction between two of the same bromocyclopentadienone molecules. It is a remarkable two-step sequence to generate the highly complex key synthetic intermediate. For the question of cubane's NMR spectra, this molecule is fully symmetrical and yield only one peak in both the 1H and 13C NMR spectra. | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/01%3A_Pericyclic_Reactions/1.02%3A_Cycloaddition_Reactions.txt |
Objectives
After completing this section, you should be able to:
1. Identify electrocyclic reactions including 4 pi, 6 pi, and 8 pi ring opening and closing reactions
2. Understand the orbital analysis of electrocyclic reactions
3. Draw curved arrows to illustrate electron flow for electrocyclic reactions
4. Given electrocylic starting materials, accurately predict the product including stereochemistry
5. Use retrosynthetic analysis to determine starting materials given an electrocyclic target
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Conrotatory
• Disrotatory
• Nazarov reaction
Study Notes
Electrocyclic reactions are often reversible ring opening and closing reactions that cleave and form 4-, 6-, and 8-membered rings with excellent stereocontrol. (The ring opened products must contain a p orbital at every atom.) We will also learn about cationic and anionic electrocyclic reactions that enable formation of 5-membered rings. Your goal is to understand the theoretical basis of these reactions using a molecular orbital analysis and to apply these reactions in synthesis. You should focus on accurately predicting products with a focus on stereochemistry and, given a complicated product, to accurately predict the starting material used to generate it.
Chapter 1.1 provided an overview of pericyclic reactions, including electrocyclic reactions. Electrocyclic reactions are intramolecular reactions involved in the formation of rings starting from a conjugated pi system or the formation of a conjugated pi system by cleaving a ring. In ring forming reactions, a new sigma bond (shown below in magenta) is formed at the terminus of a conjugated pi system, while ring opening reactions break a sigma bond (in magenta below) to generate a fully conjugated pi system. These are important reactions but they are not nearly as synthetically useful as cycloadditions. We will begin by looking at standard ring opening and closing reactions of neutral carbon-containing compounds. We will then explore applications of ionic electrocyclic reactions including the cationic Nazarov reaction for the synthesis of 5-membered rings.
Example Electrocyclic Reactions
Building on the introduction to electrocyclic reactions presented in Chapter 1.1, we must understand the following results for the formation and cleavage of 6-membered rings. (Note: Electrocyclic reactions are only possible for 1,3-cyclohexadiene derivatives or similar heterocycles and not other 6-membered rings lacking 2 pi bonds in the correct orientation.) We can easily draw curved arrows to illustrate electron flow for these reactions, but without molecular orbital analysis, we are unable to explain the resulting stereochemistry. In the next section, we will perform this molecular orbital analysis. The following scheme also illustrates the results for 4-membered ring systems. We will see an application of electrocyclic reactions in 8-membered ring systems in one of the summary problems at the end of this chapter. (Similar reactions of larger ring systems are possible but are rare for synthetically useful applications.)
Molecular Orbital Analysis of Electrocyclic Reactions
The molecular orbital analysis for electrocyclic reactions is best understood by focusing on the molecule in its acyclic fully conjugated form. (It doesn't matter whether this is the reactant or the product.) For the 6 pi system, we must first generate the pi molecular orbitals, add pi electrons, and identify the Highest Occupied Molecular Orbital (HOMO). This is similar to the strategy we used to understand the molecular orbital basis for cycloadditions in the previous chapter; however, the only reactive orbital for electrocyclic reactions is the HOMO. The ground state and excited state molecular orbital (MO) diagrams are shown below and highlight that the ground state HOMO is psi 3 while the excited state HOMO is psi 4*. Thus, our analysis of the ground state reaction will focus exclusively on psi 3 and the excited state on psi 4*.
We will begin our analysis of the reactions shown above by looking only at the thermal conditions for the 6 pi system. How can the psi 3 MO explain the results? We must draw out the structure with the psi 3 orbital and see how it will rotate to generate positive overlap for the formation of a new sigma bond in the ring at the terminal carbons. The orbitals must rotate in opposite directions, one clockwise and one counterclockwise, to enable positive overlap and bond formation. This process is called disrotatory and provides the insight we need to explain the stereochemical outcome. Also note that the same analysis works when thinking about the ring opening reaction. The question to answer for this process is, "How will the sigma bonding orbitals rotate to yield psi 3 for the acyclic product MO?". Since this is a reversible process, the rotation must be the same for the forward or reverse reaction.
Now lets look at the photochemical reactions. As we've seen in previous chapters, light promotes a ground state electron from the HOMO to the Lowest Unoccupied Molecular Orbital (LUMO). This generates a new HOMO (the previous LUMO) in the excited state molecule. So, when analyzing the 6 pi system, we will look to psi 4* to understand the stereochemical outcomes. Contrary to the thermal reactions, the orbitals now rotate in the same direction, both clockwise or both counterclockwise, to provide positive overlap for the formation of the new sigma bond in the ring forming reaction. This is called conrotatory. For the ring opening reaction, the orbitals forming the sigma bond rotate in the same direction to generate psi 4* in the acyclic molecule. We are now able to explain the stereochemical results. Thermal and photochemical reactions yield different stereochemical results because reactions happen from different HOMOs. These different orbitals rotate by either a disrotatory or conrotatory process. If the thermal reaction is disrotatory, the photoreaction must be conrotatory, and vice versa.
Exercise \(1\)
Using the same strategy illustrated above, draw out the molecular orbitals for the 4 pi reactions to explain the stereochemistry for both the thermal and photochemical reactions.
Answer
To begin, we must draw out the 4 atom pi MO diagrams to determine the ground state and excited state HOMOs.
This shows that we will be analyzing HOMO psi 2 for the ground state thermal reactions and HOMO psi 3* for the excited state photoreactions. The thermal reactions are shown below and demonstrate that they are proceeding by conrotatory processes to provide the indicated stereochemistry.
This means that the photochemical reactions must be proceeding by disrotatory processes. This is illustrated below and accounts for the indicated stereochemistry.
Generalized Statement of Woodward-Hoffmann Rules for Electrocyclic Reactions
The table below provides a useful summary for electrocyclic reactions. Drawing out the key orbitals will always provide you with the correct mode of orbital rotation. However, you can use the number of electrons involved (just count your curved arrows and multiply by two) and the table below to quickly determine whether the reaction is con- or disrotatory.
Number of Electrons Thermal Photochemical
4n + 2 Disrotatory Conrotatory
4n Conrotatory Disrotatory
Exercise \(2\)
What are the products of the following two reactions? Pay careful attention to stereochemistry.
Answer
For the first problem, this is a six pi electrocyclic ring opening. Three curved arrows show that 6 electrons are involved. The central ring is the ring that reacts since it is the only ring that produces a fully conjugated pi system when it opens. The other rings are unreactive. Using the chart above, we can quickly see that this is a conrotatory process (4n+2 = 6, light). So, what is the alkene geometry in the product? We need to do a rotation analysis as illustrated in the box below. (Many students find this easiest if the bond that is breaking/forming is always on the bottom of the structure, so we have rotated the structure of the starting material and the product to make this possible.) The blue rotation arrows are going in the same direction (conrotatory). This puts the H pointing left and the methyl pointing right. When doing this analysis, we can ignore the details of the other rings which are abbreviated by the curved lines. An interesting follow up quesiton is what would happen if the starting material was heated? This would be a 6 pi disrotatory process. What would the product structure look like? This reaction doesn't happen. Why not? (This is a good question to discuss with classmates or your instructor.)
For the second problem, this is a four pi electrocyclic ring opening. Two curved arrows show that 4 electrons are involved. The four membered ring is the ring that reacts since it is the only ring that produces a fully conjugated pi system when it opens. The five membered ring is unreactive. Using the chart above, we see that this is a conrotatory process (4n = 4, heat). What is the alkene geometry in the product? Here's where this problem gets interesting. The result would be one cis alkene and one trans alkene in a 7-membered ring. What happens if you try to build this structure with your model kit? Exactly. You can't do it. There's too much strain, and the molecule doesn't exist. So, this is a trick question. (Sorry.) There is no reaction. What happens if you shine light on this molecule? Another trick question. (Again, sorry.) An isolated alkene doesn't absorb light, so a photoreaction isn't possible.
Electrocyclic Reactions of Cations (Nazarov Reaction) and Anions for the Synthesis of 5-Membered Rings
One of the most synthetically useful electrocyclic reactions is the Nazarov reaction. This is a cationic 4 pi thermal reaction (thus, conrotatory) beginning with a conjugated dienone that generates a cyclopentenone product. When thinking retrosynthetically, a cyclopentenone target is the Nazarov retron. An example of the reaction is shown below along with its mechanism. Strong acid is present to protonate the ketone in the first step. This generates a resonance stabilized cation that undergoes the requisite 4 pi ring closing reaction. This generates the cyclopentene ring that contains a resonance stabilized positive charge. The ring closing reaction initially generates two chiral centers, one of which will return to an achiral carbon upon the following E1 reaction. The mechanism ends with an acid catalyzed enol to ketone tautomerization. So, the final product is a racemic cyclopentenone where one new chiral center is present in the product.
Though much less common, anions can also participate in electrocyclic reactions. For example, deprotonating the most acidic proton in the 8-membered ring shown below with n-butyllithium results in a delocalized carbanion that undergoes a thermal 6 pi (disrotatory) ring closing reaction to form a new 5-membered ring as part of an interesting bicyclic product.
Exercise \(3\)
What happens when the starting dienone for the Nazarov reaction is not symmetrical? We explore that in this problem. Draw the two possible Nazarov reaction products and predict which will be the major product.
Answer
Following the mechanism above, we can identify the two cyclopentenone products. One contains a trisubstituted alkene while the other is a tetrasubstituted alkene. Further, working through the mechanism, the tetrasubstituted alkene product is generated from the more substituted carbocation resonance contributor. So, the major product is the more stable tetrasubstituted alkene that is generated from the more stable resonance contributor of the carbocation.
Summary Problems
Exercise \(4\)
In 2005, Dirk Trauner's lab published an application of electrocyclic reactions for the synthesis of the natural product SNF4435 C. This process involves two consecutive electrocyclic reactions. Propose a mechanism for these two reactions. Clearly state what type of electrocyclic reaction is occuring, draw curved arrows, indicate if it is conrotatory or disrotatory, and draw an MO diagram for each step to explain the observed stereochemistry. The starting material is drawn exactly as depicted in the paper. The first step in your analysis is to redraw this tetraene in a reactive conformation (perform bond rotations being careful not to alter the geometry of an alkenes) so that it is more obvious what electrocyclic reactions are possible.
Answer
DOI - 10.1021/ol051790q (Organic Letters 2005)
Redrawing the starting material to place carbons 1 and 8 near each other enables you to see that the first step is a conrotatory (thermal) 8 pi electrocyclic ring closing reaction. Drawing the HOMO for this process (psi 4) illustrates that this must be conrotatory. The second reaction is a disrotatory (thermal) 6 pi ring closing reaction that occurs from HOMO psi 3. The stereochemistry agrees with these rotations.
Exercise \(5\)
Carolyn Bertozzi is a pioneer of bioortogonal reactions. These are highly selective non-natural reactions that happen spontaneously in biological systems to label biomolecules. An interesting introduction to the work of the Bertozzi lab is described in a 2010 review entitled "Cu-free click cycloaddition reactions in chemical biology" in Chemical Society Reviews. A transformation mentioned in that review is pictured below. Propose a mechanism for this reaction. Does it happen thermally or photochemically? Justify your answer with MO diagrams. Hints: This is a two-step reaction. Both are pericyclic reactions; one is a cycloaddition and the other is an electrocyclic reaction.
DOI - 10.1039/B901970G
Answer
This problem can be approached in either the forward or reverse direction. Either way, there are two reasonable possibilities for the mechanism. The cycloaddition must be a Diels-Alder reaction which could happen in the first or second step. If the Diels-Alder happens first, the second step is a 4 pi electrocyclic ring closing reaction (Option #1). If the Diels-Alder happens second, the first step must be a 6 pi electrocyclic ring closing reaction (Option #2). Does it matter? The curved arrows look fine either way. However, as we've seen above, it is critical to analyze the stereochemistry to see what happens with reactions that are con- or disrotatory. The key piece of stereochemical information is that the Hs on carbons 5 and 8 in the product must be syn. This means that the 4 pi ring closing reaction in Option #1 must be disrotatory which requires a photochemical reaction. In Option #2, with the 6 pi ring closing happening first, it would also have to occur in a disrotatory fashion which happens with a thermal reaction. So, Option #1 is a thermal reaction (remember, Diels-Alder reactions must be thermal) followed by a photochemical reaction. Option #2 is two sequential thermal reactions. Both are possible; however, Option #2 is much more reasonable and is what actually happens in the Bertozzi lab.
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/01%3A_Pericyclic_Reactions/1.03%3A_Electrocyclic_Reactions.txt |
Objectives
After completing this section, you should be able to:
1. Identify sigmatropic rearrangements including hyrdride shifts and Cope, Claisen, and Wittig rearrangements
2. Understand the orbital analysis of sigmatropic rearrangements
3. Draw curved arrows to illustrate electron flow for sigmatropic rearrangements
4. Given a sigmatropic rearrangement starting material, accurately predict the product including stereochemistry
5. Use retrosynthetic analysis to determine starting materials given a sigmatropic rearrangement product
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Suprafacial
• Antarafacial
• Hydride shift
• Cope rearrangement
• Claisen rearrangement
• Wittig rearrangement
Study Notes
Sigmatropic rearrangements are pericyclic reactions that, no surprise, provide rearranged products. The most common include hydrogen shifts across pi systems and formation of new carbon-carbon bonds across allyl-type structural fragments. The most synthetically useful are the Cope and Claisen rearrangements which are formally classified as [3,3] rearrangements. (See Chapter 1.1: Introduction to Pericyclic Reactions for an explanation of naming sigmatropic rearrangements.) Your goal is to understand the theoretical basis of these reactions using a molecular orbital analysis and to apply these reactions in synthesis. You should focus on accurately predicting products with a focus on stereochemistry and, given a complicated product, to accurately predict the starting material used to generate it.
Chapter 1.1 provided an overview of pericyclic reactions, including sigmatropic rearrangements. Sigmatropic rearrangements are intramolecular reactions involving the migration of a sigma bond across a pi system. Examples include hydrogen atoms migrating across dienes upon heating (some even do this spontaneously at room temperature) and hydrogen atoms migrating across alkenes with light. Double allyl-type systems also commonly react via sigmatropic rearrangements, with 1,5-dienes participating in Cope rearrangements while ally vinyl ethers produce 1,4-enones after Claisen rearrangements. Most of our attention will focus on the applications of Cope and Claisen reactions in synthesis.
Thermal and Photochemical Hydride Shifts
Building on the introduction to pericyclic reactions presented in Chapter 1.1, we must understand the following results: [1,5] hydrogen shifts occur thermally while [1,3] hydrogen shifts happen photochemically. Labeling substrates with deuterium makes it possible to see the outcomes of these reactions which could otherwise often be invisible. Examples are shown below. As we have seen in previous sections, we need to use molecular orbital diagrams to explain these results.
Molecular Orbital Explanation for Sigmatropic Rearrangements
If we think about the transition states for the hydride shift reactions shown above, we can depict the reactions as a hydrogen (or deuterium) atom (a radical) moving across either a pentadienyl radical (heat) or an allyl radical (light).
This means that we need to draw a 5-atom and a 3-atom pi molecular orbital diagram so that we can determine the HOMO for each reaction. We can then use those diagrams to explain the experimental results. As shown below, the excited state HOMO for the 3 atom system is psi 3* and the ground state HOMO for the 5 atom system is psi 3.
What does it look like when a hydrogren atom moves across these two orbital systems? As shown below, the H is able to migrate across the same face of the pi system. As mentioned previously, this is called a suprafacial process. If the H moved from the bottom face to the top face (or vice versa), that would be an antarafacial process. Antarafacial hydrogen shifts are possible, but only for rings that are 7-membered or larger. So, the two reactions are explained by the MO diagrams showing that a thermal [1,5] H shift is suprafacial and a photochemical [1,3] H shift is suprafacial. Antarafacial [1,3] or [1,5] H shifts are too high in energy, so they do not occur.
A similar analysis helps explain why thermal [3,3] sigmatropic rearrangements are common. These reactions involve suprafacial overlap between two allyl systems. Looking at the reaction we introduced in Chapter 1.1, the simplest version of the Claisen rearrangement, we can see that the key sigma bond moves across two three-atom systems. For the MO diagram, this means that allyl HOMO psi 2 overlaps in a suprafacial fashion with another allyl HOMO psi 2. We can generalize the orbital analyses of sigmatropic rearrangements as shown in the following section.
Generalized Statement of Woodward-Hoffmann Rules for Sigmatropic Rearrangements
Number of Electrons Thermal Photochemical
4n + 2 Suprafacial Antarafacial
4n Antarafacial Suprafacial
Cope Rearrangement
The Cope rearrangement is a [3,3] sigmatropic rearrangement of a 1,5-diene. It was discovered at Bryn Mawr College by Elizabeth Hardy, a graduate student in Arthur Cope's research lab. As shown below in its simplist form, this is a reversible reaction. Consequently, the Cope retron is also a 1,5-diene. To make the Cope rearrangement synthetically useful, we must introduce a driving force that will favor formation of the product. The two most popular strategies are to provide relief of ring strain and to incorporate a tautomerization as an irreversible final step. The following problems provide examples of these strategies.
Exercise \(1\)
Predict the Cope rearrangement product of this reaction.
Answer
As mentioned above, this reaction is favorable because it relieves the strain in the cyclopropane ring. This common strategy is a useful way to generate a new 7-membered ring.
Exercise \(2\)
Predict the product of this reaction.
Answer
This is an example of an Oxy-Cope rearrangement. In Oxy-Cope reactions, placement of an OH on one of the sp3 carbons connecting the 1,5-diene results in an enol after the [3,3] sigmatropic rearrangement. An irreversible tautomerization yields the final product.
Oxy-Cope Rearrangement
As seen in the previous problem, the Oxy-Cope rearrangement is an important synthetic tool. The product of this reaction is a 1,5-enone. This is the Oxy-Cope retron which will be important when approaching synthesis problems. One way to promote Oxy-Cope reactions is to deprotonate the starting alcohol with a hydride base. These reactions are referred to as oxy-anion accelerated Cope rearrangements.
Cope Rearrangement Application
The most amazing application of the Cope rearrangement is the molecule known as bullvalene. This molecule was designed and synthesized in the lab of William von Eggers Doering and published in 1963. If you play around with the molecule, you see that there are several Cope rearrangements that are initially possible. If you draw those products, you see that even more become apparent. The net result is that this is a fluxional molecule; it does not have a set structure at room temperature. In fact, when heated, its NMR spectra are a singlet at 4.2 in the proton NMR and one peak at 86 in the carbon spectrum. For an excellent article about the bullvalene origin story, check out Addison Ault's paper in the Journal of Chemical Education.
Claisen Rearrangement
As mentioned above, the Claisen rearrangement is the conversion of an allyl vinyl ether into a 1,4-enone via a [3,3] sigmatropic rearrangement. The simplist Claisen rearrangement is picture below. The forward direction is favored because the product contains a carbonyl, so this is not a reversible reaction. The following two problems provide practice predicting products of Claisen rearrangements.
Exercise \(3\)
Predict the product of the following Claisen rearrangement.
Answer
First, find the allyl vinyl ether. It is highlighted in magenta below. Second, rotate the bonds so that carbons 1 and 6 are near each other. Third, draw the curved arrows. Fourth, draw the product. As with many reactions, numbering your atoms is helpful to ensure you get the correct product.
Exercise \(4\)
Predict the product of the following Claisen rearrangement. Hint: Think carefully about the most stable structure for your product.
Answer
Draw the curved arrows for the allyl vinyl ether to generate the 1,4-enone. The tricky part of this problem is recognizing that this product is not the lowest energy molecule possible. A keto to enol tautomerization yields the final product. This is another example of the aromatic stability of benzene. We are used to converting enols to ketones. In this case the enol is more stable because of the aromaticity of benzene.
The above reaction is a very common strategy to make a C-C bond at the ortho position of a phenol. Keep this is mind when thinking about synthesis problems. (The starting material above is easily synthesized by treating phenol with sodium hydride and allyl bromide.)
Often the most challenging aspect of the Claisen rearrangement is synthesizing the starting material. Allyl vinyl ethers are difficult to obtain and chemists have developed several useful strategies to make them. How would you do it? The answer to Problem #4 above mentions starting with phenol, a vinyl ether, and reacting it with allyl bromide. However, vinyl ethers are very rare since they readily tautomerize to carbonyls unless they are part of an aromatic ring. One solution is to start with an acetal or ketal in place of the vinyl ether. An example of this strategy is shown below. What is the mechanism for this reaction? This variant of the Claisen rearrangement forms ketones or aldehydes (R=H). We will see below that other common Claisen strategies form esters, carboxylic acids, and amides.
Exercise \(5\)
Propose a mechanism for the example Claisen reaction shown above. Remember, your mechanism must generate an allyl vinyl ether so that the Claisen rearrangement can occur. Don't forget that AcOH is acetic acid.
Answer
This is a good review of acid catalyzed ketal mechanistic steps that you learned in intro organic chemistry. The mechanism starts with those standard steps. In the second to last step, the acetate anion (generated in the first step) can deprotonate to form the neutral allyl vinyl ether (this deprotonation looks like the mechanistic step we use to form enamines from secondary amines plus ketones) that undergoes the Claisen rearrangement in the final step.
Johnson-Claisen Rearrangement
A reaction analogous to the one we just discussed above is the Johnson-Claisen rearrangement that features the unusual orthoester functional group as one of its starting materials and produces ester products. An orthoester is the ester equivalent of a ketal. Using the exact same mechanism as in the answer to Problem #5 above (with R = OMe), we can understand the Johnson-Claisen rearrangement shown below. Thus, the retron for a Johson-Claisen rearrangement is a 1,4-enester.
Ireland-Claisen Rearrangement
The Ireland-Claisen rearrangement results in the formation of carboxylic acid products and proceeds via a silyl enol ether generated after forming an enolate. The reaction and mechanism are shown below. Continuing our trend of starting with allyl alcohol, we treat it with acetic anhydride to form an allyl ester. Combining that with LDA yields an enolate that reacts with trimethylsilyl chloride (TMSCl) on the oxygen of the enolate (this is standard reactivity for silyl electrophiles with enolates) to yield the key silyl enol ether. This undergoes the Ireland-Claisen rearrangement to yield a silyl ester that is easily converted to the desired carboxylic acid upon workup with aqueous acid. (This step is analogous to acidic deprotection of silyl ethers to yield alcohols.)
Exercise \(6\)
What is the product of the following reaction?
Answer
This is an example of the Eschenmoser-Claisen rearrangement. Using an orthoamide in place of the orthoester in the Johnson-Claisen rearrangement results in the production of an unsaturated amide product by the same mechanism.
Claisen Rearrangement Alkene Geometry
One final point about the Claisen rearrangement relates to the alkene geometry formed in the reaction. The example below highlights that trans alkenes are formed while cis alkenes are not. Why? This is another example of the importance of chair-like transition states. (We first saw this with the ene reaction in the cycloadditions chapter.) Putting the methyl substituent in the more stable equatorial position leads to the trans product. With the methyl in the less stable axial position, the cis product would form.
Wittig Rearrangement
All of the sigmatropic rearrangements that we have seen so far are reactions of neutral molecules. That is the case for most sigmatropic rearrangements, but charged molecules also participate in these transformations. The Wittig rearrangement is an anionic [2,3] sigmatropic rearrangement of an allylic ether to yield a homoallylic alcohol. So, the retron for this reaction is the same as the oxo-ene reaction. A generic version of this reaction is shown below. The critical components of the starting material are an allyl ether containing an electron withdrawing "Z" group. Thus, treatment with a strong base results in deprotonation next to the Z group. The resulting anion can participate in a [2,3] sigmatropic rearrangement (numbering from the sigma bond broken to the sigma bond formed) to yield a homoallylic alcohol product, after quenching with acid.
An application of the Wittig reaction is shown below where the electron withdrawing group is a resonance stabilizing alkyne. As this example shows, the Wittig rearrangement can be a powerful ring contraction reaction. In this case, converting an 18-membered ring ether into a 15-membered ring alcohol.
Exercise \(7\)
What is the product of this Wittig rearrangement?
Answer
The phenyl is the electron withdrawing group in this molecule. So, draw the anion, then the rearrangement arrows, and finally add the proton to generate the product.
Summary Problems
Exercise \(8\)
Identify the retron in the following molecule then draw the starting material (go back only one step) that would yield the target.
Answer
This molecule contains a 1,5-diene which is the retron for the Cope rearrangement. Drawing the curved arrows for the reaction enables you to determine the starting material used to make the target. Note: There is another 1,5-diene in the molecule; however, the one shown is what was used in the 2010 Organic Letters paper to make this molecule.
Exercise \(9\)
The following transformation appeared in a paper focused on the synthesis of the novel antibiotic platensimycin. Propose a synthetic route for the conversion of the starting allylic alcohol into the triene synthetic intermediate. Then, propose a mechanism for the conversion of this material into the polycyclic product. Hints: One of the steps in your mechanism is a cycloaddition. It might help to think retrosynthetically, going backwards one step from the target for this mechanism.
Answer
The synthesis portion of this problem involves three reaction types: oxidation, Wittig, and deprotection. Both oxidations are primary alcohols to aldehydes, so feel free to use PCC, Swern, or Dess-Martin interchangeably. The key Wittig reagents are shown below. After adding the first alkene, then it's time for the deprotection with fluoride (any F minus reagent will work). This reveals the second primary alcohol for the final two steps.
For the mechanism part of the problem, spotting the Diels-Alder retron in the product is the key. This disconnection gets you back to a molecule that is very similar to the starting material. Performing a thermal [1,5] H shift on the starting molecule generates the intramolecular Diels-Alder substrate. This is an excellent example of the importance of hydrogen shifts and the amazing complexity that can be synthesized using intramolecular cycloadditions.
Reference: Journal of Organic Chemistry 2009
Exercise \(10\)
Propose a mechanism for the following reaction. Hints: The mechanism is 4 steps and involves two sigmatropic rearrangements.
Answer
Looking at the reactants, it's important to recognize that we are starting with an allyl vinyl ether, the starting material for the Claisen rearrangement, and a modified Wittig reagent. This is a nitrogen Wittig reagent which is commonly called an aza Wittig reagent. It reacts just like the Wittig reagents you have seen before but produces an imine instead of an alkene when reacting with a ketone or aldehyde. There is no carbonyl at the start, so our only option is to do the Claisen rearrangement. This yields a molecule with a 4-membered ring and an aldehyde! So, we can now do the aza Wittig reaction. Comparing that imine product with the target 8-membered ring shows that we can do the reverse of our first step, a retro aza (because of the nitrogen) Claisen rearrangement, to generate the target. So, in the end it's just substituting NBn for O, but there's a lot going on to make that possible.
Reference: Organic Letters 2010
Exercise \(11\)
Propose a synthesis of the following target starting with any compounds containing six carbons or fewer. Feel free to ignore any carbons that might be in the alcohol protecting groups (PG). Hint: Retrosynthetic analysis should be very helpful.
Answer
Let's focus on retrons for this problem. The target contains a 1,4-enamide which is the retron for the Eschenmoser-Claisen rearrangement. Doing that retrosynthetic step reveals a second 1,4-enamide, so we can do another Eschenmoser-Claisen rearrangement. This yields a double amino enol ether that we can disconnect back to a 6-carbon diol and the ortho amide that we have used before for Claisen rearrangements.
For the synthesis part of the problem, all we need to do is heat the diol with excess ortho amide. This yields the allyl vinyl ether we need for the Eschenmoser-Claisen rearrangement, and the entire process can happen again to yield our target. An excellent example of creative synthetic planning!
Reference: Organic Letters 2010
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/01%3A_Pericyclic_Reactions/1.04%3A_Sigmatropic_Rearrangements.txt |
• 2.1: Introduction to Transition Metals and Mechanistic Steps
This chapter introduces concepts of bonding, oxidation state, and electron counting for organometallic transition metal complexes. It also covers fundamental mechanistic steps like oxidative addition, transmetallation, reductive elimination, migratory insertion, and beta-hydrogen elimination.
• 2.2: Pd-Catalyzed Cross Coupling Reactions
This page focuses on the key carbon-carbon and carbon-heteroatom bond forming reactions catalyzed by palladium including the Stille, Suzuki, Sonogashira, Heck, Tsuji-Trost, and Buchwald-Hartwig reactions.
• 2.3: Olefin Metathesis
This page covers olefin (alkene) metathesis using catalysts developed by Schrock and Grubbs. The main reaction is ring closing metathesis, while cross metathesis and alkyne metathesis are also covered.
• 2.4: Co-Mediated Ring Forming Reactions
This page focuses on ring forming reactions promoted by cobalt, specifically the Pauson-Khand reaction to form cyclopentenones and the alkyne cyclotrimerization reaction to yield substituted benzenes.
02: Transition Metal Catalyzed Carbon-Carbon Bond Forming Reactions
Objectives
After completing this section, you should be able to:
1. Understand bonding, electron counting, and oxidation states for transition metal complexes
2. Draw and understand common transition metal mechanistic steps
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Transition metal complex
• Ligands
• 18-electron rule
• Metal oxidation state
• Pi donor ligands
• Oxidative addition
• Transmetallation
• Reductive elimination
• Migratory insertion
• Beta hydrogen elimination
Study Notes
Transition metal catalyzed carbon-carbon bond forming reactions provide powerful methods to form key bonds in ways that are often impossible with traditional main group organometallic compounds (e.g., Grignard and organolithium reagents). This chapter provides an introduction to transition metal complexes including their bonding and common reactions. The following chapters will explore the most common transition metal catalyzed reactions including Pd catalyzed cross coupling reactions, olefin metathesis, and Co mediated ring forming reactions.
Content
Before learning about reactions catalyzed by transition metal complexes, we first need to understand a little about their structure and behavior. To begin, we need to update our vocabulary. You've already seen that we refer to molecules containing transition metals as complexes. Transition metal complexes contain the metal and other molecules bonded to the metal. These molecules bound to the metal are called ligands. Bonds between transition metals and their ligands are weaker than the covalent bonds you are used to from standard organic molecules. This is one reason why transition metals make such excellent catalysts, they are able to make and break bonds to ligands frequently throughout the course of a reaction. For example, tetrakis triphenylphosphine palladium is a yellow solid in which Pd has 4 triphenylphosphine ligands. As shown below, when dissolved in a solvent, this complex exists in equilibrium with complexes containing three and two triphenylphosphine ligands.
Electron Counting
Much time in introductory organic chemistry is spent discussing the octet rule (based on filling s and p orbitals). Transition metals have ready access to d-orbitals, so the octet rule does not apply. Instead, we add 10 electrons (because of the 5 d orbitals) and come up with the 18-electron rule. Transition metal complexes containing 18 electrons are generally stable and unreactive. For active catalysts, we must have metals that are surrounded by fewer than 18 electrons. Also, we need to learn how to determine the formal charge on the metal. This is called the metal's oxidation state. Metals with high oxidation states are electron poor and are much more reactive than low oxidation state or neutral metal complexes.
We will start by analyzing metal complexes where the ligands are bound to the metal via sigma bonds, like in tetrakis triphenylphosphine palladium above. First, we must determine the oxidation state (formal charge) of the metal. To do this, we will 1) Consider the sigma bond electrons between the metal and the ligand as belonging to the ligand. 2) Assign a formal charge to the ligand atoms bound to the metal using standard organic chemistry formal charge rules. 3) Add up the charges on all of the ligands. 4) The metal oxidation state balances the ligands' summed formal charges so that the overall complex is neutral (or negative if it's an anionic complex/positive if it's a cationic complex).
Let's practice this by looking at a Pd complex (Pd(PPh3)4) and a Rh complex ((Ph3P)3RhCl). Phosphine ligands are common in transition metal complexes, and their charge is always neutral. (Just like N, which is in the same column in the periodic table, P is neutral with 3 bonds and a lone pair.) Halogens with eight electrons and no bonds have a negative one formal charge. So, in the Pd complex, the ligands have no formal charge, and Pd has a zero oxidation state: Pd(0). In the Rh complex, the net charge on the ligands is negative one. The metal must balance that charge so that the overall complex is neutral. This means Rh has a +1 oxidation state: Rh(I).
The next step is to detemine the total electron count around the metal to see if it satisfies the 18-electron rule. To do this, we must determine 1) The total electrons donated to the metal from the ligands. 2) The valence electrons from the transition metal based on its oxidation state. 3) Add the ligand electrons plus the metal electrons.
Ligand electron counting is as simple as counting the bonds from the ligands to the metal and multiplying by 2. In both of the complexes above, there are four sigma bonds from the ligands to the metals, so there are 8 electrons from the ligands. To determine the metal's valence electrons, we must look at the periodic table and factor in its oxidation state. In the Pd complex above, we determined it is Pd(0) which means Pd has all of its valence electrons. It is in group 10 in the periodic table, so it has 10 valence electrons. In Wilkinson's catalyst above, we determined it is Rh(I) which means Rh has lost one valence electron. Rh is in group 9, so it has 8 valence electrons as a +1 metal.
Thus, we can complete our analysis of the example complexes above. Tetrakis triphenylphosphine palladium has 8 ligand electrons plus 10 metal electrons for a total of 18 electrons. It satisfies the 18 electron rule and is unreactive until it loses one or two ligands in solution. Wilkinson's catalyst has 8 ligand electrons plus 8 metal electrons for a total of 16 electrons. Thus, it can form one more bond and is reactive.
Exercise \(1\)
The following Pd complex is commonly formed in Pd catalyzed cross coupling reactions. Perform an electron counting analysis on the complex to determine the oxidation state of Pd and the total Pd electron count.
Answer
Following our strategy from above, we can determine that we have two negative one ligands, the Br and the Ph. This means that Pd must balance the net -2 charge by being in its +2 oxidation state: Pd(II). For the total electron count, we again have 8 ligand electrons. Metal ligands are 8 for a +2 Pd (two fewer than the 10 electrons for the neutral Pd). The means our total electron count is 16.
We have one more class of ligands to consider before moving on. These are pi donor ligands. This class of ligands can donate pi electrons via a pi bond. A common example is a pi complex formed when an alkene or alkyne bonds with a metal via its C-C pi bond (see below, left). This results in a neutral, 2 electron ligand. Another example that we will see frequently is an allyl ligand. There are two options for how an allyl anion can bond to a metal (see below, right). It can form a sigma bond which makes it a 2 electron, -1 ligand. However, it can also bond via its alkene pi bond, making it a 4 electron, -1 ligand. This latter option is much more common since it generally helps the metal get closer to an 18-electron complex. Here's another place to expand our transition metal vocabulary. We use the Greek letter eta to describe how many atoms in the ligand are bound to the metal. So, for the first allyl option (2 electrons, -1), we would say it is eta 1 (1 carbon sigma bonded to the metal). For the second allyl option (4 electrons, -1), we would say it is eta 3 (3 carbons bonded to the metal via one sigma and one pi bond).
Exercise \(2\)
Propose a structure for the iron complex that is formed in the following reaction. Hint: This molecule is very stable.
Answer
This deceptively simple reaction was the start of a new field of chemistry. Originally published as the sigma bond structure in 1951, an alternate "sandwich" complex was published in 1952. An X-ray crystal structure proved the novel sandwich structure and a new subdiscipline of organometallic chemistry was born. For their work in this area, Geoffrey Wilkinson and Ernst Fisher were awarded the Nobel Prize in 1973. This 2001 article in Chemical and Engineering News provides an overview of the discovery. An entertaining examination of this discovery and the subsequent scientific credit appeared in Angewandte Chemie in 2000. This article was co-authored by Roald Hoffmann and examines the role of R.B. Woodward in the ferrocene story. (These are Woodward and Hoffmann from pericyclic reaction fame appearing again!)
The initial proposal is likely the first structure that you drew. This complex has the two cyclopentadienyl ligands (abbreviated Cp) sigma bonded to Fe. This yields a 10 electron Fe(II) complex that would not be very stable. Thinking about the two Cp ligands participating in bonding to the Fe via their pi electrons yields a completely different structure. (Don't forget that the cyclopentadienyl anion is an aromatic anion.) This is the correct structure with each Cp ligand contributing 6 electrons to the complex (still a -1 donor) via eta 5 bonding. This results in a stable 18 electron Fe(II) complex. This fascinating molecule is aromatic, participating like benzene in Friedel-Crafts reactions, and is known as ferrocene. Cp ligands are ubiquitous in organometallic chemistry, so be aware of this unique bonding when reading the literature and you see metal-Cp complexes.
Common Mechanistic Steps
Many transition metal catalyzed reactions share common mechanistic steps that include oxidative addition, reductive elimination, transmetallation, migratory insertion, and beta hydrogen elimination. We will explore each of these individually and then see in the next chapter how that are combined in specific Pd catalyzed cross coupling reactions.
One note about organometallic mechanisms is that chemists often don't use curved arrows to show electron flow. Why? It is often because electron flow isn't always clear. For example, some of these mechanisms might actually be radical reactions. In this chapter, we will shown how electrons could flow if they were polar reactions since this often helps students understand what is happening. However, in keeping with the conventions of transition metal mechanisms, we won't always do this.
Oxidative Addition
Many transition metal catalyzed reactions begin with a step called oxidative addition where the metal adds to a carbon-halogen bond. You have already seen this reaction type though it may not have been called an oxidative addition. Forming a Grignard reagent from Mg and an organohalogen compound is an oxidative addition. This step is oxidative with respect to the metal because it formally loses electrons when adding carbon and halogen ligands. In the examples below, a neutral metal adds two -1 ligands, meaning the metal becomes +2. We will see many examples of oxidative addition of Pd(0). Two are shown below highlighting the reactivity of halogens and triflates (OTf). (If you haven't met a triflate before, it is a more reactive version of a tosylate that can easily be made from an alcohol or a carbonyl. We will see examples of this in later sections.) A third Pd(0) example is shown to illustrate how you can think about electron flow and curved arrows for oxidative addition. As mentioned above, we generally won't draw curved arrows for this step, and the arrows likely don't depict what is actually happening in the mechanism. The most common and synthetically useful oxidative additions occur between Csp-X/OTf or Csp2-X/OTf and metals like Pd. So, we will encountered oxidative additions with vinyl, phenyl/aryl, and alkynyl substrates.
Reductive Elimination
The opposite mechanistic step to oxidative addition is reductive elimination. In this step, two groups bound to the metal, often two carbon ligands, form a new bond while formally giving electrons back to the metal. Thus, this step reduces the metal. It is often the last step in the cycle, regenerating the transition metal catalyst and forming the key carbon-carbon bond in the product. Curved arrows are show below, but this is another step where we will normally resist drawing curved arrows.
Transmetallation
A common strategy to form a new carbon bond to the transition metal catalyst is to start with a carbon group on a different metal. The stoichiometric metal reagent then transfers the carbon group to the catalytic metal during the catalytic cycle. We will see in the next section that Pd participates in transmetallation with metals like B, Sn, Cu, and Ni. As shown below, the net result is that the halogen group on the catalytic metal (Pd) switches places with the carbon group on the stoichiometric metal (Sn). Also, note that the oxidation state of Pd doesn't change, so this is neither an oxidation nor a reduction.
Exercise \(3\)
Shown below is the catalytic cycle for a Negishi reaction. Label each mechanistic step as oxidative addition, reductive elimination, or transmetallation.
Answer
As with many Pd catalyzed reactions, the steps proceed in this order: oxidative addition (making one Pd-C bond), transmetallation (making a second Pd-C bond), and reductive elimination (making the key C-C bond).
Migratory Insertion
Some transition metal catalyzed reactions involve addition of a metal-carbon bond across a pi bond, like an alkene or alkyne, that is also bound to the metal. This is a powerful way to build up complexity in a molecule, especially when it is an intramolecular reaction, and we will see numerous examples of this step when we explore the Heck reaction in the next chapter. As shown below, an alkene (or alkyne) that is complexed to the metal via its pi bond can insert into an adjacent metal-carbon sigma bond. This is a syn addition to the alkene and is similar to the alkene hydrogenation mechanism that you learned in intro organic. The new carbon-carbon bond generally forms at the least hindered/most electron poor carbon of the alkene.
Beta Hydrogen Elimination
For many transition metal complexes bound to an sp3 hybridized carbon, beta hydrogen elimination is a common mechanistic step. This is a product yielding step and demonstrates the ability of transition metal reactions to form alkene products. Unlike E2 reactions that you learned in intro organic that proceed via an anti elimination, beta hydrogen eliminations are syn eliminations. This has important implications for product alkene geometry and must be carefully considered when proposing this type of mechanistic step. In some complicated substrates, a syn arrangement of the metal and beta hydrogen are not possible. In these cases, the base included in the reaction mixture that usually regenerates the active catalyst instead does a traditional anti E2 reaction to yield an alkene product. We will see examples of this in the following chapter.
Beta hydrogen elimination reactions occur most often after migratory insertion steps. We can explore this mechanism beginning with the product of our migratory insertion step shown above. The migratory insertion product has two beta carbons; however, only the beta carbon on the left has a hydrogen. In the initial conformation, this beta hydrogen is anti to the Pd, so intramolecular elimination is not possible. Thus, our next step is a bond rotation to orient the Pd and beta H syn. Our syn beta hydrogen elimination is now possible, and the reaction occurs. This generates our alkene product with the ketone and phenyl positioned cis, due to their orientation in the reactive conformation. Note that this reaction does not regenerate our Pd(0) catalyst. Instead, we form a Pd(II) complex containing a Pd-H bond. Standard reaction conditions include a base so that deprotonation and loss of bromide regenerates the active catalyst.
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/02%3A_Transition_Metal_Catalyzed_Carbon-Carbon_Bond_Forming_Reactions/2.01%3A_Introduction_to_Transition_Metals_and_Mechanistic_Steps.txt |
Learning Objectives
After completing this section, you should be able to:
1. Identify common Pd-catalyzed carbon-carbon bond forming reactions
2. Draw and understand reaction mechanisms
3. Use reactions in synthesis problems
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Stille reaction
• Suzuki reaction
• Sonogashira reaction
• Heck reaction
• Tsuji-Trost reaction
• Buchwald-Hartwig amination
Study Notes
Palladium catalyzed reactions are the most common transition metal catalyzed cross coupling reactions. Their importance was recognized with the 2010 Nobel Prize for Richard Heck, Ei-ichi Negishi, and Akira Suzuki. These reactions provide straightforward methods for the construction of Csp2-Csp2, Csp2-Csp, and Csp2-N bonds. We will see many examples of these types of bond formations, and several others, throughout this chapter. While learning this material, don't forget to contrast Pd catalyzed reactions with popular intro orgo carbon-carbon bond forming reactions like Grignard or organolithium reactions with carbonyls and epoxides. At least one component of these reactions was sp3 hybridized, so it was impossible to generate the types of bonds that are formed most commonly by Pd catalyzed reactions. Adding Pd catalyzed reactions to your organic chemistry toolbox will greatly expand the types of molecules that you are able to synthesize.
Content
This chapter will focus on learning and applying common Pd catalyzed reactions for the formation of 1) Carbon-Carbon bonds: Stille, Suzuki, Sonogashira, and Heck 2) Carbon-Carbon or Carbon-Oxygen bonds: Tsuji-Trost and 3) Carbon-Nitrogen bonds: Buchwald-Hartwig. If you haven't already, please refer to the previous chapter for a discussion of transition metal structure and bonding along with common mechanistic steps in transition metal catalyzed reactions.
For all of the reactions in this chapter, we will depict the palladium catalyst as LnPd(0) to acknowledge that some number of ligands are bound to Pd but their actual number and structure are unimportant to the mechanism. Two things happen when a Pd complex dissolves in the reaction solvent. First, bonds to phosphine ligands break to establish an equilibrium with complexes having fewer ligands and open coordination sites for reactions to occur. Second, some reactions will begin with Pd(II) reagents. For example, most Sonogashira reactions are run with PdCl2(PPh3)2. These Pd reagents are reduced under the reaction conditions by several possible mechanisms that we won't concern ourselves with here. The net result is that Pd(0) is produced in the flask, and this complex can then catalyze the desired reactions. Even though we will generally ignore the phosphine ligands, the structure of these molecules is critically important to the success of Pd catalyzed reactions. Many of the recent advances in this field were made possible by the development of novel phosphine ligands. The Buchwald-Hartwig reaction is one such example.
Stille Reaction
The Stille reaction is the simplest Pd catalyzed cross coupling reaction where an organohalide combines with an organotin in the presence of a Pd(0) catalyst to form a new C-C bond. To avoid side reactions and production of mixtures, both components generally have Csp2 or Csp atoms bonded to the halogen and the tin. An example reaction and mechanism are shown below. Oxidative addition of bromobenzene with the Pd catalyst yields the first Pd(II) intermediate. Transmetallation transfers the alkene from Sn to Pd, generating the tributyltin bromide byproduct and the desired Pd(II) intermediate with two C-Pd bonds. Finally, reductive elimination generates the desired product and regenerates the Pd(0) catalyst.
A useful variant of the standard Stille reaction is the carbonylative Stille reaction where a carbonyl can be added between the organohalide and organotin components. This is made possible by running the reaction under an atmosphere of CO gas. An example with the carbonylative mechanism is shown below. After oxidative addition, CO associates with Pd forming a C-Pd bond. Next, the phenyl ligand can perform a carbonyl insertion step (similar to the alkene migratory insertion shown in the previous chapter) to form one new C-C bond. We now return to the regular Stille mechanism with the transmetallation and reductive elimination steps producing the final product and regenerating the catalyst.
Exercise \(1\)
Complete the following synthesis in 2 steps, one of which is a Stille reaction.
Answer
Thinking retrosynthetically, we must be forming the single bond between the alkenes using a Stille reaction. That means we need to convert the ketone into a vinyl halide or halide equivalent in the first step. This is easily done by forming an enolate and adding triflic anhydride to make a vinyl triflate. As explained in the previous chapter, vinyl triflates participate readily in transition metal catalyzed reactions. So, our two step synthesis is 1) Deprotonation with LDA (to ensure we form the least substituted kinetic enolate) followed by addition of triflic anhydride to yield the vinyl triflate. 2) Stille reaction with vinyl tributyltin to generate the target.
Exercise \(2\)
Using retrosynthetic analysis, how would you make this target in one step using a Stille reaction?
Answer
The key bond is the one between the alkene and the alkyne. This is the bond we will make in the forward direction using the Stille reaction.
Exercise \(3\)
Propose a 1 step synthesis of the target using a Stille reaction.
Answer
This is the perfect molecule to use the carbonylative Stille reaction. We can start with a vinyl bromide and an alkynyl tin. Combining them with the palladium catalyst under a CO atmosphere yields the target.
Suzuki Reaction
The Suzuki reaction is similar to the Stille reaction but with boron used instead of tin for the transmetallation step. It also differs from the Stille reaction because Suzuki reactions require the presence of a base to activate the boron reagent prior to transmetallation. As shown below, the mechanistic steps are the same for these two popular transformations with the addition of the base promoted boron activation step that forms the reactive borate (negatively charged boron) compound. This borate reagent can be formed by a variety of oxygen bases in addition to carbonate like hydroxide and alkoxides (e.g., ethoxide, t-butoxide).
There are several advantages of the Suzuki reaction. First, we are already familiar with forming C-B bonds from intro orgo using a hydroboration reaction. Previously, we always followed this step with an oxidation reaction to form an alcohol. Now we can form the organoboron compound and use it in a Suzuki reaction. We can react alkenes or alkynes in hydroboration reactions to yield organoboron reagents for transmetallation. Several examples are shown below. Hydroboration of terminal alkynes is synthetically useful (internal alkynes yield product mixtures) when using hindered boranes like disiamylborane or catecholborane. Like all hydroborations, the reactions are syn additions with the larger boron adding to the less hindered side of the alkyne resulting in trans alkene products. To generate alkylboron reagents for use in Suzuki reactions, the most common reagent is 9-BBN (pictured below). A second advantage is the Suzuki reaction provides a way to use Pd catalysis to easily make Csp2-Csp3 or Csp-Csp3 bonds which can be challenging to make with other Pd catalyzed reactions. Hydroboration of an alkene provides the requisite B-Csp3 reagent (the 9-BBN derivative shown below) that participates in transmetallation then reductive elimination to yield the desired bond. We will see an example of this in one of the problems below.
Exercise \(4\)
What is the product of the follow reaction sequence?
Answer
The first step is a hydroboration reaction with 9-BBN which is selective for the much less hindered terminal alkene. This sets up an example of a very synthetically useful intramolecular Suzuki reaction to form a 6-membered ring.
Exercise \(5\)
Propose a synthesis of the following target starting with compounds containing 6 carbons or fewer. One of your starting materials must be an alkyne.
Answer
Thinking retrosynthetically, we can split the molecule between the two alkenes into two 6-carbon fragments that can be combined using a Suzuki reaction. The boron-containing component can then be simplified to the requisite alkyne starting material. In the forward direction, we start with 1-hexyne and hydroborate it with catecholborane (a 6-carbon reagent) to yield the vinylborane that participates in a Suzuki reaction with our 5-membered ring vinyl bromide to yield the target.
Sonogashira Reaction
The Sonogashira reaction enables the combination of an unsaturated carbon-containing halide or triflate with a terminal alkyne to yield a new Csp-Csp or Csp-Csp2 bond. One interesting aspect of this reaction is that it is catalyzed by a combination of Pd (to form the C-C bond) and Cu (to activate the terminal alkyne for transmetallation). Another key reagent is an amine base that promotes formation of the copper acetylide that participates in the transmetallation step. An example Sonogashira reaction and the mechanism is shown below. This is a very useful reaction and an important addition to our synthesis toolbox.
The previous three reactions, Stille, Suzuki, and Sonogashira, share a similar mechanism that includes oxidative addition, transmetallation, and reductive elimination steps. Other reactions with different metals have a similar catalytic cycle with palladium. Although we won't discuss them here, you are well equipped to understand these reactions when you see them in the literature. Two of the most common examples are the Negishi reaction (transmetallation with zinc) and the Kumada reaction (transmetallation with magnesium).
Heck Reaction
The Heck reaction proceeds via a different mechanistic pathway than the Stille, Suzuki, or Sonogashira reactions because it does not contain transmetallation or reductive elimination steps. Instead, the Heck reaction relies on an alkene insertion step for carbon-carbon bond formation and beta hydrogen elimination to generate the product. (As a reminder, these steps were described in detail in the previous chapter.) An example of the reaction and its mechanism are shown below. A few key points about the mechanism are worth highlighting. The alkene insertion step usually places the Pd at the more substituted position and the carbon at the least substituted position. This step is a syn addition to the alkene which necessitates a bond rotation in the next step to place a beta hydrogen syn to the Pd. There are two possible Hs that can end up syn but the conformation that places the phenyl and the ketone anti is preferred. (Note that for this mechanism, only one beta carbon has Hs. The other beta carbon is the ketone. In many reactions, there are several alkenes that can form. Heck reactions generally produce the most stable possible alkene product.) This conformation yields the more stable trans alkene product in the beta hydrogen elimination step. The Pd(0) catalyst is regenerated in the final deprotonation step that explains the need for base in the mechanism.
Heck reactions are commonly run intermolecularly, between two different reactants. However, the true power of the Heck reaction for synthesis becomes apparent when studying intramolecular Heck reactions. This is a very useful technique for quickly building molecular complexity. We will see an example of this in one of the problems below.
Exercise \(6\)
Predict the product of the following intermolecular Heck reaction.
Answer
This is a tricky problem. Most students quickly predict the conjugated diene shown below that doesn't form. This is a good illustration of the importance of evaluating potential products formed from the beta hydrogen elimination step. In this reaction, there are two beta carbons with Hs that can be eliminated with Pd. One option, the beta H on the left, yields the conjugated diene. The other option, the beta H on the right, yields an enol product that tautomerizes to the observed aldehyde product. Carbonyls are more stable than alkenes, so this thermodynamic difference drives the reaction to the aldehyde.
Exercise \(7\)
Predict the product of the following intramolecular Heck reaction.
Answer
This is an excellent example of what is known as a Heck zipper reaction where multiple alkene insertions occur to form more than one ring. Working through the mechanism, we can see what is happening. After the initial oxidative addition step, the first alkene insertion occurs to form the new 6-membered ring. Normally, we would next perform a beta hydrogen elimination reaction to generate the product. However, this molecule has a quaternary carbon at the beta position, so beta hydrogen elimination is impossible. When that happens, the molecule undergoes another alkene insertion to form a second new ring. In this case, a new five membered ring. We again look to find a cis beta hydrogen. That is now possible when the Pd is on carbon #11 and the final product is formed.
Tsuji-Trost Reaction
The Tsuji-Trost reaction is a highly useful transformation that enables formation of C-O, C-N, and C-C bonds. The key reaction components are an alkene with an allylic leaving group (halide, acetate, or epoxide) and a nucleophile which is an alcohol, an amine, or a carbon with an acidic proton (think 1,3-dicarbonyl). An example with a mechanism is shown below. Reaction of allyl acetate with the 1,3-diketone nucleophile and a palladium catalyst yields the product with a new C-C bond to the allyl group. In the mechanism, which is unlike Pd catalyzed reactions we have already seen, the first step is alkene association. Next, Pd substitutes for the leaving group, generating a pi-allyl Pd complex. The acetate leaving group is now free to act as a base, deprotonating the acidic proton on the nucleophile. The resulting stabilized enolate adds to one of the terminal carbons on the pi-allyl complex, forming the new C-C bond. The final step is alkene dissociation that yields the product and regenerates the catalyst.
There are several important details to consider for the Tsuji-Trost reaction. First, it can happen inter- or intramolecularly, and we will see examples of each in the problems below. We also must consider regiochemistry and stereochemistry, depending on the structure of the reactants. One example below demonstrates that regioselectivity is governed by sterics, with the nucleophile adding to the less substituted side of the pi-allyl Pd complex. The other example below shows that the reaction is stereospecific, with substitution occurring with retention. As we saw previously in intro orgo, the only way for this to happen is for the mechanism to include two SN2 reactions. (Remember, SN1 reactions always yield racemization, equal amounts of retention and inversion.) The two SN2 reactions are when the Pd displaces the leaving group and when the nucleophile adds to the pi-allyl complex.
Exercise \(8\)
Predict the product of this intermolecular Tsuji-Trost reaction.
Answer
The pi-allyl Pd complex is formed by breaking open the lactone. Next, the deprotonated diester adds to the least hindered side to yield the product.
Exercise \(9\)
Predict the product of this intramolecular Tsuji-Trost reaction.
Answer
The Pd adds from the back to yield the pi-allyl complex. The enolate nucleophile adds from the front to form the new 6-membered ring.
Exercise \(10\)
Predict the product of this reaction and provide a mechanism to explain its formation.
Answer
This is an example of an allylic epoxide participating in the Tsuji-Trost reaction. The Pd pi-allyl complex forms by inversion. A deprotonation step generates the carbanion nucleophile that adds with inversion to the least hindered side of the pi-allyl complex.
Buchwald-Hartwig Amination
The Buchwald-Hartwig amination was developed by Steve Buchwald (MIT) and John Hartwig (Berkeley) in the late 1990s. It has evolved into one of the most popular methods for the construction of aryl amines which are common functional groups in pharmaceutical compounds and natural products. The reaction combines an aryl halide, a primary or secondary amine, and a base in the presence of a palladium catalyst. As shown in the example below, specialized phosphine ligands pioneered by the Buchwald lab are often critical to the success of these reactions. The mechanism is similar to what we saw for the Stille, Suzuki, and Sonogashira reactions in that it starts with an oxidative addition and ends with a reductive elimination. The middle steps are different and involve the amine associating with Pd before the base deprotonates the amine proton resulting in the loss of halogen from the Pd.
Exercise \(11\)
Propose a product for this intramolecular Buchwald-Hartwig amination.
Answer
This is an excellent strategy to form a new heterocyclic ring. In this case, it's a five membered ring that is part of the bicyclic dihydroindole system.
Summary Problems
Use what we have learned about Pd catalyzed reactions to solve these summary problems. You will also need to incorporate reactions from intro orgo and the pericyclic reactions chapter to complete some of the problems.
Exercise \(12\)
Complete the following synthesis using the indicated starting materials and any other compounds you would like. Hint: Start by finding the retron in the target and going backwards at least one step.
Answer
The key is to start by identifying the cyclohexene Diels-Alder retron in the target. This simplifies the molecule and reveals a Stille retron, the single bond between the alkene and benzene ring. The final disconnection is the amide bond formation. In the forward direction, the carboxylic acid is activated as an acid chloride before adding the starting primary amine and pyridine. The next step is a Pd catalyzed Stille reaction between the aryl iodide and vinyl tin compound. The resulting molecule incorporates an electron poor dienophile and a reactive diene (the furan). Heat promotes the Diels-Alder reaction and delivers the target molecule.
Reference - Organic Letters 2006
Exercise \(13\)
Propose a mechanism for the following transformation.
Answer
This mechanism combines steps from the Heck reaction, Suzuki reaction, and a carbonylation. The mechanism begins with an oxidative addition followed by an alkyne insertion (Heck reaction). There is no hydrogen on the beta carbon, so the reaction continues with CO association and insertion steps (carbonylation). At this point, transmetallation between Pd and B occurs followed by reductive elimination (Suzuki reaction) to yield the product. Predicting this as the product of the reaction would be difficult based on what we know. However, given the structure of the product, we do have the skills to propose a mechanism.
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/02%3A_Transition_Metal_Catalyzed_Carbon-Carbon_Bond_Forming_Reactions/2.02%3A_Pd-Catalyzed_Cross_Coupling_Reactions.txt |
Objectives
After completing this section, you should be able to:
1. Identify ring closing and cross metathesis reactions
2. Draw and understand reaction mechanisms
3. Use reactions in synthesis problems
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Schrock catalyst
• Grubbs 1st generation catalyst
• Grubbs 2nd generation catalyst
• Ring closing metathesis
• Olefin (alkene) metathesis
• Alkyne metathesis
• Cross metathesis
Study Notes
Olefin metathesis reactions enable formation of new carbon-carbon bonds between two starting alkenes while generating ethylene as a byproduct. (This results in the loss of two carbons from the starting material(s) to form ethylene.) Alkyne metathesis, generating acetylene as a byproduct, is also possible but is much less common. The most powerful metathesis reaction is the intramolecular version known as Ring Closing Metathesis (RCM). The intermolecular variation is also useful and is known as Cross Metathesis (CM). Thanks mainly to the work of Robert Grubbs (Caltech) and Richard Schrock (MIT) in the 1990s, these reactions have become popular tools for synthetic organic chemists. For their key research on these reactions, Grubbs and Schrock, along with Yves Chauvin, were awarded the 2005 Nobel Prize in Chemistry. RCM provides you with another highly useful method for the synthesis of rings. This reaction is especially valuable as a method for the synthesis of large rings which are often challenging to synthesize. For the purposes of retrosynthetic analysis using RCM, we look for ring sizes of 5 or larger containing an alkene. So, the cyclohexene Diels-Alder retron is also a ring closing metathesis retron.
Content
This chapter will focus on learning and applying olefin metathesis reactions for the synthesis of rings. We will meet the most common reaction catalysts, learn the reaction mechanism, and see several examples. We will also see brief examples of alkyne metathesis and cross metathesis.
Olefin metathesis catalysts (shown below) are examples of metal-carbene complexes where the reactive metal (either Ru or Mo) contains a double bond to the carbon ligand that participates in the reaction. Furthermore, these highly reactive catalysts are actually examples of what chemists call precatalysts. The compound introduced into the reaction undergoes one round of the catalytic cycle before forming the molecule that catalyzes all further reactions. The precatalysts are less reactive than the actual catalysts, enabling chemists to handle them more easily outside of the reaction flask. Most of the original research in the Grubbs and Schrock labs was conducted in glove boxes under inert atmosphere conditions since early catalysts were highly sensitive to air and moisture. The popularity of the Grubbs catalysts is their combination of stability outside the glove box, reactivity, and functional group compatibility (only reacting with alkenes and not other functional groups).
An example ring closing metathesis (RCM) reaction is shown below along with the mechanism. In this case, we are forming a six membered ring. As mentioned above, RCM reactions generate ethylene as a byproduct, meaning that the starting diene has two more carbons than the cyclic product. Don't forget this key point when planning syntheses using RCM. The initial steps of the reaction convert the precatalyst (the molecules shown above) into the actual catalyst for the reaction. Active metathesis catalysts have CH2 attached to the metal with the structure LnM=CH2. So, initially the alkene byproduct containing the R group present in the precatalyst is formed along with the active catalyst and the desired product. We can now draw out the reaction mechanism with the correct catalyst. (For all future problems, we will ignore the precatalyst reaction and draw mechanisms beginning with what we know is the active catalyst, LnM=.) The RCM mechanism is a series of [2+2] and retro [2+2] cycloaddition reactions. Transition metals can react using d orbitals making thermal [2+2] reactions allowed mechanistic steps (unlike what we learned in the cycloadditions chapter about main group [2+2] reactions). A critical point about the first step is the regiochemistry of how the catalyst reacts with the substrate. For the reaction to be productive and lead to ring formation, the metal must react at the more hindered position. This helps explain why there is a strong preference for the catalyst to react with the least hindered alkene first, in this case the monosubstituted C1-C2 alkene. This first step yields a metallocyclobutane that breaks apart by a retro [2+2] cycloaddition to generate the ethylene byproduct and a new metal carbene with the catalyst covalently bonded to our substrate. Ring formation occurs in the next step, our second [2+2] cycloaddition. The final cycloreversion regenerates the catalyst and installs the alkene in our six-membered ring product.
Cross metathesis reactions enable formation of a substituted alkene in an intermolecular reaction with a second alkene. There are significant selectivity challenges to avoid producing large product mixtures, including E/Z product isomers. One example of a relatively straightforward application of cross metathesis is shown below. Producing a new alkene with symmetrical substitution on one side obviates the potential alkene isomer problem. So, starting with isobutylene and a monosubstituted alkene leads to generation of a trisubstituted alkene product. (As we saw previously with RCM, ethylene is generated as a reaction byproduct.) Note that we have a slightly different catalyst that was developed to promote cross metathesis. This modification of the Grubbs second generation catalyst is know as the Hoveyda-Grubbs catalyst.
Alkyne metathesis is a useful variant of standard olefin metathesis. Tungsten and molybdenum catalysts have been developed that enable efficient ring closing alkyne metathesis. Not surprisingly, this strategy is only useful for the synthesis of large rings which are able to handle the significant ring strain introduced by a cyclic alkyne. A leader in this field is Alois Fürstner who published the following transformation as part of a total synthesis of epothilone C.
Exercise \(1\)
Propose a product for the following reaction sequence.
Answer
The first step is a standard ring closing metathesis reaction to form a seven-membered ring. This step highlights that heterocyclic rings can be easily made using olefin metathesis. The second step is less straightforward but should make you think about silyl ether deprotection with fluoride to yield alcohols. In this case, since the silyl group is part of the ring, it opens the ring to yield an acyclic allyl silane. This highlights the possibility of using RCM to selectively synthesize acyclic cis alkenes. You should also note that the allyl silane in the product can be oxidized to an allylic alcohol by treatment with hydrogen peroxide.
Exercise \(2\)
Starting with compounds containing 12 carbons or fewer, propose two different syntheses of the target compound. Your answers should both incorporate ring closing metathesis, but you should use two different versions of RCM.
Answer
This problem provides an opportunity to use both alkene and alkyne metathesis. Thinking retrosynthetically, the first disconnection is to break open the ring. Don't forget to add the extra carbons necessary for the alkene and alkyne metathesis reactions. Next, you should break the ester bond to yield an acid chloride and an alcohol that fit the 12 carbons or fewer limitation.
Synthesis #1 is likely the most obvious choice. Form the ester in the first step then use the Grubbs catalyst to form the 16-membered ring. One issue with this synthesis is the unfortunate limitation of alkene RCM for large rings that often generates cis/trans product mixtures. In fact, this reaction yields a 1:1 mixture of the cis and trans isomers.
Synthesis #2 enables you to address this problem while adding an extra step to your synthesis. (This is often preferred over making difficult to separate alkene isomers.) The first step generates the diyne ester that is cyclized using alkyne RCM in the second step. This cyclic alkyne (stable in such a large ring) can be hydrogenated using Lindlar's catalyst to yield exclusively the desired cis alkene.
Exercise \(3\)
The following target is a key synthetic intermediate in a published synthesis of pseudotabersonine. Propose a one-step synthesis of this tetracycle beginning with a substituted indole as your bicyclic starting material.
Answer
Reference: Organic Letters 2010
Since we are beginning with the two rings in the indole moiety, we must disconnect the other two rings in our retrosynthetic analysis. Both have an alkene making them retrons for alkene ring closing metathesis. As always when using RCM, be sure to add the carbons that will be lost in the forward direction. The resulting tetraene is the starting material that the Martin group reacted with the Grubbs catalyst in their 2010 Organic Letters paper.
Exercise \(4\)
The following synthetic sequence outlines a strategy that can be applied to the synthesis of taxol-like molecules. First, propose a synthesis of the key ring-closing metathesis precursor. Second, propose a mechanism for the ring-closing metathesis reaction.
Answer
Reference: Organic Letters 2004
The synthesis part of the question involves reactions from intro orgo. The first step is an enolate alkylation between the two starting materials. The final step is a Grignard addition to the ketone.
This is an interesting substrate for olefin metathesis. It contains two alkenes and an alkyne which all have to react to form the target molecule. The most reactive functional group is the terminal alkene which is where the mechanism begins. We do our standard [2+2] and retro [2+2] reactions to generate the ruthenium intermediate that can react with the alkyne to form the 8-membered ring. This [2+2] reaction yields a metallocyclobutene (not butane) that opens to yield the desired cyclooctene with the ruthenium still in the molecule. This molecule can complete the mechanism with another round of [2+2] and retro [2+2] reactions. Please note this is an example of a RCM reaction that does not use LnRu= as the catalyst. Because of the structure of alkene in the final [2+2] reaction, the catalyst contains an isopropyl group. This was the result of reaction optimization when synthesizing the target molecule in the Granja lab.
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/02%3A_Transition_Metal_Catalyzed_Carbon-Carbon_Bond_Forming_Reactions/2.03%3A_Olefin_Metathesis.txt |
Objectives
After completing this section, you should be able to:
1. Identify Pauson-Khand and alkyne cyclotrimerization reactions
2. Draw and understand reaction mechanisms
3. Use these reactions in synthesis problems
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Pauson-Khand reaction
• Alkyne cyclotrimerization reaction
Study Notes
In this chapter, we will learn about two reactions that are commonly promoted by cobalt complexes, the Pauson-Khand reaction for the formation of cyclopentenones and alkyne cyclotrimerization for the synthesis of substituted benzene rings. These transformations are not as common as the Pd-catalyzed bond forming reactions or olefin metathesis reactions that we studied previously. However, they provide novel and powerful strategies for the synthesis of common and important ring structures. When they do appear in organic synthesis applications, they are often intramolecular reactions that quickly build up molecular complexity.
Content
This chapter will focus on ring forming reactions commonly promoted by cobalt. Other metal catalysts or promoters can mediate the Pauson-Khand reaction and alkyne cyclotrimerization reactions. An understanding of the cobalt-mediated reactions will help when you encounter reactions promoted by other metals. Also, unlike the previous transition metal reactions that we have studied, the Pauson-Khand reaction is generally stoichiometric, not catalytic, in cobalt. So, if you end up running this reaction in the lab, pay very careful attention to the equivalents of the cobalt reagent needed to promote the reaction.
Pauson-Khand Reaction
The Pauson-Khand reaction involves combination of an alkene, an alkyne, and carbon monoxide to generate a cyclopentenone. This can happen inter- or intramolecularly, though the intramolecular version is much more popular in synthesis applications. An example of an intermolecular Pauson-Khand reaction is shown below along with its mechanism. In the reaction, trimethylsilylacetylene combines with cyclopentene and carbon monoxide (from dicobalt octacarbonyl) to form a new cyclopentenone ring. The Pauson-Khand retron is a cyclopentenone which is the same retron as the Nazarov reaction (from the electrocyclic reactions section). The first several steps in the mechanism involve reaction of dicobaltoctacarbonyl with the alkyne to form a cobalt-alkyne complex. (The mechanism below is drawn as a 2-electron process. It is also possible to draw this as a radical mechanism. It is unclear which is the actual process.) This is a very unusual structure, with each cobalt bound to both carbons of the initial alkyne. As an aside, cobalt-alkyne complexes can be used as alkyne protecting groups (removed by oxidative decomplexation with a variety of mild oxidants), and they are generally stable. These bright red molecules are stable to TLC and column chromatography. Back to the mechanism, loss of one of the CO ligands leads to alkene association at an open coordination sight with the alkene on the cobalt farthest from the large trimethylsilyl group. The next step is an alkene insertion, like we saw previously in the Pd-catalyzed Heck reaction, to generate new C-C and C-Co bonds. This is accompanied by CO association to keep the Co fully coordinated by ligands. The following step, a CO insertion, is analogous to what we saw previously in Pd-catalyzed carbonylation reactions. One of the CO ligands inserts into the C-Co bond to make another new C-C bond. The mechanism concludes with two reductive eliminations steps to form the final C-C bond in the new 5-membered ring followed by generation of the 5-membered ring alkene. The final reductive elimination step removes cobalt from the molecule thus converting the organometallic complex into the organic product.
Mechanism
Exercise \(1\)
How would you construct the target molecule in one step using a Pauson-Khand reaction?
Answer
This molecule contains a cyclopentenone, the Pauson-Khand retron. Thinking retrosynthetically, we can disconnect the molecule as shown below. Both bonds on either side of the carbonyl are disconnected along with the bond on the other side of the alkene. This reveals the starting materails: a bicyclic alkene, a linear alkyne, and carbon monoxide (from dicobalt octacarbonyl). In the forward direction, the alkene and alkyne are combined in the presence of Co2(CO)8 to yield the target.
Exercise \(2\)
Predict the product of the following reaction.
Answer
Since the alkene and alkyne are both in the starting material, this is an intramolecular Pauson-Khand reaction. It's easier to see how the alkene and alkyne will combine if we redraw the molecule with those two functional groups close to each other. It's also possible to draw a shorthand mechanism by including the CO molecule next to the reactant. (To be clear, this is not the mechanism. The actual mechanism for the Pauson-Khand reaction is shown above. This is a shortcut that enables you to illustrate what happens and quickly generate the product structure.) This allows us to number the atoms, draw curved arrows to keep track of electrons (again, not the actual mechanism), and then draw the product. This shows that while forming the key cyclopentenone, we end up with a tricyclic product.
Co-Catalyzed Cyclotrimerization
This fascinating reaction enables the combination of three alkynes in the presence of cobalt to yield a substituted benzene product. The scope of the reaction is limited due to challenges controlling regiochemistry which often necessitates the use of symmetrically disubstituted alkynes. As shown below, the most useful reactions involve the combination of a diyne with a symmetrically disubstituted alkyne like bis(trimethylsilyl)acetylene. An example reaction and the reaction mechanism are shown below. The reaction shows production of a cyclobutane fused benzene which we have seen previously is very useful in a 4 pi electrocyclic ring opening reaction to generate a highly reactive Diels-Alder diene. The mechanism begins with loss of two CO ligands on the cobalt to yield the active catalyst, CpCo. Don't forget that we previously met Cp, the cyclopentadienyl ligand, in the Introduction to Transition Metals chapter. Double alkyne association with the diyne followed by alkyne dimerization yields a cobaltacyclopentadiene intermediate. The next step is alkyne association by bis(trimethylsilyl)acetylene. At this point, there are two mechanistic pathways that can complete the catalytic cycle. One option is alkyne insertion to yield a seven-membered ring intermediate followed by reductive elimination to yield the product. The other option is a Diels-Alder reaction to yield a Co-bridged intermediate that is followed by a retro cycloaddition to regenerate the catalyst and produce the final product. The exact mechanism will depend on the nature of the substrate, but we will not worry about this level of detail.
Mechanism
Exercise \(3\)
Propose a one step synthesis of the target molecule using a cobalt catalyzed cyclotrimerization reaction.
Answer
When doing retrosynthetic analysis, look for the disilyl substituted benzene ring. This shows the two benzene carbons that originated as bis(trimethylsilyl)acetylene. Knowing that, you can disconnect at every other bond to yield the starting alkynes. Similar to our example above, we start with a diyne which combines with the bis(silyl)acetylene and the cobalt catalyst to yield the target.
Exercise \(4\)
Predict the product of the following reaction sequence that was used in a total synthesis of estrone.
Answer
This reaction sequence involves the key steps in Johnson's synthesis of estrone. It begins with a cyclotrimerization to make the cyclobutane-fused benzene. (Note: This is an oversimplified mechanism. The curved arrows do not illustrate the actual mechanism (see above for that). They just help demonstrate the structure of the product.) This undergoes an electrocyclic ring opening reaction to yield a highly reactive diene that participates in a Diels-Alder reaction to yield the complete tetracyclic steroid core. The estrone synthesis was completed in 2 more steps that we won't worry about.
Summary Problems
Exercise \(5\)
Propose one-step syntheses of the following two targets using a cobalt catalyzed or mediated reaction.
Answer
Both of these targets contain cyclopentenones, so we know they can be made using the Pauson-Khand reaction. The answers demonstrate a shorthand target disconnection that leads backwards to the starting enyne. Note that like in all Pauson-Khand reactions, the carbonyl carbon (carbon #1) does not appear in the starting material because it originates as one of the CO ligands on cobalt. Problem A shows that allenes can react successfully in Pauson-Khand reactions (literature reference Journal of Organic Chemistry 2008).
Exercise \(6\)
Beginning with the indicated starting material and any other compounds, complete a synthesis of the target (alcyopterosin).
Answer
The keys here are the alkyne in the starting material and the benzene ring in the product. This indicates we will be using a cyclotrimerization reaction as the key step. Using the dimethyl carbon as the anchor, it's possible figure out the pieces that must be added to the starting material. The second alkyne is added as an acetylide anion in the first step. In the second step, a different acetylide anion adds via a substitution reaction to yield the necessary triyne. After an alkyne deprotection step, the final step is the cyclotrimerization reaction to yield the target.
Reference: Organic Letters 2010
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/02%3A_Transition_Metal_Catalyzed_Carbon-Carbon_Bond_Forming_Reactions/2.04%3A_Co-Mediated_Ring_Forming_Reactions.txt |
• 3.1: Introduction to Neighboring Group Participation, Rearrangements, and Fragmentations
This page provides an overview of the three types of reactions covered in this chapter: neighboring group participation, rearrangements, and fragmentations. It shows basic mechanisms for these reactions and highlights structural similarities and differences.
• 3.2: Neighboring Group Participation
This page highlights the importance of intramolecular nucleophiles (including heteroatoms, aromatic rings, and pi bonds) for mechanistic understanding of several interesting reactions.
• 3.3: Rearrangements
This chapter focuses on synthetically useful rearrangements including the pinacol, Payne, benzilic acid, Favorskii, Tiffeneau-Demjanov, Wolff, Curtius, Baeyer-Villager, and Beckmann rearrangements.
• 3.4: Fragmentations
This page focuses on synthetically useful fragmentations including the Grob and Eschenmoser fragmentations.
03: Neighboring Group Participation Rearrangements and Fragmentations
Objectives
After completing this section, you should be able to:
1. Understand the general reaction types classified as neighboring group participation, rearrangements, and fragmentations
2. Spot structural fragments that favor neighboring group participation, rearrangements, and fragmentations
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Neighboring group participation
• Rearrangement
• Fragmentation
Study Notes
In Intro Orgo, you learned about common substitution and elimination reactions of alkyl halides (and halide equivalents like tosylates) and classified them as SN1, SN2, E1, and E2. In this section, we are going to introduce you to more interesting reactions that begin with standard R-X or R-OTs starting materials. Our goal in this introductory chapter is to provide an organizational framework that will help guide your more in depth study in the following three chapters. Don't worry about the lack of problems in this chapter, there will be plenty of problems to come when we talk about neighboring group participation, rearrangements, and fragmentations each in their own specific chapter.
Content
As with many topics in Intro Orgo, we only scratched the surface when introducing reactions of alkyl halides. You learned a variety of useful substitution and elimination reactions but likely haven't met reactions that involve neighboring group participation. When studying carbocations formed in SN1 and E1 reactions, you probably learned about simple carbocation rearrangements like hydride, methyl, and alkyl shifts. However, it's unlikely that you were introduced to fascinating rearrangements like the pinacol, Favorskii, or Wolff rearrangements. Finally, fragmentation reactions, where key structural carbon-carbon bonds are cleaved to generate a new carbon skeleton, were probably not mentioned. Let's spend this chapter briefly introducing each of these reactions before studying them in depth in the following chapters.
Neighboring Group Participation (NGP)
As the name implies, this class of reactions relies on the influence of a neighboring group to explain what at first seem like incomprehensible reactions. These are generally substitution reactions that have unexpected outcomes. Their mechanisms become clear, however, once you spot the participation of an intramolecular nucleophile on the reaction outcome. Don't forget that intramolecular nucleophiles can be either a lone pair on a heteroatom or an electron rich pi bond. The following pair of reactions illustrates the importance of understanding neighboring group participation. These are substitution reactions of diastereomeric tosylates that yield the identical trans product. These look like straightforward substitution reactions where an acetate nucleophile replaces a tosylate leaving group. Reaction B fits the pattern we learned in Intro Orgo for an SN2 reaction. The cis starting material reacts with complete inversion to yield the trans product. However, what's happening in Reaction A? This substitution reaction proceeds with complete retention, the trans starting material yields a trans product. This can't be an SN1 reaction since we learned that will yield a racemic mixture (1:1 mixture of cis and trans products). How do we get only retention? The most straightforward way is if two consecutive SN2 reactions occur. How is this possible? If the acetate group that starts on the molecule participates in the reaction as the neighboring group, this explains the outcome.
Looking at the mechanisms, we can understand the reactions. In Reaction A, the neighboring acetate group (always draw out the complete functional group structure to spot if NGP is possible) can participate in an intramolecular SN2 reaction resulting in the first inversion. This is followed by the second SN2 reaction which is of the intermolecular variety with acetic acid. Deprotonation yields the trans product. The mechanism demonstrates how retention is possible: an intramolecular SN2 reaction followed by an intermolecular SN2 reaction. In Reaction B, due to the cis stereochemistry, an intramolecular SN2 reaction is impossible. The cis acetate can't participate in a backside attack since it is on the same face as the leaving group. So, this reaction can only undergo a standard intermolecular SN2 reaction to yield the same trans product.
Rearrangements
You have likely already seen carbocation rearrangements in Intro Orgo. When generating a carbocation as part of an SN1 and/or E1 reaction, you were told to "beware of rearrangement". Thus, you analyzed the structure to determine the type of carbocation generated (primary, secondary, tertiary, and/or resonance stabilized) and looked to the adjacent carbons (the alpha carbons) to see if they would be a more stable carbocation (more highly substituted or resonance stabilized). If so, a group on the adjacent carbon "shifted" to make a new bond to the carbocation. The most common rearrangement being a hydride shift, but you likely also saw alkyl shifts.
Rearrangements definitely occur from carbocation intermediates (as pictured above); however, the most synthetically useful rearrangements proceed via mechanisms that don't involve true carbocation intermediates. The lack of a carbocation helps control the selectivity of the reaction toward production of only the desired target. To identify a rearrangement that doesn't involve a carbocation, the key is to spot a leaving group next to a group that promotes rearrangement. A common example is an alcohol that can turn into a carbonyl upon rearrangement. For example, the semipinacol rearrangement shown below highlights a rearrangement promoted by the loss of a tosylate and the formation of a carbonyl. Deprotonation by sodium hydroxide generates the negatively charged intermediate that undergoes rearrangement. Formation of the new carbonyl at C1 promotes rearrangement resulting from cleavage of the C1-6 bond to form the new C6-10 bond. This rearranges the original bicyclo[6,6] system into a new bicyclo[5,7] system. We will see many more examples of cationic and anionic rearrangements in an upcoming chapter.
Fragmentations
Fragmentations are related to rearrangements in that the structural fundamentals vary slightly. For cationic intermediates, rearrangements occur when the alpha carbon is a more stable carbocation. Fragmentations occur with the beta carbon is a more stable carbocation. This results in cleavage of the alpha-beta C-C bond, the beta carbon becomes the new carbocation, and formation of a new alkene.
Investigating a substrate similar to our rearrangement example above, we can see that true carbocations aren't necessary for fragmentations either. We just need the fragmentation promoting group, an alcohol in this case, one more bond away from the leaving group. With the OH on C6, formation of the new carbonyl promotes cleavage of the 1-6 bond to form a new alkene upon loss of the tosylate leaving group. This example demonstrates the power of the Grob fragmentation for the production of medium-sized rings, in this case a 10-membered ring, that are often difficult to make.
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/03%3A_Neighboring_Group_Participation_Rearrangements_and_Fragmentations/3.01%3A_Introduction_to_Neighboring_Group_Participation_Rearrangements_and_Fragmentations.txt |
Learning Objectives
After completing this section, you should be able to:
1. Understand the power of neighboring group participation to generate unexpected reaction outcomes
2. Spot structural fragments that favor neighboring group participation
3. Draw mechanisms incorporating neighboring group participation to explain reaction outcomes
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Neighboring group participation (NGP)
• Phenonium ion
• Non-classical carbocation
Study Notes
We briefly introduced neighboring group participation (NGP) in the previous chapter. In this chapter we will see more examples that highlight the impact of a nucleophilic neighboring group on reaction rate, regiochemistry, and stereochemistry. Remember, these are all substitution reactions of alkyl compounds containing leaving groups (halide, tosylate, etc.) and some type of internal nucleophile, either a heteroatom with a lone pair or a pi bond. It is good practice to start these types of problems by drawing out the full structure of any abbreviated functional groups and working your way through the mechanism starting with possible intramolecular reactions.
We will separate this chapter into three different types of neighboring groups: 1) Heteroatoms with lone pair electrons, 2) Benzene ring pi bonds, and 3) Alkene pi bonds. Lone pair electrons on heteroatoms are the easiest neighboring groups to spot and are the most common. Pi bonds, either from benzene or an alkene, create very unusual intermediates and generate very surprising results. The alkene neighboring group even created a heated chemical controversy that involved high profile chemists (including one who would later go on to win the Nobel Prize) attempting to determine the correct mechanism for one important reaction.
Heteroatom Nucleophiles as Neighboring Groups
We have already seen an example of this class of reactions in the preceding chapter. As a reminder, the diastereomeric tosylates react with acetic acid to yield the identical trans diacetate product. (For an explanation of this outcome, including the reaction mechanisms, please revisit Chapter 3.1.) We also should note that the relative rates of these reactions vary dramatically, with Reaction A proceeding 670 times faster than Reaction B. This further highlights the power of neighboring group participation; it can dramatically enhance the rate of reactions versus standard nucleophilic substitution reactions. We will see further examples of this throughout the chapter.
We can see another NGP example below with the reactions of the different primary alkyl chlorides with water. In the absence of the rate information, we would assume these are both simple SN2 reactions. However, the dramatic differences in reaction rates indicate that something unusual is happening. This is when we should look for a neighboring group, hopefully spotting the sulfur in the bottom reaction.
The S can act as an excellent internal nucleophile and promote an intramolecular reaction. As highlighted below in the mechanisms for these two reactions, the top reaction is a straightforward SN2 reaction. The bottom reaction is much faster because of NGP with the internal S nucleophile. This creates a cyclic sulfonium ion intermediate that reacts much more quickly with water.
Exercise \(1\)
Propose a mechanism to explain the outcome of the following reaction.
Answer
This is an example of a substitution reaction that proceeds with retention. As we have already seen, this is a hallmark for an NGP mechanism. In this case, the internal nucleophile is a carboxylate that forms an epoxide-type intermediate which reacts quickly with the methanol solvent. These two consecutive SN2 reactions result in overall retention of configuration.
Exercise \(2\)
Propose a mechanism to explain how both products are formed in the reaction.
Answer
If this was a simple substitution reaction, we would only form the first product. Seeing that two products are formed, including the second one that looks very strange, we should focus on neighboring group participation. The ether O is a very good internal nucleophile. It can react to form a cationic five-membered ring intermediate. This common intermediate can lead to formation of both of the products depending on which carbon in the intermediate is attacked.
Benzene Ring Pi Bonds as Neighboring Groups
As demonstrated in the following pair of reactions, benzene ring pi bonds are very effective neighboring groups. As we have already seen, without the rate data, we would not know that anything interesting is happening when comparing these two reactions. How is it possible that the second reaction is 3,000 times faster than the first one? With no heteroatom containing a lone pair, it must be the presence of the benzene ring.
The first reaction is a standard SN2 reaction while the second one results from two SN2 reactions, as shown in the mechanism below. Like when doing electrophilic aromatic substitution reactions in Intro Orgo, the phenyl ring acts as a nucleophile to displace the tosylate leaving group. This generates a resonance stabilized three-membered ring phenonium ion intermediate. Addition of the trifluoroacetate nucleophile breaks open the three membered ring and restores aromaticity to the benzene ring.
Exercise \(3\)
Propose a mechanism to explain the results of the following reaction. Note: You are starting with a single enantiomer of the starting material.
Answer
We can verify this isn't an SN2 reaction by drawing out the product that would result from inversion at the carbon bearing the tosylate. Again, we must look to neighboring group participation. When the benzene ring attacks the tosylate it yields an achiral phenonium ion. The acetate nucleophile can attack either the left or right side of the three membered ring to yield the product with a restored benzene ring. The two products formed are mirror images and flipping the product on the left demonstrates that we do form the target product mixture.
Alkene Pi Bonds as Neighboring Groups
This is, perhaps, the most interesting example of neighboring group participation. It was definitely the most controversial and involved many research groups who investigated whether the key intermediate was a classical or non-classical carbocation. The principal players in this drama were H.C. Brown (Nobel Prize winner for hydroboration reactions) who favored the classical carbocation and Saul Winstein who promoted the non-classical explanation. Winstein's view was ultimately proven correct based on a variety of investigations including NMR and X-ray crystallography. For a detailed explanation of this important historical argument, there is an excellent description in Walling's 1983 paper in Accounts of Chemical Research. So, how do we make this fascinating carbocation and what does it have to do with neighboring group participation? The scheme below illustrate the key reactions, and we will see the carbocation when we illustrate the mechanism. First, to the reactions! The second reaction is a straightforward SN2 reaction. The first reaction has an unbelievable rate enhancement of 1011! This is truly remarkable when compared to the other rate enhancements we've seen in this chapter with the largest being 3,000 for the phenonium ion reaction.
So, what is going on? This is where we wade into the classical versus non-classical carbocation debate. Clearly, the neighboring alkene pi bond is acting as the neighboring group and pushing off the tosylate leaving group. This is our first SN2 reaction. What is the structure of the cation formed when this happens? We have two options. For both the classical and non-classical carbocation options, the structures look identical. The critical difference is that we are using different types of arrows to connect the structures. For the classical carbocation, we are showing the structures as rapidly interconverting cations that are in equilibrium. They are all distinct intermediates. For the non-classical carbocation, these cations are resonance structures, so there are not intermediates at all. Instead, the intermediate is the hybrid structure (pictured in a top down view making it easier to see). This is a very strange resonance hybrid. The three dashed lines indicate that the two electrons from the original pi bond are shared over three carbon atoms, making this an unusual three center, two electron bond! In the second SN2 reaction, acetic acid attacks this resonance hybrid to yield the product.
Summary Problems
Exercise \(4\)
The reactions of the isomeric starting materials produce very different products. Propose a mechanism to explain Reaction A. Why can't the starting material in Reaction B undergo a similar reaction? Reaction B is a preview of what we will see in the fragmentation chapter. Propose a mechanism to explain the product formation.
Answer
Reaction A should look familiar. This is another example of neighboring group participation with the N attacking as the internal nucleophile. Two SN2 reactions yield the product that is formed with net retention. In Reaction B, backside attack is impossible because the N is attacking the same face where the leaving group already is. Instead, this molecule fragments, breaking the C1-C7 bond to yield the monocyclic product shown. | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/03%3A_Neighboring_Group_Participation_Rearrangements_and_Fragmentations/3.02%3A_Neighboring_Group_Participation.txt |
Objectives
After completing this section, you should be able to:
1. Spot structural fragments that favor rearrangement reactions
2. Draw mechanisms incorporating rearrangements to explain reaction outcomes
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Pinacol rearrangement
• Semipinacol rearrangement
• Payne rearrangement
• Benzilic acid rearrangement
• Favorskii rearrangement
• Tiffeneau-Demjanov rearrangement
• Wolff rearrangement
• Curtius rearrangement
• Baeyer-Villager rearrangement
• Beckmann rearrangement
Study Notes
We briefly introduced rearrangements in a previous chapter and you likely learned a little bit about them in Intro Orgo. In this chapter, we will focus exclusively on rearrangements that are synthetically useful. Rearrangements occur readily under strongly acidic conditions, but this leads to uncontrolled decomposition for most substrates. You will see that useful rearrangements occur under both acidic and basic conditions, and our goal is to introduce you to some of the most popular rearrangements for synthesis. This generally involves a leaving group on an atom adjacent to a rearrangement promoting atom, generally an O or N. It is often very challenging to think retrosynthetically for rearrangements, so our focus will be on the forward direction. We will aim to draw mechanisms that explain reaction outcomes, sometimes including interesting stereochemical issues.
Content
Our goal in this chapter is to introduce the most important rearrangements you will likely encounter in organic synthesis papers. We also know this will provide you with the skills to understand other reactions that you might encounter during your studies. Heteroatom placement in these reactions is critical, with heteroatoms playing the roles of both leaving groups and rearrangement promoting atoms. We will see that substrates rearrange to make new C-C, C-O, and C-N bonds. These reactions also provide opportunities for ring expansions and ring contractions.
Pinacol Rearrangement
The pinacol rearrangement is the acid catalyzed rearrangement of a 1,2-diol into a ketone. It is named after the molecule pinacol, pictured below, the simplest substrate to undergo the reaction. One of the alcohols is protonated to make the leaving group (water) while the other OH participates as the rearrangement promoting group. After water leaves, generating a tertiary carbocation, the remaining alcohol forms a carbonyl, thus promoting the rearrangement of a methyl group. Deprotonation completes the mechanism to form pincolone and regenerate the catalytic acid.
For unsymmetrical diols, not surprisingly, you form the most stable carbocation intermediate. You will see an example of this in the problems below. What about substrates where different groups can migrate? The following examples demonstrate that H migrates faster than R, which is consistent with what you saw previously in Intro Orgo. An H shift yields a more stable cationic intermediate than if an alkyl group migrates. What about Ph vs H? In this case, the benzene ring migrates faster. These seems strange until we remember our previous chapter on neighboring group participation. A migratory phenyl yields a phenonium ion intermediate that is favored over H migration.
Exercise \(1\)
For the following reaction, propose a product and a mechanism to explain its formation.
Answer
Using the same strategy as the pinacol rearrangements above, we quickly generate a tertiary carbocation. Formation of the carbonyl that we know will be in the product promotes rearrangement (migration) of one of the ring bonds. This yields a very interesting result. One of the original 5-membered rings expands to form a new 6-membered ring. So, we can use the pinacol rearrangement as a ring expansion reaction!
Exercise \(2\)
For the following reaction, propose a product and a mechanism to explain its formation.
Answer
In this example, the two possible carbocations that can form after water leaves are very different. The tertiary carbocation is low enough in energy to form; however, the doubly benzylic, resonance stabilized carbocation is much lower in energy. So, it forms preferentially, resulting in a methyl shift to form the product ketone.
Another interesting aspect of the pinacol rearrangement is the impact of stereochemistry on the migrating group. If you think about this statement, it seems to contradict what we have shown in all of the reactions up to now in this chapter. How can stereochemistry of the diol (cis or trans) matter when we generate a planar carbon as part of the carbocation intermediate? It turns out, like many reactions in organic chemistry, our first look was an oversimplification. The current understanding is that an achiral carbocation does not form. Instead, a chiral ion pair forms, retaining the stereochemistry of the starting material. However, when considering rearrangements of chiral diols, we will sometimes draw the carbocation and sometimes a concerted mechanism depending on what helps most with our understanding. The key is to recognize that what does form is a chiral version of the free carbocation (the ion pair) that promotes the rearrangement. Similar to what you learned in intro orgo for SN2 and E2 reactions, the migrating group must be antiperiplanar to the leaving group. The example and problem below demonstrate that stereochemistry is critically important when considering reactions of cyclic diols. We will first consider reaction of the cis cyclohexane diol. Axial leaving groups are much more reactive than equatorial leaving groups, so the reaction occurs with the protonated axial OH leaving accompanied by formation of the new carbonyl and rearrangement of the axial methyl group, the anitperiplanar migrating group. What happens if we run the same reaction on the trans isomer of this diol? See if you can come up with an answer as part of the next problem.
Exercise \(3\)
What is the product of this reaction? Also, provide a mechanism for its formation.
Answer
Both of the alcohols are equatorial because they are larger than the axial methyl groups and they can form an intramolecular hydrogen bond to stabilize this conformation. So, when a protonated equatorial alcohol leaves, which group is antiperiplanar to participate in the pinacol rearrangement? It's clearly not one of the methyl groups. Thus, the ring bond is the only alternative. As shown in the mechanism, formation of the new carbonyl promotes cleavage of the 5-6 bond to form a new 1-5 bond. The result is a ring contraction from 6- to 5-members! This demonstrates that the pinacol rearrangement can produce both ring contractions and ring expansions, as we saw previously.
Let's consider one more type of pinacol rearrangement that is very important for complex structures and often appears in total synthesis applications. It is called a semipinacol rearrangement and involves rearrangement under basic conditions where it is impossible to form a carbocation. To understand why this is a key reaction, let's first look at the reaction below. This is a standard pinacol rearrangement. As we saw above, with an unsymmetrical diol, we will protonate the tertiary carbocation to form the more stable tertiary carbocation and then do our rearrangement. In this case, a hydrogen shift to yield the ketone product. But, what if we wanted to form the carbocation at the secondary alcohol? It is impossible to do this under acidic conditions.
The solution is to move away from acidic conditions and think about reactions we know that occur selectively at secondary alcohols. If we can make the secondary alcohol into a good leaving group, we can force that alcohol to leave, reverse the position of the rearrangement promoting O and generate a different product. How do we put that into practice? The reaction sequence below shows the standard way this is accomplished. Alcohols react selectively with tosyl chloride from least hindered (primary) to most hindered (tertiary), so we can selectively tosylate the secondary alcohol in the presence of the tertiary alcohol. Next, addition of a base results in deprotonation of the alcohol to form an alkoxide that can promote the rearrangement. Formation of the ketone promotes rearrangement of the ring bond to force out the tosylate leaving group. In this fascinating example, that results in a simultaneous ring expansion and ring contraction!
Payne Rearrangement
The Payne rearrangement involves reactions of nucleophiles with epoxy alcohols under basic conditions. Like many rearrangements, at first glance it seems confusing. We can rationalize hydroxide adding to an epoxide, but how does the sulfur nucleophile replace the poor OH leaving group? The key here is to think first about deprotonating the primary alcohol and then to focus on neighboring group participation.
Deprotonating the primary alcohol generates an alkoxide nucleophile that opens the epoxide via an intramolecular SN2 reaction to form a new epoxide. The sulfur nucleophile can now add to the less hindered side of the new epoxide via an intermolecular SN2 reaction. The final protonation step yields the target trans diol.
Benzilic Acid Rearrangement
The benzilic acid rearrangement involves conversion of a 1,2-diketone into a carboxylic acid. The conditions are deceptively simple, hydroxide followed by an acid quench, and lead to the migration of a benzene ring.
This mechanism is relatively straightforward. Hydroxide adds to one of the ketones to yield a tetrahedral intermediate. Reforming the carbonyl results in rearrangement of the phenyl onto the second carbonyl. The acid quench ultimately generates the target carboxylic acid.
Favorskii Rearrangement
The Favorskii rearrangement transforms an alpha halo ketone into an ester, as shown in the example below. Upon first inspection, this seems to continue the theme we just saw in the benzilic acid rearrangement. Add the nucleophile to the carbonyl, reform the carbonyl, and have the rearrangement occur pushing out the leaving group. For this substrate, that results in a ring contraction reaction. As we will see in the problems below, this is not the only mechanistic possibility.
Exercise \(4\)
Propose a mechanism and a product for the following reaction.
Answer
This reaction appears to behave exactly like the reaction shown above. Rearrangement occurs from the tetrahedral intermediate and yields the ester product. See the answers to Exercise #5 for an alternate mechanism to get the same product.
Exercise \(5\)
Propose a mechanism and a product for the following reaction.
Answer
As shown below, our standard Favorskii mechanism fails for this reaction. We do not form the product predicted by the mechanism used previously. Instead, we form the same product observed in the reaction from Exercise #4. How is this possible? If the two starting materials form the same product, perhaps the reactions proceed through a common intermediate.
A different way to think about these reactions is to use the methoxide as a base, not a nucleophile. If we deprotonate alpha to the carbonyl on the opposite side of the halogen, something interesting can happen. As shown below, we form two different enolates that can each undergo an intramolecular alkylation reaction to yield the same cyclopropanone intermediate. Due to the ring strain, this is a much more reactive carbonyl that does react with methoxide to yield a tetrahedral intermediate. Reforming the carbonyl breaks open the three-membered ring to selectively yield the resonance stabilized anion that is subsequently protonated by the methanol formed in the first step.
So, we have seen two potential mechanisms for the Favorskii rearrangement. Depending on the substrate, either is possible. So, the best problem solving strategy is to consider both options and then see which one looks most favorable. In the example above, resonance stabilization of an anion adjacent to the benzene ring helps favor this mechanism.
Tiffeneau-Demjanov Rearrangement
In this rearrangement, a 1,2-aminoalcohol is converted into a ketone as shown in the generic example below. The reagent that promotes this transformation is nitrous (not nitric) acid. If you previously studied nucleophilic aromatic substitution reactions, you might recognize that combining an amine and nitrous acid yields a reactive diazonium intermediate. This is a key step in the Sandmeyer reaction where anilines react with nitrous acid to yield a diazo benzene intermediate that reacts with a variety of nucleophiles to make new benzene derivatives.
In the Tiffeneau-Demjanov rearrangement, formation of the diazonium intermediate promotes the key rearrangement step. Similar to previous mechanisms, formation of the carbonyl promotes the alkyl shift and loss of the leaving group.
A historical note about this reaction is to recognize the key role played by Bianka Tchoubar in its development. She conducted the critical experiments in Tiffeneau's lab to determine the scope and limitations of the reaction. Bianka was a key figure in the organic chemistry community in France from the 1930s until the 1980s, resulting in 140 publications.
Exercise \(6\)
First, starting with cyclopentanone, how would you produce the desired aminoalcohol? Second, what product is formed upon exposure of the aminoalcohol to nitrous acid?
Answer
There are two ways to convert cyclopentanone into the target aminoalcohol. In option #1, we add sodium cyanide to yield a cyanohydrin that can be reduced with either lithium aluminum hydride or hydrogen and catalytic palladium. In option #2, we use a Henry reaction to generate a nitroalcohol that can be reduced with hydrogen and catalytic palladium. Adding nitrous acid promotes the Tiffeneau-Demjanov rearrangement to yield a ring expanded ketone.
Wolff Rearrangement
The Wolff rearrangement is similar to the Tiffeneau-Demjanov rearrangement because of the key role of a diazo intermediate. The most common variation involves reaction of a ketone with a diazo compound. As shown below, treating cyclobutanone with diazomethane yields a ring expansion reaction to form cyclopentanone via a mechanistic pathway that should look familiar.
Exercise \(7\)
Propose a mechanism and the product for the following reaction.
Answer
As with many mechanisms, it helps to draw out the complete Lewis structure for reactive functional groups. There are two common resonance structures for diazo compounds and the one with the negative charge on carbon highlights that this atom is nucleophilic. Using this carbon to form a new carbon-carbon bond yields a tetrahedral intermediate. Reforming the carbonyl leads to the rearrangement, resulting in a ring expansion and the seven-membered ring product.
Wolff rearrangements are also useful for ring contraction reactions. As shown below, treating an alpha diazoketone with heat or light in methanol promotes the rearrangement to yield the contracted ester product. We will discuss the mechanism for this reaction in the next problem. Forming the starting alpha diazoketone involves a diazo transfer reaction with a ketone. We will generally not worry about this step and instead start with the alpha diazo ketone.
Exercise \(8\)
Propose a mechanism for this reaction. Hints: The key intermediate is a ketene. The role of heat or light is to promote the loss of nitrogen gas. You don't need to factor that into your mechanism.
Answer
This is a strange mechanism for us. It actually will make more sense when we study carbenes in a later chapter. For now, let's focus on the resonance structure with the negative charge on carbon. From here, nitrogen can leave, we can form a new carbon-carbon double bond between C1 and C6, and we can rearrange to form a new C2-C6 bond. This yields the ketene intermediate that reacts with methanol to form an enol. Tautomerization yields the 5-membered ring ester product.
Curtius Rearrangement
The Curtius rearrangement generally involves the conversion of a carboxylic acid into an amine with the loss of one carbon. It is similar to the Wolff rearrangement but in place of diazomethane, this reaction uses an azide nucleophile. A generic example, shown below, involves generation of an acid chloride upon treatment of the carboxylic acid with thionyl chloride followed by reaction with sodium azide to promote the rearrangement then addition of water to generate the amine product. Other products formed include nitrogen gas and carbon dioxide.
As shown below in the full mechanism, the key intermediate is an isocyanate. This forms upon rearrangement of the azide intermediate resulting in loss of nitrogen gas and formation of a new R-N bond. Addition of water to the isocyanate generates a carbamic acid that loses carbon dioxide under the reaction conditions to complete the reaction and form the product amine. Other nucleophiles can be added to the isocyanate intermediate to yield different products including a substituted amide, as shown in the problem below.
Exercise \(9\)
Beginning with the indicated starting material, how could you use a Curtius rearrangement to generate the target amide?
Answer
This reaction has been used in the synthesis of the natural product pancratistatin. Like in our general example, the starting carboxylic acid is treated with thionyl chloride followed by sodium azide. This yields the isocyanate intermediate after a Curtius rearrangement. Adding a functionalized organolithium reagent to the isocyanide yields the target amide. This sequence nicely highlights that isocyanates can participate in more than just hydrolysis reactions.
Baeyer-Villager Rearrangement
The Baeyer-Villager rearrangement is the reaction of a peracid with an aldehyde or ketone to yield a carboxylic acid or ester, respectively. There are two possible products for ketones with selectivity generally favoring migration of the larger group. The most common peracid used synthetically is meta-chloroperoxybenzoic acid (MCPBA) which also reacts with alkenes to form epoxides. Beware of this dual reactivity when planning syntheses. The Baeyer-Villager rearrangement mechanism is shown below. The peracid adds to the aldehyde or ketone to produce a tetrahedral intermediate. Reforming the carbonyl promotes the rearrangement with H migrating exclusively in the aldehyde substrate to yield the carboxylic acid product. A similar intermediate is formed in the ketone reaction. In this case, the larger alkyl group migrates (related to which group can better stabilize a partial carbocation on the carbonyl carbon in the transition state) to yield the ester product.
Exercise \(10\)
Propose a product for the following Baeyer-Villager rearrangement.
Answer
The left side of the ketone is larger than the methyl group, so this alkyl group migrates onto the new O in the mechanism to yield the product ester. Note that no bonds were formed or broken to the chiral carbon on the cyclohexane ring so this stereochemistry does not change.
Exercise \(11\)
Propose a Baeyer-Villager rearrangement that would yield the target lactone.
Answer
Thinking about the ketone that led to the formation of the six-membered ring lactone, we should realize that we just need to remove the O in the ring from the product to yield the all-carbon starting material. In this case, it's cyclopentanone. This reaction highlights the ability of the Baeyer-Villager rearrangement to promote a ring expansion reaction.
Beckmann Rearrangement
The Beckmann rearrangement converts ketones into amides and is the nitrogen equivalent of the Baeyer-Villager reaction with ketones. The rearrangement is relatively straightforward for reactions of symmetrical ketones, as shown in the example below.
Treatment of the ketone with hydroxylamine yields an oxime that rearranges upon exposure to sulfuric acid to yield a nitrilium ion. Addition of water completes the mechanism to yield the product amide after tautomerization. The details are shown in the scheme below.
What about reaction of an unsymmetrical ketone? The structure of the oxime leads directly to the rearrangement outcome. Sterically, the OH in the oxime is oriented away from the larger ketone group. This results in the major or only product resulting from rearrangement of the larger ketone substituent.
Exercise \(12\)
What is the product of the following Beckmann rearrangement?
Answer
When starting with a cyclic ketone, a Beckmann rearrangement promotes a ring expansion reaction to yield an expanded lactam as shown below.
The Beckmann rearrangement isn't always this straightforward, as the following experiment illustrates. Beginning with a mixture of the two oximes, they are treated with tosyl chloride to yield the corresponding tosylates. Heating the tosylates yields four products. Products 1 and 2 are the standard Beckmann rearrangement products. How did products 3 and 4 form? Clearly, something strange is happening here.
To explain how all four of these products formed simultaneously, we must consider a fragmentation reaction as part of the mechanism. This results when a highly stabilized carbocation intermediate can form which is definitely the case for this substrate. Fragmentation of the tosylates yields tertiary carbocations A and C. These carbocations can combine with either of the nitriles (B or D). So, these four recombination possibilities yield the observed products. This experiment highlights another mechanistic possibility for the Beckmann rearrangement and provides a preview of the mechanism we will focus on in our next section (fragmentations!).
Summary Problems
Exercise \(13\)
Propose a mechanism for the following transformation.
Answer
The key step in this mechanism is a Payne rearrangement. This occurs after the primary alcohol is deprotonated. The resulting new epoxide reacts with the sulfur ylide produced upon deprotonation. After forming the new carbon-carbon bond, the new 5-membered ring forms via a substitution reaction with dimethyl sulfoxide as the leaving group.
Reference - Journal of the American Chemical Society 2004
Exercise \(14\)
The molecule below can be used to synthesize substituted azulenes. Our goal in this problem is to propose a three-step synthesis of this bicyclic cyclopentenone. One step involves dichloroketene while another is a ring expansion reaction.
Answer
The key step in this synthesis is a Wolff rearrangement. The synthesis begins with a ketene [2+2] cycloaddition between cycloheptatriene and dichloroketene. The resulting bicyclic ketone reacts with diazomethane and then undergoes the Wolff rearrangement. This results in a ring expansion to yield the 5,7-bicycle. The final step in the synthesis is an E2 elimination of HCl promoted by triethylamine.
Reference - Angewandte Chemie International Edition 2005
Exercise \(15\)
(+)-Sparteine is a natural product that is a very popular chiral ligand for a variety of asymmetric reactions. The first asymmetric total synthesis of this molecule was published in 2002 and it involved several reactions of interest to us. Azide 1 was prepared in eight steps from commercially available norbornadiene. Propose a mechanism for the conversion of 1 into amide 2. After five steps, 2 was converted into keto iodide 3 which upon treatment with hydroxyl amine and tosyl chloride yielded amide 4. (Note: I have modified the reaction conditions for the synthesis of 4 from the literature to make them consistent with material you have learned in this chapter.) Provide a mechanism for this transformation and then indicate how you would convert 4 into (+)-sparteine in one step.
Answer
The key step here is a Schmidt reaction which is similar to a Curtius rearrangement. The Lewis acid activates the carbonyl for attack by the azide. This tetrahedral intermediate undergoes the rearrangement to form the amide product and lose nitrogen gas.
The key step in this sequence is a Beckmann rearrangement. The mechanism starts with a substitution reaction with hydroxylamine following by an intramolecular iminium ion formation. (Note: I have omitted the proton transfer steps to save space.) In the paper cited below, the second step is irradiation with ultraviolet light to promote an unusual photo-Beckmann rearrangement. For consistency with what we learned in this chapter, we will proceed as if TsCl and heat will enable this transformation. Addition of tosyl chloride and heat should promote formation of an N-tosylate that undergoes the Beckmann rearrangement. This yields a very unusual intermediate with positive charges on adjacent atoms (and likely explains why this reaction doesn't work in the lab). It should react quickly with water to yield the amide product after deprotonation.
The synthesis of (+)-sparteine concludes with a lithium aluminum hydride reduction of the amide.
Reference - Organic Letters 2002
Exercise \(16\)
Propose a mechanism for the following transformation.
Answer
The key step in this mechanism is a semi-pinacol rearrangement. In the first step, bromine reacts with the alkene to yield a cyclic bromonium ion intermediate. This undergoes a semi-pinacol rearrangement to yield the target aldehyde.
Reference - Organic Letters 2004
Exercise \(17\)
Trost discovered the following transformation while investigating the total synthesis of pseudolaric acid B. Propose a mechanism to explain this reaction.
Answer
The key step in this mechanism is a pinacol rearrangement. The epoxide adds to the Lewis acid, then opens up to form a tertiary allylic carbocation. Next, the pinacol rearrangement yields the bicyclic 5,6-spiro product.
Reference - Journal of the American Chemical Society 2008
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/03%3A_Neighboring_Group_Participation_Rearrangements_and_Fragmentations/3.03%3A_Rearrangements.txt |
Objectives
After completing this section, you should be able to:
1. Spot structural fragments that favor fragmentation reactions
2. Draw mechanisms incorporating fragmentations to explain reaction outcomes
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Grob Fragmentation
• Eschenmoser Fragmentation
Study Notes
We briefly introduced fragmentations in a previous chapter and you likely did not learn about them in Intro Orgo. We also met the Beckmann fragmentation at the end of the previous chapter on rearrangements. In this chapter, we will focus exclusively on fragmentations that are synthetically useful. Like rearrangements, fragmentations occur readily under strongly acidic conditions, but this leads to uncontrolled decomposition for most substrates. You will see that useful fragmentations occur under conditions similar to rearrangements but with a slightly different arrangement of atoms, one more carbon between the fragmentation promoting atom (generally O or N) and the leaving group. Fragmentations are much less common than rearrangements.
Content
In this chapter, we will introduce the most important fragmentations you will likely encounter in organic synthesis papers. We also know this will provide you with the skills to understand other reactions that you might encounter during your studies. Heteroatom placement in these reactions is critical, with heteroatoms playing the roles of both leaving groups and fragmentation promoting atoms. We will see that substrates fragment to make new C-C bonds and turn cyclic molecules into acyclic targets. Fragmentations also provide a novel method for the synthesis of challenging medium-sized rings upon fragmentation of bicyclic systems.
Grob Fragmentation
The Grob fragmentation is similar to the pinacol rearrangement. Substrates that are 1,2-diols (or 1,2-diol type compounds) undergo the pinacol rearrangement to yield a new ketone or aldehyde. Substrates that are 1,3-diols (or 1,3-diol type compounds) undergo the Grob fragmentation to yield both a new ketone/aldehyde and an alkene, as shown below.
The Grob fragmentation is an excellent method for constructing medium or large rings from bicyclic systems. Two examples for the synthesis of 10-membered rings are shown below. The example on the left is a standard Grob fragmentation. A deprotonated alcohol forms a carbonyl that promotes cleavage of the adjacent C-C bond (the fragmentation) to generate a new alkene and loss of the tosylate leaving group. The example on the right highlights that nitrogen can also function as a fragmentation promoting atom. In this case, fragmentation and loss of tosylate yields a new alkene and an iminium ion. Sodium borohydride reduces the iminium ion to yield the amine product. The example on the left is actually an oversimplification of this reaction because it does not show stereochemistry. Fragmentations are similar to E2 reactions and cyclic pinacol rearrangements in that the reacting groups must be antiperiplanar for constructive orbital overlap to occur. The follow problem explores the importance of stereochemistry for the Grob fragmentation.
Exercise \(1\)
Only one of the following two reactions yields a Grob fragmentation. Which reaction is it? Explain. Hint: It is very helpful to draw out these trans decalin molecules in the chair conformation.
Answer
Reaction #1 undergoes the fragmentation while reaction #2 does not. Focusing on the chair conformation for each molecule, we can explain why. In the first reaction, the fragmenting C-C bond and the bond to the leaving group are anti (both in red). In reaction #2, the fragmenting C-C bond and the leaving group (both in red) are not anti. Instead, a C-H bond is anti to the fragmenting C-C bond so the reaction does not occur.
Exercise \(2\)
Predict the product of the following reaction.
Answer
Following our standard strategy for the Grob fragmentation, we can show how the starting tricycle fragments into the product bicycle. As always, numbering helps keep track of atoms.
1,5-Fragmentations
Though much less common than Grob fragmentations, 1,5-fragmentations are also possible. These substrates start as ketones or aldehydes and the reaction is promoted by base induced enolate formation. Upon fragmentation, they yield an unsaturated ketone and an alkene, as shown below.
Exercise \(3\)
Predict the product of the following reaction.
Answer
After formation of the enolate, the molecule can fragment as shown above to yield the 11-membered ring product.
Eschenmoser Fragmentation
The Eschenomser fragmentation (sometimes called the Eschenmoser-Tanabe fragmentation) was discovered in 1967 in the lab of Albert Eschenmoser at the Swiss Federal Institute of Technology (ETH) in Zurich, Switzerland. Much of the early work on the scope of this reaction in the Eschenmoser lab was conducted by Dorothee Felix. The reaction begins with the epoxidation of a cyclic conjugated enone to yield an epoxy carbonyl. Treatment of this molecule with tosyl hydrazine promotes fragmentation to remove the ring, generating a new carbonyl and alkyne.
The mechanism of the fragmentation reaction is similar to the Wolff-Kishner reaction (reduction reaction of carbonyls with hydrazine and base) with important differences due to the presence of the epoxide and the tosylate group. After formation of the tosyl hydrazone, the important mechanistic steps begin. Movement of electrons from the nitrogen attached to the tosyl forms a double bond between the nitrogens, creates a C5-C6 alkene, and opens the epoxide. Next, the electrons flow back in the opposite direction. A carbonyl forms which cleaves the C2-C6 bond (the fragmentation) to yield the product alkyne while generating nitrogen gas (N2) and the tosyl leaving group.
Exercise \(4\)
What is the product of the following reaction sequence?
Answer
Like all Eschenmoser fragmentations, the reaction will ultimately cleave the alkene bond in the original conjugated enone. The means that the alkene carbon farthest from the carbonyl (the beta carbon) will turn into the product ketone while the other carbon from the alkene and the ketone carbon will turn into the product alkyne. Applying this along with the mechanism show above yields the target, containing one fewer ring than the starting material.
Summary Problems
Exercise \(5\)
The triterpene shown below was shown to arise biosynthetically via a Grob fragmentation. Propose a reasonable precursor that would participate in this fragmentation to yield the given target.
Answer
Returning to our introduction of the Grob fragmentation above, we can see that the product has a carbonyl at C1 and an alkene between C2 and C3. This means for our 1,3-diol type starting material, we need the alcohol at C1 and the leaving group at C3. We also need a bond between C1 and C2 that will cleave during the fragmentation reaction. That means our starting material is the bicyclic molecule shown below.
Reference - Angewandte Chemie 2006
Exercise \(6\)
The following undesired reaction occurred during John Wood’s synthesis of welwitindolinone. Propose a mechanism to explain this transformation. Hint: The final step of the mechanism is a tautomerization reaction to generate the bicyclic aromatic system from a lactam.
Answer
The mechanism starts with an intramolecular carbonyl addition of the amine to the ketone. (Base removes a proton from the amine after the addition.) The resulting tetrahedral intermediate is set up perfectly for a fragmentation due to the chloride leaving group at the 3-position. So, the carbonyl reforms, the C1-C2 bond cleaves to form the C2-C3 alkene, and chloride leaves. This yields a lactam that tautomerizes to the product quinoline structure.
Reference: Journal of the American Chemical Society 2008
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/03%3A_Neighboring_Group_Participation_Rearrangements_and_Fragmentations/3.04%3A_Fragmentations.txt |
Objectives
After completing this section, you should be able to:
1. Understand radical reactions involved in functional group conversions and carbon-carbon bond formation
2. Understand radical chain reactions and radical combination reactions
3. Draw mechanisms incorporating radicals to explain reaction outcomes
4. Plan syntheses using radical reactions
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Radical initiation
• AIBN
• Radical propagation
• Bu3SnH
• Radical termination
• Barton-McCombie deoxygenation
• Barton decarboxylation
• Hofmann-Loeffler-Freytag reaction
• Pinacol reaction
• McMurry reaction
• Acyloin reaction
Study Notes
Radical reactions are nearly always covered in Intro Orgo. Students generally learn about radical chain reactions for the bromination of alkanes (with Br2) and alkenes (with HBr and peroxides). They also often encounter allylic and benzylic bromination with NBS (a very interesting mechanism involving polar steps as part of a radical chain reaction). In this chapter, we will highlight several other useful radical chain reactions for functional group interconversions and, most importantly, for carbon-carbon bond formation. These reactions will proceed by the standard outline you have seen before: initiation with a radical initiator and propagation with a radical propagator. We will also meet synthetically useful radical reactions that proceed by a process called radical combination. These will look like termination steps from a radical chain reaction but they will be the productive pathway yielding our desired product. Radical reactions are currently a very popular research area, with many top synthetic organic labs working to develop useful transformations that are impossible using polar reactions. The information you learn in this chapter will help you understand these recent developments.
Don't forget that all mechanisms in this chapter involve the movement of single electrons (radicals!), so we will use single-headed arrows in our mechanisms.
Content
Our goal in this chapter is to introduce fundamental radical reactions that you will likely encounter in organic synthesis papers. This will build on what you have already learned in Intro Orgo about bromination reactions with bromine as the chain propagating radical by demonstrating the power of radical reactions using tributyltin radicals. These transformations will enable dehalogenation, decarboxylation, and deoxygenation reactions. Most importantly, they will demonstrate the utility of radical reactions for the synthesis of carbon-carbon bonds, especially 5- and 6-membered rings. We will also see the power of radical combination reactions for remote functionalization that can lead to the synthesis of heterocyclic ring systems.
Initiation and Propagation Steps
Before getting into the actual radical reactions, we first need to comment on two key reagents: AIBN (AzobisIsoButyroNitrile) and Bu3SnH (tributyltin hydride). We will use AIBN as our radical initiator and Bu3SnH as the precursor for our chain propagating radical, tributyltin radical. Upon heating, AIBN cleaves to form nitrogen gas and two radicals that react with Bu3SnH to yield two tributyltin radicals, as shown below. Tributyltin is our chain propagating radical, so these two radicals can each start a radical chain process (the propagation steps) resulting in the production of our desired molecule. Please remember a few key points about initiation and propagation steps: 1) We only need a small amount of the initiator to begin the chain process. Generally, 1 mol% of AIBN (0.01 equivalents) is sufficient to generate enough tributyltin radicals for the propagation steps. Although not a true catalyst, you can think about AIBN like the catalyst that must be present to make the chain reaction possible. 2) The chain propagating radical, tributyltin radical, must be a reactant in your first propagation step and a product in your last propagation step. Tributyltin radical looks a bit like a catalyst (think Pd(0) in a Stille or Suzuki reaction) but it is different in a key way. Tributyltin hydride is consumed in the reaction and must be present in a stoichiometric amount (at least 1.0 equivalents) or the chain will be broken before all of the starting material reacts. Don't forget, the propagation steps are the only productive steps in a chain reaction. This is where your product is formed. 3) Termination steps break the chain and are undesired. In radical chain reactions, combining two radicals to form a new bond is never a productive reaction. Your product will never come from a termination step. These molecules are undesired byproducts and their formation must be very limited for the radical chain reaction to be synthetically useful.
Dehalogenation Reactions
Tributyltin hydride is a useful reagent for the dehalogenation (reduction) of alkyl halides. This illustrates a radical chain mechanism using tributyltin as the chain propagating radical. (As always, AIBN and tributyltin hydride combine in the initiation steps shown above to produce tributyltin radical.) Once the tributyltin radical is formed, it reacts to form a new Sn-Br bond by cleaving the C-Br bond to generate a carbon radical. This forms tributyltin bromide, a byproduct that must be removed after the reaction, and the new carbon radical that is part of our chain process. The carbon radical reacts with tributyltin hydride (remember, we have a very small amount of AIBN, just enough to start the process, and at least 1 equivalent of Bu3SnH to react completely with the alkyl bromide starting material) to form the new C-H bond in the product and another molecule of tributyltin radical that can participate in another cycle of the chain process.
Barton-McCombie Deoxygenation
Similar to the dehalogenation reaction above, radical reactions can be used to remove an alcohol from a molecule, thus deoxygenating (reducing) the compound. We need to use some unusual chemistry to make the deoxygenation possible, specifically we need to form a xanthate, a functional group similar to a carbonate that contains two sulfur atoms. One of the sulfurs is part of a thio carbonyl (C=S) which is very reactive toward radical reactions and readily reacts with tributyltin radical to form a Sn-S bond and a carbon radical. To form the xanthate, we react the alcohol with potassium hydride to generate a negatively charged oxygen nucleophile. This reacts with carbon disulfide (the sulfur equivalent of carbon dioxide) and then methyl iodide to generate the xanthate intermediate. Once the xanthate is formed, it can react with tributyltin radical to participate in our radical chain reaction. The first step generates a new S-Sn bond and a carbon radical that decomposes to yield a dithiocarbonate ester (a byproduct that must be removed after the reaction) and a new carbon radical. At this point, the molecule has been successfully deoxygenated and the final step with tributyltin hydride completes the reaction by forming a C-H bond and regenerating the chain propagating tin radical.
Barton Decarboxylation
You likely learned about polar decarboxylation reactions in Intro Orgo. These reactions enable the removal of a carboxylic acid in a 1,3-carbonyl acid functional group. The second carbonyl positioned beta to the carboxylic acid carbonyl is critical in the mechanism (either acidic or basic conditions). These polar reactions do not work when the other carbonyl is absent. Thus, a decorboxylation reaction that doesn't require the presence of a second carbonyl at a specific position is highly valuable for synthetic chemists. This example highlights the importance of the Barton decarboxylation, a radical decarboxylation that is similar to the deoxygenation reaction shown above. It relies on a different reagent than the deoxygenation, so a xanthate ester isn't formed, but the steps are very similar, including reaction of tributyltin radical with a thiocarbonyl. In the problem below, you can propose a mechanism for this interesting reaction.
Exercise \(1\)
The reaction scheme below illustrates the multistep Barton decarboxylation reaction. Predict the product of the first two steps. Then, using that product, provide a mechanism for the radical decarboxylation step.
Answer
The first two steps are Intro Orgo reactions. Thionyl chloride converts the carboxylic acid into an acid chloride that reacts withe the hydroxyl amine (fancy alcohol) to generate the hydroxyl amine ester product. As we saw previously, the C=S bond reacts readily with a tributyltin radical. The resulting carbon radical can generate a very stable pyridine byproduct upon cleavage of the weak N-O bond. This generates an oxygen radical that can yield carbon dioxide, thus driving this step, via C-C bond cleavage. This is the key decarboxylation step. The resulting carbon radical combines with tributyltin hydride to yield the product and regenerate the chain propagating radical.
Carbon-Carbon Bond Forming Reactions
Radical chain reactions mediated by Bu3SnH can also promote the formation of carbon-carbon bonds. These are among the most important radical reactions in synthetic chemistry since they provide us with another option in our toolbox of methods to make critical molecular connections. Key components in these reactions are a carbon-halogen bond that can react with the chain propagating tributyltin radical to generate a carbon radical that combines with an alkene or alkyne to form the new carbon-carbon bond. These reactions occur both inter- and intramolecularly with the latter being favored for higher yields and better regioselectivity. Radical reactions are kinetically controlled so they provide a very useful strategy for the synthesis of 5-membered rings. This is even true for most instances when selecting between the formation of 5- versus 6-membered rings in intramolecular cyclization reactions. (The smaller ring forms faster and, thus, is favored.) In intermolecular reactions, it is often helpful to have an electron poor alkene or alkyne as the radical partner to help favor regioselectivity with the (generally) more substituted radical formed in the first step. An example reaction and mechanism are shown below for a standard intermolecular carbon-carbon bond forming reaction. An intramolecular example is shown in the following problem.
Mechanism
Exercise \(2\)
For the following intramolecular C-C bond forming reaction, predict the product and propose a mechanism for its formation.
Answer
Like in our intermolecular example above, the tributyltin radical attacks the carbon-halogen bond to yield a carbon radical. This new radical participates in an intramolecular C-C bond forming reaction to form a 5-membered ring radical that reacts in the final step to form the product and regenerate the chain propagating radical. Note, because of ring strain, we won't form the 4-membered ring resulting from addition to the other side of the alkene.
Exercise \(3\)
Predict the product of the following reaction.
Answer
This problem highlights that radicals can easily form on sp2 carbons and is a reminder that bond rotation is always important to consider. The key intermediate forms and reacts to yield the new 6-membered ring which forms faster than the 7-membered ring alternative.
Non-Chain Radical Combination Reactions
Some highly useful radical reactions involve the combination of two radicals to form an important new bond. (These look like a termination step in a radical chain mechanism.) These reactions have become increasingly relevant as new and more mild methods for Hydrogen Atom Transfer (HAT) reactions have recently been developed. We will explore some classic examples in this class of radical reactions which you can then apply to understand more contemporary transformations. The Hofmann-Loeffler-Freytag reaction enables remote functionalization of a haloamine that ultimately results in the formation of a new pyrrolidine ring (saturated 5-membered ring containing N). The reaction and mechanism are shown below. The first step in the reaction is the radical portion of the mechanism. (The chloroamine can be formed by treating the amine with t-butyl hypochlorite (tBuOCl).) Light promotes homolytic cleavage of the weak N-Cl bond to yield a Cl radical and an N radical. The next steps help explain the highly selective nature of this reaction. Bond rotation enables a HAT via a highly favored six-membered ring transition state that also transforms the reactive nitrogen radical into a more stable secondary carbon radical. The product of the first portion of the reaction forms via a radical combination reaction of the initially generated Cl radical with the newly formed C radical. This 1,4-chloroamine can undergo an intramolecular SN2 reaction to yield the pyrrolidine product. A related example of a Hoffmann-Loeffler-Freytag reaction to yield a lactam is shown in the next problem.
Mechanism
Exercise \(4\)
Propose a mechanism for this Hofmann-Loeffler-Freytag reaction.
Answer
Iodine promotes formation of the requisite haloamine via a substitution reaction. Light initiates the radical portion of the mechanism which proceeds via the key six-membered ring transition state as shown in the prior example. After the radical combination reaction to yield the iodo amide, hydroxide deprotonates the nitrogen making the amide nucleophilic. Attack by the more reactive O yields the iminium ion-type intermediate that undergoes hydrolysis to ultimately yield the target lactone.
A related reaction promoted by lead tetraacetate (Pb(OAc)4) that proceeds via a key oxygen radical provides a synthetically useful method for the synthesis of tetrahydrofurans. As shown below, treatment of an alcohol with Pb(OAc)4 yields a substituted tetrahydrofuran (THF). The mechanism does not have a radical combination step, instead relying on a Single Electron Transfer (SET) process between a carbon radical and lead to yield a carbocation. The mechanism begins with a substitution reaction on lead with the starting alcohol. The new O-Pb bond is weak and can homolytically cleave to yield Pb(III) and an oxygen radical. Like in the Hofmann-Loefflear-Freytag reaction, the oxygen radical does a Hydrogen Atom Transfer (HAT) step via a 6-membered ring transition state to yield a carbon radical that reacts with a formally positive Pb(II) via SET. Thus, the lead is reduced and the carbon is oxidized to a carbocation that readily reacts with the intramolecular alcohol to yield the desired THF.
Mechanism
Exercise \(5\)
Propose a product for the following reaction.
Answer
Like in the mechanism above, the lead reagent will promote formation of an oxygen radical from the alcohol. This will participate in a HAT reaction with the methyl group, followed by a SET reaction, and finally cyclization to form the new 5-membered ring.
Other useful radical combination reactions involve dimerization of carbonyl starting materials. The three most popular are the pinacol reaction, the McMurry reaction, and the acyloin reaction. These involve starting with ketones (pinacol and McMurry) or esters (acyloin) and adding a strong metal reducing agent like sodium, magnesium, or titanium to generate ketyl radicals that dimerize and ultimately produce diol (pinacol), alkene (McMurry), or keto alcohol (acyloin) products. Intermolecular examples of each are shown below. Intramolecular reactions are also possible and will occasionally show up in total synthesis papers.
Pinacol Reaction
Two equivalents of acetone react with magnesium metal via Single Electron Transfer (SET) to yield two radical anions (ketyl radical anions). These two anions react with cationic magnesium (+2) to yield the neutral diradical that undergoes a radical combination reaction to form a new carbon-carbon bond. The resulting 5-membered ring breaks down in the acidic workup to yield the diol product. In this case, the product is pinacol, the starting point for the pinacol rearrangement that we saw in Chapter 3.
McMurry Reaction
The McMurry reaction is nearly identical to the pinacol reaction except for the final step. The key difference is the use of titanium metal, generated in situ from titanium trichloride and lithium aluminum hydride. Titanium reacts with two ketones to yield two ketal radical anions that react with cationic titanium to yield a neutral diradical. The diradical participates in the radical combination reaction to yield a new 5-membered ring include a new C-C bond. At this point, the McMurry reaction diverges from the pinacol reaction. The titanocycle is not stable, instead it undergoes a deoxygenation reaction to yield an alkene product. Overall, the McMurry reaction acts like a reverse ozonolysis reaction by combining two ketones to yield an alkene.
Acyloin Reaction
The acyloin reaction begins just like the pinacol reaction with SET reactions to yield two radical anions. Dimerization (radical combination) yields the new carbon-carbon bond and an intermediate that looks like a double tetrahedral intermediate from the carbonyl addition section of Intro Orgo. Accordingly, ethoxide is a good leaving group which promotes formation of a 1,2-diketone. This compound is highly reactive toward reduction, so it will accept two more electrons from sodium to yield a diradical dianion. Radical combination generates a new pi bond. This diradical is stable and is quenched in the workup to produce a diol. Since this is an enol, it will tautomerize to yield the keto alcohol product.
Summary Problems
Exercise \(6\)
One of the most powerful examples of the utility of carbon-carbon bond forming radical reactions in synthesis is Curran's synthesis of hirsutene published in 1985. The reaction below is the final step in the synthesis. Propose the structure of hirsutene and a mechanism for its formation.
Answer
This follows the pattern we observed previously. The tin radical reacts with the iodide to form a carbon radical that reacts with the alkene to form a new 5-membered ring. The resulting tertiary radical reacts with the alkyne to form the third 5-membered ring in hirsutene. Note that the stereochemistry in the starting material determines the stereochemistry in the product. The alkyl chain coming out forms a new bond from above the initial 5-membered ring. The alkyl chain going back forms a new bond from beneath the initial 5-membered ring.
Exercise \(7\)
Though much less common, it is possible to use tin reagents other than tributyltin hydride to promote carbon-carbon bond forming reactions. One example is allyl tributyltin. An example of this reaction is shown below. Propose a mechanism for the initiation and propagation steps. Hint: Like in all of our other reactions with tin, tributyltin radical is the chain propagating radical.
Answer
Without a Sn-H bond, radicals that react with the allyl tin reagent must add to the alkene, like we have seen in other radical reactions in this section. So, for the initiation steps, the initiating radical adds to the alkene then the resulting carbon radical decomposes to yield an allyl group on the initiating group plus the chain propagating tin radical. In the propagation steps, the first step is our standard Sn + Br to yield the first carbon radical. The only option for the new carbon radical is to add to the alkene of the tin allyl group. This generates a secondary carbon radical than decomposes just like in the initiation steps, leaving the allyl group on the product and regenerating the chain propagating tin radical. Note: alkenes are much more reactive that C-C or C-Sn single bonds.
Exercise \(8\)
Provide a mechanism for the reaction shown below. Hints: 1) Don't forget about the ground state structure/behavior of oxygen. 2) t-BuSH reacts similarly to Bu3SnH.
Answer
The mechanism begins with standard initiation steps. AIBN generates the initiating carbon radical that reacts with t-butyl thiol to yield the chain propagating S radical. Since this behaves like the tributyltin radical, it will add to the weak C=S making a new S-S bond and carbon radical. This new radical quickly reacts to generate a stable pyridine, gaseous carbon dioxide, and a new secondary carbon radical. Remembering that oxygen exists as a diradical, we see that the next step is a radical combination between the carbon radical and O2. In the final step, the remaining O radical reacts with the beginning thiol to generate the product and regenerate the chain propagating radical.
Literature Reference: Barton Tetrahedron 1985
Exercise \(9\)
Provide a mechanism for the following transformation. Remember, SmI2 is an excellent single electron donor. Hint: At some point in the mechanism you will need to form an alpha-N radical (a radical on a carbon next to the N).
Answer
Samarium diiodide starts the reaction by donating an electron (Single Electron Transfer (SET)), to the carbonyl. This yields a radical anion that undergoes a radical cyclization reaction to generate the new 5-membered ring and a tertiary radical. In the next step, this radical adds to the adjacent pi bond to form a transient 3-membered ring and a secondary radical. Regeneration of the alkene enables radical cleavage of the unstable cyclopropane to yield a stabilized alpha-N radical. Another SET mediated by samarium converts the radical into a carbanion. Both anions are quenched in the acidic workup to yield the target molecule.
Literature Reference: Wood Chemical Science 2020
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/04%3A_Radical_Reactions.txt |
Objectives
After completing this section, you should be able to:
1. Understand the structure of singlet and triplet carbenes
2. Understand how to generate carbenes and carbenoids
3. Understand carbene reactions involved in cyclopropantion and C-H insertion reactions
4. Draw mechanisms incorporating carbenes to explain reaction outcomes
5. Plan syntheses using carbene reactions
Key Terms
Make certain that you can define, and use in context, the key terms below.
• Singlet carbene
• Triplet carbene
• Carbenoid
• Alpha Elimination
• Cyclopropanation
• Simmons-Smith reaction
• C-H Insertion
Study Notes
Carbenes are the final member in the four reactive intermediates of carbon, joining carbanions, carbocations, and radicals. It is unusual to meet carbenes in Intro Orgo, so it is often a shock to realize there is a brand new version of reactive carbon that wasn't encountered earlier. Carbenes are much less important that the other three intermediates mentioned above, and we will learn only a few fundamental reactions to make them and two reactions that carbenes participate in. They provide us with one critical synthetic transformation, cyclopropanation. This reaction is the most important transformation of carbenes, and it functions as the carbon equivalent of MCPBA epoxidation. The reactivity of carbenes can be tuned by coordination with a variety of metals to form metal carbenoids. We will mention this briefly and encourage you to learn more about this fascinating research topic if you are interested.
Content
Our goals in this chapter are three-fold: 1) Understand the structure of carbenes. 2) Understand how to make carbenes and carbenoids. 3) Understand carbene cyclopropanation and carbene C-H insertion reactions. These topics cover the fundamentals of carbenes and carbene reactions and will prepare you to use and understand carbenes in the context of organic synthesis.
Carbene Structure
A carbene is a neutral form of carbon that has two bonds and two additional electrons. It is highly reactive because it has an incomplete octet with only six electrons around carbon. Carbenes can exist with their nonbonding electrons as either a lone pair or two radicals. The diradical form is known as a triplet carbene, while the structure with a lone pair is called a singlet carbene. Carbene structure is not always obvious and can depend on the method used to make the carbene. However, since we are often more comfortable thinking about carbocations and carbanions than radicals, we will simplify our analysis of carbenes and assume that they are singlet carbenes. Again, this is an oversimplification, and you are encouraged to learn more about carbenes and their structure if you are interested. Getting back to singlet carbenes, since they contain both a lone pair in a hybrid orbital (sp2) and an empty p orbital, they behave as if they are simultaneously a carbanion and a carbocation. This strange combination leads to unique reactivity. To help us recognize this type of reactivity, we will draw carbenes with both a positive and negative formal charge, thus indicating that they are neutral but highly reactive.
One relatively stable and popular class of carbenes are N-heterocyclic carbenes. These molecules are very useful as organometallic ligands, and they are seen frequently in organic synthesis.
Carbene Synthesis
The two most popular methods for the synthesis of carbenes are from alpha-elimination of a halo compound and from decomposition of a diazo compound. A common example of the alpha-elimination reaction is the deprotonation reaction of chloroform with hydroxide to yield dichlorocarbene.
Diazo decomposition is the preferred method for carbene and carbenoid generation and occurs upon exposure to heat, light, or a metal promoter (like Rh or Cu). A standard carbene can be generated from any diazo compound with heat or light with loss of nitrogen gas. Treating a diazo compound with a metal converts the diazo into a metal carbenoid. In the example below, rhodium acetate enables formation of a rhodium carbenoid. (Note: the "Ln" in the structure means some number of acetate ligands are attached to rhodium.) Rhodium and copper carbenoids have very similar reactivity to carbenes; however, the metal carbenoids are often more selective and their reactivity can be tuned based on the ligands present on the metal.
Before continuing on to carbene reactions, it is important to comment on how diazo compounds are generated. The most popular methods start with an acid chloride, an aldehyde, or a 1,3-dicarbonyl. Treating an acid chloride with two equivalents of diazomethane yields an alpha-diazo ketone. The first equivalent of diazomethane participates in a carbonyl substitution reaction. The second equivalent is the base that deprotonates the diazonium intermediate to yield the diazo product.
Treating an aldehyde with tosylhydrazine and base yields a diazo compound. As expected, tosylhydrazine reacts with the aldehyde to yield a tosylhydrazone. Methoxide removes the acidic N-H proton. The resulting anion undergoes loss of tosyl to yield the neutral diazo product.
1,3-Dicarbonyl compounds readily undergo diazo transfer reactions where an azide donates two nitrogens to form the diazo product. An example using tosyl azide is shown below. A variety of azides can participate in these reactions.
Carbene Reactions
Cyclopropanation Reactions
The most important carbene reactions are ones that enable formation of cyclopropanes. Depending on the structure of the target cyclopropane, different carbene starting materials are employed. If a dihalocyclopropane is the target, then a standard alpha-elimination reaction is used. As shown below, treating the starting alkene with chloroform and a base readily yields a dichlorocyclopropane product. The reaction mechanism is a stereospecific, concerted process.
When the goal is synthesis of a cyclopropane bearing a new CH2 group, the best option is the Simmons-Smith reaction. As shown below, combination of an alkene with diiodomethane and a Zn/Cu metallic couple yields the target cyclopropane. In the mechanism of this reaction, a free carbene is not formed. Instead, zinc first does an oxidative addition with one of the C-I bonds. The resulting organozinc intermediate behaves like a carbene, participating in a concerted reaction with the alkene to yield the cyclopropane and zinc(II) iodide.
Generating more highly substituted cyclopropanes involves the use of diazo compounds in both inter- and intramolecular reactions. Synthesis applications often take advantage of the increase in structural complexity available with intramolecular cyclopropanation reactions. An intermolecular example is shown below while an intramolecular reaction is featured in the following problem. The example problem below highlights a synthesis that proceeds via a copper carbenoid.
Exercise \(1\)
Propose the product of the following reaction.
Answer
Copper reacts with the diazo compound to form a copper carbenoid. This reacts like a carbene to enable an intramolecular cyclopropanation reaction. In addition to the 3-membered ring, the reaction is selective for the production of the favored 6-membered ring as part of the bicyclic product.
C-H Insertion Reactions
C-H insertion reactions provide another reaction pathway for carbenes. (As we will see shortly, C-H insertion reactions adjacent to the carbene carbon to yield an alkene are a major limitation in carbene applications.) Intramolecular C-H insertion reactions provide another method for the synthesis of 5-membered rings. As we have seen previously, this results from a favorable 6-membered ring transition state. In the example below, a rhodium carbenoid forms first. There is no alkene for a potential cyclopropanation reaction, so a C-H insertion reaction occurs. The most favorable 6-membered ring transition state yields the 5-membered ring product.
Limitations of Carbene Reactions
When planning carbene reactions, there are several limitations to be aware of. First, don't forget that alpha-diazo ketones also participate in Wolff rearrangements (as we saw in Chapter 3 and mentioned in Exercise 3.3.8). The mechanism introduced in Chapter 3 did not involve a carbene. However, it is possible to draw an alternate mechanism that proceeds via a carbene. Compare the following mechanism to the one provided in the answer to Exercise 3.3.8. Both proceed to a ketene, they just get there in different ways. The preference for the Wolff rearrangement is to draw it as a carbene mechanism.
Second, carbenes with an adjacent C-H bond often undergo C-H insertions to yield an alkene before participating in cyclopropanation or C-H insertion to form a ring. This can be mitigated by some metal carbenoids, but it is always important to be wary of this undesired side reaction.
Third, nucleophilic atoms, like oxygens in ethers, can add directly to the carbene and produce rearrangement products. This is called the Stevens rearrangement and an example is provided below. The rhodium carbenoid reacts by the expected C-H insertion reaction to yield a new 6-membered ring (the ether O blocks formation of a 5-membered ring since there are no C-H bonds at this position), and it also reacts via a Stevens rearrangement to yield a 5-membered ring cyclic ether. In this mechanism, the ether O adds to the empty p orbital of the carbene then the resulting carbanion participates in a substitution reaction with the oxygen as the neutral leaving group.
Summary Problems
Exercise \(2\)
The Arndt-Eistert reaction is useful method for the homologation (extension by one carbon) of carboxylic acids. Propose a mechanism for the second and third steps in this reaction.
Answer
Thionyl chloride converts the carboxylic acid into an acid chloride. As outlined previously in the chapter, conversion of an acid chloride into a diazo compound happens readily with 2 equivalents of diazomethane. Heat or light will generate the carbene that undergoes a Wolff rearrangement to yield a ketene. Hydration of the ketene yields a carboxylic acid having one more carbon than the starting material.
Exercise \(3\)
We mentioned the molecule bullvalene in the section on sigmatropic rearrangements in Chapter 1.4. In this problem, we will see several steps in the Doering group's synthesis of bullvalene. A key intermediate in the bullvalene synthesis is barbaralone (named for Barbara Ferrier, the first person to make it). Starting with benzene, how would you make barbaralone? Hint: One reaction in your synthesis should be the Buchner reaction.
Answer
This synthesis utilizes two carbene steps. In the Buchner reaction, a carbene reacts with benzene to form a cyclopropane that quickly opens via a [3,3] sigmatropic rearrangement to yield a cycloheptatriene. This ethyl ester is hydrolyzed to the carboxylic acid and then the acid chloride. Treatment with 2 equivalents of diazomethane yields an alpha diazo ketone that reacts with copper to furnish a carbenoid that undergoes another cyclopropanation reaction to yield barbaralone.
Exercise \(4\)
Propose a mechanism for the following transformation.
Answer
Carbenoid formation leads to an intramolecular cyclopropanation that also generates a new 5-membered ring. The cyclopropane fragments via a Cope rearrangement to form the 7-membered ring in the product. The final two steps are a base catalyzed isomerization to yield the desired target.
Literature reference - Stoltz Chemical Science 2017
Contributors
• Prof. Kevin Shea (Smith College) | textbooks/chem/Organic_Chemistry/Organic_Synthesis_(Shea)/05%3A_Carbene_Reactions.txt |
• 1.1: Difunctional Carboxyloids
The first synthetic polymer to really seize the public's consciousness -- the first one to change the course of world events -- was nylon. Not long after its discovery in the 1930s, it had supplanted silk as the material of choice for stockings, having transformed a luxury item into a widely available commodity. Nylon came just in time for the Second World War. Soon, parachutes, too, went from being made of silk to being fabricated from nylon.
• 1.2: Cyclic Carboxyloids
Nylon 6 is not an alternating co-polymer like nylon 66. It's just a polymer. And it isn't made from a difunctional monomer like nylon 66 or proteins. Instead, it's made from a cyclic amide, sometimes called a lactam. To polymerize, the lactam has to break open into a linear form, and the lactam monomers end up enchained head-to-tail. This process is called ring-opening polymerization.
• 1.3: Olefins
Polyolefins are made from "olefins", which you may know as alkenes. Olefin is an older term for an alkene that is still commonly used in the industry. These compounds make up a significant fraction of commercially-used polymers today. If you think of the common recyclable plastics, polyethylene (#2 and #4, depending on how the material is made), poly(vinyl chloride) (#3), polypropylene (#5) and polystyrene (#6) are all examples of polyolefins.
• 1.4: Cyclic Olefins
Polyolefins are usually made from olefins by tethering one alkene unit to the next, trading in a pi bond within a monomer for a single bond between two repeat units. However, there is another approach that converts cyclic alkenes into polymers. This approach is reminiscent of the ring-opening of cyclic esters and amides.
• 1.5: Coordination Polymers
Polymers are long-chain molecules formed from individual molecular building blocks. Typically, the building blocks are organic molecules held together via covalent bonds. What other kinds of building blocks are available? The formation of coordination compounds is one of many important aspects of inorganic chemistry. In a coordination compound, an electron pair donor, called a ligand, shares its electron pair with a metal atom.
• 1.6: Supramolecular Assemblies
So far, we have seen how covalent bonds can be used to bind monomers together into longer chains, forming polymers. There has also been a great deal of interest recently in using intermolecular attractions to make similar structures. Of course, intermolecular attractions are very important in forming large, organized structures in biology. Think about the twin helices held together in DNA, or the secondary and higher-order structures in proteins.
• 1.7: Other Polymers
The principle of functionality means that almost any type of organic reaction could potentially be used to make polymers. For instance, if a compound has two functional groups of the same kind, it could undergo reaction at two different sites, forming new bonds with two neighbors. The compound thereby becomes enchained in a trio of formerly independent molecules. If the neighboring molecules are also difunctional, then this pattern can repeat, forming a polymer.
• 1.8: Polymer Topology
Polymers are very large molecules made from smaller ones. How those smaller units are arranged within the polymer is an issue we haven't addressed very closely yet. Topology is the study of three-dimensional shapes and relationships, or of how individual parts are arranged within a whole.
• 1.E: Solutions for Selected Problems
Thumbnail: Space-filling model of a section of the polyethylene terephthalate polymer, also known as PET and PETE, a polyester used in most plastic bottles. Color code: Carbon, C (black), Hydrogen, H (white), and Oxygen, O (red). (Public Domain; Jynto).
01: Monomers and Polymers
The first synthetic polymer to really seize the public's consciousness -- the first one to change the course of world events -- was nylon. Not long after its discovery in the 1930s, it had supplanted silk as the material of choice for stockings, having transformed a luxury item into a widely available commodity. Nylon came just in time for the Second World War. Soon, parachutes, too, went from being made of silk to being fabricated from nylon.
The development of nylon took place at Du Pont's Experimental Research Station in Delaware. For much of the twentieth century, Du Pont and other large companies relied on the pursuit of basic science as an engine of future development. Knowledge was the infrastructure that made future products possible, and so scientists were asked to explore the unknown; the applications of this work would follow naturally.
When Wallace Carrothers of Du Pont developed nylon, he had been trying to explore some ideas proposed by Hermann Staudinger at ETH Zurich. These ideas, initially controversial, held that many materials around us, including things like cotton and spider silk, are made of small molecular units covalently bonded to each other. These little molecules, monomers, bind together to form immensely long chains. In so doing, the monomers lose their individual identities and simply become repeating units in a long chain. They come together to form polymers.
Nylon, in particular, is an example of a polyamide. It is composed of very long chains that contain regularly repeating amide bonds, N-C=O. That's the same motif found in proteins in biology, and it's the same pattern found in spider's silk, which is composed of protein.
Conceptually, we are taught in biology class that proteins are made of amino acids. An amide bond (or peptide bond) occurs when the nitrogen of an amine is brought together with the carbon of a carboxylic acid. A molecule of water is lost, and the nitrogen takes the place of the oxygen on the carboxylic acid, forming an amide, instead. The loss of water leads to the term "condensation reaction" because early studies of these reactions led to water forming droplets on laboratory glassware as it bubbled out of the reaction. In practice, industrial nylon production is really no more complicated than that.
Of course, not just any old amide can become a protein. What makes amino acids special is the fact that they are difunctional. They contain not just one functional group (an amine, say, or a carboxylic acid), but two. So, when the amine group of an amino acid bonds with the carbonyl of a neighbour, the carboxyl group can bond to the amine of a different neighbour. It's like it has two hands; it can hold onto a friend with each hand, and each of those friends can hold onto another, and so on, forming a chain.
This functionality is a key part of how polymers can form. Because difunctional molecules can form bonds in two directions, a simple coupling reaction (an amine plus a carboxylic acid making an amide, two small molecules making another small molecule) becomes a polymerization (many small molecules forming an enormous molecule).
Well, nylon isn't made from an amino acid, although the coupling reaction is similar. Instead, it is made from two different molecules, both of which are difunctional: a diamine and a dicarboxylic acid. The diamine bonds to a neighbouring carbonyl through each end and the dicarboxylic acid bonds to a neighbouring amine through each end. For the most common material, each of the two reactants is six carbons long, giving rise to the term "nylon 66".
Nylon 66 is an example of something called an alternating co-polymer. It isn't just one unit repeating over and over along a chain, but two. The two monomers make the polymer together. We sometimes say that the monomers are enchained when they become part of a polymer. And of course, they have to alternate along the chain, so that one can bind to the complementary other.
In the drawings above, the dashed lines are meant to suggest continuity: the pattern shown in the drawing keeps repeating to the right and to the left. More commonly, polymers are drawn using parenthetical notation. Below, the part shown in the parentheses is what keeps repeating. If we could make a stamp of that picture, we could construct the polymer chain by simply stamping that image over and over across a sheet of paper. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.01%3A_Difunctional_Carboxyloids.txt |
Carrothers' development of nylon at Du Pont led indirectly to a very similar polymer that was made in a very different way. Across the Atlantic Ocean, Paul Schlack at IG Farben in Germany was looking for a way to make a similar material that would not be subject to Du Pont's patent. His efforts led to a material called perlon, sometimes referred to as nylon 6.
Nylon 6 is not an alternating co-polymer like nylon 66. It's just a polymer. And it isn't made from a difunctional monomer like nylon 66 or proteins. Instead, it's made from a cyclic amide, sometimes called a lactam. To polymerize, the lactam has to break open into a linear form, and the lactam monomers end up enchained head-to-tail. This process is called ring-opening polymerization.
There are a number of things that are different about these two ways of making nylon. One of them turns on the whole concept of synthesis. Synthesis is the act of making things. It's a little like cooking. You gather the ingredients. You take the right amount of this and the right amount of that. You mix them together and you get something wonderful. But that's the difference here: in making nylon 66, Carrothers mixed two different compounds together. He poured one compound out of one bottle and another compound out of another bottle. He mixed them together and he got a polymer.
In contrast, Schlack didn't mix anything together, at least as far as we have seen so far. It all came out of one bottle. So why wasn't it already nylon 6 in the bottle? If the monomers just react with themselves, couldn't they have just gone ahead and done that in the bottle?
Ring-opening polymerization, at least in this context, is an example of a chain reaction. Chain reactions don't just happen by themselves; they need a jump start.
To understand why, you have to appreciate that these two materials come about through two very different classes of polymerization reactions. Nylon 66 is the product of a condensation reaction. An amine is mixed with a carboxylic acid, water is released, and an amide bond is formed. Even if the pair of reactants is chosen to be more synthetically efficient -- say, an amine and an acid chloride instead of a carboxylic acid -- a condensation reaction still results, in this case releasing hydrogen chloride.
Nylon 6, on the other hand, is not the product of a condensation reaction -- at least at first glance. Look carefully at the monomer and the polymer. Count the atoms. There's nothing missing. No molecule of water or hydrogen chloride or anything else was released. The nitrogen at one end of the chain simply attaches to the carbonyl of the next.
But if you walk down to the end of that chain, until you reach the very last carbonyl, what do you find there? There's no nitrogen attached. That nitrogen was the nucleophile that bound to the carbonyl in the next monomer. There has to be something, though, because the carbon cannot be sitting there with only three bonds. So there must have been some nucleophile that added to that carbonyl, springing loose the amine that added to the next carbonyl, springing the next amine, and so on. There had to be an original nucleophile.
This is the jump start that the chain reaction needed. A nucleophile had to be added to get things going. Once it reacted with the first monomer, the amine became the nucleophile for the next monomer, and that ring-opening produced another amine nucleophile, and so on. Because there is always a new nucleophile produced when the next ring opens, the reaction just keeps going. That first nucleophile, the one that got everything started, is called an initiator. The need for an initiator is a hallmark of chain reactions.
In addition to the initiator, ring-opening polymerizations frequently employ catalysts to accelerate the reaction, just as in some condensation polymerizations. The catalyst may be a Lewis acid that activates the carbonyl or an "organocatalyst" that does the same thing via hydrogen bonding. Both catalytic approaches can also make use of nucleophilic catalysis. In that case, the nucleophilicity of the nucleophile may be enhanced, or else a temporary nucleophile may add to the carbonyl until it is replaced by the nucleophilic group resulting in enchainment.
Other polyamides can also be made by ring-opening cyclic amides. The same approach is also used to prepare polyesters from cyclic esters, also called lactones. In that case, the reaction is sometimes called ring-opening trans-esterification polymerization, or ROTEP for short. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.02%3A_Cyclic_Carboxyloids.txt |
Polyolefins are made from "olefins", which you may know as alkenes. Olefin is an older term for an alkene that is still commonly used in the industry.
These compounds make up a significant fraction of commercially-used polymers today. If you think of the common recyclable plastics, polyethylene (#2 and #4, depending on how the material is made), poly(vinyl chloride) (#3), polypropylene (#5) and polystyrene (#6) are all examples of polyolefins.
Polyolefins have been known for some time, although it took about a century from the time that they were first documented until they were recognized as polymers. German chemists in the mid-1800s were aware that certain tree resins gave rise to hard materials over time, and these materials included polystyrene.
If you compare the repeating structure of polystyrene to the structure of styrene, you can imagine that polystyrene is made from a series of styrene molecules that have been strung together. Essentially, the double bond of styrene has moved to attach to the next molecule, and the double bond there has done the same thing, and so on.
Polyolefins are formed in the same way from a wide variety of alkenes, leading to an array of different materials with properties that are suited to unique applications.
Note that, like ring-opening polymerization, olefin polymerization depends on molecules reacting with other molecules that are just like them. That isn't likely to happen; they will need something to come along and induce reaction between them. Olefin polymerization is thus another case in which monomers are tied together through a chain reaction. The reaction will require an initiator to get things started. That initiator is likely to become an end group, hanging from one end of the polymer chain or the other.
1.04: Cyclic Olefins
Polyolefins are usually made from olefins by tethering one alkene unit to the next, trading in a pi bond within a monomer for a single bond between two repeat units. However, there is another approach that converts cyclic alkenes into polymers. This approach is reminiscent of the ring-opening of cyclic esters and amides.
For example, imagine a cyclopentene ring opening up at the double bond and reaching out to join with other rings on either side of it.
A series of cyclopentene rings that joined together in a row would look something like this:
We would probably draw it in the usual zig-zag conformation.
This act of taking a cyclic alkene, splitting open its double bond, and knitting it together with other such alkenes in a long chain is called ring-opening metathesis polymerization. Sometimes it is called ROMP for short. It has some things in common with other ring-opening polymerizations, such as ROTEP. In both ROTEP and ROMP, like molecules react together to form a polymer. In ROTEP, it would be two cyclic esters. In ROMP, it would be two cyclic alkenes. That was different from condensation polymerization, which required two complementary molecules, such as a difunctional amine and a difunctional acid chloride.
Partly as a consequence of like molecules reacting together, ROTEP and ROMP are both chain reactions. In order to get the molecules to react with themselves, they need an initiator. The initiator jump-starts the reaction.
Other olefin polymerizations followed this pattern as well. Alkenes don't normally react together (there are some circumstances when they will, but we needn't get into that now, as those events don't usually have anything to do with polymerization). Olefin polymerizations, in general, go through chain reactions that require an initiator to get started.
Despite these similarities, ROTEP and ROMP reactions are actually quite different in how they occur, as are regular olefin polymerizations, and the conditions required to initiate polymerization are unique to each case. ROMP requires something called an olefin metathesis catalyst. An olefin metathesis catalyst is a transition metal compound that is capable of splitting the double bond of an alkene in half and putting the two pieces together with other alkenes. The key part of an olefin metathesis catalyst is a metal-carbon double bond. That is the group that is capable of switching the ends of alkenes around with different partners. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.03%3A_Olefins.txt |
Polymers are long-chain molecules formed from individual molecular building blocks. Typically, the building blocks are organic molecules held together via covalent bonds. What other kinds of building blocks are available?
The formation of coordination compounds is one of many important aspects of inorganic chemistry. In a coordination compound, an electron pair donor, called a ligand, shares its electron pair with a metal atom. Frequently, the metal atom is a transition metal, and very commonly it is a transition metal ion, but there are other examples as well.
For example, the nitrogen in a pyridine ring has a lone pair. Pyridine can act as a ligand if its lone pair is shared with a metal center, such as the vanadium in trichloro(oxo)vanadium. The lone pair becomes a nitrogen-vanadium bond. Sometimes, this bond is drawn as a short arrow from the lone pair to the vanadium, emphasizing its origin, but more often it is simply drawn as a line, like any other bond.
One of the interesting things about metal atoms is their capacity to form variable numbers of bonds. Although palladium dichloride could form a coordination complex by binding with one pyridine ligand, it can also do so with two pyridines. In the former case, it would form a three-coordinate complex, but the latter case would lead to a four-coordinate palladium compound.
Remember, the ability to bond with two neighbouring groups, rather than just one, is an important feature that can allow a small molecule to become enchained in a polymer. The palladium here is a link in a chain that is three units long. What if the ligand also had this capacity to bind to two things? What if, instead of pyridine, the ligand were 4,4'-bipyridine?
That molecule could bind a palladium atom on either end. Once a bpy (that's an abbreviation for bipyridine, pronounced "bippy") bound to a palladium atom, it would still have a second nitrogen lone pair that it could use to bind another. The palladium, too, would be free to bind a second bpy. As a result, these two monomeric units are able to form an alternating chain.
Why would people want to do that? The general idea is to exploit the properties of these metal atoms in new ways. What are metals good at? They are good at conducting electricity, and sometimes they hold useful magnetic properties. If these properties can be incorporated into a material that is more like organic polymers, which tend to be inexpensive and very lightweight, then maybe they can be used to make materials that will have all sorts of applications in everyday life. Metals are also very good catalysts for a wide range of reactions. A coordination polymer serves to space metal atoms out evenly within a structure that has a lot of surface area, which could promote catalytic efficiency. All of these potential uses have driven a great deal of research into coordination polymers in recent years (as well as related "metal-organic frameworks").
Now, can all of the bonds along this metal-ligand chain really hold together to form a polymer? One key difference between the covalent bonds you have seen in organic polymers and the dative bonds here is that dative bonds are reversible. There is always an equilibrium between the metal-ligand complex and the free metal and ligand. Of course, this bond might be very strong, in which case the equilibrium lies toward the metal-ligand complex. In other words, a large fraction of the material would form the metal-ligand complex. How large a fraction? The real value isn't that important at the moment. For our purposes, let's just say we put a metal and a ligand together and 90% of the molecules form a complex. And suppose that's also true in the next step, bringing another ligand into the picture to bind to the metal, and in the step after that, bringing another metal in to bind to the other end of one of the ligands, and so on. So, suppose each of these events leads to 90% product formation.
Just getting to that four-unit chain relies on three different equilibria. If, as we say, each step proceeds about 90% of the way (which sounds pretty good; you would be happy with a 90% yield on a reaction in the lab), then the entire three-step process would yield 0.90 x 0.90 x 0.90 = 0.73, or 73% product. Not bad. But useful polymer chains might be a thousand units long or more, in which case the amount of polymer actually formed of the proper chain length would be about (0.90)999 = 1.94 x 10-46 or 1.94 x 10-44 %, and that's a ridiculously small amount.
Now, a real calculation of the equilibrium concentration is of course more sophisticated than that, but this quick exercise underscores an important point: in order to form a stable coordination polymer of an appreciable length, the metal must bind the ligand very tightly. For that reason, many approaches to coordination polymers have employed multidentate ligands, which of course bind more tightly than monodentate ligands. For example, the polymer formed using the bidentate binding shown below would be much more stable with respect to depolymerization (i.e. falling apart into monomers again) than the monodentate example shown earlier. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.05%3A_Coordination_Polymers.txt |
So far, we have seen how covalent bonds can be used to bind monomers together into longer chains, forming polymers. There has also been a great deal of interest recently in using intermolecular attractions to make similar structures. Of course, intermolecular attractions are very important in forming large, organized structures in biology. Think about the twin helices held together in DNA, or the secondary and higher-order structures in proteins. Clearly, intermolecular attractions can be used to form large, stable structures.
If the goal is to form a chain, analogous to the chains of monomers we have seen in other polymers, then it would be useful to have two sites within a monomer that can interact with other monomers. Such a molecule would be able to hold hands with two other partners, so to speak. If we take hydrogen bonding as an example of a strong intermolecular force, then we can look at a molecule such as ethylene glycol (also called 1,2-dihydroxyethane).
Ethylene glycol looks kind of like ethanol (CH3CH2OH) with an extra OH group. A physical sample of ethylene glycol is a colorless liquid, just like ethanol, but there are noticeable differences between the two. The boiling point of ethylene glycol is around 198°C, which is over a hundred degrees higher than that of ethanol. Furthermore, ethanol isn't very viscous; it pours quite easily, but if you tried to stir or pour ethylene glycol you would notice that it seems much thicker. Something is holding those ethylene glycol molecules very tightly together. It isn't hard to imagine the formation of chain structures involving groups of these molecules.
If we look at diols containing longer carbon chains, it's even easier to imagine something that looks kind of like a polymer, with monomers enchained together via hydrogen bonding interactions.
Hydrogen bonds are strong, but they are still just intermolecular attractions, rather than true covalent bonds. They are readily reversible, so any chains that formed via interaction between these diols could depolymerize very easily. Individual diols could readily fall of the end of the chain, and diol chains could just as easily break somewhere in the middle.
What about a tetraol? If there were two hydroxy groups at either end of the chain, then the chain would be much less likely to come apart. Even if one hydrogen bond broke at one of the linkages, there would still be a second one holding two neighbors together. A molecule like that could form a much more stable supramolecular assembly.
The dynamic nature of supramolecular assemblies is part of what makes them so interesting to researchers. Stable structures that can easily form under one set of circumstances, but gently come apart under another, could have lots of applications. For example, some researchers are interested in developing novel materials, such as liquid crystals, that might be converted from one form into another using different stimuli. Others are trying to develop materials that have useful medicinal and therapeutic applications.
1.07: Other Polymers
The principle of functionality means that almost any type of organic reaction could potentially be used to make polymers. For instance, if a compound has two functional groups of the same kind, it could undergo reaction at two different sites, forming new bonds with two neighbors. The compound thereby becomes enchained in a trio of formerly independent molecules. If the neighboring molecules are also difunctional, then this pattern can repeat, forming a polymer.
Take the Diels Alder reaction as an illustrative example. On paper, the reaction is fairly straightforward, even if it isn't all that common. A molecule with a pair of conjugated alkenes, the diene, cyclizes with another alkene, the dienophile, to form a new six-membered ring, a cyclohexene.
The diene needs two double bonds but the dienophile only needs one. Nevertheless, what happens if the dienophile has an extra double bond? The Diels Alder adduct that forms would end up with two double bonds: one formed from the original diene, as always, and the other leftover from the dienophile.
The resulting compound is a difunctional alkene. Each end of this molecule could potentially undergo another Diels Alder reaction with an additional diene. Each time that reaction occurs, a new alkene is left behind where the diene used to be, preserving that difunctionality for another step. That difunctionality forms the basis for potential polymer chemistry.
In fact, Diels Alder reactions have been exploited by researchers in a number of ways to make materials with useful properties. The fact that six-membered rings are introduced along the backbone, rather than a chain of single bonds, means that these materials display varying levels of conformational rigidity, resulting in some unique properties. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.06%3A_Supramolecular_Assemblies.txt |
Polymers are very large molecules made from smaller ones. How those smaller units are arranged within the polymer is an issue we haven't addressed very closely yet. Topology is the study of three-dimensional shapes and relationships, or of how individual parts are arranged within a whole. Let's take a look at the topology of polymers.
At the simplest level, we have been thinking of polymers as chains of monomers strung together like beads on a string. We even use the verb "enchain" to describe the act of taking a monomer and tying it into a larger polymer. The chain is the most basic of polymer structures. This topology is often referred to simply as a chain or, to underscore the structure, a linear polymer.
Even a simple chain, in this context, can have additional structural features worth considering. These features are connected to how a polymer chain grows up from individual monomers. If the polymer results from a chain reaction, typically used for the polymerization of alkenes, then the growing chain usually has two distinct ends. One end, sometimes called the tail, is the site of the first monomer to be incorporated into the polymer, as well as some remnant of the initiator to begin the polymerization process. The other end, sometimes called the head but more commonly just the growing end, is the active site, the place where new monomers are about to be enchained into the polymer.
That's what happens in most cases when the initiator is capable of reacting with one monomer and starting the chain reaction. However, sometimes initiators are difunctional. Difunctional initiators are capable of getting two monomers to start growing a polymer chain: one in each direction. In that case, the initiator fragment is left behind in the middle of this new growing chain, which grows outward from the middles. Both ends of the chain are growing ends. In this type of growth, the chain is referred to as "telechelic".
Sometimes, polymer chains do not have simple, linear structures. Instead, their chains branch out here and there. This topology is called "branched". Instead of looking like a snake or a piece of spaghetti, this structure looks more like a length of seaweed. Branching is sometimes an artifact of how the polymer was made, and so sometimes the same monomers may lead to a more linear polymer or a more branched one. The prime example if polyethylene, which can form high-density polyethylene (HDPE) or low-density polyethylene (LDPE) depending on the conditions under which the ethylene is enchained.
In a branched polymer, smaller chains grow out like limbs along a tree trunk or leaves along a stem. We can take that arrangement a step further into something that looks more like a net. In what we call a crosslinked structure, branches connect one main chain to the next, tying them together into one big piece. Sometimes, this type of structure is called a thermoset.
The word "thermoset" is really a description of a physical property of a polymer, in contrast to "thermoplastic". A thermoplastic is a polymer that can be melted and reformed into new shapes after polymerization. In contrast, once a thermoset has been polymerized, it retains its shape even when heated; it doesn't melt. These terms have connotations about the topology of the material, nonetheless.
The reason thermoplastics can be melted and formed into new shapes is that they are made of separate molecules. The molecules may be very long, and they may even be branched, but at high enough temperatures these molecules can move completely independently of each other. They can melt and so the material that is comprised of these separate molecules take on new shapes.
In a thermoset, crosslinks connect the different chains in the material, forming bridges that span from chain to chain to chain, essentially uniting the material into one big molecule. If it is one big molecule, the chains can never move completely independently of each other, and the material cannot form a new shape. Looked at in another way, those crosslinks tie the main chains in place. They may be able to move around some, but they can never get very far. If the amount of crosslinking is sufficient, they will always hold the material in the same basic shape.
Of course, just a little bit of crosslinking may not have the same effect. You may have two or three chains tied together to form one big molecule, but they behave more like highly branched chains than like extended nets.
It's also worth noting that the term "crosslink" can actually mean different things when used in different ways. Sometimes, crosslinks refer to true covalent bonds connecting two chains together. These connections are called "chemical crosslinks"; the word chemical refers to the covalent bonds. Alternatively, chains can be connected to each other through strong intermolecular forces. That's not at all the same thing, of course, because these intermolecular forces can be overcome with sufficient energy, and so at some point, the chains may no longer be tied together. These connections are called "physical crosslinks" to distinguish them from permanent bonds.
To use a biology analogy, sulfide bonds in proteins would be an example of chemical crosslinks; they hold the protein firmly in one shape, and a chemical reaction is required to break that connection. The ordinary hydrogen bonds that are so prevalent in the protein are physical crosslinks. Because they can be overcome by adding heat, proteins are very sensitive to temperature.
Dendrimers are another type of branched structure; the term comes from the Greek dendron or tree. Dendrimers differ from regular branched polymers in that they have a much more regular branching pattern. A dendrimer will actually grow outward from the center, branching out at regular intervals.
Many dendrimers are polyamides, although there are other types as well. Generally, at least one of the monomers is trifunctional, which introduces the branching in a predictable way.
As dendrimers grow outward, layer by layer, they are often described in terms of generations. Suppose you start with a trifunctional monomer in the middle. That's generation 0. If the monomer is polymerized outward until there is another set of trifunctional monomers attached on the edge, we have a first generation dendrimer. If we keep going and add another layer of trifunctional monomers, we have a second generation dendrimer, and so on.
These dendrimers start to look circular on paper, like pancakes, but steric interactions between the groups forces things into three dimensions. As a result, dendrimers are roughly spherical in shape.
In some of the examples of polymers we have seen, the chain is actually composed of two different monomers. That's true in the case of polyamides such as nylon-6,6. In that example, the chain is composed of difunctional amines alternating with difunctional carboxyloids (such as carboxylic acids or acid chlorides). We can think of a polymer like that as being composed of two different monomers. Of course, because of their complementary reactivity they have to alternate: an amine and then a carboxyloid, to form an amide, and so on. We can think of these polymers as "co-polymers", meaning they are formed from more than one kind of monomer. We can go further and say that they are "alternating co-polymers" because the two different monomers alternate with each other along the chain.
In some cases, there is no need for the two different monomers to alternate the way they do in nylon-6,6. If you take a mixture of alkenes that are capable of forming polymers and you polymerize them together, you may well get them randomly enchained into a growing polymer. This arrangement is called a "random co-polymer" or sometimes a "statistical co-polymer". For example, maybe you manage to get a random sequence of propene and vinyl chloride units along with the polymer.
Notice that in the above example we still have the same zig-zag main chain in the polymer; all that changes between one monomer and the other is the group attached to that main chain. It is sometimes useful to think about this main chain or "backbone" separately from the attached or "pendant groups". In this case, we have what looks like a polyethylene backbone with pendant chlorides or methyls randomly attached.
A random arrangement isn't the only possibility. Maybe all of the vinyl chlorides polymerized in a row, and then all of the propenes were incorporated after them. The result would be a diblock co-polymer; there is a solid block of poly(vinyl chloride) at one end of the chain and a solid block of polypropylene at the other end.
There are a couple of ways that could happen. Maybe you waited until all of the vinyl chlorides were enchained before adding any propene so that all of the vinyl chlorides were enchained at the tail end and the propenes were added at the growing end later. Alternatively, maybe you added them all at once but the vinyl chlorides just underwent polymerization a whole lot faster than the propenes; the vinyl chlorides all became enchained before the propenes had a chance.
Those possibilities assume that the monomers are all enchained linearly. There are other possibilities. Maybe one set of monomers form the polymer backbone and the other set form pendant branches along the chain. That arrangement is called a "graft co-polymer", as though we have little apple trees grafted onto the trunk of another breed.
1.E: Solutions for Selected Problems
Problem MP1.1.
Problem MP1.2.
Problem MP1.3.
Problem MP2.1.
Problem MP2.2.
Problem MP2.3.
Problem MP3.1.
Problem MP3.2.
Problem MP4.1.
Problem MP4.2.
Problem MP4.3.
Problem MP5.1.
Problem MP5.2.
Problem MP5.3.
Problem MP6.1.
Problem MP7.1.
Problem MP7.2. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/01%3A_Monomers_and_Polymers/1.08%3A_Polymer_Topology.txt |
• 2.1: Ziegler-Natta Polymerization
There are alkenes that do not give rise to cations stable enough for cationic polymerization, and don't form stable enough anions for anionic polymerization, either. The two most prominent examples are ethene and propene. These monomers are instead polymerized via an organometallic cycle of alkene associations and 1,2-insertions, so that the last alkene becomes part of a metal alkyl group ready to undergo insertion with the next alkene that binds the metal.
• 2.2: Solutions to Selected Problems
• 2.3: Step Growth and Chain Growth
Step growth and chain growth are two broad classes of polymerization methods. They use monomers with distinct characteristics and display some different growth patterns.
• 2.4: Cationic Polymerization
Alkenes, or olefins, are probably the most common polymer feedstock. The pi bonds of alkenes are inherently nucleophilic. The addition of an electrophile to an alkene results in cleavage of the pi bond, with the electrons drawn to the electrophile, and consequently a cation can be left on one end of the former double bond.
• 2.5: Living Cationic Polymerization
Alkenes are a common polymer feedstock, used to make a range of very familiar plastics in everyday use. Because their properties depend strongly on their molecular weights and molecular weight distributions, it is very important to be able to control the growth of these long-chain polymers from their alkene monomers. Termination events contribute to a widening disperity between chains that have undergone these events and those that continue to grow.
• 2.6: Anionic Polymerization
It seems obvious that cationic methods would be employed for the polymerisation of alkenes. After all, alkenes are nucleophilic, and so they should react readily with cationic initiators. It might be surprising that there are a number of alkene polymerisations that work very well using anionic initiators. One of the most common scenarios for this approach is in the polymerisation of conjugated hydrocarbons, such as styrene
• 2.7: Living Anionic Polymerization
• 2.8: Ring-Opening Polymerization
On an earlier page, we saw that polyesters and polyamides are typically obtained via condensation polymerization. Ring-opening polymerization (ROP) offers a second approach to these kinds of materials. Using this method, cyclic esters (lactones) or amides (lactams) are opened up to make extended chain structures. The reaction is typically driven by the release of ring strain.
• 2.9: Radical Polymerization
Radical polymerization is a very common approach to making polymers. There are a number of very reliable methods of carrying out radical polymerization, leading to high molecular weight materials.
• 2.10: Living Radical Polymerization- RAFT
• 2.11: Living Radical Polymerization- ATRP
02: Synthetic Methods in Polymer Chemistry
There are alkenes that do not give rise to cations stable enough for cationic polymerization, and don't form stable enough anions for anionic polymerization, either. The two most prominent examples are ethene and propene. Nevertheless, ethene and propene are the most common feedstocks for polymers, with about 100 millions tons of global polyethylene production annually, and around 50 million of polypropylene. So, how do we make such huge quantities of polyethylene and polypropylene?
These monomers are instead polymerized via an organometallic cycle of alkene associations and 1,2-insertions, so that the last alkene becomes part of a metal alkyl group ready to undergo insertion with the next alkene that binds the metal. This process was developed in 1952 by Karl Ziegler at the Max Planck Institute for Coal Research in Mulheim, Germany. The method was later improved by Giulio Natta at Milan Polytechnic.
Commonly used industrial conditions for Ziegler-Natta polymerization employ a multi-component system. There is always an early transition metal catalyst, from the left-hand side of the d-block of the periodic table. Typically this catalyst is TiCl4 or TiCl3; under the reaction conditions, TiCl4 is probably reduced to TiCl3 anyway. This catalyst is supported on other metal halides such as MgCl2. In heterogeneous catalysis, a support is a layer which strongly interacts with the catalyst, tuning its properties; exactly how that works is somewhat complicated. The catalyst and support are deposited on a solid carrier, such as silica, SiO2, which distributes the catalyst, increasing the usable surface area. In addition, the carrier may modify the three-dimensional space around the catalyst, providing a binding pocket for the catalyst and the substrate, for example. Finally, the system needs an activator, which is a p-block metal alkyl such as triethylaluminum. This compound transfers a starting alkyl group to the catalyst and may also reduce the transition metal species.
A Ti(III) rather than Ti(IV) catalyst has the advantage of being better able to bind the alkene substrate. Ti(IV) is a d0 species, and d0 metals are typically incapable of binding alkenes. The relatively weak π-bond donation to a metal is normally augmented by back donation from an occupied metal d orbital into the alkene π* orbital.
Exercise 2.1.1
Let's take a look at some alternative Ziegler-Natta reagents. Identify each component as a possible catalyst, support, carrier or activator.
1. ZrCl4
2. EtLi
3. Mg(OEt)2
4. Al2O3
5. AlMe3
6. CrCl6
7. VCl5
8. MgO
As indicated previously, once an alkene binds at the transition metal, an initial alkyl that was previously transfered from the aluminum is able to participate in a 1,2-insertion reaction. As a result, the alkene is incorporated into the alkyl chain. Subsequent alkene association and 1,2-insertion steps serve to extend the chain, eventually into oligomers and then polymers.
Homogeneous catalysts have also been used commercially for these polymerizations. These systems have been developed more recently by Walter Kaminsky at University of Hamburg, Germany and Hans-Herbert Brintzinger at University of Konstanz, Germany. These soluble catalysts have expanded the scope of stereochemical control over the polymerization of propene and other terminal alkenes or 1-olefins.
When propene is enchained into a polymer, a new chiral center is formed at every position where a methyl group branches from the backbone. Rather than trying to assign each of these chiral centers with stereochemical configurations (R) or (S), we instead describe the relative stereochemical relationships along the backbone. The term used to describe these relationships is "tacticity". If there is no apparent relationship between the projection of the methyl groups along the backbone, the polymer is termed "atactic". If the methyl groups alternate, pointing first one direction, then the other, all the way along the chain, then the polymer is termed "syndiotactic". If the methyl groups all project the same direction, the polymer is described as "isotactic".
Tacticity in polymers is frequently determined by NMR spectroscopy. Because diastereomers have different physical properties, they display slightly different shifts in the 1H or 13C NMR spectrum. In a polymer chain, we think about chiral centers in pairs, which we call "diads", because every pair of chiral centers has two possible diastereochemical relationships and therefore two possible chemical shifts. If the methyl groups are on the same side, the chiral centers have a "meso" relationship, whereas if they are on opposite sides, they have a "racemo" relationship.
In principle, these are two different stereochemical relationships that could give rise to two different NMR signals. If the polymer is truly isotactic, we might only see the signal that comes from the meso diad, because all of the stereocenters would have that same relationship along the chain. If the polymer is truly syndiotactic, we might see only the signal from the racemo diad, at a slightly different chemcial shift as the meso. On the other hand, we might see two signals, and the relative integration of the two signals might indicate that the polymer is mostly syndiotactic, mostly isotactic, or completely atactic.
Exercise 2.1.2
Label the following diads.
We can expand from there to think about triads, a sequence of three chiral centers, considering the relationship between the first pair of chiral centers in the triad and then the next pair. We would then describe these relationships as mm, for two meso relationships in a row; rr, for two racemic relationships in a row; or mr, for a meso and a racemo relationship. We might be able to detect different signals for all of these relationships.
Given a big enough magnetic field strength an an NMR spectrometer, we may even be able to detect the difference between diads and tetrads, allowing us to see more detail along different segments of the polymer backbone.
Exercise 2.1.1
Label the following triads and tetrads.
Control over polymer stereochemistry typically occurs in one of two ways. The first way is called chain end control. Given an existing stereocenter at the end of the chain, the approach of the next monomer will be influenced by how that existing chiral center projects in space. As a result, the last chiral center influences how the next one is formed. Conceptually, this idea is the same as saying that in a simple organic reaction, an existing chiral center may force an incoming group to one face of a molecule rather than another, leading to preferential formation of a particular stereochemistry where the new bond is made. In the case of polymers, it means the very first insertion reaction, when the first chiral center forms, determines all of the subsequent stereochemistries along the chain. Of course, there isn't a simple way to predict what that stereochemistry will be: syndiotactic or isotactic. However, the drawing below shows how such an interaction could lead to a syndiotactic polymer.
The second mechanism is called site control. In site control, the specific shape of the reagent or catalyst that facilitates the reaction also determines the stereochemictry. This idea is also seen in organic reactions. Chiral reagents or catalysts, whether for hydrogenations, hydroborations, or other reactions, will fit with the substrate in one way more easily than another, leading to preferential formation of one stereochemistry and not the other.
Site control is a topic of interest in homegeneous catalysis, so let's look at how it would work in a titanocene. The first basic lesson in titanocene chemistry is that the two cyclopentadienyls form an angle with each other, so that the molecule takes on a little bit of a wedge shape. One end of the titanocenes are closed together more, and the other end is more open. It's kind of like a clam shell.
If another molecule, such as a propene, approaches to bind to the titanium, it will have to negotiate the wedge shape to minimize steric interactions. For example, keeping the methyl group away from the titanium, toward the opening on the wedge, would cause less crowding than if the methyl spun around to project deeper inside the wedge.
When discussing the formation of a new chiral center, the key idea is whether a new group will bind to one face of a molecule or the other face. We are going to show one face of the propene in blue and the other in red to emphasize this difference. So far, the methyl of the propene will remain further away from the titanium, but there is nothing to prevent the propene from flipping over; the methyl group could be either up or down in the picture.
Much of the focus in these homogeneous catalysts has been on developing some kind of asymmetry in the catalyst. The catalyst shown below has Cs symmetry. The top is now different from the bottom. Specifically, the bottom is much more crowed, so that it is no longer very likely that the propene would approach with its methyl group down. It would be much easier to come in with its methyl group pointed up in this picture.
To add one more wrinkle, the propene isn't going to bind in the middle of the wedge, but will be off to one side or the other. That's because there should already be an alkyl group on the metal, which we have shown in this case as a simple methyl group. The propene could approach from the near side, in front of the methyl (shown on the left), or from the far side, behind the methyl (shown on the right).
If the propene is on the near side from our perspective, it is actually presenting its blue face to the alkyl chain. When the 1,2-insertion occurs, the alkyl will bond to the blue face, pushing the methyl and hydrogen of the propene towards us (seen in the top picture). However, if the propene is on the far side from our perspective, it really is presenting its red face to the alkyl chain. When the 1,2-insertion occurs, the alkyl will bond to the red face, pushing the methyl and hydrogen of the propene away from us.
Which one really happens? Both. During the insertion, the alkyl chain shifts over to the coordinated olefin, opening up a coordination site. Notice in the top picture above, the propene is on a wedge and the alkyl is on a dash. After the insertion, the alkyl chain is on a wedge and a new coordination site opens up behind it. A new propene moves into that site. As a result, the new propene approaches first from one side, then the other, alternative every time. This windshield washer mechanism results in a syndiotactive stereochemistry.
What is we change the shape of the catalyst? In the picture below, we have a catalyst with C2 symmetry. With this symmetry, the propene can approach with its methyl up only if it comes from the front. With its methyl down, it can only approach from the rear.
If you think about it, that means the propene is always presenting the same face to the metal. It only binds to the titanium through the blue face. That means the alkyl chain always binds to the blue face. If the alkyl always binds to the same face, it is always forming a new chiral center with the same configuration. It's making an isotactic polymer. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.01%3A_Ziegler-Natta_Polymerization.txt |
SM11. Solutions to Selected Problems
Problem SM1.1.
Problem SM1.2.
Problem SM1.3.
Problem SM1.4.
Problem SM1.5.
a) After seven steps, the polymer will be 27 = 128 units long; MW = 128 x 150 g/mol = 19,200 g/mol.
b) After sixteen steps, the polymer will be 216 = 65,536 units long; MW = [(120 + 130)/2] x 65,360 g/mol = 8,192,000 g/mol.
c) After ten steps, the polymer will be 210 = 1,024 units long; MW = [(100 + 105)/2] x 1,024 g/mol = 104,960 g/mol. One condensation will occur at the two ends of each monomer, except for the ones on the ends, but that's a loss of approximately 1,023 x 15 = 15,345 g/mol. The net molecular weight is 104,960 - 15,345 g/mol = 89,615 g/mol.
d) Chain growth occurs linearly, so the molecular weight is 30 x 105 g/mol + 65 g/mol (end group) = 3,215 g/mol.
Problem SM2.1.
Problem SM2.2.
Problem SM2.3.
Problem SM2.4.
Problem SM2.5.
These large, highly charged anions are not very soluble, limiting their interaction with the growing polymer chains.
Problem SM3.1.
a) GaCl3
b) SnCl4
c) ZnCl2
d) FeCl3
Problem SM3.2.
Problem SM3.3.
Problem SM4.1.
The additional resonance stabilization of the anion by the nitrile group makes anionic polymerization proceed smoothly.
Problem SM4.2.
a) Vinyl chloride's electronegative chlorine might stabilize an anion on the adjacent carbon. However, that inductive effect may be offset by lone pair repulsion between carbon and chlorine.
b) Under these strongly basic conditions, 1,2-elimination may result from the growing polymer chain. Very basic conditions may even result in 1,2-elimination from viyl chloride to give acetylene (ethyne).
Problem SM5.1.
a) Li+, Na+, K+
b) Be2+, Mg2+, Ca2+
Problem SM5.2.
a) K+, Na+, Li+
b) Ca2+, Mg2+, Be2+
Problem SM5.3.
a) Et3Al
b) Et2Zn
c) Ph3B
d) (CH3O)2AlCH3
Problem SM6.1.
Problem SM6.2.
Problem SM6.3.
Problem SM7.1.
Problem SM7.2.
Problem SM7.3.
Problem SM8.1.
The more stable the radical, the more rapidly it will form from the monomer, leading to faster polymerization.
Problem SM8.2.
The more stable the radical generated upon fragmentation of R, the more the equilibrium will shift toward growing phase.
Problem SM8.3.
The more stable the radical formed, the more the equilibrium will shift toward the dormant phase.
Problem SM8.4.
a) With a fast-polymerizing monomer like this one, we need a chain control agent that will provide a similarly stable radical, so that there is an appreciable equilibrium allowing chains to move into the dormant phase; hence the phenyl substituent. However, to get into a new growing phase, we need an R group that fragments very easily, such as this disubstituted benzyl.
b) With a monomer on the lower end of radical stability, we need a chain transfer agent that isn't terribly stable as a radical, or else the chains will shift completely into the dormant phase. We also need an R group that will fragment at a rate comparable to the original radical chain, so we have one that will form a modestly stable radical.
Problem SM9.1.
Problem SM9.2.
a)
b)
Problem SM9.3.
Problem SM9.4.
a)
b) The radical generated in the first step may initiate new polymer chains instead. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.02%3A_Solutions_to_Selected_Problems.txt |
Step growth and chain growth are two broad classes of polymerization methods. They use monomers with distinct characteristics and display some different growth patterns.
Condensation polymerizations take place via a step-growth process. In a classis condensation reaction, a carboxylic acid reacts with a neutral nucleophile, such as an alcohol or an amine. Substitution leads to formation of either an ester (from an alcohol nucleophile) ar an amide (from an amine nucleophile), with water as a side-product. For example, nylon-6,6 is a condensation polymer.
Nylon-6,6 is produced via a condensation reaction between 1,6-hexanedioic acid (or adipic acid) and 1,6-hexanediamine (or hexamethylenediamine). The amine end group of the nucleophile loses a proton and the carboxylic acid loses an OH group, for the formal loss of one water molecule each time an amide bond forms.
Poly(ethyleneterephthalate) or PETE is a very common example of a polyester.
PETE is formed from the reaction of dimethyl terephthalate with 1,2-ethanediol (or ethylene glycol). Once again, a condensation reaction takes place, with a small molecule formed as a side product. An OMe group is displaced fron the carboxyl electrophile and a proton is lost from the nucleophilic alcohol group, adding up to the formal loss of methanol.
It's worth talking about the conditions used for these reactions. On paper, we might think about using more reactive monomers, such as acyl halide electrophiles, in order to have much faster reactions. However, these reactions are either exothermic in the case of the polyamide or thermoneutral in the case of the polyester, which is formed through a trans-esterification. Both reactions occur in equilibrium. Although they may be relatively slow, they should still get to product eventually, especially if the equilibrium can be shifted to the right. In both of these cases, that shift can be acheived through removal of the water or methanol side-products, which are relatively volatile compared to the other reactants.
Note also that the neutral nucleophile and the relatively poor leaving groups in these cases suggest the reaction will be pretty slow. To speed things up, a Lewis acid may be added (such as the antimony oxide, Sb2O3, employed in the second example).
The monomers used in step-growth polymerization usually have a couple of characteristics that we can see in these examples. Most often, there are two different monomers. Neither monomer would undergo polymerization on its own, but the two monomers are complementary to each other, so that each provides the other with an appropriate reactive partner.
More importantly, each of the monomers we have seen here is mutlifunctional. In this context, functionality describes the number of reactive functional groups. Each of the monomers we have seen so far contains two functional groups; they each have a functionality of two, or they are each difunctional. Once a functional group has reacted on a difunctional monomer, it still has another one that can react. Every monomer can therefore make bonds with two others, allowing for chain formation.
This higher functionality is essential for step-growth polymerization. It isn't limited to difunctional monomers. Monomers could be trifunctional, tetrafunctional, and so on, leading to more highly branched or chemically crosslinked polymers, rather than simple chains.
Problem SM1.1.
Identify the monomers that would be used to produce these polymers.
Because of difunctionality in the monomers, condensation polymers can grow in two directions at once. Two complementary monomers connect with each other to form a dimer. Now, suppose all the monomers are reacting at roughly similar rates (not quite true if you know about polymer molecular weight distribution, but close enough). When one end of the dimer goes to react again, chances are it will find another dimer, because most of the monomers have already reacted. That reaction will form a tetramer. The typical tetramer reacting another time will find another tetramer, forming an octamer, and so on.
Most of the subsequent chapters in this text deal with a variety of ways of carrying out chain polymerizations. In chain polymerization, we usually start with just one kind of monomer, which is again unreactive by itself. This time, instead of adding a complementary monomer, we add an initiator. The initiator converts one monomer molecule into a reactive intermediate, capable of reacting with another monomer to form a new reactive intermediate, and so on. This is a chain reaction.
Notice the difference from step growth. In chain growth, the polymer chain always grows one monomer at a time. In step growth, the polymer chain doubles with each step. As a result, the rate of growth of the polymer chain is very different in these two cases.
Chain growth results in a steady increase in chain length with every coupling step. Growth is linear.
Chain growth is exponential, with chain length doubling at every coupling step. Notice that the y axis on this graph is logarithmic rather than linear.
What are the consequences of these two growth patterns? Let's take a look at how many steps it would take to form a high molecular weight polymer, maybe with a molecular weight of one million g/mol (sometimes expressed as one million Da in polymer chemistry; Da stands for Dalton). Starting with a monomer of molecular weight 100 g/mol and using step growth, in which chain length doubles every time a reaction takes place, it will take only thirteen or fourteen steps to reach a molecular weight of a million Da, because 213 = 8,192 repeat units. Using chain growth, it would take ten thousand steps to reach the same molecular weight, assuming a monomer of similar size.
That's a very approximate calculation, and it neglects the weight of side products during condensation polymerization, but we are looking at differences that are orders of magnitude apart.
Why is this difference significant? For one thing, it doesn't matter that much if a condensation reaction is slow. If it only takes twenty (or even thirty) steps to reach a molecular weight of a million (not unheard of in a commercial polymer, but still pretty high), then there's no hurry. In chain polymerization, there is a much higher premium on speed. We need very fast reactions or sometimes very fast catalysts that can turn around and enchain monomer after monomer reliably.
If we take these two broad methods and look at them in a different way, we get new information. If, instead of thinking about the number of bonding events or reaction steps, we focus on the percent conversion, we get a graph that looks like this.
Looked at in this way, we see the linear nature of chain growth. Every time a new monomer is added, the polymer chain grows by the same amount (the weight or length of one new repeat unit, depending on whether we are thiniing in terms of molecular weight or chain length). Step growth works exponentially: doubling, quadrupling, and so on. Both modes of polymerization proceed, ideally, until all of the monomer has completely reacted; in other words, to 100% conversion. But what happens if growth is suddenly arrested because of some unwanted side reaction? In step growth, such an event happening even in the very late stages of polymerization might result in a chain length that is only half what we had expected. Such disastrous results in chain growth would result only if growth stopped after only consuming half the monomer. Arresting chain growth at 80 or 90% conversion still results in pretty long polymer chains, even if they aren't the expected length. Because of this feature, step growth polymerization requires very reliable chemistry. Otherwise, step growth would always result in relatively short chains.
In either case, it's useful to know that polymerization is often followed by a quenching reaction to arrest the process once the polymer has reached the desired chain length. Usually that results in a polymer with two different groups at the end of the chain. The choice of those groups can occasionally be useful in modifying the properties of the polymer.
Imagine a case, however, in which this termination happens accidentally. In a chain growth, the polymer has a reactive intermediate at one end only. If something caps that end of the polymer chain, it stops growing altogether. Unexpected terminations can be a significant problems in chain growth.
Of course, they are also a problem in step growth, but remember that in step growth the polymer is growing from two ends. If one end gets capped, the other end can keep growing. The polymer will grow more slowly, but it won't stop altogether.
As a result, there is a great deal of interest in developing more reliable ways of carrying out chain growth. That's especially important because one wrong step and chain growth stops altogether. Chain growth is, in fact, a crucial way of making polymers, despite what may seem like some limitations here. A key reason is that step growth is mostly limited to condensation polymerization of carboxyloid compounds such as polyamides and polyesters, as well as a few related materials such as polyurethanes. Chain growth is used to polmerize a wide array of compounds that contain carbob-carbon double bonds (alkenes or olefins) as well as a variety of cyclic monomers. Both approaches are needed to make a range of useful materials.
Problem SM1.2.
Provide structures of the polyesters made from the monomers as indicated.
Problem SM1.3.
The following difunctional monomers would undergo polymerization together; in some cases the reaction happens without condensation of a side product. Show the product structures that result.
Problem SM1.4.
A few polymers are obtained from difunctional monomers that contain both nucleophilic and electrophilic sides. Show the polymers that result in these cases.
Problem SM1.5.
Calculate the molecular weight of the polymer in each of the following cases.
a) Step polymerization involving monomers of molecular weight 150 g/mol, allowed to undergo seven polymerization steps.
b) Step polymerization involving monomers of 120 and 130 g/mol, allowed to undergo sixteen polymerization steps.
c) Step polymerization involving monomers of 100 and 105 g/mol, allowed to undergo ten polymerization steps, producing a condensation product with molecular weight 15 g/mol.
d) Chain polymerization of a monomer with molecular weight 105 g/mol, allowed to undergo thirty polymerization steps, using an initiator with molecular weight 65. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.03%3A_Step_Growth_and_Chain_Growth.txt |
Alkenes, or olefins, are probably the most common polymer feedstock. The pi bonds of alkenes are inherently nucleophilic. The addition of an electrophile to an alkene results in cleavage of the pi bond, with the electrons drawn to the electrophile, and consequently a cation can be left on one end of the former double bond.
If the resulting cation interacts with another alkene, another electrophilic reaction occurs, and so on, leading to polymerization. Because of the presence of a key cationic intermediate along the reaction pathway, this method of polymerization is termed cationic polymerization. It is an example of a chain reaction that is initiated by a cationic species.
It stands to reason that alkenes that form stable cations would be particularly susceptible to polymerization via cationic methods. Cations are high-energy intermediates, so if a cation is relatively easy to form then the barrier for the process is lowered considerably. Isobutylene, (CH3)2C=CH2, is a good example of an alkene that would form a stable cation upon protonation. The relative stability of more-substituted cations dictates that the positive charge is found at a tertiary position.
Degree of substitution is just one way that carbocations can be stabilized. Conjugation is also a major factor in cation stability. Alkenes that give rise to delocalized cations are also good candidates for cationic polymerizaton. Styrene, PhCH=CH2, as well as butadiene, CH2=CHCH=CH2, and the related natural product isoprene, CH2=C(CH3)CH=CH2, can all be polymerized using cationic methods.
These examples are stabilized by allylic delocalization: the cation is part of a system that contains a C=C-C+ feature, which is delocalized as +C-C=C. Cations can also be stabilized via pi-donation. Oxygen and nitrogen are particularly good pi-donors towards carbocations, so oxygen and nitrogen substituents on alkenes render these compounds susceptible to cationic polymerization. Vinyl ethers, for example, can be polymerized this way.
The obvious way to initiate these polymerizations is through addition of a strong acid. Protonation of the alkene should certainly result in formation of a carbocation. However, most common strong acids are mineral acids, typically found in aqueous media. The presence of nucleophilic water would quickly cap the carbocation before polymerization could take place.
Of course, hydrogen chloride does not require water as a solvent. The addition of dry HCl gas to the olefin of interest should result in cation formation without subsequent hydration. Unfortunately, just the presence of the chloride ion, which can also act as a nucleophile, dooms polymerization to an early death.
There are other, less nucleophilic acids that are not as strongly prone to classical addition across an alkene compared to the extent that we associate with HCl. Sulfuric acid and various sulfonic acids come to mind. Even in these cases, however, polymerization is limited by eventual collapse of the nucleophile with the cationic chain end.
Instead of using strong Brønsted acids to inititate polymerization, it is actually very common to use Lewis acids. Lewis acids cannot initiate cationic polymerization directly, but they can co-initiate the reaction in the presence of a water impurity, for example. The Lewis acid activates the water towards release of a proton. The proton then initiates the reaction. Traces of HCl can also act as the initiator when a Lewis acid co-initiator is employed, because the Lewis acid ties up the chloride ion so that it cannot connect with the growing cationic chain end.
Protic initiators are not strictly required for cationic polymerization. Strong Lewis acids are also capable of activating alkyl halides. The carbocation that results can then initiate the polymerization chain reaction.
Even in these systems, there are chain-termination events that destroy the reactive intermediate and stop polymerization. As indicated previously, combination of a nucleophilic anion with the cationic chain end is a common fate of growing polymers. Halide ions, released in equilibrium from the Lewis acid-base complex, are still a possible problem, even though this event is much less likely in the presence of Lewis acidic co-initiators. Other, similar anions, such as hydroxide anions in water-initiated polymerizations, can play a similar role.
Anionic species can also carry out beta-eliminations on cationic chain ends. Beta-elimination results in the loss of the cationic intermediate, replacing it with a cain-end elkene. That event results in the production of a new initiator, and although it could be argued that the initiator might just re-initiate the same polymer chain, given relative concentrations it is much more likely that the initiator transfers the proton to a new monomer. In that case, a new growing chain starts from scratch and begins to accumulate monomers.
Surprisingly, chain-transfer events are another common mode of termination. In these cases, a hydride must be transferred from one chain to another, or from one position along the backbone to the growing chain end. This unusual hydride transfer is probably facilitated by very close contact distance between two chains, or between two segments of one chain, in a concentrated polymerization milieu. As a result, one chain end stops growing, and the chain begins growing from a new position along its backbone. This event results in chain branching, changing the morphology and properties of the polymer.
Normally, assuming a relatively rapid initiation step (formation of a cation from the first monomer), growth of tandem polymer chains will proceed at roughly the same rate, so that all of the chains reach similar lengths by the time all monomers have been consumed. What happens if some of those chains die prematurely? Other chains continue to grow to longer lengths. As a result, there is a wider distribution of chain lengths, or molecular weights, than there would be otherwise. A wider molecular weight distribution means that physical properties of the material may be less reproducible.
In a typical polymerization process, the reaction is followed by quenching, in which some agent is added that will stop further polymerization. In cationic polymerizations, quenching might be accomplished through addition of aqueous acid, because either the water or the conjugate of the acid will destroy the cation. The material is usually precipitated in a solvent that can dissolve any remaining monomer, but not polymer. Because some chains may have died very early, they might also remain in solution, leaving only larger chains in the precipitate.
Exercise 2.4.1
Choose the most stable cation from each of the following groups. Provide a reason for the stability of that cation.
Exercise 2.4.2
In each group, select the alkene most suitable for cationic polymerization.
Exercise 2.4.3
Provide a mechanism for the formation of a protic initiator from the interaction of boron trifluoride with water.
Exercise 2.4.4
Anethole in a naturally-occuring compound that has been used in cationic polymerizations. Show why anethole should be a good monomer for this method.
Exercise 2.4.5
Phosphotungstates such as H3PW12O40 have been used as cationic initiators and do not show much termination by the counterion, below. Propose a reason for the low nucleophilicity of this counterion for the growing cationic chain. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.04%3A_Cationic_Polymerization.txt |
Alkenes are a common polymer feedstock, used to make a range of very familiar plastics in everyday use. Because their properties depend strongly on their molecular weights and molecular weight distributions, it is very important to be able to control the growth of these long-chain polymers from their alkene monomers. Termination events, such as combination of the cation with a nucleophile, elimination, or chain transfer, all contribute to a widening disperity between chains that have undergone these events and those that continue to grow.
In any polymerization, as in any other reaction, we are dealing with thousands of reactions involving thousands of different molecules all happening at the same time. Several reactive chains growing in tandem would be expected to grow at similar rates, and ultimately they would reach similar degrees of polymerization. That means they would all have the same number of monomers incorporated into the final polymer chain and so they would all have similar molecular weight.
What happens when one of these growing chains undergoes some kind of termination event? Chain death, as this problem is commonly called, has a couple of direct consequences. Obviously, the chain that stopped growing does not keep up with the others, so its molecular weight is lower than the rest. That fact alone might lead us to believe the average molecular weight would be lower than we had expected in this polymerization. If we added more and more polymer to the polymerization over time, we wouldn't expect the molecular weight to keep increasing, because the reactive sites keep getting quenched.
More subtly, that growing chain had been slated to enchain a certain number of monomers, some of which are still left in the system when it dies. What happens to them? Of course, the other growing chains will each get some extra monomers. That means the dispersity problem gets even worse; one chain stops growing and is shorter than anticipated, but others gobble up the extra monomers and become longer than anticipated. Dispersity, the distribution of molecular weights, gets wider.
The problem can be even worse when a termination occurs through elimination, resulting in the formation of a new alkene. Because that new alkene is already found at the end of a polymer chain, when it combines with a cationic chain end the polymer molecular weight effectively doubles.
In other words, although we would reasonably expect molecular weight to stop increasing as chains die, resulting in lower molecular weight than expected, the opposite can also happen. We can observe an increase in molecular weight, because the chains that are left are consuming all of the monomer, becoming much bigger, and occasionally two chains are combining to dramatically increase molecular weight. Molecular weight increase is proceeding in an uncontrolled and somewhat unpredictable way.
Furthermore, the reaction of a cationic growing chain with a macroalkene results in a sudden shift in polymer morphology. Instead of a straight chain, we now have a dramatically branched polymer. That change in morphology has a severe impact of polymer properties.
All of these factors mean that the distribution of molecular weights gets broader when chain reactions undergo unexpected termination. The term for the width of molecular weight distribution is dispersity (D), sometimes called polydispersity index (PDI). In the graph below, the red line describes a sample from a polymerization that was not as well controlled as the one described by the blue line. There are a lot more polymer chains that are both longer and shorter than the average.
In an ideal polymerization, growing chains wouldn't die unexpectedly. They would continue to grow, and we could easily predict how long each chain would become based on how much monomer and initiator we added. Each initiator molecule would start one growing chain, and each growing chain would enchain its portion of the monomers. The molecular weight would be predicted easily because the polymer chains would grow linearly with added monomer. A polymerization in which there is a linear relationship between added monomer and molecular weight, even as the molecular weight becomes very high, is called a "living polymerization".
A second feature of living polymerization is that dispersity (or PDI) stays relatively constant throughout the course of a reaction. Because there are no chain terminations, the reactive chains all continue to grow at the same pace, and the polymers that result are of uniform molecular weight.
In order to achieve a living cationic polymerization, we would have to prevent unwanted terminations steps. There are a variety of systems that accomplish this goal, but they share some common features. Essentially, these methods intentionally allow a nucleophile to combine with the cationic chain end, but use a Lewis acid to reactivate the resulting compound and regenerate the growing chain end.
Like other cationic polymerizations, these processes could be initiated by addition of a protic acid to an alkene to generate a cation. That initial cation would then begin reacting with the nucleophilic alkenes around it, generating subsequent cations that sustain a chain reaction. In practice, it's instead very common to add an ionizable compound, such as a tertiary alkyl halide, along with a Lewis acid co-initiator. The Lewis acid promotes ionization of the initiator, forming a cation.
Once that cation has formed, regardless of how it got there, it can begin to react with the monomers around it. The chain reaction will continue, consuming more and more monomers, resulting in a longer growing chain with higher molecular weight.
Lewis acid-base adducts always form reversibly, however. At some point, the adduct can give back to the cationic chain end the same anionic group that it once extracted from the initiator. The chain stops growing, because it no longer contains a reactive site. However, just as the Lewis acid once extracted an anionic group from the initiator, it can extract it from this polymer chain. When it does, the cationic chain end will start reacting with more monomers, and the chain will grow again.
Living cationic polymerization depends on an equilibrium between a reactive, growing phase and an unreactive, dormant phase. There is an inherent trade-off here. The concentration of reactive cationic chain ends is lowered because a large fraction of the chains are always dormant. That low concentration of cations means there is less opportunity for chain termination events; that's helpful in keeping dispersity low. On the other hand, that low concentration of cations also means the polymer grows much more slowly than it would otherwise.
This trade-off works to our advantage because of kinetic conditions that hold true throughout most of the duration of the polymerization. Termination steps depend on the collision between two species that are both held at relatively low concentration, whether it is a cationic chain end and an anionic base or two different growing chains. Propagation depends on the concentration of reactive chain ends and the concentration of monomer. Although one of those concentrations is very low, the other one is very high all the way up until nearly the end of the polymerization. That means propagation remains fast relative to termination.
For example, if your target is a polymer of molecular weight 100,000 Da and the monomers have molecular weight of 100 Da, then you would conduct the polymerization with one initiator for every thousand monomers. Each time a monomer was consumed, the relative monomer concentration would change very, very little, and it would remain high enough to keep the polymerization reaction moving forward at a reasonable pace. Only near the end of the polymerization would the monomer concentration drop significantly enough to slow the reaction down drastically.
Exercise 2.5.1
Choose the best Lewis acid from each of the following pairs.
1. NCl3 GaCl3
2. SnCl4 Cl2
3. CCl4 ZnCl2
4. FeCl3 XeCl2
Exercise 2.5.2
Provide a mechanism for this polymerization. Show chain control using equilibriu between a growing phase and a dormant phase.
Exercise 2.5.3
Living cationic polymerizations are sometimes promoted when a weakly basic molecule is added to the reaction mixture. Although different reasons have been proposed for the role of the Lewis base, it is well-established that adding a neutral Lewis base to a Lewis acid can result in a more active, cationic Lewis acid. Show how this stronger Lewis acid forms using triethylamine and gallium chloride. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.05%3A_Living_Cationic_Polymerization.txt |
If you have studied some organic chemistry, then it seems obvious that cationic methods would be employed for the polymerisation of alkenes. After all, alkenes are nucleophilic, and so they should react readily with cationic initiators. It might be surprising that there are a number of alkene polymerisations that work very well using anionic initiators.
One of the most common scenarios for this approach is in the polymerisation of conjugated hydrocarbons, such as styrene. From the point of view of Lewis structures, this seems like a funny combination. There is no obvious electrophile in an alkene; not even in a conjugated one. Why would it interact with an anionic initiator?
A molecular orbital treatment of conjugated systems provides an answer. When alkenes are conjugated, several things happen. There is at least one extended π-bonding orbital that moves to lower energy because of conjugation, and we think of that factor to explain the relative stability of conjugated alkenes toward electrophiles, compared to the reactivity of isolated alkenes. At the same time, there are also π-bonding orbitals that are raised a little in energy as well as corresponding π*- or π-antibonding orbitals that are lowered in energy by conjugation. This feature explains the well-known shift in optical absorption of conjugated alkenes toward the visible region of the spectrum; we see that in brightly coloured compounds such as carotene and xanthophylls in autumn leaves. It is partly that lowering of the Lowest Unnocupied Molecular Orbital (LUMO) that leads to the increased electophilicity of conjugated alkenes compared to regular alkenes.
Once an anionic initiator has added to a conjugated alkene, the conjugated alkene itself becomes an anion. That leads to the second part of the explanation. If you look at the anion that results, you will find resonance stabilisation. That delocalisation is always a powerful factor that allows ions to form more easily. In fact, we can easily choose an anionic initiator that is more reactive than that delocalised one, so that there is a driving force toward polymer initiation. Alkyllithiums, such as butyllithium, are good candidates.
There may be other alkenes that come to mind as being potentially suitable for anionic polymerisation. Again, if you have studied organic chemistry, you might thing of 1,4-addition of nucleophiles to conjugated enones. Anionnic initiation of these kinds of monomers is also possible. Acrylate polymerisation, involving monomers that contain a C=C-CO2 unit, is commercially very important.
The 1,4-addition of anionic initiators does not proceed as smoothly as we might expect, however. This approach is prone to a number of side reactions. For example, Claisen condensations result in the loss of an alkoxide leaving group and chain death.
On the other hand, the presence of α-hydrogens along the polymer can lead to a different problem, with equilibrium proton transfer resulting from "back-biting". This phenomenon leads to uncontrolled branching of the polymer chain.
The potential for side reactions means that anionic polymerisations are prone to poor molecular weight control. Chains that die early will have lower molecular weight than those that continue to grow, leading to broad molecular weight distributions. Like cationic polymerisations, anionic polymerisations can be greatly enhanced through the use of living methods.
Problem SM4.1.
Anionic polymerization of p-substituted styrene proceeds very well when the substituent is an electron-withdrawing group such as nitrile. Explain the reason for the success of this approach.
Problem SM4.2.
Vinyl chloride (CH2=CHCl) might be expected to be a good candidate for anionic polymerization, but side reactions make it unsuitable.
a) Explain why the resulting anion might be expected to show some stability.
b) Show the side reaction that would result under anionic polymerization conditions. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.06%3A_Anionic_Polymerisation.txt |
Unwanted side reactions in anionic polymerization, such as back-biting or Claisen reactions with acrylate chains, lead to early chain death and a broadening of the molecular weight distribution. This problem is intrinsic to polymer growth. Because reactive chain ends are needed to enchain additional monomers, there is always the potential that these relatively high-energy species will go off track and lead to different products.
Living polymerization describes any system in which early chain death is limited, so that polymer chains can continue to grow uniformly. In these systems, the molecular weight increases linearly with the percent conversion of monomer to polymer. In addition, dispersity remains low even at high percent conversion.
The reactive chain ends in anionic polymerizations are nucleophilic carbon anions. If you have studied these kinds of compounds before, the idea of covalency might come to mind. Carbon anions are easier to work with if they are not really anions, but instead share their electrons with their counterions to some degree. So, for example, we might choose to employ lithium counterions with these anionic chain ends, rather than sodium or potassium. The smaller, more electronegative lithium (at least compared to sodium or potassium) can form a polar covalent bond with carbon, stabilizing the nucleophile.
Of course, even an alkyllithium is a strong enough nucleophile to initiate anionic polymerization, provided the resulting anion is more stable than the initial one. In general, it can initiate the formation of growing chains if the resulting anion is delocalized.
We can think of the growing chains as being in equilibrium between having covalent lithium-carbon bonds and forming ion pairs. The ion pair would be more ready to react with the next monomer. That equilibrium could form a basis for a dormant state and a growing state. Just as in living cationic polymerization, the growing state is necessary for polymer chain growth but is susceptible to unwanted side reactions. The dormant state protects the growing chain by limiting the growing chain concentration, consequently limiting the degree of side reactions.
So, just using a lithium counterion, for instance, might be expected to promote living polymerization, keeping dispersity low. For that reason, it may be surprising that one of the strategies used for chain control in anionic polymerizations is to add potassium alkoxides along with the alkyllithium initiator. If lithium bases have greater covalency and offer greater control, why would you add potassium bases?
That question has even more merit if you explore the history of mixed-metal bases. Schlosser's base is a well-precedented example. Typically, it's a mixture of butylllithium and potassium tert-butoxide. Developed by Manfred Schlosser at EPF (ETH) Lausanne in Switzerland, mixtures of alkylithiums and potassium alkoxides form powerful bases capable of deprotonating hydrocarbons such as toluene. The mechanism of achieving such high base strength is believed to involve transfer of an alkyl anion from lithium to potassium. From the point of view of making growing chains more covalent, providing a dormant state, this doesn't seem like a good idea. Nevertheless, it works. How?
One of the other features of these mixtures (Schlosser called them LiCKOR bases, noting the mix of lithium and potassium components) is a high level of aggregation. Aggregates are clusters of molecules that stick together. For Schlosser's base, the simplext aggregate would be one alkylithium molecule bound to one potassium tert-butoxide molecule.
What holds aggregates like this together? The anions can bridge between alkali metals. With the alkoxide ion, that's easy to imagine: the oxygen atom has more than one lone pair, so it can donate one to lithium and one to potassium. It's a little harder to see how the alkyl anion, with only one lone pair, could do that. However, that sort of interaction in which one lone pair is shared between two or more lithium ions, although rare, is pretty well-documented in some alkyllithiums. It's like the alkyl anion has been caught midway between two lithiums, transferring from one to another.
Bigger aggregates could form if additional molecules stuck together. We can easily picture this happening if one alkyllithium combined with two potassium alkoxides.
You can probably imagine even larger aggregates. Maybe two alkoxides come together with one alkyllithium, held together by bridging oxygens. In fact, these structures seem to be very dynamic. They can come apart in solution, and they can come together to make even bigger structures. In reality, a number of different aggregation states will exist in equilibrium with each other, and some might contain eight or twelve alkali cations together with their accompanying anions.
So, what is the role of aggregation in producing a dormant state? It may cap the end of the growing chain temporarily, so that the anionic chain end is less likely to interact with monomers. Reaction would occur only when the aggregate broke up, freeing an anionic chain end.
Aggregate formation can also be promoted by other anions, including simple halides such as chloride and fluoride. As a result, the addition of simple lithium salts can be effectine in promoting living anionic polymerization. The alkoxide base needn't play a role.
An alternative strategy for living anionic polymerization involves the addition of Lewis acidic compounds as chain control agents. In these cases, the equilibrium between dormant and growing chains would involve coordination of the anionic chain end to the Lewis acidic atom. Because Lewis acid-base complexes occur in equilibrium, some fraction of the polymers would always exist in the growing phase, but a greater fraction would always be found in the dormant phase.
Exercise 2.7.1
Rank the following ions in terms of covalency with oxygen (most covalent to least covalent).
1. Na+, Li+, K+
2. Mg2+, Ca2+, Be2+
Exercise 2.7.2
Coordination number can vary with the size of a cation. Rank the following ions from largest to smallest.
1. Na+, Li+, K+
2. Mg2+, Ca2+, Be2+
Exercise 2.7.3
Which compounds would be expected to stabilize growing anionic chains?
1. Et3N or Et3Al
2. Et2Zn or Et2O
3. Ph3B or Ph3N
4. (CH3O)2AlCH3 or (CH3O)2CHCH3 | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.07%3A_Living_Anionic_Polymerization.txt |
On an earlier page, we saw that polyesters and polyamides are typically obtained via condensation polymerization. Ring-opening polymerization (ROP) offers a second approach to these kinds of materials. Using this method, cyclic esters (lactones) or amides (lactams) are opened up to make extended chain structures. The reaction is typically driven by the release of ring strain. For example, caprolactone is ring-opened to produce polycaprolactone.
Caprolactone and other ROP monomers are different from the monomers used in condensation polymerization in a couple of ways. First of all, ROP monomers contain only one functional group, not two. Furthermore, they don't need a complementary monomer. These monomers just react with each other. Also, when they react, they don't form a new functional group, as was sometimes the case in condensation reactions. The monomers were esters and the products are esters, too. This swapping of esters leads to the more specific term "ring-opening trans-esterification polymerization" (ROTEP) when lactone monomers are used.
Let's take a look at how these monomers can be enchained together without the need for complementary monomers. The transformation is instead carried out using a chain reaction. An initiator is needed. In this case, an alkoxide nucleophile can be added to open the first monomer, but when that happens, the substitution reaction at the carboxyloid carbon displaces an alkoxide leaving group. The new alkoxide can then add to the next monomer, propagating the chain reaction.
In this drawing, you can explicitly see the new ester linkage formed between two monomers, replacing the ester linkage in the monomer.
Ring-opening polymerizations like this one are typically quenched with acid to convert the alkoxide nucleophile into a less reactive alcohol, arresting polymerization.
Caprolactone and other cyclic esters can be polymerized via anionic methods, but that isn't the only way. It's very common for these polymerizations to be performed via Lewis acid catalysis instead.
A range of Lewis acids can be used for this purpose, and common examples include iron, zinc, tin and aluminum compounds. This example uses the commercially available tin octoate as a catalyst.
The role of the Lews acid can be twofold. It can accelerate the reaction via electrophilic activation, in which it binds the carbonyl oxygen of the ester, making it more electrophilic.
In addition, when neutral alcohols bind to Lewis acids such as tin octoate, they are more easily deprotonated, becoming better nucleophiles. Both of these effects probably add up to very rapid ring-opening polymerization under these conditions.
Usually, ring-opened polymers grown either through anionic or Lewis acid conditions have a single growing end, but that doesn't have to be the case. If a difunctional initiator is used instead of a simple alcohol, two monomers would be initiated, allowing growth in two different directions. For example, ring-opening polymerization of ethylene oxide is often accomplished with ethylene glycol as an initiator. The ethylene glycol unit becomes enchained in the middle of the polymer chain, rather than at the end.
This type of structure, with two growing ends, is sometimes called a "telechelic" polymer. It can be very useful in the synthesis of triblock copolymers, because a different kind of monomer can be added to both ends of an existing polymer, leading to an ABA structure.
Problem SM6.1.
Some catalysts for ring-opening polymerization contain both a Lewis acid and a nucleophile for initiation of chain growth. Identify these two components in the following catalysts.
Problem SM6.2.
Some organocatalysts promote ring-opening polymerization of cyclic esters through nucleophilic catalysis, acting as an initial nucleophile. Show the mechanism for the following reaction.
Problem SM6.3.
Some organocatalysts can activate both the nucleophile and the electrophile, and also use a catalytic approximation strategy. Provide a structure that illustrates these features in the following system. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.08%3A_Ring-Opening_Polymerization.txt |
Radical polymerization is a very common approach to making polymers. There are a number of very reliable methods of carrying out radical polymerization, leading to high molecular weight materials. A typical example is the formation of polystyrene under the influence of the radical initiator, azoisobutyronitrile (AIBN).
Radical reactions start with an initiation reaction, in which radicals first come into being. When AIBN is used as an initiator, the compound decomposes partly because of the strong N-N triple bond that is formed, and partly because of the relatively stable radical that results. This radical is both tertiary and delocalized.
Other initiators can also be used in radical polymerizations. Benzoyl peroxide (BPO) is one common example. In this case, the initiation reaction is sparked by the weak O-O single bond. The benzoate radical that forms from that homolysis further decomposes via decarboxylation, rapidly giving rise to a phenyl radical.
Phenyl radicals and isobutyronitrile radicals are both able to initiate the polymerization of styrene. Not just any radical will peform that task. Simple peroxides are not very good at initiating radical polymerization, despite the fact that they also contain O-O bonds that are readily cleaved.
Oxygen-centered radicals do not typically add to double bonds. It's much more common to see them abstract hydrogen atoms. Accordingly, tert-butyl peroxides and other such peroxides would not be used to initiate the polymerization of styrene.
The radical formed via initiation induces cleavage of the pi bond in styrene, resulting in a benzylic radical. This is an example of a propagation step, because old radicals have disappeared and new radicals have appeared.
Note that this step takes place with regioselectivity. There are two possible radicals that could result, but only one actually forms. That's the more stable one. It is stabilized by delocalization because of the benzene ring.
Once that benzylic radical forms, it can continue to propagate, enchaining more styrene molecules. Each time a new styrene molecule is enchained, it forms a new benzylic radical.
Eventually, radical reactions undergo termination, in which two radicals combine in some way so that neither one is a radical anymore. The most obvious way for that to happen is if two growing chains meet head-to-head, with two benzylic radicals connecting to make a bond.
Other termination events are also prevalent in radical polymerizations. Hydrogen atom abstraction is pretty common, with one growing radical stealing a hydrogen atom alpha to the radilca at the head of another chain. The reason that may happen is that it results in the formation of two new bonds: a C-H bond and a C-C pi bond. In that case, both chains stop growing. The one that contains the new pi bond could eventually be re-initiated, of course, but in the meantime it will lag behind most of the other chains that will continue to grow without it. That lag will lead to a broadening molecular weight disctribution.
Partly because of the immense size of the molecules involved, hydrogen atom abstraction need to necessarily take place alpha to another radical. One growing chain may pass another in such a way that its benzylic radical grazes a hydrogen atom along the backbone of another. Hydrogen atom abstraction may occur more randomly in this case, although it may still result in a new benzylic radical.
In this case, one chain becomes dead; it stops growing altogether, just like one of the chains in the previous case. However, the other chain continues to grow. In fact, with two radical sites, it will begin to grow twice as quickly as the rest of thegrowing chains. Once again, this development leads to a dramatic broadening of the molecular weight distribution. What's more, this "back-biting" event leads to a change in morphology of the growing chain. Because it is now growing from two different sites, this chain becomes branched, unlike all of the other chains that are growing in a linear fashion. Not only do we now have chains of vastly different lengths, but we also have chains of vastly different shapes, and their properties are becoming less well-defined.
These termination steps are important because they result in the death of growing chains and therefore have an impact on the polymers that result. In later sections, we will see some strategies that limit these termination steps, leading to more uniform molecular weight distribution and more reliable polymer properties.
So far, we have focused on polystyrene as an example of radical polymerization, because of the obvious stability of the benzylic radical. Other monomers can also be polymerized under similar conditions. Like styrene, they may give rise to resonance-delocalized radicals. In other cases, the radical stability may be less obvious, such as in the case of vinyl chloride. Vinyl chloride (CH2=CHCl) is a pretty candidate for radical polymerization because radicals are stabilized by adjacent halogen atoms.
That may come as a surprise. This stability results from a hyperconjugation effect, in which overlap between the radical electron and a nonbonding pair on the halogen results in a net stabilization. This idea is similar to the one behind the stability of highly substituted radicals, such as radicals on tertiary carbons.
Problem SM7.1.
Show the polymers that would result from radical polymerization of the following monomers.
Problem SM7.2.
Show why radicals formed from the following monomers are relatively stable:
a) acrylonitrile, CH2=CHCN
b) methyl acrylate, CH2=CHCO2Me
Problem SM7.3.
Apart from resonance delocalization, degree of substitution is also a modest stabilizing factor for radicals. Indicate the order of stability of the following radicals: | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.09%3A_Radical_Polymerization.txt |
The termination of radical polymerizations can lead to broadening molecular weight distributions and divergent morphology, with branched structures eventually adding in to the mixture of linear chains that were already growing. There are several methods commonly used to prevent these random terminations. All of them work by keeping the concentration of growing chains low. That strategy limits the chance that radical chain ends encounter other chains and undergo termination reactions.
The underlying argument for this strategy is that the kinetics of chain growth and chain termination steps depend differently on the concentration of growing chains. A propagation step requires only one growing chain to encounter a highly abundant monomer molecule. A termination typically requires two growing chains to encounter each other. Consequently,
Rate propagation = k [polymer.][monomer]
Rate termination = k [polymer.][polymer.] = [polymer.]2
There is a trade-off. Limiting the concentration of growing chains results in slower polymer growth. At the same time, it results in dramatically slower rates of chain termination.
Reversible addition-fragmentation chain transfer (RAFT) polymerization is a common strategy used to limit the concentration of polymer chains at any given time, while ensuring a steady supply of these growing chains in the long run. The strategy has been described as putting growing chains in suspended animation, waking them up periodically to continue polymerization. The key component of this strategy is the chain transfer agent, sometimes called a chain control agent. Usually, these chain transfer agents are thiocarbonylthio compounds like the one shown below.
Note that the conditions are otherwise very similar to a regular radical polymerization. The process starts with initiation of radicals, in this case through decomposition of the AIBN.
The isobutyronitrile radical then propagates the radical to the first styrene molecule. So far, nothing different has happened compared to a regular radical polymerization. The resulting benzylic radical propagates with additional styrene monomers, one at a time, enchaining them into the polymer.
After a few of those propagation events, the growing chain encounters a molecule of the chain transfer agent. Insetad of adding to the pi bond of the styrene, the radical adds to the pi bond of the thiocarbonyl. The new radical is centered on the carbon between the sulfur atoms. If you have encountered sulfur and phosphorus chemistry before, you will know that these large main group atoms have a knack for stabilizing species next to them that would otherwise be very reactive. That's true with radicals, too. The result is that the radical has found a stable resting state. While it sits there, no polymerization can occur, but no terminations will happen, either.
This step happens reversibly, so the growing chain can be released again. It is useful to think of this equilibrium as partitioning the polymers between a growing phase, in which chains can grow but they may also die, and a dormant phase, in which chains are safely sleeping but will not grow.
The structure of these particular chain transfer agents leads to a second possible equilibrium. Rather than releasing the original growing chain from the dormant phase, a radical could be released by cleaving the other sulfur-carbon bond, forming a new thiocarbonyl. The radical that is released is then free to propagate, forming a new polymer chain.
As a result, we can think of this chain control agent as participating in multiple equilibria, with two chains that can each be either dormant or growing. The quilibrium constants are typically chosen so that the chains spend more of their time in a dromant state, but at any given time one of them may be released and start growing again.
As a result, the rate of termination is cut down drastically. Molecular weight distribution (or dispersity) is much narrower and consequently the polymer has much more well-defined properties. Furthermore, the low chance of termination means that the polymer chain can continue to grow until it reaches very high molecular weight.
Problem SM8.1.
Relative rates are very important in controlling polymerization via RAFT. Rationalize the relative radical polymerization rates for the monomers below.
Problem SM8.2.
Relative rates of fragmentation of the chain control agent are important in releasing the growing phase during polymerization. Rationalize the relative fragmentation rates for the chain control agents below.
Problem SM8.3.
Relative rates of addition of radical to the chain control agent are important in controlling the concentration of the growing phase during polymerization. Rationalize the relative addition rates for the chain control agents below.
Problem SM8.4.
Effective chain control by RAFT requires selective matching between monomer and chain control agent so that there is an appreciable equilibrium between growing phase and dormant phase. Explain the choice of reagents here. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.10%3A_Living_Radical_Polymerization-_RAFT.txt |
SM9. Living Radical Polymerization: ATRP
Atom transfer radical polymerization (ATRP) is another common method of carrying out living polymerization. This method, like RAFT and other methods, sets up an equilibrium between a dormant phase of polymer chains and a growing phase. In the dormant phase, radicals have been trapped and are protected from termination, but are also unable to participate in polymerization. In the growing phase, radical chains have been released and can continue to grow by adding monomers in a radical mechanism. The idea is to release just a few radical chains at a time, recapturing them before they have a chance to undergo chain termination events.
ATRP is initiated in a very different way from other radical polymerizations. It does not depend on a typical initiator such as AIBN or benzoyl peroxide. Instead, the initiator is an alkyl halide. On its own, the alkyl halide seems more suited for polar reactions such as nucleophilic substitutions than radical reactions. The second component of the chain control agent, usually a copper(I) halide, is required in order to activate the initiator, turning it into a radical. Polymerization ensues.
In the initiation step, a halogen atom is transferred from the alkyl halide to the copper(I) ion, leaving behind a radical. In formal terms, the copper has become copper(II), so mechanistically this step appears to involve an electron transfer.
Because the halogen atom on the alkyl halide has lone pairs, it is likely that the electron transfer takes place via an inner-sphere mechanism after the alkyl halide has coordinated to the copper. However, if something prevents the alkyl halide from coordinating to the copper, an outer-sphere electron transfer would still be possible.
There is always an additional component to the chain control system. The copper salts themselves are pretty insoluble. A chelating ligand is added to bind the copper ion and help get it into solution, where it can much more readily participate in the polymerization reaction. A few examples of these ligands are shown below.
Problem SM9.1.
Show CuBr binding with each of the ligands shown above.
Problem SM9.2.
Show a mechanism of initiator activation via
a) outer-sphere electron transfer
b) inner-sphere electron transfer
Once the initiator has been activated, it can add to the first styrene molecule. The radical continues to propagate, building up a long polymer chain. In the final polymer, both end groups come from the initiator.
Chain control once again relies on an equilibrium between a dormant phase and a growing phase. In the dormant phase, the radical chain end has been capped with a bromine atom. In the growing phase, the radical chain end is free and able to add to additional monomers.
One problem that can arise in living polymerizations is that at some point a termination may still occur. Of course, that results in a dead chain that will no longer grow. An additional consequence is that the chain can no longer go back into the dormant phase, and if the chain cannot go back, then neither can the copper(II).
Thus there is a buildup of copper(II) in the system.
What happens when there is a buildup of copper(II)? The job of copper(II) is to deactivate the radical, sending it back into the dormant phase. If copper(II) starts building up, the equilibrium shifts, driving all of the polymer chains into the dormant phase. Polymerization grinds to a stop.
There are a couple of clever solutions to this problem. Perhaps the simplest method uses copper metal to reduce the copper(II) to copper(I). This method has been coined "supplemental activation reducing agent" or SARA.
Problem SM9.3.
In addition to SARA, there are other methods of converting deactivated copper back into its active form in order to shift chains back into the growing phase. Initiators for continuous activator regeneration (ICAR) uses organic reducing agents to convert CuII back into CuI. Provide a mechanism for this reaction using phenylhydrazine (PhNHNH2) as the reducing agent.
Problem SM9.4.
Activators regenerated by electron transfer (ARGET) is a similar method to SARA and ICAR, but relies on the addition of the thermal initiator, AIBN, to the polymerization.
a) Provide a mechanism for how AIBN regenerates CuI ion.
b) Explain the disadvantageous side reaction that can result under these conditions. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/02%3A_Synthetic_Methods_in_Polymer_Chemistry/2.11%3A_Living_Radical_Polymerization-_ATRP.txt |
KP1. Thermodynamics of Polymerization
Polymerization is the process of taking individual monomers and enchaining them into a macromolecule. Thermodynamically, there are always two factors to consider in a reaction: enthalpic contributions and entropic contributions.
ΔG = ΔH - TΔS
Entropic changes in chain polymerizations tend to be qualitatively similar in different cases. Because a large number of monomers are being enchained together in a macromolecule, the number of degrees of freedom is always decreasing, regardless of the specific polymerization process under consideration. Consequently, the entropy change associated with polymerization is always negative.
ΔS = kBln(Wf/Wi)
in which kB is Boltzmann's constant and W is the number of microstates available in the final or initial state. If Wf < Wi, then ΔS < 0 because the natural log of a fraction is a negative number.
Free energy correlates positively with enthalpy and negatively with entropy. That means that a negative entropy change translates into a positive, and unfavorable, free energy change. Polymerization is always entropically disfavored, regardless of the circumstances. However, because entropy changes are generally rather small compared to enthalpy changes, polymerization proceeds despite this handicap. The entropy factor begins to dominate the equation only at higher temperatures, because the entropy contribution scales with temperature (in Kelvin) in the expression for free energy.
As a result, thermodynamically, chain polymerizations should be run under cooler conditions in order to maximize the degree of polymerization. Nevertheless, there are certain polymerizations (such as ring-opening polymerization of cyclic esters) that are frequently run at elevated temperatures because of a significant barrier to the reaction. Of course, in these cases there tends to be a maximum extent of polymerization before the process bottoms out in an equilibrium between polymer and monomer.
In general, this entropic problem in entropy gives rise to the term "ceiling temperature", which is the temperature above which polymerization of a particular monomer will not occur.
In condensation polymerizations, the entropy of reaction is more subtle. The formation of condensation side products, such as water or methanol, means that there are not overall changes in the number of molecules before and after polymerization. For example, n co-monomers are converted into 1 polymer chain and n-1 water molecules. There may still be entropy changes in the reaction, but the sign of the entropy change is not as easy to predict compared to chain polymerization.
Since the entropy for chain polymerizations is always positive, then all chain polymerizations must be exothermic, because otherwise they would never proceed. The reasons for these enthalpically-favored reactions varies from one case to another.
In olefin polymerization (alkene polymerization), the enthalpy of reaction has been measured in a wide range of cases; it is usually around -20 kcal/mol. This negative ΔH value can be partly attributed to the exchange of one C=C π bond for one C-C σ bond for every alkene monomer enchained. The strength of an average C=C double bond is approximately 147 kcal/mol, whereas the strength of an average C-C bond is about 83 kcal/mol. By difference, the strength of a C=C π bond is about 64 kcal/mol. Thus, it costs 64 kcal/mol to break the π bond but we drop 83 kcal/mol for each σ bond made, for a net decrease of around 20 kcal/mol.
Bond Bond Energy (kcal/mol) Bond Bond Energy (kcal/mol) Bond Bond Energy (kcal/mol)
C-C 83 C-N 73 C=C 147
C-O 86 C-Cl 81 C=O 178
N-H 93 Cl-H 102 O-H 111
To be sure, there are other factors that contribute to the enthalpy change. C-H bond strengths change significantly with hybridization changes. Intramolecular steric interactions are drastically different in standard monomers compared to polymers. In alkene polymerization in particular, phase changes between gas phase or liquid phase starting material and solid product is appreciable, although still small compared to the bond strength issues. Nevertheless, at first glance we would expect a moderately exothermic reaction, and that's what has been measured experimentally.
Problem KP1.1.
Diels Alder polymerization depends on a [4+2] cycloaddition. Thermodynamic studies on the parent reaction indicate ΔH = -40 kcal/mol. Explain this observation in terms of bond strengths.
Taking a second example, the ring-opening polymerization of cyclic esters is considered to be exothermic because of ring strain. Values of ΔH are strongly dependent upon ring size, which is intimately connected with ring strain. For γ-butyrolactone, a five-membered ring containing a carbonyl, ΔH = -1 kcal/mol; that's not enough to compensate for a negative entropy of reaction, so that this monomer can only be polymerized at low temperatures. On the other hand, for δ-valerolactone, ΔH = -3 kcal/mol, whereas for ε-caprolactone, ΔH = -7 kcal/mol. These six- and seven-membered lactones are much more readily polymerized because of increasing ring strain.
Problem KP1.2.
Ceiling temperature can be calculated from an assumption that ΔG = 0, at which point polymerization is readily reversible. A kinetic approach to equilibrium gives the relation:
Tc = ΔH / (ΔS + Rlog[M])
in which R is the ideal gas constant, 1.98 cal K-1 mol-1 (or 8.314 J K-1 mol-1)
Calculate the ceiling temperature for caprolactone polymerization if ΔH = -7 kcal mol-1 and ΔS = -8.6 cal K-1 mol-1. The concentration of neat caprolactone is 8.7 M.
(Data from: Olsen, P.; Odelius, K.; Albertsson, A.-C. Biomacromolecules 2016, 17, 699-719.)
Given a less pronounced entropic bias against condensation polymerization, endothermicity becomes less crucial. Indeed, some condensation polymerizations are effected through simple trans-esterifications, which ought to be close to thermoneutral. As pointed out in the condensation section, these reactions are frequently driven to completion by removal of the condensation side product, small molecules such as methanol or water.
Problem KP1.3.
Some polyamides are made industrially from the corresponding difunctional amines and acid chlorides, releasing HCl. Calculate the enthalpy change for this reaction. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/03%3A_Kinetics_and_Thermodynamics_of_Polymerization/3.01%3A_Thermodynamics_of_Polymerization.txt |
KP2. Kinetics of Step-Growth Polymerization
It is important to understand how reactions proceed over time. This information can tell us how long it will take for a polymer to reach an optimum length. It can also provide insight into how the polymerization occurs, just as kinetics can provide insight into other reaction mechanisms.
Condensation polymers grow through a a step-growth process. The reaction generally involves substitution at a carboxyloid. For example, polyester formation often involves the substitution of an alcohol functional group at a carboxylic acid, making an ester. The drawing below illustrates that action using ethylene glycol and oxalic acid, to use common names prevalent in industry. Oxalic acid is not typically used in making polyesters, but it is drawn here as a very simple example structure.
If you know some basic kinetics, you know that we can express the rate of a reaction as the rate of change of product concentration per change in time. In this case, the product is the polymer, abbreviated P. Alternatively, we could express the rate of the reaction as the rate of change of monomer concentration per change in time. Just by looking at the equation of reaction, we expect that every time a polymer molecule is made, a monomer molecule is used up. In this case, we have a co-polymer, made of two different monomers, so that's why we could express the reaction rate as the rate of change of either monomer concentration. The negative sign in the rate suggests we are looking at the rate of disappearance of monomer; as polymer appears, monomer disappears.
If you know something about organic reaction mechanisms, you may know that the very first step in an esterification is the donation of a lone pair from an alcohol oxygen on the diol to a carbonyl carbon on the dicarboxylic acid. (It could actually be the second step, but we'll get to that later.) That first step in the mechanism is an elementary reaction. All of the other steps you would draw in a mechanism are also elementary reactions. The nice thing about elementary reactions is that we always know the rate law of each one: it's a rate constant times the concentrations of all of the species leading into that step. In this case, this first step is also the rate-determining step, so the entire reaction is governed by the rate law of that first step. It requires both monomers, so we get a simple rate law.
Note that, based on the stoichiometry (the ratio of monomers in the equation of reaction), we would probably run this polymerization with equal amounts of diol and dicarboxylic acid. Instead of worrying about M1 and M2 separately, we can call them both [M].
So, what we have here is a simple second-order rate law. This form of the rate law is called the differential rate law, because it is expressed in terms of the rate of change of one thing with the rate of change of the other. If that isn't exactly what you are interested in, and you just want to see what happens to concentrations over time, you may want to integrate this expression to obtain the integrated from of the rate law.
If we were to measure monomer concentration periodically during the polymerization and plot its inverse over time, we would obtain a straight line. The slope of the line would be the rate constant, and the intercept would reflect the inverse of the monomer concentration with which we had started the reaction.
At this point, we need to pause and talk about what we actually mean by monomer concentration. Because of a very specific feature of condensation polymerizations, it isn't really the concentration of ethylene glycol or oxalic acid. Take a look at this drawing of this step-growth process and we will talk about why that is.
Remember that, in step-growth polymerization, the majority of molecules are probably reacting at roughly similar rates. Two monomers link together to form dimers, two dimers form tetramers, two tetramers form octamers, and so on. It won't go as perfectly as we see in this picture; there will inevitably be some stragglers that throw the averages off. But notice what this means: the monomers disappear right away. Nobody is really interested in how quickly the monomers turn into dimers. They want to know how the gowth of the polymer chain is progressing.
So, in the above expression, [M] doesn't refer to the concentrations of those two monomers at all. It refers to the concentration of the reactive functional groups in those monomers. After all, once the monomers are gone, the same esterification reaction keeps happening, and it's really the concentration of alcohol and carboxylic acid groups that matters, whether they are found on monomers, dimers, tertramers, or whatever. In these rate expressions, sometimes people will explicitly use [OH] and [CO2H] in the rate expression, underscoring that they are really following the disappearance of those functional groups.
And notice what's happening to those functional groups. Their concentration is smoothly dropping over time. Furthermore, ester concentration, representing the new functional group in the polymer, is growing in at the same rate. [OH] and [CO2H] drop by 8 (or in half) in the first step; [CO2R] increases by 8, and so on. SO the rate of consumption of reactant functional groups equals the rate of appearance of product functional groups.
The other thing we are interested in here is the degree of polymerization: the number of monomers incorporated into the chain. The polymer is growing at a certain rate, but how long is it getting? Just looking at the simple scheme of polymer growth above, we see the average degree of polymerization increasing (written as DP or sometimes as Xn; a bar over the symbol signals that we are looking at an average). If you look carefully, you will see that the degree of polymerization is always the original monomer concentration divided by the remaining monomer concentration. That turns out to be a very important relationship.
Note that this relationship is just the inverse of the fraction of monomer remaining (or, really, the fraction of those monomer functional groups left over).
The opposite quantity is the fraction of monomer converted. That describes the functional groups already converted into esters. Because we are dealing with fractions, the fraction converted is just one minus the fraction remaining. If we do a little algebra, we get a variation on that piece of information.
Why did we do that last part? Because now we have arrived back at the expression for degree of polymerization. So, now we have an equivalent way of expressing degree of polymerization in terms of the fraction converted. This relationship is called the Carothers equation, in honor of Wallace Carothers, the DuPont chemist who invented nylon.
Problem KP2.1.
Predict the degree of polymerization if 75% of monomer end groups have been converted to polymer.
So, what does that relationship look like? Here's some data from Paul Flory, another DuPont chemist and Nobel laureate, obtained during the period between the World Wars when this kind of chemistry was just getting started (Flory, P. J. J. Am. Chem. Soc. 1937, 59, 466-470).
That's fraction converted on the x axis and the expression for degree of polymerization on the y axis. This is exponential growth. The degree of polymerization is starting to shoot upward dramatically. We already knew that from our introduction to step growth, but what stands out from the data is how a higher degree of polymerization, and high molecular weight, doesn't really occur until very late in the reaction. If we need high molecular weight condensation polymers, the chemistry better be pretty reliable; otherwise, if things go wrong, we'll be left with some short-chain goo instead of long-chain materials.
Now let's take these new relationships and look at them in the context of our second-order kinetics. We're starting with that same integrated rate law, but we have mulitplied everything on both sides of the equation by the starting monomer concentration. That's fair game in algebra, remember. The reason for doing that is to get a term that resembles our relationships that describe degree of polymerization.
So far, we have taken the ratio we obtained in the first line and worked out its relationship to fraction converted, which we substituted in the last line. Now we bring in the Carothers equation to recall that this is really telling us how degree of polymerization changes over time. This relationship indicates the degree of polymerization should increase linearly over time.
This relationship indicates the degree of polymerization should increase linearly over time. Once again, we're going to use Flory's data to illustrate the relationship, using the fraction conversion data. He could measure that data easily, and knew how it was related to degree of polymerization.
It was supposed to be linear. But don't worry; Flory knew what went wrong. You might know, too, if you remember a few things about substitution at carboxloids. The trouble is, we are dealing with a neutral nucleophile and a fairly unreactive electrophile. That reaction should be really slow. It isn't that slow, thoough, because it is self-catalyzed. One carboxylic acid monomer can activate another, making it a better electrophile.
That means the coupling of the alcohol with the carboxylic acid isn't the first elementary step of the reaction, but the second. Before the alcohol and carboxylic acid can come together, two carboxylic acids need to have collided already. That means we have a third order reaction. As it happens, the integrated form of a third order rate law isn't all that much more complicated, and we can do some similar algebra with it to see how the degree of polymerization would vary with time given this new information.
We can take Flory's data again and give it this new treatment.
This time, we get a linear relationship. That correction illustrates one of the fundamental uses of kinetics, which is a tool that lets us probe how reactions happen. In this very specific case, kinetics showed that there was an extra (pre-rate determining) step in the reaction. In addition, we found that it is the square of the degree of polymerization that is increasing linearly with time.
Problem KP2.2.
Given an initial alcohol group concentration [M]0 = 17 M, determine the rate constant from Flory's data in L2mol-2s-1.
Flory also established a statistical basis for predicting molecular weights and molecular weight distributions in polymers. For example, once we have the degree of polymerization, it is a simple step to calculate the approximate molecular weight. That's just the degree of polymerization times the molecular weight of the monomer; in other words, the number of monomers in the chain times the weight of each one. (This approach neglects any small molecules such as water that are lost during the condensation reaction.)
Mn = DP x M0
Mn = M0 / (1 - p)
This is called the number average molecular weight. It's based on taking one average value for the chain length and converting that into the chain molecular weight. Of course, some chains may be much shorter (Flory actually shows, based on probability, that there will be a number of monomers left over) and as such they contribute less to the overall sample.
A second way to approach the average chain molecular weight is to assess the fraction of the total weight contributed by each chain length. The average chain molecular weight would then be given by:
Mw = Σ πxxM0
in which πx is the fraction of chains of length x.
What is the fraction of chains of length x? The approach starts with considering the probability that a chain will be x units long. As noted above, a chain x units long has x-1 ester linkages and 2 unreacted chain ends. We know the probability of a chain end being converted to a polymer linkage is p, and in a chain x units long that will have to happen x-1 times, for a probability px-1. We also know that the probability of a functional group being left unreacted is 1-p, and that has to happen twice here, for a probability (1-p)2. The probability of having a chain of length x is:
πx = x px-1(1-p)2
That means:
Mw = Σ x px-1(1-p)2 xM0
Mw = (1-p)2M0 Σ x2 px-1
The series on the right is reducible to:
Σ x2 px-1 = (1 + p) / (1 - p)3
So that:
Mw = (1-p)2M0(1 + p) / (1 - p)3
Mw = M0(1 + p) / (1 - p)
That provides a straightforward prediction of the weight-average molecular weight based on the fraction converted in condensation polymerization. Although Mn and Mw presented here are essentially theoretical predictions of molecular weight, these quantities can also be measured experimentally using different methods.
The ratio between these two quantities has always been used as an expression of the breadth of the molecular weight distribution, called the polydispersity index or simply the dispersity:
D = Mw/Mn
In the case of condensation polymerization,
D = [M0(1 + p) / (1 - p)]/[M0 / (1 - p)]
D = 1 + p
Because p is always a fraction, the result is that the dispersity of a condensation polymerization is statistically expected to be less than 2.0.
Problem KP2.3.
Given 99% conversion of monomers with average molecular weight 120 g/mol, calculate Mw, Mn and D. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/03%3A_Kinetics_and_Thermodynamics_of_Polymerization/3.02%3A_Kinetics_of_Step-Growth_Polymerization.txt |
KP3. Kinetics of Chain Polymerization
The polymerization of alkenes occurs in a very different way than monomers that undergo condensation reactions. Whether it occurs through an anionic, cationic, or radical mechanism, polymerization of alkenes involves a chain reaction. Chain reactions are classically illustrated by radical chemistry, so we will look at a treatment of radical chain polymerization to see some of the factors that influence polymerization rates.
A typical radical polymerization starts with the thermal decomposition of a radical initiator to provide two radicals. The rate of decomposition depends only on the decomposition rate constant and the concentration of the initiator.
Once the radicals have been generated, they are able to undergo radical addition to a monomer double bond. Of course, radical addition to a double bond results in a new radical where the double bond used to be. Although this is formally a radical propagation step, in polymer chemistry it is termed the initiation step, because it is the first time a monomer has undergone radical addition. This step consumes the first monomer and produces a new radical species which will become the growing radical chain. It requires a collision between the radical and a monomer, so the rate of initiation depends on those two concentrations and the chain initiation rate constant, ki.
That rate law depends on a reactive intermediate. It's not a very helpful rate law, because the reactive intermediate isn't something that we have directly measured out and added to the reaction, and it might not even occur at high enough levels that we can measure its concentration as the reaction progresses. We usually look for ways to express the rate law in ways that do not include reactive intermediates. In this particular situation, the way of getting around this situation is to assume that the decomposition of the initiator is the rate determining step. Making the radicals in the first place is probably the slow part because it is heavily dependent on bond breaking, which is energy intensive. Once we have radical, it probably undergoes addition to a monomer fairly quickly, initiating chain growth. If that's true, we can assume that the chain initiation step proceeds very quickly afterward, so that the rate really depends only on the rate of the decomposition step.
In practice, polymer chemists add another factor to the rate law. This factor, f, takes into account the fact that only some of the radicals from the initiator actually react with monomers to initiate growing chains. The rest decay through some other side reactions. Usually, f is assumed to be around 0.5.
Once the first monomer has been initiated into a radical, it can react with another monomer to enchain it and make a new radical. This is the principal propagation step of the chain reaction. That step will keep repeating, adding more monomers into the chain. The rate constant for this step, kprop or kp, is identical no matter how many monomers have been enchained, but is distinct from ki because of the different nature of the radical intermediates in the two different steps. For example, the radical obtained from addition to styrene looks pretty similar whether it is the first in the chain or the tenth. Each time, the rate of consumption of monomer depends on the propagation rate constant, the concentration of the monomer, and the concentration of the propagating radical. Notice that in the rate law we just write that propagating radical the same way (M dot) no matter how long the chain grows.
There is one last process, or group of processes, to complete the chain reaction cycle. In termination, two radicals combine in some way to form closed-shell products. There are a variety of ways that can happen in a radical polymerization. The simplest event conceptually is coupling, in which two radical chains come together and form a bond. That idea is shown below.
Of course, that rate depends on the termination rate constant and the concentrations of both growing chains. The same would be true if the reaction terminated by conproportionation, in which one radical abstracted an alpha hydrogen from the other, forming one saturated and one unsaturated product.
Once again, these last two rates -- of propagation and of termination -- depend on concentrations of reactive intermediates, which we do not typically know. This time we will use a very standard assumption, which is that the concentration of this reactive species remains constant, being consumed as soon as it is generated. The usual way that we apply the steady state approximation is to assume zero change in concentration of the reactive intermediate. That means that the sum of all the rates for processes generating the intermediate equal the sum of all the rates consuming the intermediate. In polymer chemistry, we take a slight shortcut, and just assume that the rate of appearance of the radical in the first place equals its rate of disappearance. We already have expressions for both of those rates.
By rearranging, we can get an expression for the reactive chain end concentration. Then we can just substitute the result into our expression for propagation rate:
The result sums up the factors that control the growth rate for the polymer. The growth rate increases linearly with the concentration of monomer, and as the square root of the initiator concentration. The rest of the factors are just constants, so we can think of the rate law as one combined constant and those two concentration dependences. Sometimes an aggregate constant of this type is just designated k'.
Note that the propagation rate varies linearly with the concentration of growing chains, but that the termination rate varies with square of the concentration of growing chains. That difference is significant, and it underlies the strategy of living polymerization, which is to keep the concentration of growing chains low. Doing so has the adverse affect of slowing the rate of propagation, but the effect on the termination rate is much greater, making this trade-off worth it to achieve enhanced molecular weight control.
Problem KP3.1.
Suppose the following data were obtained for a radical polymerization conducted at various starting concentrations of monomer. If all of the data were conducted at an initiator concentration of 0.25 mmol L-1, calculate k'.
Kinetics can also be used to establish a theoretical polymer chain length. This quantity is called kinetic chain length, represented by "vee bar". In principle, it's just the ratio of the chain propagation rate to the chain initiation rate. That should tell us the number of monomers per each chain initiated. Because of our earlier assumption about the initiation rate equalling the termination rate, we can subsitute the terms for those two rates and simplify the result.
Substituting our earlier expression for the concentration of the growing radical chain ends gives a new expression.
That means that we can theoretically control the length of polymer chains, because the average length should vary linearly with the concentration of monomer and inversely with the square root of initiator concentration.
As chain reactions, cationic and anionic polymerizations have some characteristics in common with radical polymerizations. One key difference is that the initiator does not fragment into two pieces, each capable of initiating a chain; instead, one added initiator starts polymerization of one chain, without the prior decompositions step seen in radical reactions. Also, termination steps in radical reactions are dominated by radical recombinations (as well as other variations), but in anionic and cationic terminations it is more likely that the growing chain encounters an impurity and reacts with that.
Problem KP3.2.
Given that for a cationic polymerization:
Rateinit = ki [M][I] and Rateterm = kt[M+]
Provide a steady state expression for the concentration of growing cationic chain.
Problem KP3.3.
If the propagation step in a cationic polymerization is written as follows:
M+ + M → M-M+
Provide the rate expression for cationic polymerization.
Problem KP3.4.
By analogy to radical polymerization, the kinetic chain length in cationic polymerization can be expressed as the ratio of propagation rate to initiation rate. Provide an expression in terms of constants and known concentrations.
Problem KP3.5.
In a living polymerization, chain length simplifies to [M]0/[I]0 because all initiation leads to polymerization and there are no terminations steps. Given a radical polymerization with [M]0 = 4.5 M and [I]0 = 1.25 mM, compare chain length in the following situations:
a) a living polymerization
b) a non-living polymerization with f = 0.5; kp = kt = 0.003 s-1; kd = 0.0001 s-1.
c) a non-living polymerization with f = 0.5; kp = 0.003 s-1; kt = 0.3 s-1; kd = 0.0001 s-1.
d) a non-living polymerization with f = 0.5; kp = 0.003 s-1; kt = 0.003 s-1; kd = 0.1 s-1. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/03%3A_Kinetics_and_Thermodynamics_of_Polymerization/3.03%3A_Kinetics_of_Chain_Polymerization.txt |
KP4. Kinetics of Catalytic Polymerization
The Ziegler-Natta polymerization of alkenes is conducted under catalytic conditions. In most cases, the process involves heterogeneous catalysis, in which the reaction takes place on the surface of a solid. The kinetics of catalyzed reactions have some features that are different from other reactions and that are worth exploring.
The fact that the reaction is taking place on the surface of a solid is a key feature that must be stressed in the kinetics of heterogeneous catalysis. The treatment of rates therefore uses an approach developed by Irving Langmuir, a long-time scientist at General Electric who was awarded the Nobel Prize for the study of surfaces. One of the things that makes metal catalysts so useful is their ability to adsorb molecules on their surfaces (think of the precious metals used in catalytic hydrogenation, adsorbing alkenes and hydrogen; or the iron in the Haber-Bosch synthesis of ammonia, adsorbing dinitrogen and hydrogen, to cite two important examples). "Adsorption" refers to the adhesion of molecules to a surface. Langmuir thought of this adsorption process as a dynamic one, with molecules landing on and sticking to open spots on a surface even as other molecules lifted off to create vacancies.
Desorption, on the other hand, refers to the process of molecules leaving the surface.
In terms of kinetics, the rate of desorption of a molecule would depend on some rate constant, kd, and the fraction of the surface covered by these molecules, θ (that's the Greek letter, theta). The greater the surface fraction covered by molecules, the greater the chance that you will encounter one desorbing.
Rated = kd θ
On the other hand, the rate os adsorption of a molecule onto a surface will depend on some rate constant ka, the concentration of the molecule to be adsorbed, and the fraction of the surface still available. That last part is 1 - θ, because the fraction covered plus the fraction uncovered would equal the whole. Note that Langmuir was interested in gas phase molecules adsorbing onto a surface, and so he expressed things in terms of pressure rather than concentration.
Ratea = ka (1 - θ) [M]
At equilibrium, these two rates will equal each other:
kd θ = ka [M] - ka θ [M]
θ (kd + ka [M]) = ka [M]
θ = ka [M] / (kd + ka [M])
Now we have an expression for the fraction of the surface covered by the substrate. This term is called the "Langmuir isotherm" and it shows up in various surface and catalytic phenomena. It is usually expressed slightly differently, in terms of an equilibrium constant for adsorption:
The rate of a reaction catalyzed on that surface will depend on the catalyst concentration as well as a rate constant for enchainment, kprop or simply kp, and the amount of surface covered by monomer (unbound monomer will not undergo propagation). An additional factor, x*, takes into account the fact that only a fraction of the catalyst is active.
Superficially, the form of the rate law has something in common with the Michaelis-Menten equation, with which you may already be familiar. The Michaelis-Menten equation relates the rate of an enzyme-catalyzed reaction to enzyme concentration, the rate constant for the catalytic reaction, and rate constants for reversible substrate binding with the enzyme.
Maybe that resemblance shouldn't be too surprising. After all, both equations describe catalytic processes, in which either the surface or the binding site must accomodate the substrate so as to carry out a subsequent series of reactions. Both equations take the form of a saturation curve, indicating that the rate of reaction will level out if the surface or binding site becomes fully occupied.
Problem KP4.1.
One way to evaluate multi-term relationships is to consider what happens under different conditions. What happens to the rate law for catalytic polymerization if monomer concentration is very low, so that 1 >> Keq[M]?
Problem KP4.2.
What happens to the rate law for catalytic polymerization if monomer concentration is very high, so that Keq[M] >> 1?
3.05: Solutions to Selected Problems
KP5. Solutions to Selected Problems
Problem KP1.1.
Two C-C σ bonds and one new π bond are made. Three old π bonds are lost.
ΔH = bonds broken - bonds made
ΔH = (3 x 64) - ((2 x 83) + 64) kcal/mol
ΔH = -38 kcal/mol
Problem KP1.2.
Tc = ΔH / (ΔS + Rlog[M])
Tc = - 7,000 cal mol-1 (-8.6 cal K-1 mol-1 + 1.98 cal K-1 mol-1 log(8.7))
Tc = - 7,000 (-8.6 + 1.98 (0.939)) K
Tc = - 7,000 (-6.74) K
Tc = 1038 K = 765 °C
Problem KP1.3.
For each amide bond, a C-N bond and a H-Cl bond are made. A C-Cl and a N-H bond are lost.
ΔH = bonds broken - bonds made
= (C-Cl + N-H) - (C-N + H-Cl)
ΔH = (81 + 93) - (73 + 102) kcal/mol
ΔH = -1 kcal/mol
Problem KP2.1.
75% conversion means 0.75 in terms of fractions.
DP = 1 / (1 - p)
DP = 1 / (1 - 0.75)
DP = 1 / 0.25
DP = 4
Problem KP2.2.
slope = 2 [M]0 k = 0.717 s-1
k = 0.717 s-1 / (2 x 17 mol L-1)
k = 0.021 L mol-1s-1
Problem KP2.3.
Mn = M0 / (1 - p)
= 120 g/mol / (1 - 0.99)
= 120 g/mol / 0.01
= 12,000 g/mol
Mw = M0(1 + p) / (1 - p)
= 120 g/mol (1 + 0.99) / ( 1 - 0.99)
= 120 g/mol (1.99) / 0.01
= 23,800 g/mol
D = 1 + p
= 1 + 0.99
= 1.99
Problem KP3.1.
Rate = k' [M][I]1/2
slope = k' [I]1/2
0.0024136 s = k' (0.00025)1/2
k' = 0.015 s-1
Problem KP3.2.
At steady state:
Rateinit = Rateterm
ki [M][I] = kt[M+]
Rearranging:
[M+] = (ki/kt) [M][I]
Problem KP3.3.
Rateprop = kp [M+][M]
Substituting the steady state expression for [M+]:
Rate = (kikp/kt) [M]2[I]
Problem KP3.4.
v = Rateprop/Rateinit
v = kp [M+][M] / ki [M][I] = (kp / ki)[M+] / [I]
but [M+] is not a known quantity. Alternatively, at steady state, Rateinit = Rateterm
v = Rateprop/Rateterm
v = kp [M+][M] / kt [M+] = (kp / kt)[M]
Problem KP3.5.
a) v = [M]0/[I]0 = 4.5 / 1.25 x 10-3 = 3,400
b) v = (kp/2f kt kd)([M]/[I]1/2)
v = (0.003 / 2(0.5)(0.003)(0.0001))(4.5/(1.25 x 10-3)1/2) = (1/0.0001)(4.5/0.035) = 128/0.0001 = 1,280,000
c) v = (kp/2f kt kd)([M]/[I]1/2)
v = (0.003 / 2(0.5)(0.03)(0.0001))(4.5/(1.25 x 10-3)1/2) = (0.01/0.0001)(4.5/0.035) = 128/0.01 = 12,800
d) v = (kp/2f kt kd)([M]/[I]1/2)
v = (0.003 / 2(0.5)(0.003)(0.1))(4.5/(1.25 x 10-3)1/2) = (1/0.1)(4.5/0.035) = 128/0.1 = 1,280
Problem KP4.1.
The key point is that, when two terms are added together and one is much larger than the other, the sum is approximately the same as the larger of the two terms. You can ignore the smaller one.
Problem KP4.2. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/03%3A_Kinetics_and_Thermodynamics_of_Polymerization/3.04%3A_Kinetics_of_Catalytic_Polymerization.txt |
• 4.1: Molecular Weight of Polymers
It might seem obvious that molecular weight is an essential property of any molecular compound. In polymers, molecular weight takes on added significance. That's because a polymer is a large molecule made up of repeating units, but how many repeating units? Thirty? A thousand? A million? Any of those possibilities might still be considered a representative of the same material, but their molecular weights would be very different, and so would their properties.
• 4.2: Viscosity of Polymers
Viscosity is a term we use to describe the "thickness" of different liquids.
• 4.3: Rheology
Viscosity measurements are the realm of a field of science called rheology. Rheology is, literally, the study of flow.
• 4.4: Glass Transition
The glass transition is probably the most commonly-cited characteristic of a polymeric material. At the glass transition temperature, the physical nature of the material changes subtly. It goes from being a rubbery, flexible material at higher temperature, above the glass transition, to a glassy, harder material at a lower temperature. The material is still a solid either way, but there is a definite change in how it responds to stimuli.
• 4.5: Crystallinity in Polymers
Polymers are different from many other such solids. Metals, for instance, have crystalline structures, in which atoms form a regularly repeating pattern, row upon row. Polymers, in contrast, are generally somewhat amorphous. Think of a strainer filled with cooked spaghetti, the long chains of pasta looping over each other randomly. That's how the molecules of amorphous or "shapeless" polymers tend to distribute themselves.
• 4.6: Microphase Separation
Sometimes, more rigid segments of a polymer are deliberately built into the structure. If the block lengths are the right size, the two segments are able to separate into two phases. As a result of stronger intermolecular attractions, lengths of chains containing hard segments cluster together, pushing out the soft segments that would otherwise get in the way of these intermolecular attractions. This phenomenon is called microphase separation.
• 4.7: Stress-Strain Relationships
• 4.8: Storage and Loss Modulus
• 4.9: Modulus, Temperature, Time
• 4.10: Chapter Solutions
04: Polymer Properties
Molecular weight is one of the most central aspects of polymer properties. Of course, all molecules have molecular weights of their own. It might seem obvious that molecular weight is an essential property of any molecular compound. In polymers, molecular weight takes on added significance. That's because a polymer is a large molecule made up of repeating units, but how many repeating units? Thirty? A thousand? A million? Any of those possibilities might still be considered a representative of the same material, but their molecular weights would be very different, and so would their properties.
That variation introduces some unique aspects of polymer molecular weight. Because polymers are assembled from smaller molecules, the length (and consequently the molecular weight) of a polymer chain depends on the number of monomers that have been enchained into the polymer. The number of enchained monomers in an average polymer chain in a material is called the degree of polymerization (DP).
Notice that key point: it is just an average. In any given material, there will be some chains that have added more monomers and some chains that have added fewer. Why the difference? First of all, polymer growth is a dynamic process. It requires monomers to come together and react. What if one monomer starts reacting, forming a growing chain, before any of the others get started? With its head start, this chain will become longer than the rest. What if something goes wrong with one of the growing chains, and it can no longer add new monomers? That chain experienced an early death, and it will never grow as long as the others.
As a result, when we are speaking about the molecular weight of a polymer, we are always talking about an average value. Some chains in the material will be longer (and heavier) and some chains in the material will be shorter (and lighter). As with any group of measurements, it's helpful to know how widely distributed the individual values really are. In polymer chemistry, the width of the distribution of molecular weights is described by the dispersity (Ð, also called, in older texts, the polydispersity or the polydispersity index, PDI). The dispersity of a polymer sample if often between 1 and 2 (although it can be even higher than 2). The closer it is to 1, the narrower the distribution. That is, a dispersity of 1.0 would mean that all of the chains in a sample are exactly the same length, with the same molecular weight.
The original idea of dispersity was based on alternative methods of measuring the molecular weight (or the chain length) of a polymer sample. One set of methods gave something called the number average molecular weight (symbol Mn). These methods essentially took the weight of a sample, counted the molecules in a sample, and therefore found the average weight of each molecule in that sample. A classic example of this approach is a colligative properties experiment, such as a freezing point depression. You know that impurities in a liquid tend to disrupt intermolecular interactions and lower the freezing point of the liquid. You may also know that the amount by which the freezing gets lowered depends on the number of molecules or ions that get dissolved. Hence, if you weigh a sample of polymer, dissolve it in a solvent, and measure the freezing point, you could figure out the number of molecules dissolved and consequently arrive at Mn.
That's not so easy in practice; freezing point depressions are very small. They're not used very often anymore. A very common example of the kind of measurement widely used to determine Mn today is end group analysis. In end group analysis, we use 1H NMR measurements to determine the ratio of a specific proton in the repeat units to a specific proton in the end group. Remember, the end group might be something like the initiator, which only added onto the first monomer to get the polymerization going. By the end of the polymerization, it is still found at the end of the polymer chain, so it is an end group. There is only one of them per chain, whereas there are lots of monomers enchained in the polymer, so the ratio of those enchained monomers to the end group tells us how long the chain is.
The other set of methods upon which dispersity was based gave something called the weight average molecular weight (symbol Mw). The classic example was a light-scattering experiment. In this experiment, a solution of polymer was exposed to a beam of light and the resulting scattered light -- coming from the sample in different directions -- was analyzed to determine the size of the polymer chains in the solution. The results were more heavily influenced by the larger molecules in solution. As a result, this measurement of molecular weight was always higher than measurements based on counting every single molecule.
The resulting ratio, Ð = Mw / Mn, became known as the polydispersity index or, more recently, the dispersity. Because Mw was always more strongly influenced by longer chains, it was a little bigger than Mn and therefore the dispersity was always bigger than 1.0.
Nowadays, both molecular weight and dispersity are most commonly measured using gel permeation chromatography (GPC), synonymous with size-exclusion chromatography (SEC). This method is a high-performance liquid chromatography (HPLC) technique. Solvent containing a sample of polymer is pumped through a specialized chromatography column capable of separating molecules based on their size differences. As sample emerges from the column, it is detected and recorded. Most commonly, the presence of sample in the solvent emerging from the column causes a slight change in the refractive index. A graph of refractive index versus time presents a record of the amount of sample emerging from the column at a given time. Because the column separated molecules based on size, the time axis corresponds indirectly with chain length of molecular weight.
How can the column separate molecules based on size? The column is packed with a porous material, usually insoluble polymer beads. The pore sizes vary. These pores a crucial to separation because molecules flowing through the column may tarry in the pores. Smaller molecules could become delayed in any of the pores in the material, whereas larger molecules will only be delayed in the very largest pores. Consequently, a longer elution time corresponds to a lower molecular weight.
If you injected a series of different polymers into a GPC, each having a different molecular weight distribution, you would observe each one eluting at a different time. What's more, each peak may be broader or narrower, depending on the dispersity of that particular sample.
The wider the peak in GPC, the broader the distribution of molecular weights; the narrower the peaks, the more uniform are the chains. Normally a software package analyzes the curve to determine the dispersity.
Note that the x axis on a GPC trace is most commonly labeled as "elution time" and it normally runs left to right. However, often the x axis is labeled "molecular wright" because that is really the quantity we are interested in. In fact, sometimes the axis is reversed, so that peaks with higher molecular weights appear to the right, because it can feel more natural to look at it that way. You need to look carefully at the data to see how it is displayed.
There are some problems with relying on GPC for molecular weight measurements. The main difficulty is that polymers in solution tend to coil into balls, and those coils will contain greater or lesser amounts of solvent, depending on how strongly the polymer and solvent interact with each other. If it interacts more strongly with the solvent, it will pull lots more solvent molecules inside its coils. The coil has to get bigger to make room for those internal solvent molecules. If it doesn't interact strongly with the solvent, it will mostly just stick to itself, blocking the solvent molecules out. There is a broad range of behaviors in between.
As a result, different polymers may swell to different extents in different solvents. That matters because GPC is really using size of the polymer coil as an index of its molecular weight, so comparing GPC traces of two different kinds of polymers has to be done with caution.
Problem CP1.1.
In each of the following cases, state which polymer has the higher molecular weight, and which one has a narrower dispersity
Problem CP1.2.
Calculate the molecular weight of the following samples.
Problem CP1.3.
Use NMR end group analysis to determine the degrees of polymerization in the following samples. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.01%3A_Molecular_Weight.txt |
Polymers have come to occupy a very important niche in the materials we use every day. What makes them unique? Why do they have properties that aren't easily replicated by other materials?
Let's start with a property that you are probably familiar with. Viscosity is a term we use to describe the "thickness" of different liquids. For example, we say that honey is more viscous than water. Motor oil is more viscous than gasoline. When we say that, we mean that water is much easier to stir or to pour than honey. The honey moves more slowly. It resists the movement of the spoon when we stir it. Viscosity is often described in very general terms as "resistance to flow". The honey doesn't flow very easily, especially compared to something like water.
Now, honey isn't a polymer. It's a very concentrated solution. It contains a little bit of water and a whole lot of sugars, plus other small molecules produced by the plants from which the bees gathered the nectar to make the honey. The sugars aren't polymers, either; they are simple monosaccharides such as glucose and fructose. Nevertheless, this high viscosity is something honey has in common with polymer solutions, as well as with some oligomers, which are short-chain polymers that can be liquids instead of solids.
Why is honey so viscous? Partly it's just that the sugar molecules in the water are much larger than the water molecules, so they experience a lot more drag as they move through the solution compared to just plain water.
The strong intermolecular attractions between the individual sugar molecules are also a major factor. Sugars are covered in OH or hydroxyl groups that are capable of strong hydrogen bonds. As these molecules move past each other in the very concentrated solution, they cling to each other, slowing down the flow of the liquid.
The motor oil example works in a similar way. Motor oil and gasoline are composed of the same class of compounds, hydrocarbons, composed of carbon chains covered in hydrogen atoms. They're very similar to each other. The intermolecular attractions between the molecules are much weaker than those between sugar molecules; they're just London dispersion forces. The major difference between motor oil and gasoline is that the motor oil contains much longer hydrocarbon chains. The molecules are bigger, and so they experience more drag as they move through the liquid. The amount of intermolecular interactions is also crucial in determining how tightly two molecules hold onto each other, and this factor is especially important in very weak London dispersion forces, where a small advantage goes a long way. Those longer hydrocarbon chains in the motor oil cling to each other much more strongly than the shorter ones in the gasoline, so gasoline flows much more easily than motor oil.
For similar reasons, oligomers and polymers, if they are liquids, also display high viscosity. The extended contact between their long molecular chains leads to enhanced intermolecular attractions that contribute to a resistance to flow. Increased drag is also a factor; even in solution, these very large molecules encounter more resistance as they move past solvent molecules compared to the resistance smaller molecules would experience.
In fact, viscosity measurements of polymer solutions are another way to determine the size of the polymer -- leading to the chain length and the molecular weight. The larger the polymer, the more drag and also the more intermolecular attraction, and so the higher the viscosity.
So, the viscosity increases with the molecular weight, but not necessarily in a linear way. That can make it more difficult to use a graph like this to predict molecular weight from a viscosity measurement. One common thing to do in this situation is to take the log of the values. That approach will often give a straight line when two things are related.
Often in nature one physical property that depends on another will follow a "power law". That means that one quantity depends on the other raised to some exponent. In this case, the relationship is described by the Mark-Houwink equation:
\[ [η] = KM^α\]
In this equation, [η] is the intrinsic viscosity; that's the viscosity attributed to the solute, rather than the viscosity of the solvent itself. M is the molecular weight. K and α are constants for a particular polymer.
If you take the log of both sides of the equation -- that's OK to do mathematically, just like you could multiply both sides of an equation by four or subtract three from both sides and still have an equivalent expression -- you get:
\[\log[η] = log(KM^a)\]
If you take the log of two things multiplied together, it's the same as adding the individual logs of both things.
\[\log[η] = \log K + \log(M^a)\]
And if you take the log of something raised to an exponent, the exponent just comes down to become a coefficient.
\[\log[η] = \log K + α\log M\]
or
\[\log[η] = α \log M + \log K\]
That looks like an equation for a straight line:
\[y = mx + b.\]
Here, you plot \(\log[η]\) on the y axis, \(\log M\) on the x axis, and you get a straight line with those constants, \(α\) and \(K\), as the slope and y-intercept, respectively.
This approach does not yield an absolute measurement. You couldn't take two completely different polymers, take viscosity measurements, and make conclusions about which one had a higher average molecular weight. However, if you took a series of polymers of the same kind and measured their viscosities in solution, you would be able to deduce which ones were longer and which ones were shorter.
In fact, the dependence of viscosity on the molecular weight of polymers is more complicated than this graph suggests. If you deal with large enough polymers, the graph starts to look like this one:
Why does it bend? Remember, polymers in solution are coiled up into balls. A longer chain just corresponds to a bigger coil. At some point, these chains become big enough that they aren't likely to stay coiled independently of each other. They become more and more likely to interact with each other. They become entangled with each other.
There are plenty of other factors you would have to take into account in using viscosity to assess molecular weight. One of them is concentration, for example. The more concentrated a polymer solution, the more large molecules you have exerting drag and interacting with each other. Higher concentration leads to a higher viscosity measurement. Consequently, you would have to make sure you accounted for the polymer concentrations if you used this method.
Problem PP2.1.
Estimate the values of α and K for the polymer that provided the plot below.
Problem PP2.2.
Suppose you have a polymer described by the plot below. If your sample has a molecular weight of 1 million g/mol, what would you expect for intrinsic viscosity under the conditions of this experiment?
Problem PP2.3.
Suppose you have a calibration plot, below, for a particular polymer. If you measure the intrinsic viscosity of a sample and find that the value is 800 ml / g, what is the molecular weight of the sample?
Problem PP2.4.
Ethanol (CH3CH2OH) has an "absolute viscosity" of 1.095 centipoise, whereas ethylene glycol (HOCH2CH2OH) has an absolute viscosity of 16.2 centipoise. Explain why the two values are so different. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.02%3A_Viscosity.txt |
Viscosity measurements are the realm of a field of science called rheology. Rheology is, literally, the study of flow. Another very simple definition, attributed to chemical engineer Chris Macosko at University of Minnesota, is the study of "what happens when you squish stuff". There's an element of force or pressure that comes into play here, too.
One of the common ways of assessing properties in rheology is to place a sample between two parallel plates and move one plate with respect to the other. One plate says still and the other one moves. What happens to the liquid between the plates? There ought to be some friction between the stationary plate and the liquid that will keep the liquid still. There also ought to be some friction between the moving plate and the liquid that will make the liquid move along at the same speed as that plate. So at one extreme, the liquid is moving along with the sliding plate and at the other extreme the liquid is perfectly still. If we imagine that the liquid in between these two extremes is divided into very thin layers, then each layer must be moving at a slightly different speed than the next.
In the diagram, the symbol, u, stands for the speed of the layer of liquid. The arrow beside the layer is meant to convey its relative speed: the top layer is moving the fastest, the next layer is a little slower, and so on; the bottom layer isn't moving at all. There is an important quantity, called the strain rate or shear rate (given as a symbol the Greek letter gamma, γ, with a dot on top) that describes how the speed of the liquid moving along the x-axis (left to right) changes, layer by layer, in the y-direction.
What does this picture have to do with viscosity? Well, in order to get that top plate to move, we have to apply some sort of force to it. In rheology the force is given per unit area (like pressure); this force per unit area is described as the stress (given as a symbol the Greek letter sigma, σ).
What if we wanted to slide the plate even faster? What would happen to the liquid? Well, the shear rate would go up. The top layer would be moving even faster, and the bottom layer would still be stationary. And what would we have to do to get that faster shear rate? We would have to push the top plate a little harder. The result is a graph that looks like this:
The graph says that shear stress and shear rate are directly related. The stress is called "shear stress" because of the direction of the force parallel to the liquid, causing shear strain. When we increase the shear rate, the shear stress also increases proportionally. So, if we did this experiment using water as the liquid between the plates, we would get a linear relationship between shear stress and shear rate. That's how Sir Isaac Newton described the behavior of liquids, so when we see that linear relationship between stress and strain, we describe the liquid as a Newtonian liquid.
What happens if we do the experiment with honey instead of water? What would happen, for example, if we exerted a certain stress to get a certain shear rate with the water, and we wanted to get the same shear rate using honey? Well, the honey is thicker than the water. It's not going to move as easily. We will have to apply a greater stress in order to achieve the same shear rate that we measured with water.
That would be true for every possible shear rate. If we were to graph shear stress vs. shear strain with honey, we would get a linear relationship, just like we did with water, but the value of shear stress would always be higher than for water. The slope of the line for honey would be greater than the slope of the line for water.
Viscosity is formally defined as the slope of that line. Honey, being thicker than water, requires higher and higher stresses to achieve the same strain rates that you would get using lower stresses with water. In the simple kitchen analog of the experiment, you have to push the spoon harder to stir the honey than you do if you are just stirring water.
Honey and water are Newtonian liquids. The relationship is simple: increase the shear stress and you increase the shear rate. Not all liquids behave that way. Polymer solutions don't, for example. They display a non-linear relationship, in which the shear stresses needed to get higher shear rates are just not as high as you would expect.
That behavior is described as shear thinning. What makes polymer solutions do that? Remember that, in solution, an individual polymer molecule tends to coil up into a ball. Normally, those balls are spherical. At high shear rates, they distort into ellipses. That reduces the amount of drag, making the solution a little less viscous than you would expect. The shear stress still increases with higher shear rates, but not as much as if it were a Newtonian fluid.
That's not all. Remember, polymers display chain entanglement, especially when they are very large. Chain entanglement increases the viscosity of the solution. These chains are mobile because the polymer is constantly undergoing conformational changes. As the solution gets sheared, some of the polymer chains may wiggle loose from their neighbors, but random conformational changes will always result in new entanglements right away.
But what happens at a very high rate of shear strain? The polymer chains still become untangled, but they don't have time to form new entanglements, because everything is whizzing by too quickly. Without those entanglements, viscosity doesn't increase as sharply as you would expect.
Some liquids behave just like water, but polymer solutions do not. That's a consequence of the enormously long chain structures of the polymers. It's worth pointing out that there are other things that don't behave in the "normal" way. Colloids often display shear thickening. You can see examples of this in videos of people running across wading pools filled with corn starch and water, a colloidal mixture in which the solid cornstarch particles are just suspended in the water, not dissolved. If you walk across the pool, you sink. If you run, you stay on top. That's shear thickening, and it has to do with how those solid particles move during shearing. When you walk, they just slide past each other, but when you run, they all collide, and come to a stop.
A Bingham plastic, on the other hand, actually appears to be a solid until you give it a good shove; then it flows like a liquid. Peanut butter behaves sort of like that.
Exercise 4.3.1
Honey is a Newtonian fluid, but molasses undergoes shear thinning. What might you deduce about the composition of molasses and honey? | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.03%3A_Rheology.txt |
The glass transition is probably the most commonly-cited characteristic of a polymeric material. At the glass transition temperature, the physical nature of the material changes subtly. It goes from being a rubbery, flexible material at higher temperature, above the glass transition, to a glassy, harder material at a lower temperature. The material is still a solid either way, but there is a definite change in how it responds to stimuli.
A material could certainly be useful in either state. We might want a plastic to be more rigid, like a water bottle or a rod for a shower curtain. Alternatively, we might want it to be softer, like a seat cushion. Either way, it might be helpful to know the temperature at which the material will change from one type to the other.
The glass transition is a little like what happens to gummy bears when you put them in ice cream. Straight from the bag, gummy bears are chewy, but they don't feel like they will break your teeth. Put them on ice cream and that changes. They become much harder to chew.
The classic explanation for the glass transition is based on the idea of chain flow. Polymers are long-chain molecules and, given a little energy, the chains move around. They wiggle. They undergo bond rotations, switching from one conformation to another. A block of material contains piles of chains, like a nest of snakes. Chain ends and loops squirm past each other continuously.
Chain flow allows a material to adapt when forces are exerted on it. We can bend an eraser because the chains in the rubber slide over each other and adopt a new shape. At least, that's what happens at room temperature. It might not work if the eraser were accidentally dropped in some liquid nitrogen, which is very, very cold.
As a block of material cools down, it contracts. The molecules become packed more closely together. At some point, the free volume -- that's the amount of volume in the material that is not actually taken up by the molecules -- becomes too small to allow chains to move past each other. All of those chains need a little room around them if they are going to undergo conformational changes, and conformational changes are how polymer chains move. Without that extra room, the material suddenly becomes less flexible.
If we start in the glassy state and increase the temperature, the volume of the material is expanding, and the free volume is increasing, too. At some point, the free volume becomes great enough that chains can slip past each other. The material becomes more flexible. It becomes rubbery.
Now, this transition is not the same thing as melting. The chains are not completely overcoming their interactions with each other and gaining freedom of movement in any direction. The chains are still highly entangled. Portions of the chain are sliding past each other but overall things have not changed that drastically. The block of polymer does not turn into a gooey puddle of liquid.
Well, if a material can change from glassy to rubbery at a certain temperature, there will be consequences in how the material behaves. For example, the tires on your car are meant to be rubbery and flexible; that factor helps them grip the road. If the weather gets too cold and your tires become glassy, the tires no longer have the same amount of flexibility, and you don't have as much traction. "All-weather tires" are composed of a rubber that has a very low glass transition temperature, helping to avoid this problem. ("Snow tires" are a different thing; they have patterns in the treads that help channel away the snow in order to improve traction.) Knowing when this change will occur would be very useful. So, how do we determine the glass transition?
Phase changes, such as melting points, are measured using calorimetry, so let's start there. When a solid is heated, its temperature increases. That seems simple enough. Temperature is basically a measure of heat content, so as heat flows in, the measured temperature rises. This simple relationship breaks down at the melting point. At that point, heat flowing into the material is consumed by the breaking of intermolecular forces. Overcoming these attractions costs extra energy. Consequently, there is a point at which the temperature rise in the material stalls out temporarily while it melts.
This extra heat needed to melt the material is called the heat of melting or, more commonly, the heat of fusion. Heat of fusion actually refers to the opposite process as the material is cooled and frozen. As heat is carried away from a cooling material, it cools down gradually, but there comes a point at which those strong intermolecular interactions form, giving off some extra heat. The fusion temperature is the same as the melting temperature, and the heat of fusion is the same as the heat of melting, but in one case the heat is added and in another case the heat is given off. In principle, if we just heat something up and look for that stalling point in the temperature, we can find the temperature of the phase transition.
Differential scanning calorimetry
Differential scanning calorimetry (DSC) is a technique that is commonly used to measure phase transitions, including the glass transition temperature. It is based on the same ideas but the experiment is run in a slightly different way. In DSC, we have two tiny sample chambers side by side. One contains the material we are interested in and the other (empty) one is used as a reference. The instrument heats both samples at a constant rate, all the while maintaining both of them at the same temperature. Consequently, it may actually add more heat to one chamber than the other, so that they both reach 30.5 °C at the same time, then 30.6 °C, and so on. Once we reach the melting point of the sample of interest, heat flow into that sample must be increased so that it can keep up with the reference. Past that point, the heat flow drops back to a more normal level.
If we look at a DSC scan, we see a graph of heat flow on the y-axis and temperature on the x-axis. The heat flow usually stays pretty constant as the temperature increases. At the melting point, heat flow increases, but then it drops back down once the melting point has been overcome. It may not drop back to the same level as before, however, because the heat flow that is recorded is related to the heat capacity of the material. The solid and the liquid usually respond differently to heat. In general, because the molecules in a liquid can freely rotate, they have something else to do with added heat. Liquids thus have a slightly higher heat capacity than their corresponding solids.
All of that explanation hopefully prepares us for a DSC study of the glass transition temperature. It works the same way, but only up to a point. If we look at a DSC scan, we still see heat flow on the y-axis and temperature on the x-axis. At the glass transition, heat flow bumps up slightly -- and remains there. It looks very different from a melting point. A DSC scan of a melting point shows a "peak" at the transition temperature. A DSC scan of a glass transition point shows only a little step, like a riverbank
Why is it different? Well, the glass transition is not really a phase change like the melting point or boiling point. It does not involve a physical change of state. It was a solid before and it is a solid afterward. No intermolecular interactions must suddenly be overcome to free the molecules from each other. Instead, it's just a volume change. The free volume became great enough that the chains can slip past each other, but the chains are still clinging together in a solid-state. Now, that additional movement does have consequences. The material becomes more flexible. If heat flows in, there is more freedom of motion into which the heat can be distributed. In other words, there is a slight increase in heat capacity, and that's what we observe in DSC.
One practical note: DSC scans can actually be displayed in two different ways. The y-axis can either display heat flowing in or heat flowing out. That means melting points can look like peaks or like valleys, depending on how the data is displayed. Often the data are labeled with an arrow that says "endo" to tell you which direction along the y-axis means more heat is flowing in (or possibly "exo", meaning which direction means heat is flowing out). You need to look carefully if things seem backward.
Problem CP4.1.
For each DSC trace, state what sort of transition is occurring at what temperature.
Why does the glass transition temperature vary from one kind of polymer to another? What structural factors influence the glass transition temperature? This structure-property relationship isn't a straightforward one, as there seem to be a number of different variables involved. However, the simplest of these factors is just molecular weight. The higher the molecular weight of a polymer, the higher its glass transition temperature. This relationship is true only up to a certain point, however. The non-linear dependence of the glass transition temperature of molecular weight is described by the Flory-Fox equation:
\[T_g = T_{g(∞)} - \dfrac{K}{ M_n}\]
Here, Tg(∞) refers to the glass transition temperature of an infinitely long chain of the polymer. K is a constant for a particular polymer, such as polystyrene or polyethylene.
A plot of Tg vs. Mn resembles a saturation curve; the line rises sharply, gradually stalling out and continuing parallel to the x-axis. In other words, although this relationship of increasing glass transition temperature with increasing molecular weight holds true at relatively low molecular weights, the glass transition temperature remains constant once a threshold molecular weight has been reached. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.04%3A_Glass_Transition.txt |
We often think of polymers in the form of plastics: solid materials that serve some structural function, like a water bottle or some Venetian blinds. Polymers are different from many other such solids. Metals, for instance, have crystalline structures, in which atoms form a regularly repeating pattern, row upon row. Polymers, in contrast, are generally somewhat amorphous. Think of a strainer filled with cooked spaghetti, the long chains of pasta looping over each other randomly. That's how the molecules of amorphous or "shapeless" polymers tend to distribute themselves.
Polymers are capable of forming more ordered structures. If they aligned instead with their long chains parallel to each other, they would be able to get much closer together, and intermolecular attractions between the chains would be much stronger. The energy of the system would decrease, so this more crystalline structure should be inherently favored.
So why doesn't that always happen? Think about a polymer cooling from a completely melted state, in which case the structure is certainly amorphous. As it cools, the material becomes less like a liquid, and then less soft. With small molecules, that transition is relatively abrupt, as the molecules slide into place, guided by their strengthening intermolecular attractions. The molecules just have to rotate a little to face the correct direction, or maybe budge a little to the side. With polymers, the transition is much more gradual, because those great, long chains have to slide over each other and uncoil in order to lie parallel. At some point, there just isn't enough energy for them to keep working their way into the optimal alignment. As a result, many polymers are semi-crystalline, with regions called lamellae where portions of chains have aligned parallel to each other, but also with large amorphous areas that are much more randomly oriented.
As a result, a polymer sample might be 80% amorphous with only 20% of its chain lengths aligned in crystalline lamellae. Even so, those crystalline domains exert a strong influence on the properties of the polymer. Because of the stronger intermolecular attractions between these closer chains, the chains are much less able to slide past each other. The result is a material that is stronger and more rigid, and that can be very important for reliable structural materials.
How do we know whether a polymer sample contains crystalline domains? We can use differential scanning calorimetry to observe the transition between an ordered crystalline phase and a disordered melt phase with added heat. As we saw when we were looking at glass transition, a melting point shows up as a peak on a DSC trace. You can look for it at a temperature above Tg, which you will recall looks more like a step in the baseline. Below Tg, the chains are not mobile enough to move out of alignment with each other. In the example below, you can see Tg at about 128 °C as well as a melting point, Tm, when attractions between chains in the lamellae are overcome, at 155-160 °C.
If this were a regular molecular or elemental solid, the melting point would be the same as the fusion point; the material would melt at the same temperature during warming as it froze during cooling. In polymers, it usually doesn't work out that way, and a hysteresis is observed. In hysteresis, a sample has changed because of previous events, leading to results that don't seem to be reproducible in the way that we would normally expect. In general, the crystallization temperature, Tc, is lower than the melting temperature, Tm. That's because, as the material melts, the chains move out of alignment with each other and, because of chain entanglement, they are hindered from getting back into alignment with each other again, leading to a delay in crystallization.
Example 4.5.1
Problem PP5.1.
Describe what is observed in each of the following DSC traces.
Crystallinity can also be probed in other ways. The classic approach is through X-ray diffraction. When X-rays pass through ordered materials, they give rise to diffraction patterns, bright spots in space where ricocheting X-rays have constructively interfered with each other, shining starkly amid the blackness where the X-rays have undergone destructive interference.
The drawing below is a cartoon of an X-ray diffraction pattern. It might be captured by a digital camera or, in earlier days, a sheet of photographic film. The bright spots result from X-rays scattering out from the middle, where they have encountered the sample. In the middle, the black circle blocks the original X-ray beam, which would otherwise be too bright to allow observation of those spots around it.
The data from an X-ray diffraction experiment is fed into a program that can mathematically deconvolute why the spots showed up where they did. The process has been compared to throwing a stone over your shoulder into a pond, then watching the ripples to decide where the stone must be. With one stone, that should be pretty easy. You would find the stone at the center of a circle of ripples. With two or more stones, interference patterns make the ripples more complex, and so it may take more work to determine where each stone lies in the pond.
The picture below illustrates the results of a relatively simple X-ray diffraction experiment commonly used with crystalline polymer samples. The technique is called wide-angle X-ray scattering, or WAXS. It simply reports the angle at which scattering intensity is observed relative to the initial X-ray beam. That angle corresponds to a distance between atoms in the sample, which in this case usually corresponds to a distance between parallel polymer chains. Sometimes there can be more than one peak, but this example is a simple one to start with.
The relationship between this scattering angle and an interatomic distance is given by Bragg's Law, a fundamental starting point in X-ray crystallography first expressed by Sir William Bragg and his son, Sir Lawrence Bragg. The law states that:
$nλ = 2d\sin θ \label{bragg1}$
in which n is an integer (typically assumed to be 1), lambda is the wavelength of the X-ray used (commonly 1.541 Å, although others are possible), d is the distance between regularly repeating atoms and θ is the scattering angle.
Rearranging Equation \ref{bragg1}, we find:
$d = \dfrac{1.541}{2\sin θ}$
In other words, the distance between atoms is inversely proportional to the scattering angle. The greater the angle, the smaller the distance.
Example 4.5.1
In each of the following cases, determine which sample (A or B) has a higher crystalline content. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.05%3A_Crystallinity_in_Polymers.txt |
Crystalline domains provide additional strength to polymer materials. The strong attraction possible between closely-aligned chains results in long segments of the polymer being held more firmly in position. Consequently, chain flow is more limited, and the material becomes more rigid.
Sometimes, more rigid segments of a polymer are deliberately built into the structure. For example, in block co-polymers, softer, more flexible blocks are often paired with harder, more rigid blocks. The soft segments may have greater conformational flexibility, or weaker intermolecular attractions between themselves, or both. The hard segments may be more conformationally rigid or they may have stronger intermolecular attractions, such as strong dipoles or hydrogen bonds.
If the block lengths are the right size, the two segments are able to separate into two phases. As a result of stronger intermolecular attractions, lengths of chains containing hard segments cluster together, pushing out the soft segments that would otherwise get in the way of these intermolecular attractions. This phenomenon is called microphase separation. The result is that the material contains islands of strength and rigidity in a matrix of flexible polymer chains. That can be a very useful combination. The flexible chains of the soft segments allow the polymer to be distorted, bent or compressed, but the hard segments put limits on that flexibility, keeping the material firmly together.
Because we are usually dealing with very large numbers of enchained monomers, the difference between the two kinds of segments need not even be dramatic. A copolymer of butadiene and styrene, both hydrocarbons, can form microphase separated materials. In this case, intermolecular attractions are dominated by weak London dispersion forces, but the aromatic groups of the styrene, with their delocalized pi systems, have London dispersion forces that are slightly stronger. As a result, the polystyrene blocks can cluster together, surrounded by the softer polybutadiene blocks.
Problem PP6.1.
Identify the hard segment and the soft segment in each of the following block-co-polymers.
Sometimes, the separation between these phases can be directly observed via microscopy. Tunneling electron microscopy (TEM) is a technique that can generate images of a cross-sectional slice of the material. The material is generally stained with a heavy metal, such as osmium, that binds preferentially to one phase or the other. The stained phase shows up darker under TEM than the phase that isn't stained.
X-ray diffraction techniques can often be used to measure distances between hard segments. Small-angle X-ray scattering (SAXS) is very similar to wide-angle X-ray scattering (WAXS). Because of the inverse relationship between scattering angle and distance, SAXS is used to probe regularly repeating structures at greater distances than those seen in WAXS. That makes it possible to see peaks if the hard segments are distributed regularly enough within the soft matrix.
Note that, in SAXS, the x-axis is usually labeled as q, the scattering vector:
q = 4πsinθ / λ
But since d = 2sinθ/λ then q = 2π/d or d = 2π/q. That gives us a pretty straightforward way of calculating distances between regularly-spaced hard segments (or any other regularly-spaced objects). Once again, just as in WAXS, there is an inverse relationship between the quantity shown on the x-axis and distances through space.
Problem PP6.2.
Calculate the approximate distances revealed in the following SAXS results. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.06%3A_Microphase_Separation.txt |
If you think about materials made from polymers, a couple of things might readily come to mind. You might think about the ubiquitous use of plastics in structures around us, such as automobile bumpers. You might think about a rubber band that you use to fasten something more firmly, either at home or in the lab. In either case, the function of the material relies on its response to stimuli. Do we want it to be flexible, to change shape, but then snap back to where it came from? Exactly how much flexibility do we need? Where is the trade-off between flexibility and strength?
These questions are important in polymer chemistry. Consequently, we often need to probe how materials behave under different conditions so that we know how they can be employed most effectively.
Tensile testing is one of the simplest ways to probe how a material responds to stress. Remember, the stress in this context means the force exerted on the material per unit of cross-sectional area. As such, it has units of pressure, such as Pascals (Pa). To do the experiment, the two ends of a sample are attached to two clamps, one of which is movable. The movable clamp is then pulled so that the sample becomes stretched. The force required to pull the sample is recorded and, given the cross-sectional area of the sample perpendicular to the force, the force is converted into units of strain.
The experiment also measures the distance that the sample is stretched (the strain). That distance is usually expressed as a ratio, comparing the change in length to the original length of the sample. It can be written either as a fraction or as a percent.
In the rough sketch below, we can see what typically happens in such an experiment. The sample stretches (the strain increases), and it gets harder to stretch as seen by the increasing force (and therefore increasing stress) that is needed to keep stretching it. Eventually, the stress plummets, because of the sample breaks.
Already in this graph, we can see a couple of important pieces of information we can learn from tensile testing. Number one, how far can we stretch this material before it breaks? That quantity is called "strain at break". In this sample the strain at break looks like about 1.1 or 110%; that means the sample was stretched to twice its original length, and then some. Number two, how much stress can the sample support without breaking? That quantity is called "ultimate tensile strength". In this sample, the ultimate tensile strength is just over 750 Pa.
If we look more closely at the graph from another sample, we will get a third important quantity, and see some additional features. The important quantity can be derived from the first part of the curve (A), in which stress increases linearly with strain. In this linear region, the material is behaving as a "Hookean solid", meaning it obeys Hooke's Law. Hooke's Law says that stress and strain should be directly proportional. In his original words, the extension is proportional to the force:
F = kx
in which F is the force, x is the extension of the solid, and k is the proportionality constant.
Hooke's Law is commonly applied to the behavior of mechanical springs, but it also holds for other solid materials. The slope in the linear region of the graph (A) would equal that proportionality constant, k, because the graph shows the ratio of F/x. In materials science, this slope is more commonly called "Young's modulus". It is a measure of the inherent stiffness of the material.
Young's modulus: E = σ/ε
in which σ = stress and ε = extensional strain.
The extensional strain is just the strain observed by stretching the material. In the graph shown above, Young's modulus is around:
250 Pa / 0.1 = 2,500 Pa.
This initial region of the curve, in which Hooke's Law is obeyed, is sometimes called the "linear elastic region". The word "elastic" does have an immediate connotation in everyday English, bringing to mind a rubber band that can be stretched, so we think of the stretching part when we hear the word "elastic". However, the returning motion is an essential feature of elastic behavior. The rubber band always comes back to its original shape. Likewise, within the linear elastic region, any solid material returns to its original shape after it is deformed under stress.
With most solids, such as aluminum or concrete, the linear elastic region spans a very narrow range of strains. Just by looking, we wouldn't notice these materials being deformed. However, this linear stress-strain relationship is typical of solids. In rheology, that behavior is described as elastic. Of course, with many polymers, a sample can be stretched so far that you can see the change with the naked eye, and it still returns to that original shape.
What makes polymers different? The long-chain structure of polymers does make them behave differently from other materials. These chains can undergo conformational change: each bond along the chain can rotate, converting the polymer chain into a slightly different shape. That ability gives a "soft material" a great deal of flexibility. The conformations of chains can adapt to accommodate stress, moving the chains into new shapes that offer lower-energy packing between each other. When the stress is removed, the chains eventually slide back into their original conformations. They return to their equilibrium shapes.
The presence of physical or chemical crosslinks help the material return to its original shape, functioning as anchor points so that the chains don't stray too far. Those interactions might be found in hard-phase interactions in a microphase separated material, as illustrated here, although they could also be found in a homogeneous material.
The linear elastic region isn't all we see in the stress-strain curve above. At point (B), the linear relationship is suddenly lost. The stress might even drop, as seen in this particular case. This feature on the graph is called the "yield point". The stress being experienced by the material, and the resulting strain, has become sufficient to overcome the natural elastic behavior of the solid. As noted previously, physical crosslinks such as hydrogen bonds help to reinforce the elastic behavior of a polymer sample. If, at some point, those interactions are overcome, the chains will start to slide more readily past each other.
As a result, the material loses its elasticity. When the stress is released, the material will still spring back as the chains settle into a new conformational equilibrium. However, that equilibrium will not be the same as the one before. New physical crosslinks will form as groups form intermolecular interactions with the nearest neighbors they encounter. It isn't likely that these will always be the same groups that they were interacting with before. As a result, the material will settle into a slightly different shape. You have probably seen this happen before when a rubber band has been stretched too far, too often, or for too long. Chains have dropped their old interactions and picked up new ones that formed more easily in the extended shape.
At point (C) of the example graph, the slope of the curve starts to increase. The same change in stress results in smaller and smaller changes in strain; the material is getting stiffer. This phenomenon is called "strain hardening". This feature would not always be observed, but if it did happen, what would explain it? In polymers, one explanation may lay in the fact that the volume of the material should remain constant as it is stretched. If the sample is getting longer, that means it is also getting narrower.
As a consequence of that narrowing cross-section, chains become compressed together. At some point, physical crosslinks begin occurring between neighboring chains. These crosslinks don't occur in equilibrium positions, with polymer chains coiled around each other like they were in the original sample. These crosslinks occur when chains are extended, lying parallel to each other, at closer contact distances than normal.
We can see the opposite sort of thing happening at (D), when the slope of the curve is decreasing instead of increasing. This phenomenon is called "strain-softening". Again, it might not be observed in all samples. When it does happen, what causes it? In this case, the answer is simpler. Having overcome the interactions that held the chains together, there is nothing left to resist further deformation. As the chains begin to disentangle from each other, it becomes even easier to pull them apart, facilitating the extension or stretching of the sample.
Eventually, at the breaking point (E), the chains start to lose contact with each other at some location in the sample, resulting in a catastrophic rupture of the sample.
Problem PP7.1.
In each of the following curves, estimate ultimate tensile strength and strain at break.
Problem PP7.2.
In each of the following curves, calculate Young's modulus.
Problem PP7.3.
In each of the following curves, identify any diagnostic features. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.07%3A_Stress-Strain_Relationships.txt |
We saw earlier that the inherent stiffness of a material can be assessed by its Young's modulus. The Young's modulus is the ratio of the stress-induced in a material under an applied strain. The strain is the amount of deformation in the material, such as the change in length in an extensional experiment, expressed as a fraction of the beginning length. The stress is the force exerted on the sample divided by the cross-sectional area of the sample. If the strain is limited to a very small deformation, then it varies linearly with stress. If we graph the relationship, then the slope of the line gives us Young's modulus, E. That's the proportionality constant between stress and strain in Hooke's Law.
E = σ / ε
Hooke's Law is sometimes used to describe the behavior of mechanical springs. The modulus can be thought of the resistance to stretching a spring; the more resistance the spring offers, the greater the force needed to stretch it. The same force is what snaps the spring back into place once you let it go.
In the experiments we saw earlier, we didn't let go. We continued to stretch the material farther and farther, applying generally increasing stress until the material finally broke. Now we will look at a much more limited approach. Instead of stretching the material as far as we can, we will only stretch it a tiny bit, then release the stress so that it snaps back to its original length. We can then stress it again and release it again. We can keep repeating. Instead of a continuously increasing strain, this sample is subjected to an oscillatory strain, one that repeats in a cycle.
This approach is called dynamic mechanical analysis. We can use dynamic mechanical analysis to measure the modulus of the material. Instead of continuously moving all the way through the linear elastic region, beyond which Hooke's law breaks down, we carefully keep the sample in the Hookean region for the entire experiment. Now, one experiment should be good enough to extract the modulus, but we are letting go and doing it over again. Why?
The principle reason for running the experiment this way is to get some additional information. We can get this information because polymers don't quite follow Hooke's Law perfectly. In reality, even within the linear elastic region, the stress-strain curve is not quite linear. In the picture below, the curvature is exaggerated quite a bit, just for illustrative purposes.
Even if the relationship is not quite linear, then as we release the strain, the stress in the material should simply follow the curve back down to zero. It does not. Instead, there is a phenomenon called hysteresis at work. Hysteresis just means that a property of the material depends on how the material came to be in its current situation. In this case, Hooke's Law seems to imply that a specific sample subjected to a specific strain would experience a specific stress (or vice versa). However, it depends whether we are stretching the sample or letting it relax again. As we let the sample relax back to its initial length, the strain is different from what we saw when we were stretching it. Typically, it's lower. Again, we can see this in the curve below, where the curvature has been exaggerated.
The difference between the loading curve (when the stress was first applied) and the unloading curve (when the stress was removed) represents an energy loss. A force was applied to move a sample or a portion of a sample, some distance. When the sample snapped back the same distance, the force was unequal to the one that was initially applied. Some energy was therefore lost.
The slope of the loading curve, analogous to Young's modulus in a tensile testing experiment, is called the storage modulus, E'. The storage modulus is a measure of how much energy must be put into the sample in order to distort it. The difference between the loading and unloading curves is called the loss modulus, E". It measures energy lost during that cycling strain.
Why would energy be lost in this experiment? In a polymer, it has to do chiefly with chain flow. The resistance to deformation in a polymer comes from entanglement, including both physical crosslinks and more general occlusions as chains encounter each other while undergoing conformational changes to accommodate the new shape of the material. Once the stress is removed, the material springs back to its equilibrium shape, but there is no reason chains would have to follow the exact same conformational pathway to return to their equilibrium conformations. Because they have moved out of their original positions, they are able to follow a lower-energy pathway back to their starting point, a pathway in which there is less resistance between neighboring chains.
For that reason, stretching a polymer is not quite the same as stretching a mechanical spring. A "spring-and-dashpot" analogy is often invoked to describe soft materials. Whereas a spring simply bounces back to its original shape after being pulled, a dashpot does not. If you don't know what a dashpot is, picture the hydraulic arms that support the hatchback on a car when you open it upward. There is some resistance to opening the hatchback because a piston is being pulled through a hydraulic fluid as the arm stretches. When we stop lifting, the arms stay at that length, because the hydraulic fluid also resists the movement of the piston back to its original position. The dashpot has a tendency to stay put rather than spring back.
Polymers display a little of both properties. They have an elastic element, rooted in entanglement, that makes them resist deformation and return to their original shapes. They also have a viscous element, rooted in chain flow. That viscous element means that, when we distort polymeric materials, they might not return to exactly the same form as when they started out. Taken together, these behaviors are described as viscoelastic properties. Many materials have viscoelastic properties, meaning they display some aspects of elastic solids and some aspects of viscous liquids.
So far, we have concentrated on extensional deformations of materials: we have been looking at what happens when we stretch them. It's worth looking at another type of deformation because it is very commonly used in materials testing. This second approach uses shear instead of an extension to probe how the material will respond. A shear force is applied unevenly to a material so that it tilts or twists rather than stretching.
One of the reasons this approach is used so often is because it is very easy to do. A sample is sandwiched between two plates. The bottom plate is held in place while the top plate is twisted, shearing the material held in between.
If we take a closer look at a layer of the sample, maybe at the surface, along the edge of the sandwich, we can imagine breaking it down into individual layers. Under shear strain, those layers move different amounts. The top layer, right beneath that top plate moves the most. The bottom layer, sitting on the stationary lower plate, doesn't move at all. In between, each layer moves a little further than the one beneath it. This gradation of deformation across the sample is very much like what we saw in the analysis of the viscosity of liquids. The difference is that viscosity looks at the variation of strain with time. Nevertheless, modulus in solids is roughly analogous to viscosity in liquids.
We can use this parallel plate geometry to obtain values for storage modulus and loss modulus, just like we can via an extensional geometry. The values we get are not quite the same. For this reason, modulus obtained from shear experiments is given a different symbol than modulus obtained from extensional experiments. In a shear experiment,
G = σ / ε
That means storage modulus is given the symbol G' and loss modulus is given the symbol G". Apart from providing a little more information about how the experiment was actually conducted, this distinction between shear modulus and extension modulus is important because the resulting values are quite different. In general, the value of the storage modulus obtained from an extensional experiment is about three times larger than the value of storage modulus obtained from a shear experiment.
E' = 3 G'
The reason for the difference is that extension actually involves deformation of the material in three directions. As the material is stretched in one direction (let's say it's the y-direction), in order to preserve the constant volume of the material (there is still the same amount of stuff before and after stretching), the material compresses in both the other two directions (x and z).
Problem PP8.1.
Metric prefixes are frequently encountered when reading about modulus. Rank the following units of stress from smallest to largest, and in each case provide a conversion factor to Pa.
GPa kPa MPa Pa | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.08%3A_Storage_and_Loss_Modulus.txt |
The storage modulus measures the resistance to deformation in an elastic solid. It's related to the proportionality constant between stress and strain in Hooke's Law, which states that extension increases with force. In the dynamic mechanical analysis, we look at the stress (σ), which is the force per cross-sectional unit area, needed to cause an extension in the sample, or the strain (ε).
\[E = \dfrac{σ}{ ε}\]
Alternatively, in a shear experiment:
\[G = \dfrac{σ}{ε}\]
The dynamic mechanical analysis differs from simple tensile testing by performing the experiment cyclically. The sample is stretched and released (or sheared and released). It can then be subjected to additional stress and released again. There is an element of time involved here. Specifically, because the experiment is cyclic, it can be carried out at different frequencies. When you do that, and you plot the resulting modulus against frequency, you can get additional information about the sample. The results would typically be presented in a graph like this one:
What the graph tells us is that frequency clearly matters. When the experiment is run at higher frequencies, the storage modulus is higher. The material appears to be stiffer. In contrast, the loss modulus is lower at those high frequencies; the material behaves much less like a viscous liquid. In particular, the sharp drop in loss modulus is related to the relaxation time of the material. In this context, that's the time it takes the chains to flow into new conformations in response to the applied stress. If they don't have time to flow, then that viscous response of the material is lost. The material behaves much more like an elastic solid when subjected to high-frequency cyclic deformation. That's important to know, because a material might be subjected to vibrations or other stimuli during everyday use, and its properties might change accordingly.
Another variation on this kind of experiment is called dynamic mechanical thermal analysis. Instead of changing the frequency of the stimulus throughout the experiment, the frequency is held constant and the temperature is changed instead. As a result, we can again see how the material responds under different conditions, which might tell us how it will behave in everyday applications. The result of the experiment might be a graph like the one below:
At this point, you are already familiar with the glass transition. It shouldn't be surprising that the properties are dependent on temperature. At the glass transition temperature, the expanding volume of the material with increasing temperature becomes sufficient to allow chain flow. As a result, the material suddenly behaves much more like a viscous liquid. Loss modulus increases. The stiffness of the material drops as the entangled chains not longer resist deformation as strongly. Storage modulus decreases.
The dynamic mechanical thermal analysis thus provides an alternative way to determine the glass transition temperature. Because it is actually measuring a different physical phenomenon than differential scanning calorimetry, the Tg obtained from a DMTA experiment may not agree exactly with one obtained from a DSC experiment. Nevertheless, it is often useful to have different ways of assessing properties.
In order to facilitate the analysis of the Tg in this experiment, a different quantity is usually displayed. Tan delta is just the ratio of the loss modulus to the storage modulus. It peaks at the glass transition temperature.
The term "tan delta" refers to a mathematical treatment of storage modulus; it's what happens in-phase with (or at the same time as) the application of stress, whereas loss modulus happens out-of-phase with the application of stress. Because it would take some time for the chains to move into new confirmation when they are subjected to stress, the strain actually lags behind the stress in these experiments.
Delta refers to the phase lag, the amount of time between application of stress and the observation of maximum strain. You may remember that a sine curve and cosine curve are out of phase with each other. Storage modulus is described as being proportional to cosδ whereas loss modulus is proportional to sinδ. The ratio of cosδ to sinδ is just tanδ.
Why does tanδ peak at the glass transition temperature? Clearly, as chains begin to move more freely, loss modulus increases. Consequently, the material also becomes less stiff and more rubbery. The storage modulus drops. If tan delta is the ratio of loss modulus to storage modulus, it should increase at that point -- and it does. Why does it drop again? That's because loss modulus refers to an energy loss, but because the material has gotten softer, less stress (and less energy) is put into the sample in the first place, so the energy loss also gets smaller. As a result, tan delta goes up at the glass transition but drops again shortly beyond that point.
Problem PP9.1.
Estimate the storage and loss, modulus in the glassy phase and rubbery phase in each of the following cases.
Problem PP9.2.
Estimate Tg in each of the following cases. | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.09%3A_Modulus_Temperature_Time.txt |
Problem PP1.1.
a) With a shorter elution time, I has a higher molecular weight than II. The narrower peak means that I has a narrower dispersity than II.
b) With a shorter elution time, III has a higher molecular weight than IV. The wider peak means that III has a broader dispersity than IV.
c) With a higher elution time, V has a lower molecular weight than VI. The wider peak means that V has a broader dispersity than VI.
Problem PP1.2.
a) repeat unit: 109 x (C4H6O2) = 109 x 86.09 g/mol = 9,383.81 g/mol
end groups: C9H11 + Br = 119.19 + 79.9 = 199.09 g/mol
total: 9,383.81 + 199.09 = 9,582.90 g/mol
b) repeat unit: 763 x (C8H8) = 763 x 104.15 g/mol = 79,466.45 g/mol
end groups: C4H9 + H = 57.12 + 1.008 = 58.13 g/mol
total: 79,466.45 + 58.13 = 79,524.58 g/mol
c) repeat unit: 48 x (C4H6O2) = 48 x 86.09 g/mol = 4,132.32 g/mol
end groups: C4H9 + C7H5S2 = 57.12 + 153.26 = 210.38 g/mol
total: 4,132.32 + 210.38 = 4,342.70 g/mol
Problem PP1.3.
The ratio of the repeat unit integral per proton to the end group proton per integral gives the degree of polymerization. We could take the entire integration of the end group and divide it by the entire number of protons in that group, or select one position to represent the end group. Similarly, we can select one position to represent the repeat unit.
a) repeat unit integral per proton = 36.0 / 2H = 18
end group integral per proton = 0.32 / 2H = 0.16
degree of polymerization = 18/0.16 = 112
b) repeat unit integral per proton = 26.0 / 1H = 26
end group integral per proton = 1.32 / 9H = 0.15
degree of polymerization = 26/0.15 = 173
c) repeat unit integral per proton = 54.0 / 4H = 13.5
end group integral per proton = 0.49 / 1H = 0.49
degree of polymerization = 13.5/0.49 = 28
Problem PP2.1.
α is the slope, which is [rise]/[run]. That's approximately [4.0-2.4]/[6.6-4.4] = 1.6/2.2 = 0.73.
K is the y-intercept. The equation for a straight line is y = mx +b; in this case, y = 0.73x + b. If we choose a point on the line, such as (x,y) = (4.9, 2.0), we can substitute those values on for x and y to get b.
So 2.0 = 0.73(4.9) + b, or b = 2.0 - 3.56 = -1.56.
Problem PP2.2.
If the molecular weight is a million g/mol, then log(Mw) = 6. Interpolating, log([η]) = 4, or [η] = 10,000 ml/g.
Problem PP2.3.
If the intrinsic viscosity, [η] = 800 ml/g, then log([η]) = 2.9. Interpolating, log(Mw) = 5.1, or Mw = 126,000 g/mol.
Problem PP2.4.
Ethylene glycol can form hydrogen bonds at either end of the molecule, forming a supramolecular assembly much like a polymer. As a result, it has a much greater drag in solution, higher viscosity.
Problem PP3.1.
Honey is a concentrated solution of simple sugars, which are small molecules. Molasses, although similar to honey in some ways, also contain starches, which are polymers. This polymeric content leads to shear-thinning behavior.
Problem PP4.1.
a) There is a glass transition at around -18°C.
b) There is a melting point at around 125°C.
c) There is a glass transition at around -4°C.
d) There is a glass transition at around 117°C and a melting point at around 146°C.
Problem PP5.1.
a) Tg is observed at around 78°C, Tm is observed at around 117°C and Tc is observed at around 104°C.
b) Tg is observed at around 134°C and Tm is observed at around 167°C, but Tc is not observed; the sample failed to crystallize, but remained an amorphous solid.
c) Tm is observed at around 194°C and Tc is observed at around 187°C. Tg is not observed, and probably occurs below 150°C
d) Tg is observed at around 123°C, but Tm is not observed. The experiment checked much higher than Tg (over a hundred degrees), so the material may be an amorphous solid.
Problem PP6.1.
Problem PP6.2.
a) d = (2 x 3.14) / 0.40 = 16 Å; d = (2 x 3.14) / 0.70 = 9.0 Å
b) d = (2 x 3.14) / 0.25 = 25 Å; d = (2 x 3.14) / 0.85 = 7.4 Å
c) d = (2 x 3.14) / 0.25 = 25 Å; d = (2 x 3.14) / 0.52 = 12 Å; d = (2 x 3.14) / 0.66 = 10 Å
Problem PP7.1.
a) ultimate tensile strength = 800 Pa; strain at break = 55%
b) ultimate tensile strength = 750 Pa; strain at break = 215%
c) ultimate tensile strength = 220 Pa; strain at break = 120%
Problem PP7.2.
a) E = σ / ε = 180 Pa / 0.30 = 600 Pa
b) E = σ / ε = 450 Pa / 0.15 = 3,000 Pa
c) E = σ / ε = 50 Pa / 0.25 = 200 Pa
d) E = σ / ε = 75 Pa / 0.30 = 250 Pa
Problem PP7.3.
Problem PP8.1.
1 Pa = 1 Pa
1 kPa = 1,000 Pa
1 MPa = 1,000,000 Pa
1GPa = 1,000,000,000 Pa
Problem PP9.1.
a) glassy: storage modulus = 15 MPa; loss modulus = 70 kPa
rubbery: storage modulus = 7 MPa; loss modulus = 80 kPa
b) glassy: storage modulus = 600 kPa; loss modulus = 140 kPa
rubbery: storage modulus = 130 kPa; loss modulus = 150 kPa
c) glassy: storage modulus = 320 kPa; loss modulus = 80 kPa
rubbery: storage modulus = 70 kPa; loss modulus = 70 kPa
Problem PP9.2.
a) 89 °C
b) 170 °C
c) 124 °C | textbooks/chem/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.10%3A_Chapter_Solutions.txt |
A free radical is an atom or group of atoms containing an unpaired electron. A free-radical reaction then is any chemical reaction in which a species with an unpaired electron is involved at some stage along the reaction pathway. Although a free radical conceivably can be a starting material or a product in a reaction, in practice, free radicals are reaction intermediates in nearly every instance.
02: Chain Reactions
A. The Initiation Phase
1. Thermal Initiation
Radical chain reactions usually occur in solutions maintained at 110 oC or less. Most compounds cannot initiate a reaction under these conditions because most molecules do not undergo bond homolysis rapidly enough at or below 110 oC to provide the supply of radicals needed for a productive chain reaction.3 Compounds that can initiate reaction under these conditions often contain a weak σ-bond (e.g., the O–O bond in benzoyl peroxide) that cleaves thermally to generate a pair of radicals (eq 1).
In addition to simple heating of a reaction mixture, thermal initiation also can be brought about sonochemically. Irradiation of homogeneous liquids with high-intensity ultrasound creates localized, superheated, sonochemical cavities in which radicals are generated by thermal reaction.4,5 Sonochemically initiated reactions can be conducted in solutions for which the temperature outside the sonochemical cavities is well below the 80-110 0C that is typical for many radical reactions; for example, sonochemical reaction of tri-n-butyltin hydride in a solution held at 22 oC initiates a chain reaction by homolytically cleaving a tin–hydrogen bond (eq 2).4
An appropriate initiator should provide a steady supply of radicals during the entire reaction. The needed supply will exist if the initiator has a lifetime comparable to the time required for completion of the reaction being conducted.3 [The lifetime of an initiator is often described in terms of the amount of time required for 50% of the material to react, that is, its half-life (t½).] Continuous formation of radicals is necessary because the time of existence of a typical radical chain is short, usually less than one second; consequently, new chains must be started regularly.3 If the half-life of an initiator is too short to provide the necessary supply of radicals during the entire reaction, either an initiator with a longer half-life can be used, or the initiator can be added to the reaction mixture continuously (or at regular intervals) during the reaction.
2. Thermal Initiators
a. 2,2'-Azobis(isobutyronitrile)
2,2'-Azobis(isobutyronitrile) (AIBN) is easily the most widely used initiator in radical reactions of carbohydrates. There are compelling reasons for this status. AIBN has a half-life of one hour at 85 oC (five hours at 70 oC);3,6,7 consequently, it can continuously supply sufficient initiating radicals at moderate temperatures for reactions requiring several hours to reach completion. Other advantages of AIBN are that it is easily handled, generates good yields of radicals (yields of radicals available for chain initiation are not 100% because some radicals combine before they can escape from the solvent cage), and has a rate of decomposition that is almost independent of the solvent.3
The 2‑cyano-2-propyl radical (3) generated from AIBN (eq 3) is relatively stable and usually reacts with bonds that are weaker than most C–H bonds; thus, hydrogen-atom abstraction from the solvent or carbohydrate reactant generally is not a complicating factor. Since the 2-cyano-2-propyl radical readily abstracts a hydrogen atom from a compound with a tin–hydrogen bond, AIBN is an excellent initiator for the frequently encountered reactions in which tri-n-butyltin hydride is the hydrogen-atom source.3
Another advantage of AIBN is that it does not experience induced decomposition; that is, the kinetics of its reaction are first order, regardless of the solvent or initiator concentration.6 (Induced decomposition in this case refers to the more rapid decomposition that sometimes occurs when an initiator undergoes bimolecular reaction in addition to the normal unimolecular one.)
b. 4,4'-Azobis(4-cyanovaleric Acid) (ACBA); 1,1'Azobis-(cyclohexanecarbonitrile) (ABCN); 2,2'-Azobis(2,4-dimethyl-4-methoxyvaleronitrile) (V-70); and Di-tert-butylhyponitrite (TBHN)
Azo compounds that are known to initiate radical reactions of carbohydrates are shown in Table 1. Even though the vast majority of reactions are initiated by AIBN, each of the other azo compounds listed in Table 1 has a characteristic that is useful in certain situations. The water soluble 4,4'-azobis(4-cyanovaleric acid) (ACBA) initiates reactions run in aqueous solution.15 1,1'-Azobis(cyclohexanecarbonitrile) (ABCN or ABC) has a longer half-life than does AIBN and, thus, is better suited for reactions that require higher temperature or extended reaction times.16,17 2,2'-Azobis(2,4-dimethyl-4-methoxyvaleronitrile) (V-70), in contrast, reacts rapidly enough in solution that it initiates reactions run at or near room termperature.10–13 Di-tert-butyl hyponitrite (TBHN) differs from other azo initiators in that it forms oxygen-centered radicals.14
c. Peroxides
The peroxides shown in Table 2 also are initiators for radical reactions of carbohydrates. These compounds are useful when the 2-cyano-2-propyl radical 3 (eq 3), or any carbon-centered radical generated from other initiators listed in Table 1, does not have the necessary reactivity to cause a chain reaction. Such a situation arises when an initiating radical is required to abstract a hydrogen atom from a carbon–hydrogen bond. The tert-butoxy radicals formed from di-tert-butyl peroxide (4) or 2,2-di-tert-butylperoxybutane18,20–23 (5) are effective at this type of hydrogen-atom abstraction. Sometimes reaction initiation requires a carbon-centered radical that can add to an O-thiocarbonyl group. One radical reactive enough to make this addition is CH3(CH2)9CH2· formed from dilauroyl peroxide (6) (Scheme 2).24–29 Benzoyl peroxide (7) and di-tert-butyl peroxide30–32 (4) also produce radicals that add to an O-thiocarbonyl group.
The ability of radicals formed from peroxides to abstract hydrogen atoms from carbon–hydrogen bonds can be both an advantage and a disadvantage. It is an advantage when such abstraction is necessary for reaction to proceed but a disadvantage when hydrogen-atom abstraction from the carbohydrate or the solvent causes undesired reaction. One undesired reaction is induced decomposition, which can be illustrated by considering the reactions of benzoyl peroxide. At low concentrations in an inert solvent the reaction of benzoyl peroxide follows the first-order kinetics expected for unimolecular reaction,6 but at higher concentrations or in the presence of reactive solvents benzoyl peroxide undergoes faster reaction.6,33–35 The half-life of this peroxide in ethyl ether in a sealed tube at 80 oC is five minutes rather than the one hour observed in benzene at 95 oC.6 The enhanced rate of reaction in ethyl ether is attributed to the induced decomposition that occurs when the radical created by abstraction of a hydrogen atom from the solvent by either C6H5· or C6H5CO2· reacts with benzoyl peroxide (Scheme 3).6
3. Photochemical Initiation
Photochemical initiation requires a compound to absorb a photon of light and then use this energy to cleave a bond homolytically and, in so doing, create an initiating radical (or radicals). It is desirable for the initiator to absorb visible or long-wavelength ultraviolet (UV) radiation because being able to absorb this type of light minimizes the possibility that another chromophore in a reactant molecule will be excited and undergo an unwanted photochemical reaction. Photochemical initiation is particularly useful for thermally labile substrates because reaction can be conducted at low temperatures.
Carbohydrates that form radicals as a result of light absorption include iodides, bromides, azides, selenides, hypoiodites, and esters of N-hydroxy-pyridine-2-thione.36 For hypoiodites (eq 4) and esters of N-hydroxypyridine-2-thione (eq 5) photolysis with visible light causes radical formation. Since very few functional groups in carbohydrates react as a result of absorbing visible light, there is little danger that this type of radiation will cause a competing photochemical reaction. For compounds that react with visible light, care must be exercised to protect them from premature reaction due to inadvertent light exposure. Since UV radiation is required for bond breaking in carbohydrate iodides, bromides, azides, and selenides, the possibility for undesired photochemical reaction due to exciting a different chromophore in the substrate is greater when irradiating one of these compounds.
Although radicals form from photolysis of all the compounds mentioned in the previous paragraph, the extent to which a chain reaction takes place depends on both the structure of the substrate and the reaction conditions. In the case of esters of N-hydroxypyridine-2-thione, for example, the bond homolysis shown in eq 5 initiates a chain reaction with a quantum yield that ranges from 6 to 35 when R=(CH2)14CH3 and from 19 to 34 when R=C6H11.37 The range of values for each compound is due to quantum yields being determined under different reaction conditions. (The quantum yield for a reaction is the number of molecules reacted for each photon absorbed; thus, for a radical chain reaction the quantum yield is a measure of the number of propagation cycles produced by each initiating radical.)
Bond homolysis is only one of the possible reactions that can take place when a molecule absorbs a photon of light. Other photochemical reactions (e.g., cycloaddition and geometric isomerization) do not involve radical formation. Reference 36 contains a discussion of both the radical-forming and nonradical-forming photochemical reactions of carbohydrates.
4. Photochemical Initiators
Direct photolysis of carbohydrates is not the only way to initiate their radical reactions photochemically. Irradiation of noncarbohydrates also produces radicals that cause carbohydrate reaction; thus, photolysis of azo compounds, ketones, and hexaalkylditins all generate radicals that initiate chain reactions.
a. 2,2'-Azobis(isobutyronitrile)
2,2'-Azobis(isobutyronitrile) (AIBN) has a maximum absorption at 345 nm in its UV spectrum and fragments when irradiated with light of this wavelength to produce nitrogen and two 2-cyano-2-propyl radicals (eq 3).3,7 Because many compounds are transparent to 345-nm light, it is often possible to generate 2-cyano-2-propyl radicals photochemically in a reaction mixture in which AIBN is the only light-absorbing compound. (In contrast to thermal reaction, photolysis of AIBN has the advantage that it can initiate reactions at or below room temperature.)
b. Acetone and Benzophenone
A characteristic of ketones, such as acetone and benzophenone, is that absorption of UV radiation produces an excited state (n,π*) that has considerable radical character on the carbonyl oxygen atom.38 As a result, many excited ketones have reactivity similar to that of alkoxy radicals; in particular, once a photon is absorbed, the excited ketone can abstract a hydrogen atom to begin a chain reaction.38,39 An example of the way in which this type of reaction takes place is shown in Scheme 4, where excited acetone abstracts a hydrogen atom from the S–H bond in ethanethiol to initiate an addition reaction.40
Benzophenone has a longer wavelength and more intense (n,π*) absorption than does acetone; consequently, it is easier to form excited benzophenone without having a carbohydrate reactant absorb the incident light. The presence of the photoproducts benzhydrol and benzpinacol, as well as unreacted benzophenone, can make product purification difficult.41
c. Hexaalkylditins
The chain-carrying radical in many reactions of carbohydrates is the tri-n-butyltin radical (8). This radical usually is generated during the initiation phase of a reaction when the 2‑cyano-2-propyl radical 3 abstracts a hydrogen atom from Bu3SnH (eq 6). Sometimes it is necessary to generate Bu3Sn· without Bu3SnH being present in the reaction mixture. In such a situation Bu3Sn· can be formed by photolysis of hexabutylditin with ultraviolet light (eq 7).42
5. Chemical Initiation
Thermal and photochemical initiation draw energy for radical formation from heating (vibrational excitation) and photon absorption (electronic excitation), respectively. Chemical initiation is different in that it uses energy stored in bonds to form initiating radicals; for example, triethylboron reacts with molecular oxygen to produce ethyl radicals (eq 8) that then initiate radical reactions.43–46 The driving force behind radical formation is the greater strength of the B–O bond [BDE = 519 kJ/mol (124 kcal/mol) in (EtO)3B] when compared to the B–C bond [BDE = 344 kJ/mol (82.2 kcal/mol) in Et3B].47 The activation energy for this reaction (eq 8) must be exceptionally low because Et3B–O2 can initiate reactions at temperatures as low as - 78 oC.47,48
Even though ethyl radicals are generated in a reaction involving molecular oxygen (eq 8), they also can react with oxygen molecules to form peroxy radicals (eq 9). Forming peroxy radicals diverts the ethyl radicals from their role as initiators, but this diversion only temporarily interrupts the initiation process because peroxy radicals react with triethylboron to generate new ethyl radicals (eq 10).47 As long as the amount of oxygen added to the system is kept well below the amount of triethylboron present, the ethyl radicals needed to initiate reaction will continue to be produced as oxygen is introduced.
Initiation of a reaction by triethylboron–oxygen shares an important feature with photochemical initiation; namely, both are able to generate radicals well below room temperature. The mild conditions for reactions initiated in this manner can lead to fewer side reactions and higher product yields.49 Triethylboron–oxygen initiation can be used in a broad range of situations that include reaction in aqueous solution.47
The ability of triethylboron–oxygen to initiate reactions at low temperature indicates that sometimes the sole purpose of heating a reaction mixture is to decompose the initiator. The reaction shown in eq 11, for example, proceeds well at room temperature, when initiated by triethylboron–oxygen. The only apparent reason the reaction must be conducted at higher temperature when AIBN replaces triethylboron–oxygen is to enable fragmentation of the initiator (eq 11).50
The hydrogen-atom donor in most radical reactions is either a tin or silicon hydride. The most frequently used tin hydride is Bu3SnH and the most common silicon hydride is (Me3Si)3SiH. The silicon hydride used in the reaction shown in eq 11 is a less common but effective hydrogen-atom donor.
B. The Propagation Phase
1. General Characteristics
The propagation phase for a chain reaction consists of a group of elementary reactions (also called reaction steps) that are repeated a number of times (Scheme 1). (Elementary reactions are discussed in Chapter 4.) Each repetition represents one reaction cycle, and the number of cycles completed for each initiating radical is the chain length. The final step in a propagation sequence completes the current cycle and generates the radical needed to begin a new cycle. The products from any radical chain reaction are determined by the identity and reactivity of the molecules and radicals participating in the propagation phase.
2. Examples of Propagation Sequences
The basic characteristics of a propagation sequence can be seen by examining two specific reactions. One of these is dehalogenation of a D-glucopyranosyl bromide with tri-n-butyltin deuteride51 (Scheme 5), and the other is addition of a D-glucopyranos-1-yl radical to acrylonitrile in the presence of tri-n-butyltin hydride (Scheme 6).52,53
a. Substitution
Reductive dehalogenation with tri-n-butyltin deuteride, a substitution reaction with two propagation steps, is a relatively simple chain process (Scheme 5).51 In the propagation phase for this reaction R· and Bu3Sn· alternate in carrying the chain forward by each reacting only with a particular type of reactant molecule; that is, Bu3Sn· reacts only with the glucosyl halide (RX), and the carbohydrate radical (R·) reacts exclusively with Bu3SnD. This discriminating reactivity has been called "disciplined behavior".54
b. Addition
Radical addition to an unsaturated compound is a more complex process than atom substitution because radical addition typically involves at least three propagation steps (Scheme 6, desired sequence). Introducing another step (in this case, addition of R· to CH2=CHCN) to the propagation sequence complicates the sequence not just by the existence of a third elementary reaction but also by the constraints this new reaction places on the entire process. These constraints are discussed in the next several paragraphs.
One requirement for success in the addition process shown in Scheme 6 is that R· must be reactive enough to add to acrylonitrile. Carbon-centered, carbohydrate radicals (R·) add rapidly to compounds with electron-deficient multiple bonds but these same radicals add only slowly (too slowly for observable reaction) to compounds with multiple bonds that are not electron-deficient. The reasons behind different rates of addition of R· to compounds with electron-deficient and electron-rich multiple bonds are discussed in Chapter 7. [If an addition reaction is internal (i.e., a radical cyclization), it sometimes will take place even if the multiple bond is not electron-deficient.]
A second requirement for a successful reaction is that R· exist long enough in solution to add to an unsaturated reactant. One reaction that could prevent this addition from taking place is chain termination (R· + R· $\rightarrow$ RR), but chain termination is unlikely to do so because the very low concentration of R· causes the rate of radical combination to be much slower than the rate of addition of R· to a compound with an electron-deficient multiple bond. A more probable reason for R· not adding to an unsaturated compound is that this radical undergoes hydrogen-atom abstraction from tri-n-butyltin hydride before addition can occur (Scheme 6, diverting sequence). Such an abstraction diverts reaction away from the desired product. The rate of addition of R· to an unsaturated compound, therefore, must be faster than its rate of hydrogen-atom abstraction from tri-n-butyltin hydride in order to keep the chain reaction from being directed into an unwanted propagation sequence (Scheme 6).
A third constraint placed on the overall addition process by the reaction of R· with acrylonitrile is that the adduct radical must not add to a second molecule of the nitrile before hydrogen-atom abstraction takes place. Although this competing addition sometimes does occur, in most instances hydrogen-atom abstraction is more rapid than a second radical addition.
Another way of describing the constraints placed on the desired sequence in Scheme 6 by an additional propagation step is in terms of chain-transfer reactions. In each propagation sequence the chain-transfer step ends the existing cycle and creates the radical that begins a new cycle. In the reactions shown in Scheme 6 the third step in the desired sequence and the second step in the diverting sequence are the chain-transfer steps for their respective reactions. Radical addition cannot be successful if the rate of the chain-transfer step in the diverting sequence is greater than the rate of the radical-transforming step (i.e., addition of R· to CH2=CHCN) in the desired sequence. In other words, the addition reaction shown in Scheme 6 will be a minor process if most of the time R· reacts with tri-n-butyltin hydride to produce RH (the simple-reduction product) before it adds to acrylonitrile.
3. Rate Constants and Reaction Rates
Analysis of the reaction shown in Scheme 6 can be done in a more quantitative manner using the rate constants shown in Scheme 7 and the information given in eq 12.52 (The rate constants in Scheme 7 are a few of the many listed in the tables located in Chapter 8.) The first step in the formation of any product in this reaction is bromine-atom abstraction by the tri-n-butyltin radical (eq 13). (Equations 13-17 are found in Scheme 7.) After bromine-atom abstraction, the major reaction product (10) is formed by a combination of two steps, addition of the D-glucopyranos-1-yl radical 13 to acrylonitrile to give the adduct radical 14 (eq 14) and abstraction by 14 of a hydrogen atom from tri-n-butyltin hydride (eq 15). These reaction steps (equations 14 and 15) need to take place in preference to those that produce the side products 11 and 12 (eq 12). Which product will be the major one and which will be the minor ones in this reaction is determined by reactant concentrations and rate constants for the various, possible elementary reactions.
Unfortunately, not all of the rate constant information needed to analyze the reaction shown in eq 12 is available; consequently, in order to proceed it is necessary to estimate some rate constants based upon information from reaction of model radicals. For example, the rate constant for addition of the pyranos-1-yl radical 13 to acrylonitrile (eq 14) is assumed to be similar to that for addition of the model radical ·CH2OH to this nitrile.55 The rate constant for this reaction (eq 14) is smaller than that for hydrogen-atom abstraction by a typical carbon-centered radical from tri-n-butyltin hydride (TBTH) (eq 16);56 therefore, if acrylonitrile and TBTH are present in equal amounts, the major reaction product should be compound 11. The desired reaction (eq 14) will be favored, however, if acrylonitrile is present in much greater amount than TBTH. In the reaction shown in eq 12 the ratio of acrylonitrile to TBTH (5/1) is large enough to make 10 the major product (58%).52 The 21% yield of 11, however, shows that even with an excess of acrylonitrile there still is substantial reaction of the radical 13 with tri-n-butyltin hydride (eq 16).
There is a limit to how large the ratio of acrylonitrile to tri-n-butyltin hydride can become before adding more nitrile to the reaction mixture is counterproductive. This limit will be reached when reaction of 14 with acrylonitrile (eq 17, k = 1 x 103 M‑1 s-1 )57 competes effectively with reaction of this radical (14) with TBTH (eq 15, k = 2 x 106 M-1 s-1).56 The difference in rate constants for these two reactions is so great, however, that having acrylonitrile present in fivefold excess is possible without significant formation of the radical 15 (eq 17) and its hydrogen-atom abstraction product 12 (eq 12). The considerable difference in the rate constants for the reactions shown in eq 14 and eq 17 is due primarily to a difference in radical philicity; that is, 13 is more nucleophilic than 14. Radical philicity and its effect on radical reactivity are discussed in Chapter 7.
4. Solvent Effects
Although solvent selection often plays a decisive role in ionic reactions, strong solvent effects in radical reactions are less common because radical centers are relatively nonpolar and, thus, are not highly solvated.3 A similar statement applies to transition states in radical reactions because most reactions do not develop much, if any, separation of charge at the transition state; hence, stabilization by polar solvents is usually not a significant factor.
There are some situations in which solvent effects are significant in radical reactions.58 One of these occurs during phosphatoxy group migration.59 The rate determining step in this reaction involves formation of a polarized transition state on its way to becoming an ion pair (Scheme 8); consequently, reaction is faster in solvents that stabilize charge separation. Acyloxy and phosphatoxy group migrations are similar in many ways, including more rapid reaction in polar solvents.
C. The Termination Phase
Any process that stops the participation of a radical in a propagation sequence terminates the reaction chain. In typical chain reactions, such as those described in Schemes 1, 4, 5, and 6, radical combination and disproportionation are terminating steps. Chain termination in this way is limited primarily by how rapidly radicals diffuse through a solution.60,61 Successful chain reactions have concentrations of radicals low enough (approximately 1 x 10-7 M)57,62 to prevent even diffusion controlled processes from competing effectively with chain propagation. Because low radical concentrations make chain-terminating reactions less competitive, adding only a limited amount of an initiator with an appropriate half-life to a reaction mixture generates the small but steady supply of radicals required to maintain reaction and avoid premature chain termination.
Chain termination can occur when a participating radical is transformed into a radical that is not involved in the reaction sequence.63 A common way for this to happen is to have a carbon-centered radical (R·) combine with a molecule of oxygen to give a peroxy radical (ROO·) (eq 18). Formation of ROO· would terminate propagation sequences such as those shown in Schemes 5 and 6 because a peroxy radical is not a participant in either sequence. (Forming a peroxy radical by reaction with oxygen not only terminates a desired propagation sequence, but it also creates a reactive radical that is capable of becoming part of a different reaction sequence.) Because reaction of carbon-centered radicals with molecular oxygen occurs at or near diffusion control (k = 109-1010 M-1s-1),64 minimizing chain termination by oxygen, requires radical reactions to be conducted in an inert atmosphere such as that provided by argon or nitrogen.
Even when a reaction is conducted in an inert atmosphere, it is nearly impossible to eliminate all traces of oxygen from a reaction mixture. A small amount of oxygen can cause initial reaction chains to have short lengths. Once the oxygen concentration falls to a low enough level that premature chain termination is no longer significant, reaction can proceed without noticeable inhibition by oxygen. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/02%3A_Chain_Reactions/II._Basic_Stages_of_a_Radical_Chain_Reaction.txt |
Reaction efficiency is a relative concept. Reference to efficiency can take the form of a statement that one reaction is more efficient than another or that a reaction is an efficient process. Establishing a dividing line between efficient and inefficient reactions involves an arbitrary decision; nevertheless, it is sometimes helpful in discussing chain reactions to use the term efficiency and to associate a numerical value with it. Choosing chain length to measure efficiency can provide this number; thus, one way to define an efficient reaction is as one that has a chain length greater than 100.65
The efficiency of a chain reaction is determined by its relative rates of propagation (rp) and termination (rt). A reaction becomes more efficient as the ratio rp/rt increases; thus, the chain length in a reaction differs significantly when rp /rt is 10/1 as opposed to when it is equal to 1/1 (Figure 1). When rp /rt is 1/1, only 12.5% of the initiating radicals begin a chain destined to have a length greater than 2, but if the ratio of rpto rt is raised to 10/1, more than 75% of the initiating radicals produce chains with lengths greater than 2 (Figure 1). Even when rp/rt is 10/1, the reaction would not be described as an efficient one because a chain length of 100 in such a reaction would be a rare event. If, on the other hand, rp/rt is equal to 1000/1, nearly every initiated chain will complete two cycles (Figure 1) and most chains will have a length greater than 100; thus, the reaction is an efficient one. One desirable characteristic of a reaction with a long chain length is that converting all the starting material into product requires only a small amount of initiator.
It is informative to consider some actual numbers for rates of propagation (rp) and termination (rt) reactions to better appreciate how these rates determine chain length. One way to do this is to analyze a typical reaction such as the dehalogenation process shown in Scheme 9. Since the rate determining propagation step in this type of reaction is hydrogen-atom abstraction by R· from Bu3SnH,66,67 the rate of propagation (kp) is given by eq 19. A typical value for the rate constant (kp) for this reaction is 2 x 106 M-1 s-1.56 Since combination of two R· radicals is assumed to be the only significant termination process, the rate of termination is given by eq 20. The rate constant for termination (kt) will depend upon how rapidly these radicals come together in solution, that is, upon their rate of diffusion. The rate constant for diffusion is 1 x 1010 M-1 s-1 in benzene,68 a common solvent for such reactions. A typical radical concentration in a chain reaction is less than 1 x 10-7 M.57,62 A normal concentration for a hydrogen-atom donor is one molar. Based upon these numbers and the assumptions made about this reaction, dehalogenation with tri-n-butyltin hydride as the hydrogen-atom donor would be quite efficient because rp /rt would be approximately 2000 (eq 21).
Although discussion of chain reactions naturally focuses on how to increase efficiency, it is informative to examine the other end of the efficiency spectrum to determine the problems associated with inefficient reactions. As reactions become less efficient, more chains must be started in order for reaction to reach completion; consequently, more initiator must be added to the reaction mixture to compensate for the decrease in the average chain length. Increasing the initiator concentration increases the possibility of side reactions involving the initiator and radicals present in solution. As efficiency drops, chain termination products become more abundant. When this happens, product purification sometimes is more difficult. There are clear disadvantages to inefficient reactions.
A guideline for deciding when a chain reaction becomes too inefficient to be synthetically useful can be formulated in terms of rates of propagation and termination reactions.57 A criterion for usefulness is that the rate of propagation for a reaction (eq 19) should be greater than its rate of termination (eq 20).57 (This means that the first possibility pictured in Figure 1 would be just on the “wrong side” of the line for synthetic usefulness.) This dividing line is a reasonable one because any reaction in which an undesired product is formed at a faster rate than the desired product (rt > rp) is unlikely to be effective in synthesis.
Using the guideline that rp /rt should be greater than unity in a synthetically useful reaction sets a lower limit on the rate constant for propagation (kp).57 This limit is based on the following assumptions: (a) chain termination is a diffusion controlled process for which the rate constant (kt) is between 1 x 109 and 1 x 1010 M-1 s-1;68 (b) the radical concentration in a typical reaction is approximately 1 x 10-7 M;57,62 and (c) the concentration of the hydrogen-atom donor Bu3SnH is 1 M. Based on these assumptions, the rate constant for propagation (kp) for a synthetically useful reaction should be greater than 1 x 102 M-1 s-1 (eq 22 and eq 23). This limiting value for kp would hold for any reaction that satisfies the assumptions (a)-(c).
IV. Summary
Free-radical processes can be divided into chain and nonchain reactions. Chain reactions consist of initiation, propagation, and termination phases. A similar set of reactions (i.e., radical formation, transformation of one radical into another, and radical disappearance) also occurs in nonchain processes. The difference in these two types of reaction is that in a chain reaction the transformation of one radical into another also creates the radical needed to start the transformation process anew, but for nonchain reactions each radical formed causes only one “trip” through the transformation cycle. The most widely used initiator in chain reaction is 2,2'-azobis(isobutyronitrile), a compound that provides the continuous supply of radicals needed to sustain a typical reaction; that is, a reaction that takes place over a period of several hours at 80-110 oC. Peroxides also are thermal initiators, but they are less commonly used because they produce reactive radicals that can cause undesired side reactions. Triethylboron–oxygen, ultrasound, and light all initiate radical reactions and have the added advantage that they can be used in reactions that are conducted at or below room temperature.
At the core of a chain reaction is the propagation phase, the part of the reaction where reactant molecules are converted into products. Each propagation sequence consists of a group of elementary reactions. Successful propagation depends upon the ability of each participating radical to react selectively with only one type of molecule present in the reaction mixture.
The final phase in a chain reaction is termination. Chain reactions are terminated by any process, such as radical combination, that removes a participating radical from the propagation sequence. Reaction efficiency is a measure of how long a typical chain reaction continues before termination takes place. An efficient reaction is generally regarded as one with a chain length greater than 100. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/02%3A_Chain_Reactions/III._Reaction_Efficiency.txt |
Although many radical reactions in carbohydrate chemistry are chain processes, nonchain reactions also play a significant role in the chemistry of these compounds. As mentioned at the beginning of Chapter 2, chain and nonchain reactions each involve radical formation, transformation, and disappearance. The difference is that for chain reactions the transformation cycle typically is repeated many times for each initiating radical, but for nonchain reactions each radical formed causes transformation to take place only once. The radicals that participate in nonchain reactions sometimes are formed by bond homolysis but more often are produced by electron transfer. Bond homolysis is usually a photochemical reaction. Electron transfer typically involves transition-metal-generated radicals.
03: Nonchain Reactions
Reactions of radicals generated from transition-metal complexes can be divided into two types based on the direction of electron flow. In some of these reactions the transition metal accepts an electron during radical formation (oxidative electron transfer) and in others it donates an electron during this process (reductive electron transfer). The compounds that most often participate in oxidative electron transfer are manganese(III) acetate [Mn(OAc)3] and ammonium cerium(IV) nitrate [(NH4)2Ce(NO3)6], while those frequently involved in reductive electron transfer are bis(cyclopentadienyl)titanium(III) chloride (Cp2TiCl), and samarium(II) iodide (SmI2). Carbohydrates that are bonded to a cobalt-containing complex by a C–Co bond form radicals by oxidative electron transfer and then frequently reform a C–Co bond by reductive electron transfer.
Coenzyme B12 (5, Figure 1) is one of a group of biologically active molecules that have similar structures.7 Each member of this group has a cobalt atom surrounded by a macrocyclic ligand (a corrin ring) that bears various substituents. In addition to the corrin ring the cobalt atom in each of these compounds also is coordinated with a ligand that contains a phosphate group, a sugar moiety, and a nitrogenous base. Compounds related to 5 differ from each other in the structure of the R group attached to cobalt. R represents the 5'-deoxyadenosyl group in coenzyme B12 (5), but for related compounds R can be as structurally simple as a methyl or hydroxyl group.7
The original stimulus for study of carbon–cobalt bond homolysis as a pathway for forming carbon-centered radicals came from investigation of the reactions of coenzyme B12 (5).8–10 In biological systems enzyme-induced homolysis of the carbon–cobalt bond in 5 produces the 5'-deoxyadenosyl radical 6 and the cobalt-centered radical 7 (B12r, eq 1).8–10 In experiments outside biological settings the 5'-deoxyadenosyl radical (6) is produced from coenzyme B12 (5) by photolysis with visible light.11 When photolysis is conducted in the absence of an effective hydrogen-atom donor or other radical trap, cyclization follows homolysis of the carbon–cobalt bond (Scheme 5).8-10,12
b. Cobaloxime Complexes
The discovery that carbon–cobalt bond homolysis in coenzyme B12 (5) produced the carbon-centered radical 6 (eq 1), led to investigation of simpler molecules that could model this behavior. Cobaloximes are one of several types of compounds found to be effective choices for this role.13–16 Carbohydrate cobaloximes 8 and 9 produce radicals 10 and 4, which recombine in the absence of radical traps (Scheme 6).13 In the presence of compounds that react with radicals, 10 and 4 undergo characteristic radical reactions; thus, the D-mannopyranos-1-yl radical 10 adds to acrylonitrile (11) to give the adduct radical 12, which then combines with ·Co(dmgH)2py (4) to form the addition product 13 (Scheme 7).13
A necessary condition for the reaction shown in Scheme 7 is that 4 [Co(dmgH)2py] be stable enough to remain unchanged while the addition of 10 to 11 is taking place. The needed stability of 4 derives from protection of its radical center by the attached ligands; thus, 4 can be viewed as a persistent radical.
c. The Persistent-Radical Effect
Persistent radicals, such as ·Co(dmgH)2py (4), are responsible for a type of reactivity known as the persistent-radical effect.17–19 This effect causes a reaction that generates a persistent radical (R1·) and a transient radical (R2·) in equal amounts to give a higher yield of the cross-coupling product (R1R2) than would be expected from random radical coupling. The explanation for greater cross-coupling product formation begins with the recognition that although persistent and transient radicals are formed in equal amounts, this equality is short lived. Due to the reactive nature of transient radicals their concentration decreases more rapidly in the early stages of a reaction than does the concentration of persistent radicals. (Transient radicals combine, disproportionate, and undergo other reactions much more rapidly than persistent radicals.) The rapidly developed, higher concentration of persistent radicals in the early stages of reaction means that any newly formed, transient radical is more likely to encounter and combine with a persistent radical than with another transient one; in other words, the cross-coupling product R1R2 becomes the major coupling product.
An example of the persistent radical effect is shown in the reaction given in Scheme 4, where carbon–cobalt bond homolysis in 1 or 2 produces the persistent radical 4 and the transient radical 3. Even with the extended heating or photolysis needed to reach equilibrium, there was no evidence of formation of a coupling product other than the cross-coupling products 1 and 2. The persistent radical effect also is operative in the addition reaction shown in Scheme 7. In this case the transient radical 12, produced by addition of 10 to acrylonitrile (11) , and the persistent radical 4 combine to form the only radical-coupling product isolated.
2. Carbon-Mercury Bond Homolysis
There are similarities in reactivity among compounds with C–Co and C–Hg bonds. Both bonds are strong enough to exist in stable structures at room temperature but both readily cleave upon photolysis. The result in each case is formation of a metal-centered and a carbon-centered radical. Carbon-centered radicals produced by carbon–mercury bond homolysis undergo typical radical reactions, such as the hydrogen-atom abstraction shown in Scheme 8.20
3. Manganese(III) Acetate [Mn(OAc)3] Reactions
Carbon-centered radicals can be generated by reaction of manganese(III) acetate with CH-acidic compounds such as the β-diketone shown in Scheme 9.21–24 The first step in this process is formation of the enolate 15.23 In the presence of an unsaturated compound two mechanisms for reaction of 15 are considered to be possible. In the first of these electron transfer forms manganese(II) acetate and the resonance-stabilized radical 16, which then adds to an unsaturated compound. A second possible pathway for addition is a concerted process in which the enolate 15 reacts directly with the unsaturated compound to produce the adduct radical 17 (Scheme 9).23 Reaction by either of these pathways is believed to take place by inner-sphere electron transfer.
Since radical centers with two, attached carbonyl groups are electrophilic, radicals such as 16 (Scheme 9) add most easily to unsaturated compounds with electron-rich multiple bonds.22 This is the point at which carbohydrates typically become involved in reactions begun by manganese(III) acetate because glycals have electron-rich π systems that are attractive targets for addition of electrophilic radicals; for example, the radical 19, formed by reaction of dimethylmalonate (18) with manganese(III) acetate (eq 2), adds to the tri-O-acetyl-D-glucal 20 to produce the stereoisomeric radicals 21a and 21b (Scheme 10).25,26 This addition, which occurs regioselectively at C-2, is followed by oxidation of the resulting radicals with a second molecule of manganese(III) acetate to give the corresponding cations 22a and 22b. These cations react with the solvent (acetic acid) to yield the final products (23a, 23b, 24a, and 24b). Manganese(III) acetate, therefore, is involved in both the formation and disappearance of the radicals in this reaction. (Electrophilic radicals and other aspects of radical philicity are discussed in Chapter 7.)
Manganese(III) acetate has a more complicated structure than the formula Mn(OAc)3 indicates. It is an oxo-centered trimer of three manganese ions held together by six bridging acetates.27–29 Three representations for this structure are shown in Figure 2. It is often convenient in discussing reactions of this compound to use the abbreviated formula Mn(OAc)3.
4. Ammonium Cerium(IV) Nitrate [(NH4)2Ce(NO3)6] Reactions
Reaction of CH-acidic compounds with ammonium cerium(IV) nitrate generates electrophilic, resonance-stabilized radicals in a manner similar to reaction with manganese(III) acetate.30,31 As mentioned in the previous section, these radicals add readily to the electron-rich double bonds such those found in glycals (eq 3).30 Oxidation of CH-acidic compounds with ammonium cerium(IV) nitrate to produce electrophilic radicals has the advantage, when compared to reactions with manganese(III) acetate, of being able to be conducted at or below room temperature. [The reactions of manganese(III) acetate and ammonium cerium(IV) nitrate are discussed further in Chapter 21 of Volume II.]
III. Photochemically Generated Radicals
Although, as described in Chapter 2, photolysis sometimes initiates chain reactions, it also can produce radicals that undergo nonchain reactions. In a photochemically initiated chain reaction the number of photons that must be absorbed to cause complete reaction typically is far smaller than the number of molecules reacted. (A radical formed by absorption of one photon can begin a chain that produces many product molecules.) In a nonchain reaction the number of photons absorbed typically must be at least equal to the number of molecules reacted. Actually, it is rare that each, absorbed photon causes a reaction to take place because reaction is only one of the ways an excited molecule dissipates its energy; consequently, for complete reaction to occur in a nonchain process the number of photons absorbed often greatly exceeds the number of molecules reacted.
Although cleavage of weak carbon–metal bonds (e.g., carbon–cobalt bonds) tends to occur readily upon photolysis, photochemical processes do not require a reactant to have a weak bond in order for bond homolysis to take place. When ultraviolet light is absorbed by a compound, enough energy is present in the excited system to break even strong bonds.
IV. Thermally Generated Radicals
Heating of carbohydrates has a limited role in causing useful radical reactions. Few carbohydrates or their derivatives have bonds weak enough to generate radicals at temperatures that avoid general structural decomposition. As described earlier in this chapter, a carbohydrate derivative with a carbon–cobalt or oxygen-iodine bond can generate radicals by thermal reaction, but even for such compounds radical formation usually takes place photochemically.
V. Summary
Transition-metal-generated radicals are involved in most nonchain, radical reactions of carbohydrates. In some of these reactions the transition metal accepts an electron, and in others it is an electron donor. The carbohydrate radicals thus produced undergo typical radical reactions, such as addition to a double bond and hydrogen-atom abstraction. Manganese(III) acetate and ammonium cerium(IV) nitrate both react with CH-acidic compounds, such as those with β-dicarbonyl substituents, to produce electrophilic radicals that add readily to electron-rich double bonds (e.g., those present in glycals). Bis(cyclopentadienyl)titanium chloride (Cp2TiCl) reacts with glycosyl halides to produce pyranos-1-yl radicals. In the absence of a radical trap these radicals generate anomeric mixtures of glycosyl titanium compounds that undergo β‑elimination to form glycals. Radical intermediates also are produced when Cp2TiCl causes reductive opening of epoxide rings. The samarium(II) iodide–hexamethylphosphoramide (SmI2–HMPA) complex often serves as an electron donor in radical-forming reactions where a carbohydrate sulfone or halide is the electron acceptor.
Organocobalt and organomercury compounds generate radicals by carbon–cobalt and carbon–mercury bond homolysis, respectively. These compounds form carbon-centered radicals by both thermal and photochemical reaction. Carbon–cobalt bonds also undergo enzymatic cleavage, but in nonbiological settings photochemical bond homolysis is most common.
Photolysis of a variety of carbohydrates produces radicals that participate in nonchain reactions. Excited carbonyl compounds generate radicals by hydrogen-atom abstraction and by C–C bond fragmentation. Oxygen–iodine bonds cleave homolytically upon photolysis to produce highly reactive, alkoxy radicals. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/03%3A_Nonchain_Reactions/II%3A_Transition-Metal-Generated_Radicals.txt |
Radical reactions are composed of sequences of elementary reactions. An elementary radical reaction is one that does not produce an intermediate. Most elementary, radical reactions consist of transformation of one radical into another, but those that create radicals from nonradicals or cause radicals to disappear also qualify. Where radicals are concerned, an elementary reaction either creates a radical from a nonradical, transforms one radical into another, or causes a radical to disappear.
04: Elementary Reactions
An elementary reaction is one that has no intermediates. Every reaction that forms an intermediate actually is a combination of two or more elementary reactions. Where radicals are concerned, an elementary reaction either creates a radical from a nonradical, transforms one radical into another, or causes a radical to disappear.1,2
For chain reactions the propagation phase always contains at least two elementary reactions (Scheme 1). Nonchain reactions are similar in that they also contain at least two elementary reactions (Scheme 2). The elementary reactions upon which the free-radical chemistry of carbohydrates is based are listed in a general form in Table 1.1,2 Specific examples are given in the discussion of each reaction that takes place in this chapter. In describing these reactions the term “carbohydrate radical” (CARB·) refers to a radical centered on one of the atoms, usually carbon, in a carbohydrate.
02. Atom Abstraction
A. Halogen-Atom Abstraction
Halogen-atom abstraction can take place both in forming a halogenated carbohydrate and in removing a halogen atom from such a compound. When abstraction is from a halogenated carbohydrate, it produces a carbohydrate radical (eq 1). The abstracting radical typically is tin-centered or silicon-centered. In the reaction shown in eq 2, for example, a carbohydrate radical forms when a tin-centered abstracts an iodine atom from a deoxyiodo sugar.3 Abstraction that generates a halogenated carbohydrate takes place when I2, Br2, or another halogen donor reacts with a carbohydrate radical (eq 3). Equation 4 describes a reaction of this type.4
B. Hydrogen-Atom Abstraction
Hydrogen-atom abstraction is an elementary reaction that permeates the free-radical chemistry of carbohydrates. Because it is the final propagation step in many chain reactions, hydrogen-atom abstraction often converts a carbon-centered radical into a stable product. The hydrogen-atom donor in such reactions usually is a tin or silicon hydride, but sometimes a thiol or selenol serves in this role (eq 5). The final step in the simple reduction shown in eq 6 is a typical, hydrogen-atom-abstraction reaction.3
Carbohydrates also can serve as hydrogen atom donors (eq 7). A radical centered on a bromine, chlorine, or oxygen atom (and, sometimes, on a sulfur or carbon atom) is able to abstract a hydrogen atom from a carbohydrate in an elementary reaction that can be highly regioselective. For intermolecular reactions this selectivity is due to radicals preferentially abstracting the hydrogen atoms that produce the most stable carbon-centered radicals. In the reaction shown in eq 8, for example, the bromine atom abstracts only the hydrogen atom that produces the highly resonance-stabilized radical 1.4
If a radical is centered on an oxygen or carbon atom in a carbohydrate, internal abstraction becomes a possibility. Such abstraction is regioselective not only because a more stable radical is being produced but also because the radical center is able easily to come within bonding distance of a limited number of hydrogen atoms (sometimes only one). In the reaction shown in eq 9, the only hydrogen atom abstracted is the one that is 1,6-related to the radical center.5 Although an oxygen-centered radical (e.g., 2 in eq 9) is reactive enough to abstract a hydrogen atom from any carbon-hydrogen bond in a carbohydrate,6 only the most reactive carbon-centered radicals (e.g., primary and vinylic ones) are capable of such reaction (eq 10).7 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/04%3A_Elementary_Reactions/01._Introduction.txt |
Group abstraction is possible when a group is attached to a carbon atom in a carbohydrate by a bond to a sulfur, selenium, or tellurium atom (eq 11). The abstracting radical in nearly every instance is tin-centered (eq 12)8 or silicon-centered (eq 13).9 Although group abstraction reactions can be elementary ones, they cease to be so if an intermediate with a hypervalent atom forms (Scheme 3). (Generating a hypervalent atom causes abstraction to become a combination of radical addition and α-fragmentation.) Only rarely is a hypervalent atom believed to be involved if a sulfur or selenium atom provides the link to the carbohydrate framework.10,11 (An example of a noncarbohydrate, selenide reaction that appears to involve a hypervalent selenium atom is described in Chapter 8, Section III.B.) Where the connection is to a tellurium atom, computational investigations indicate that a radical with a hypervalent atom is likely form.10
04. Radical Addition
A. Intermolecular Reaction
1. Addition to a Multiple Bond
Addition of a carbon-centered radical to a multiple bond in an unsaturated compound is an elementary reaction that forms a new carbon–carbon bond (eq 14 and eq 15). One way for this to happen is for a radical centered on one of the carbon atoms in a pyranoid or furanoid ring to add to an unsaturated noncarbohydrate. Examples of this type of addition are found in the reactions shown in eq 16,12 where a radical is centered on C-2 in a pyranoid ring, and eq 17,13 where the radical center is on C-3' in a furanoid ring. It is also possible to have the radical center located on a carbon atom that is in an open-chain structure (eq 18).14 Addition can involve a noncarbohydrate radical or a carbohydrate radical adding to an unsaturated carbohydrate (eq 1915 and eq 2016, respectively). When the radical and the compound to which it is adding are both carbohydrates, reaction creates complex structures quickly (eq 20).
Heteroatoms play a role in radical addition when a radical centered on a nitrogen, phosphorous, silicon, sulfur, or tin atom adds to an unsaturated carbohydrate; for example, a nitrogen-centered radical is involved in the reaction shown in eq 19,15 and the adding radical in the reaction pictured in eq 21 is sulfur-centered.17 Radicals also add to unsaturated carbohydrates in which the multiple bond contains one or two heteroatoms. An example of this type of reaction is given in eq 22, where addition is to a carbon–sulfur double bond.18 Similar radical addition reactions occur with carbohydrates containing carbon–oxygen, carbon–nitrogen, and nitrogen–oxygen multiple bonds.
2. Addition That Forms a Radical with a Hypervalent Atom
Although most addition reactions consist of a radical adding to a multiple bond, reaction that does not involve a double or triple bond also can take place. This happens when addition produces a radical in which an atom has an expanded octet (eq 23). Such a reaction is thought to take place when a telluride, such as 3, reacts with a methyl radical (eq 24).19
B. Intramolecular Reaction (Radical Cyclization)
Radical cyclization (eq 25) is an intramolecular version of radical addition that merits special mention due to the synthetic importance of new ring formation. Five- and six-membered rings are created most often, but larger rings also can be produced. A typical radical cyclization reaction is shown in eq 26.20 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/04%3A_Elementary_Reactions/03._Group_Abstraction.txt |
A. β-Fragmentation
Homolytic β-fragmentation of a radical is an elementary reaction that cleaves a bond to one of the atoms adjacent to a radical center. (Other names for β-fragmentation are β-cleavage and β-scission.) This type of reaction sometimes produces an unsaturated carbohydrate by expelling a noncarbohydrate radical (eq 27), and other times it gives a carbohydrate radical and an unsaturated noncarbohydrate (eq 28). Equation 29 illustrates the first of these possibilities with a reaction in which the radical 4 fragments to give C6H5SO2· and the unsaturated carbohydrate 5.21 Being able to form a stabilized radical such as C6H5SO2· is an essential factor in this reaction.
A β-fragmentation reaction producing a carbohydrate radical and an unsaturated noncarbohydrate is one driven by formation of a compound with a thermodynamically stabilized multiple bond (usually a carbon–oxygen double bond) and a radical that also is stabilized (usually by an oxygen atom attached to the radical center). The reaction shown in eq 30 fits this pattern because it produces formaldehyde and the oxygen stabilized radical 6.22 Forming an aromatic ring is another way for providing a substantial driving force for β‑fragmentation (eq 31).23 A further option for β-fragmentation is ring opening, a possibility that presents itself when a radical is centered on an atom attached to the ring (eq 32).24
B. Heterolytic β-Fragmentation
When a radical is centered on a carbon atom that has an effective nucleofuge attached to a neighboring carbon atom, the possibility exists for formation of a radical cation (eq 33). The bond from the neighboring carbon atom to the leaving group needs to be one that does not cleave homolytically with ease; otherwise, β-fragmentation producing ionic intermediates could be preempted by homolytic fragmentation. Heterolytic β-fragmentation occurs in the reaction shown in eq 34.25
C. α-Fragmentation
α-Fragmentation is an elementary reaction in which a bond attached to a radical center cleaves homolytically. This reaction is rare because it requires the energy-demanding step of bond breaking without the energetic compensation of bond formation. One situation in which α-fragmentation takes place is in the formation of the isonitrile and stabilized, sulfur-centered radical shown in eq 35.26 A second occurs in the fragmentation of the hypervalent radical shown in eq 36.19
D. Bond Homolysis
Bond homolysis either produces a pair of radicals (eq 37), or if the bond being broken is part of a ring system, a diradical. Thermal reaction cleaves the weakest bond in a molecule; thus, when the cobaloxime 7 is heated, the carbon–cobalt bond, one of the weakest covalent bonds known, breaks homolytically at temperatures well below those necessary for cleavage of other bonds in the molecule (eq 38).27 This bond homolysis involves electron transfer with cobalt acting as the electron acceptor.
Photochemical reaction offers a range of possibilities for bond homolysis (eq 39). Success depends both upon a compound being able to absorb the incident light and on this light supplying sufficient energy for bond breaking. Absorption of visible light provides the energy needed to cleave weaker covalent bonds, such as the iodine–oxygen bond in the reaction shown in eq 40.28 UV radiation is energetic enough to break stronger bonds, such as the carbon–carbon bonds in the reactions pictured in eq 41.29
Unlike thermal reaction, bond breaking during a photochemical process does not necessarily cleave the weakest bond in a molecule. Selectivity in bond breaking during photolysis results from a combination of factors that control the reactivity of electronically excited molecules. In the reaction shown in eq 41, for instance, excitation energy is quickly localized in the keto group in the substrate. This localization leads to one of the characteristic reactions of an excited aldehyde or ketone, namely, breaking the bond between the carbonyl carbon atom and one of its attached carbon atoms.30 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/04%3A_Elementary_Reactions/05._Fragmentation_Reactions.txt |
A. Reactions of Carbohydrate Radicals
Transition-metal complexes can act as electron transfer agents when reacting with carbohydrate radicals (eq 42 and 43). In the reaction shown in eq 44, titanium donates an electron during formation of the carbon-titanium bond between titanocene(III) chloride (Cp2TiCl) and the pyranos-l‑yl radical 8.31,32 In the reaction shown in eq 45, ammonium cerium nitrate [(NH4)2Ce(NO3)6] accepts an electron from the pyranos-1-yl radical 9 to convert it into the corresponding cation.33,34 Other compounds that serve as electron donors in reactions with carbohydrate radicals are SmI2, Cr(EDTA)2-, [Ru(bpy)3]2+, and Co(dmgH)2py. (The structures of the ligands in these compounds are pictured in Figure 1). In addition to (NH4)2Ce(NO3)6, Mn(OAc)3 also acts as an electron acceptor in reactions with carbohydrate radicals.
B. Formation of Carbohydrate Radicals
When an electron is transferred to a carbohydrate chloride, bromide, iodide, or sulfone, the resulting radical anion reacts to form a carbohydrate radical (Scheme 4). The two electron donors participating in most of these reactions are SmI2 (eq 46)35 and Cp2TiCl (eq 47),36 but other transition-metal complexes [e.g., Cr(EDTA)2-] are able to function in this capacity. Although the radical anion 10 is pictured in Scheme 4 as a discrete intermediate, in some instances cleavage of the RX bond may be simultaneous with electron transfer.
Solvated electrons, which are more reactive as electron donors than transition-metal complexes, combine with esterified carbohydrates to produce radical anions. These radical anions then expel carboxylate anions to form carbohydrate radicals (eq 48).37
Reaction of an aldehyde or ketone with samarium(II) iodide produces a samarium ketyl (eq 49),38 an intermediate considered to be a hybrid of structures 11-13 (Figure 2). These ketyls exhibit reactivity characteristic of carbon-centered radicals.
07. Radical Combination
Radical combination involving carbohydrates takes place either by reaction between two carbohydrate radicals or between a carbohydrate and a noncarbohydrate radical (eq 50). Successful radical combination requires that the rates of competing reactions (e.g., hydrogen-atom abstraction and radical addition) be reduced to the point that radicals exist long enough in solution to combine. Under most conditions the lifetimes of typical carbohydrate radicals are too short for two of them to diffuse through solution and react. If conditions are selected to minimize competing reactions, pyranos-1-yl radicals, which are among the most stable carbohydrate radicals, exist in solution long enough to come into contact with each other and thus form dimers, although in low yield (eq 51).39 If conditions are chosen that produce large numbers of radicals in a short period of time, radical concentration can be raised to the point where substantial combination takes place (Scheme 5).40
The noncarbohydrate radicals taking part in radical combination have a range of possible structures. The basic requirement in most instances is that the noncarbohydrate radical be sufficiently stable that its concentration in solution builds to the point that it will combine quickly with the more reactive carbohydrate radicals as they are produced. [Radicals with considerable stability are described as being either persistent or stable (Chapter 2, Section I). The presence of such radicals provides the basis for the “persistent radical effect” discussed in Chapter 3 (Section II.B.1.c.).] Noncarbohydrate participants in radical combination range from stable compounds, such as nitric oxide (Scheme 6),41 to resonance-stabilized radicals, such as the 2‑pyridylthiyl radical (eq 52).23 Electrolysis is different from most reactions because it can produce locally high enough concentrations of radicals to allow even reactive ones to combine (eq 53).42
Some electron-transfer reactions between carbohydrate radicals and transition-metal complexes have a similarity to radical combination. In the reaction shown in eq 54, for instance, a change in oxidation states accompanies the combination between the carbohydrate radical and the cobalt complex.27
08. Disproportionation
Disproportionation involving carbohydrate radicals (eq 55 and eq 56) is similar to radical combination in that most such radicals do not exist long enough in solution to come into contact with each other before another reaction takes place. As mentioned in the previous section, one situation in which two radicals can interact is when locally high concentrations are created by electrolysis; thus, when the radicals shown in eq 57 come within bonding distance, both disproportionation (10%-20%) and radical combination (69%) take place.42 In contrast, disproportionation is the exclusive process in the reaction shown in eq 58 because there is considerable thermodynamic gain from forming a highly resonance-stabilized ketone while avoiding the hindrance inherent in the combination of sterically demanding radicals.43
09. Group Migration
Acyl, silyl, phosphatoxy, phenyl, and cyano groups all are capable of undergoing group migration. Because in most reactions these groups are stable substituents and often act as protecting groups, migration is an event that depends not only on reaction conditions but also on the substrate having a particular type of structure. Sometimes it is unclear whether group migration is an elementary reaction or a combination of elementary reactions. The silyl group migration pictured in Scheme 7 is thought to be an elementary reaction, but it is possible that this process consists of a pair of reactions, β‑fragmentation and radical cyclization (Scheme 8).44 The group migration shown in eq 5945 appeared at first to be an elementary reaction, but extensive investigation has shown that it consisted of heterolytic fragmentation followed by ionic recombination (Scheme 9).46–48
10. Summary
Radical reactions are composed of sequences of elementary reactions. An elementary radical reaction is one that does not produce an intermediate. Most elementary, radical reactions consist of transformation of one radical into another, but those that create radicals from nonradicals or cause radicals to disappear also qualify. Elementary reactions are listed in a general form in Table 1. Examples of carbohydrates that undergo each type of reaction are discussed. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/04%3A_Elementary_Reactions/06._Electron_Transfer.txt |
The propagation phase in a chain reaction is composed of a sequence of elementary reactions. The simple reduction pictured in Scheme 1 consists of two such reactions, and the radical addition shown in Scheme 2 has three elementary reactions in its propagation phase. Even though the overall reaction in each case (Schemes 1 and 2) consists of two or more elementary reactions occurring in sequence, neither overall reaction is described as a sequential one because to merit this designation, radical formation from the substrate and radical conversion to the product are not included.1 With these beginning and ending steps removed, the reactions shown in Schemes 1 and 2 do not qualify as sequential. [Cascade, domino, and tandem are other names used to describe sequential reactions.]
Elementary radical reactions readily occur in sequence because most produce a new radical poised for further reaction.1,2 Since all participants in a sequential reaction, including the intermediate radicals, are present in the reaction mixture at the same time, achieving the selectivity necessary to have each radical react in the desired manner at the correct time is a challenging task, one that is critical to the success of the overall process. This selectivity is controlled by the rates of the various, possible propagation steps.
Successful, sequential reactions have several common characteristics. Propagation steps are faster than termination steps. Intermediate radicals react rapidly with the correct nonradicals, but avoid reaction with the solvent and initiator. The final radical, but not the intermediate ones, is converted into a stable product.
05: Sequential Reactions
When two or more reactions occur in sequence, there is a savings in time, effort, and chemicals, compared to performing each reaction individually; therefore, sequential reactions can increase synthetic efficiency and provide a positive environmental impact by reducing chemical use.3–5 An example of this efficiency is seen in the process pictured in Scheme 3, where two new rings form in a single reaction.6 There is a price to be paid for this efficiency. It comes in the form of the additional effort that usually is necessary in preparing the starting materials and establishing the reaction conditions so that each step in this more complicated process proceeds in the desired direction.4,5 Not only may more synthetic work be required in substrate preparation, but controlling product regio- and stereochemistry also may be more challenging in a process where multiple structural changes occur in a single reaction.
III. Two-Step Sequential Reactions
When planning a sequential reaction, it is natural to gravitate toward elementary reactions that are rapid and occur in a highly predictable fashion; also, because a more complex structure is often the target in a synthetic reaction, forming a carbon–carbon bond is usually one of the goals of a sequential process. Since internal addition of a radical to a multiple bond often is both rapid and predictable and usually involves new carbon–carbon bond formation, many sequential reactions include at least one radical cyclization step; some contain more. Cyclization-cyclization, the first sequential reaction to be discussed, involves forming of two rings.
IV. Three-Step Sequential Reactions
Three-step sequential reactions often consist primarily of combinations of cyclization and β-fragmentation steps.56,73–82 One group of such reactions involves ring opening of cyclic acetals.73–77 Since these reactions contain more fragmentation than cyclization, they tend to produce less complicated structures (e.g., compounds with fewer rings). In the reaction shown in Scheme 12, for example, internal radical addition to a cyano group creates a nitrogen-centered radical. This cyclization then is followed by successive β‑fragmentation steps. Together these elementary reactions cause acetal ring opening.73
Balancing the advantage gained against additional effort needed to prepare the starting materials and adjust the reaction conditions is always a consideration when conducting a sequential reaction. Such consideration motivated the preparation of the acetal 6 (Scheme 12),73–75 when the effort necessary to synthesize 7, the first acetal used in this type of reaction, significantly offset the advantage derived from the sequential process.73,76,77
The synthetic potential of three-step sequential reactions changes when there is more cyclization than fragmentation. When such a change occurs, reactions can be used to build more complex structures.78–81 Specifically, the process shown in Scheme 13, which contains only cyclization steps, generates three new rings in a single reaction.81
V. Related Reactions
Parallel reactions and sequences of reactions cause multiple structural changes to occur in molecules under a single set of conditions. These reactions are different from sequential reactions, however, in that neither parallel reactions nor sequences of reactions are a series of conversions of one radical into another; rather, they each consist of two or more distinct, complete reactions taking place under a single set of conditions.
VI. Summary
Elementary radical reactions readily occur in sequence because many of these reactions produce new radicals ready for further reaction. Such reactions provide an opportunity for multiple structural change under a single set of reaction conditions; consequently, these reactions, when properly chosen, represent an increase in synthetic efficiency. Two-step sequential reactions often involve radical cyclization in combination with radical addition, β-fragmentation, or hydrogen-atom abstraction. Three-step reactions usually consist of a combination of cyclization and β-fragmentation steps. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/05%3A_Sequential_Reactions/II._Advantages_and_Disadvantages_of_Sequen.txt |
Establishing the structure of a free radical is a prerequisite for understanding its reactivity. Structural determination for a radical requires the same type of information needed to establish the structure of any reactive intermediate or stable molecule. The process begins by identifying the constituent atoms and their connectivity. Since radicals normally are generated from known compounds, connectivity information usually comes directly from the substrate structure. The configuration at ever carbon atom, except the one where the radical is centered, ordinarily is unchanged from that in the substrate. With this basic, structural information in hand, one can turn to investigating the remaining unknowns, that is, radical-center configuration and radical conformation. Although reactive intermediates, such as radicals, present special problems in structural determination due to their transient nature, the basic information needed is the same for both intermediates and stable molecules.
06: Radical Structure
The structural formula for a radical often, but not always, can be deduced from a combination of different types of information. This information includes the structure of the radical precursor, the method of radical formation, and the identity of the reaction products. For example, in the reaction shown in Scheme l the structure of the deoxyiodo sugar 1, the known reactivity of alkyl iodides with the tri-n-butyltin radical, and the structure of the product 3 together provide enough information to assign the basic structural formula 2 to the intermediate radical.1 At this point, the configuration at the radical center in 2 and the conformation of this radical remain to be determined. The way in which radical configuration and conformation are assigned is discussed in Sections III and IV, respectively, in this chapter.
The same type of information that effectively establishes the structure of the radical 2 (Scheme 1) is insufficient for determining the structures of the radicals produced by hydrogen-atom abstraction from simple sugars. Due to the large number of hydrogen atoms present in even simple sugars, knowing the structure of the starting material has limited value in establishing the identity of any particular intermediate radical. Product structures also are of limited usefulness due to the large number of compounds generated by hydrogen-atom abstraction (at least twenty-five from D-glucose2), and the probability that molecular rearrangement has occurred during formation of some of these products.2,3 Proposing structures for the radicals generated by hydrogen-atom abstraction from even a simple sugar, such as D‑glucose, can involve a good deal of speculation, but such speculation can be reduced by using electron spin resonance (ESR) to observe radicals directly.
It is possible to identify six, first-formed radicals in the ESR spectrum of the mixture produced by reaction of α-D-glucopyranose (4) with the hydroxyl radical (eq 1).3 These six radicals are the ones generated by hydrogen-atom abstraction from of the six carbon atoms present in 4. (Hydrogen-atom abstraction from the oxygen atoms is too slow to be competitive.) Identification of first-formed radicals is possible because when the structure of the radical precursor is combined with information from its ESR spectrum, the combination provides a basis for assigning a structural formula to each radical.
Whenever a radical reaction is encountered for the first time, the structure of any intermediate radical is naturally a topic of primary interest. Once the basic structure of a radical has been established, the unknowns that usually remain are the configuration at a radical center and conformation of the radical. Establishing this configuration and determining radical conformation often involve both experimental findings and molecular-orbital calculations. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/06%3A_Radical_Structure/II._Structural_Formulas.txt |
A. Planar and Pyramidal Structures
The configuration of a radical defines the location in space of the atoms directly attached to the radical center. When three such atoms are bonded to the carbon atom upon which a radical is centered, the configuration is either planar or pyramidal.4 A planar configuration is one in which the atoms directly attached to the radical center and the center itself all exist in the same plane (Figure 1). For pyramidal radicals the plane defined by these directly attached atoms no longer includes the central atom (Figure 1).
Nearly every carbon-centered radical has a pyramidal configuration at the radical center, but these radicals vary widely in how close their configurations are to being planar.5 A terminology has arisen that is designed to indicate approximate radical configuration. If a radical center has a nearly planar arrangement of attached atoms, the radical is described as being π‑type.6 (Since in a π-type radical the orbital in which the electron is centered is close to being a p orbital, this orbital is often referred to as being p‑type.) If a radical has a decidedly pyramidal configuration (i.e., one approaching that corresponding to sp3 hybridization), the radical is described as being σ‑type.
Pyramidal, carbon-centered radicals with no electronegative substituents attached to the radical center tend to have a small distortion from planarity;5 that is, they usually are considered to be π-type radicals. (It is worth noting that the magnitude of the angle of distortion can be deceptive. The relatively small 6.2o distortion from planarity reported for the ethyl radical means that this radical is actually about 1/3 of the way to being sp3 hybridized.7) The distortion from planar arrangement increases as electron-withdrawing substituents replace other groups attached to the radical center. The change in configuration that accompanies replacement of the hydrogen atoms in the methyl radical by fluorine atoms provides a clear example of the effect of electronegative substituents on radical geometry.5,8,9 The methyl radical is either planar, or nearly so,10 but progressive replacement of hydrogen atoms with fluorine atoms produces pyramidal radicals with structures increasingly further from planarity until the trifluoromethyl radical is reached, in which case the F–C–F bond angles are similar to those found in tetrahedral structures.11
B. Configurational Determination from α-13C Hyperfine Coupling Constants
Information about radical configuration can be obtained from analysis of α-13C hyperfine coupling constants. These coupling constants, obtained from the ESR spectra of 13C-enriched radicals, provide a sensitive measure of the hybridization at a radical center.12,13 The configuration of a pyranos-1‑yl radical is naturally of considerable interest due to the unique role of the anomeric carbon atom in carbohydrate chemistry. The α-13C hyperfine coupling constants, obtained from the ESR spectrum of the 13C-enriched D-glucopyranos-1-yl radical 5, show the deviation from planarity for this radical to be 3.9o (Figure 2).6,14 Since an sp3‑hybridized σ radical would have a deviation 19.5o, the D‑glucopyranos-1‑yl radical 5 is considered to be π‑type.6 Radicals centered at C-2 (6), C-3 (7), and C-4 (8) in pyranoid rings also have π‑type configurations.15
Since organic radicals with no electronegative substituents attached to the radical center have π-type configurations, finding that radicals 6-8 have this type of configuration is not surprising. Because radical centers with electronegative atoms attached become more pyramidal, it is surprising to discover that the radical 5, which has an oxygen atom bonded to the radical center, also has a π-type configuration. To understand why this configuration is adopted, it is helpful to analyze of the stability of 5 as determined by frontier-orbital interactions.
C. Theoretical Explanation of Observed Configurations
1. Frontier-Orbital Interactions
Frontier-orbital interactions are based on an approximate, quantum-mechanical method that assumes that all interactions between occupied orbitals in a bimolecular reaction can be neglected and that the only interactions that need to be considered are between the highest occupied molecular orbital (HOMO) of one reactant and the lowest unoccupied molecular orbital (LUMO) of the other. A small energy difference between the HOMO and the LUMO (the frontier orbitals) translates into a large stabilizing interaction. In radical reactions the singly occupied molecular orbital (SOMO) can be either an HOMO or a LUMO.16 Although frontier-orbital interactions are intended to be applied to bimolecular reactions, they can be used for understanding radical structure. In making such an application the radical is formally split into two fragments and fragment recombination is treated as a bimolecular reaction.16 This approach, which has enjoyed widespread application and success in explaining radical structure,17 will be used to rationalize the π-type configuration at C-1 adopted by the D-glucopyranos-1‑yl radical 5 (Figure 2).
2. pc /po Orbital Interaction
Experimental and theoretical studies show that the two unshared pairs of electrons on an oxygen atom in a pyranoid ring, do not have the same energy.18,19 The higher energy pair exists in a p‑type orbital while the lower energy pair is in a hybrid orbital that has considerable s character. As pictured in Figure 3, stabilization should result from interaction of the electrons in a p-type orbital on a ring oxygen atom (po) with the electron in the singly occupied, p-type orbital on an adjacent carbon atom (pc). The increase in energy of the electron in the singly occupied molecular orbital (SOMO) is more than offset by the combined decrease in energy of the two electrons in the doubly occupied orbital. Because a nonparallel alignment of orbitals would exist in a pyranos-1-yl radical with a σ‑type configuration, stabilizing orbital interaction for such a radical would be less than that for a radical with a π-type configuration, thus, there is a gain in radical stabilization to be had from having a p-type orbital at C-1 even though this atom has an electronegative oxygen atom attached. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/06%3A_Radical_Structure/III._Radical-Center_Configuration.txt |
Conformations of a molecule are usually viewed as arrangements of atoms that differ only by rotation about one or more single bonds. This view needs to be expanded where radicals are concerned because in some instances radicals with pyramidal configurations change from one conformation to another by inversion of configuration at the radical center. Consider the case of the ethyl radical (Figure 4). This radical is reported to have a pyramidal configuration with a 6.2o distortion from planarity and an extremely low (0.15 kcal/mol) energy barrier between conformations.20–22 For the ethyl radical bond rotation and inversion of configuration both can contribute to conversion of one conformation into another (Figure 4).20
The intriguing complications associated with changes in conformation of alkyl radicals are not a major focus for conformational analysis of carbohydrates. The primary objective where carbohydrate radicals are concerned is to determine why a particular conformation is preferred and then to understand how conformation influences reactivity. ESR spectroscopy is an essential tool in achieving these objectives because it enables direct observation of radicals and, in so doing, provides valuable information about their conformation.
V. Quasi-Anomeric Radical Stabilization
Radicals centered at various carbon atoms in a pyranoid ring can be divided, on the basis of their stability, into two groups. The first group includes the pyranos-1-yl and pyranos-5-yl radicals, intermediates that are stable enough to be generated and observed in toluene or tetrahydrofuran. The second group, pyranosyl radicals centered at C-2, C-3, and C-4, cannot be observed in toluene or tetrahydrofuran because they abstract hydrogen atoms from these solvents too rapidly.15 Since only the pyranos-1-yl and pyranos-5-yl radicals are capable of experiencing stabilization from the quasi-anomeric effect (Figure 8), the special stability of these radicals provides further support for the existence of quasi-anomeric stabilization.
VI. Summary
Determining the structure of a radical is essential to understanding its reactivity. The process begins by establishing the structural formula for the radical, that is, by identifying the constituent atoms, their connectivity, and elements of stereochemistry. Remaining unknown at this point typically are radical-center configuration and radical conformation.
The structural formula of a radical often can be determined reliably from knowledge of the structure of the radical precursor, the method of radical formation, and the reaction products. In instances where this information is insufficient, direct observation of the radical by ESR spectroscopy sometimes is possible and can provide the additional information needed to establish a structural formula.
The configuration at a radical center defines the location in space of the atoms directly attached to this central atom. Nearly every carbon-centered radical has a pyramidal configuration, but these radicals vary widely in how close their configurations are to being planar. If a radical is nearly planar, it is described as being π-type. If, on the other hand, a radical is much more pyramidal, it is considered to be a σ-type radical. Information about radical structure is obtained from molecular-orbital calculations and from observation of α-13C hyperfine coupling constants (determined from ESR spectra of the 13C-enriched radicals).
A conformation of a radical is one of the arrangements of atoms that can be formed by rotation about one or more single bonds. Pyranos-1-yl radicals have been extensively studied and some have been found to favor unexpected conformations. Perhaps most striking among these is the 2,3,4,6-tetra-O-acetyl-D-glucopyranos-1-yl radical, which exists in a distorted B2,5 boat conformation.
Information about radical conformations is derived from both experimental and theoretical studies. Experimental information comes from analysis of ESR spectra. Study of pyranos-1-yl radicals has led to the identification of the quasi-anomeric effect as a general, controlling influence in determining conformations in many radicals. Understanding of radical conformation comes both from simple and complex applications of molecular-orbital theory. Frontier-orbital interactions offer a simple, theoretical means for rationalizing radical conformation. The far more sophisticated ab initio molecular-orbital calculations also provide understanding of the reasons for a radical adopting a particular conformation. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/06%3A_Radical_Structure/IV._Radical_Conformation.txt |
A. The Evans-Polanyi Relation
For many radical reactions there is a simple relation between the energy of activation for the reaction and its enthalpy. This relation, which is referred to by several, similar names1–3 (Evans-Polanyi being a common one), is given in eq 1. Equation 1 expresses in a quantitative fashion the notion that in a group of closely related reactions the enthalpy for a particular reaction should be related to its energy of activation; specifically, energies of activation should decrease in a linear fashion as reactions become more exothermic.
Once the two constants in eq 1 have been determined, it is possible to predict the energy of activation for reaction of any member of the group from knowledge of the reaction enthalpy. The numerical value of the constant α represents the fraction of the overall enthalpy change that exists at the transition state. The value of α can be viewed as a measure of how far a reaction has proceeded along the reaction coordinate when the transition state is reached. The later a transition state occurs in a reaction the closer α will be to unity.
B. Nucleophilic and Electrophilic Radicals
Although radicals are neutral species, they often exhibit behavior characteristic of either nucleophilic or electrophilic intermediates.4,5 This behavior facilitates certain types of reaction; for example, in the addition reactions shown in eq 2, the carbon-centered, cyclohexyl radical behaves as a nucleophile by adding more rapidly to compounds with more electron-deficient double bonds than to ones in which the double bonds are less electron-deficient.6 In contrast, the malonyl radical 1 can be viewed as electrophilic because it adds to electron-rich double bonds such as that in the D-glucal 2 (eq 3).7 A good beginning point for discussing radical philicity is examining some hydrogen-abstraction reactions.
II. Bond Polarities, Bond-Dissociation Ener
Bond polarities, bond-dissociation energies, and rate constants for abstraction of hydrogen atoms bonded to tin, silicon, sulfur, selenium, and carbon all are given in Table 1. Based on Pauling’s electronegativity values, hydrogen has a small negative charge when bonded to tin or silicon and a small positive charge when bonded to sulfur, selenium, or carbon. The information in Table 2 shows that for each type of bond, the rate constant for hydrogen-atom abstraction by simple primary, secondary, and tertiary, carbon-centered radicals is nearly the same.
When the bond dissociation energies in Table 1 are used to calculate reaction enthalpies, they show that the reaction in eq 4 is more exothermic than that in eq 5. If the Evans-Polanyi relation is obeyed, the first reaction (eq 4) should have a lower energy of activation than the second (eq 5), but the opposite appears to be true. The rate constants for these reactions, when related to activation energies through the Arrhenius equation (eq 6), show that, unless the frequency factors for these two reactions are quite different, the reaction given in eq 4 actually has a higher energy of activation. Clearly, something in addition to reaction enthalpies must have a significant role in determining energies of activation for these two reactions.
The identification of a likely candidate for this additional factor can be made by returning to the reactions pictured in equations 2 and 3 and recalling that these reactions show some carbon-centered radicals to be nucleophilic and others electrophilic. If one assumes that the tert-butyl radical is similar in its philicity to the nucleophilic cyclohexyl radical, then in the reaction in eq 5 there is a polarity match between the nucleophilic radical and the electron-deficient hydrogen atom being abstracted. Since a similar match does not exist in the slower reaction (eq 4), radical philicity becomes a prime candidate for the factor that joins with reaction enthalpy in explaining the rate constants for hydrogen-abstraction reactions.
III. Determining Radical Philicity
Since radical philicity appears to have an important role in radical reactivity, it is valuable to be able to determine easily whether a particular radical is electrophilic or nucleophilic. A procedure that accomplishes this task would be especially useful if it could be implemented simply by inspecting the structure of the radical in question. Fortunately, two such procedures exist. One is based on atom electronegativity16 and the other on cation and anion stability.17 More sophisticated techniques for determining radical philicity also are known. One of these, principal component analysis,18,19 extracts information about nucleophilicity and electrophilicity of radicals from experimentally determined rate constants. A second, based on ab initio molecular orbital calculations, determines radical philicity from the extent and direction of charge transfer between a radical and an alkene at the transition state for an addition reaction.3,20–23 These different approaches to determining radical philicity are described in the next four sections. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/07%3A_Radical_Philicity/I._Introduction.txt |
A. Valence Bond Theory
One way to explain the existence of radical philicity begins with the addition reaction shown in eq 7. The transition-state structure in this reaction can be represented as a hybrid of valance-bond structures. If there is no separation of charge, the transition state can be represented by the contributors 3 and 4 shown in Figure 1. Unequal electron distribution at the transition state can be taken into account by including additional resonance contributors. If in the transition state there is a transfer of electron density from R· to the C–C double bond, this transfer can be represented by adding contributors 5 and 6 (Figure 2). If electron transfer is in the other direction, resonance contributors 7 and 8 (but not 5 and 6) make a significant contribution to the transition state structure (Figure 2).
1. Nucleophilic Radicals
To illustrate the way in which valance bond theory explains radical nucleophilicity, it is instructive to examine the reaction shown in eq 8. In this reaction the D-glucopyranos-1-yl radical 9 is considered to be nucleophilic because it adds to the electron-deficient double bond in acrylonitrile.24 The valence-bond structures 10-14 (Figure 3) all potentially contribute to the transition-state structure in this reaction. Structures 10 and 11 are major contributors that have no separation of charge. Structures 12 and 13 are minor but significant contributors, minor because they involve separation of charge but significant because they stabilize either negative charge (12) or positive and negative (13) charges effectively. Structure 14 is not significant because it has charge-separation and the charges are not effectively stabilized. Since structures 12 and 13 make a greater contribution than does 14 to the transition state in this reaction (eq 8), the radical 9 becomes a net electron-donor at the transition state and, thus, is considered to be nucleophilic.
2. Electrophilic Radicals
If a radical center has a sufficiently strongly electron-withdrawing substituent (or substituents) attached, the inherently nucleophilic character of a carbon-centered radical is reduced to the point that the radical becomes electrophilic.18 For example, the malonyl radical 1, which has two electron-withdrawing groups attached to the radical center, is considered to be electrophilic because it adds to the electron-rich double bond in the D‑glucal 2 (eq 3).7 The electrophilicity of 1 can be understood primarily in terms of the importance to the transition-state structure of the charge-transfer contributor 15, a structure in which the electron-withdrawing methoxycarbonyl groups stabilize the negative charge and the ring oxygen atom stabilizes the positive charge (Figure 4). Together these stabilizing interactions increase the contribution at the transition state from a structure in which the radical 1 is acting as an electrophile by accepting electron density from the D-glucal 2. The resonance structure 16 is not an important contributor at the transition state because within 16 there is a destabilizing shift of electron density away from a radical center that contains electron-withdrawing, methoxycarbonyl substituents.
3. Ambiphilic Radicals
Inherent in defining radical philicity in terms of electron-transfer is the idea that the philicity of a radical is a function of the reaction in question. This means that instead of describing a radical as nucleophilic, it should be described as nucleophilic in a particular reaction. It is fair to say, however, that radicals that are moderately or strongly electrophilic or nucleophilic in one reaction are likely to have the same philicity in all reactions, but radicals that are weakly nucleophilic or electrophilic in one reaction are better candidates for a philicity change in a different reaction. Radicals that are nucleophilic in one reaction but electrophilic in another are classified as ambiphilic.
B. Molecular Orbital Theory: Frontier-Orbital Interactions
When a transition-state structure for a reaction resembles the structure of the starting materials, frontier-orbital interactions provide qualitative information about energy changes taking place at the transition state. (Since frontier orbitals are based on the structures of the starting materials, the further the transition state is along the reaction pathway the less reliable frontier-orbital interactions will be in predicting or rationalizing reactivity.) According to Hammond’s postulate,25 an exothermic reaction should have an early transition state with a structure resembling that of the starting materials; therefore, such a reaction should be suitable for analysis by frontier-orbital interactions. A reaction involving addition of a carbon-centered radical to a carbon–carbon double bond is a prime candidate for this type of analysis because such a reaction replaces a π bond with a more stable σ bond, a change that should produce a decidedly exothermic reaction.1,3,6,26
1. Nucleophilic Radicals
A beginning point for explaining radical nucleophilicity in terms of frontier-orbital interactions is found in Figure 5, which pictures the singly occupied molecular orbital (SOMO) in the radical 17 interacting with both the π* (LUMO) and the π (HOMO) orbitals of the alkene 18. Identifying the most important interaction is critical to determining the nucleophilicity of the adding radical. When the SOMO of 17 interacts with the alkene 18, the greater interaction is with the π* orbital of the alkene (Figure 5).6 Convincing evidence supporting this position comes from plotting calculated HOMO and LUMO energies of substituted alkenes against the natural logarithm of the relative rate constants (ln krel) for addition of a carbon-centered radical (the tert-butyl radical was used) to these alkenes.27 A linear correlation exists between ln krel and LUMO energies, but no such correlation exists between ln krel and HOMO energies. The correlation with LUMO energies then is consistent with the dominant frontier-orbital interaction being between the SOMO of the radical 17 and the π* orbital of the alkene 18 (Figure 5).
The next step in understanding how frontier-orbital interactions can explain radical nucleophilicity involves the addition of the radical 17 to the alkene 19, a compound in which the double bond contains the electron-withdrawing substituent Z (Figure 6). When Z replaces one of the hydrogen atoms attached to a doubly bonded carbon atom, the π* orbital is stabilized and the associated energy level moves closer to that of the SOMO of 17.28 This change in energy level position increases the interaction between the SOMO and the π* orbital (Figure 6). Greater interaction translates into a lower transition-state barrier for reaction; therefore, the radical 17 will add more rapidly to the alkene containing the electron-withdrawing Z group than to an unsubstituted alkene. This preferential reaction with electron-deficient alkenes makes the radical 17 nucleophilic.
It is possible to increase the nucleophilicity of a carbon-centered radical still further if its SOMO energy level moves even closer to that of the π* orbital of an alkene. This type of change occurs when an oxygen atom is attached directly to the radical center because interaction between the p‑type orbital on the carbon atom and the p-type orbital on the adjacent oxygen atom raises the SOMO energy level in the resulting radical (20) (Figure 7). This higher energy level places the SOMO closer energetically to the π* orbital of the reactant alkene. Such a change further increases orbital interaction and in so doing causes greater transition-state stabilization. The enhanced reactivity, due to the presence of the attached oxygen atom, means that the radical 20 will be even more nucleophilic than 17; thus, this oxygen-substituted radical (20) is considered to be strongly nucleophilic.
There is an additional way of viewing the frontier-orbital interaction between an alkene and a carbon-centered radical. Understanding this alternative view begins by recalling that the major, frontier-orbital interaction between a carbon-centered radical and an unsaturated compound is between the SOMO of the radical and the π* orbital (LUMO) of the alkene (Figures 5-7). Since SOMO-LUMO interaction is the most important and since any electron donation at the transition state resulting from this interaction must involve electron transfer from the SOMO (the LUMO has no electrons to transfer), the radical is acting as an electron donor and, therefore, is behaving as a nucleophile.29
2. Electrophilic Radicals
If a hydrogen atom attached to a carbon-centered radical is replaced by an electron-withdrawing substituent (e.g., a cyano or carbonyl group), the resulting radical becomes more electrophilic.6,18,30,31 Additional substitution of this type further increases radical electrophilicity (Figure 8). The electron-withdrawing group causes the energy level associated with the singly occupied molecular orbital of the substituted radical to move to a position lower in energy; that is, the radical becomes more stable.6,32 When the energy level of an SOMO in a carbon-centered radical becomes sufficiently low, the major, frontier-orbital interaction with an alkene changes; that is, the primary interaction is no longer with the π* orbital of the alkene but rather with its π orbital (Figure 9). When this change occurs, the primary shift in electron density at the transition state is away from the filled HOMO toward the partially filled SOMO; thus, the radical is electrophilic.
Figure 10 pictures the frontier-orbital interaction of the radical 21 with an alkene that has an electron-donating substituent. Since the HOMO for the substituted alkene (Figure 10) is higher in energy than the HOMO of the unsubstituted alkene (Figure 9),28 transition-state stabilization from SOMO-HOMO interaction will be greater for reaction involving the substituted alkene (Figure 10). Due to this greater stabilization, the radical 21 reacts more rapidly with the more electron-rich alkene, a behavior expected from an electrophilic intermediate.
C. Balancing Polar and Enthalpy Effects
The discussion at the beginning of this chapter focused on groups of similar reactions that obey the Evans-Polanyi relation, that is, reactions in which the energies of activation can be determined from reaction enthalpies using eq 1. Attention then turned to reactions where this simple relation (eq 1) does not hold. The energies of activation for reactions that do not obey the Evans-Polanyi relation are influenced by polar effects operative at the transition state. Since some reactions are more subject to enthalpy effects and others to polar effects, the question naturally arises as to what the balance is between these two. Principal component analysis answers this question with the finding that “the dominant factors influencing radical addition reactions are polar effects alone for strongly nucleophilic or strongly electrophilic radicals...and enthalpy effects alone for weakly nucleophilic or weakly electrophilic radicals”.18 For moderately nucleophilic or moderately electrophilic radicals both polar and enthalpy effects are important.18 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/07%3A_Radical_Philicity/IV._Explaining_Radical_Philicity.txt |
A. Hydrogen-atom abstraction
The reactions shown in equations 4 and 5 illustrate the importance of radical philicity in hydrogen-abstraction reactions by showing that the nucleophilic radical R· abstracts the electron-deficient hydrogen atom attached to sulfur (eq 5) more rapidly than the electron-rich hydrogen atom bonded to tin (eq 4).33–35 The differences in rate constants and enthalpies for these two reactions underscore the fact that radical philicity affects the stability of the transition-state structure in a reaction but not the overall energy changes due to bond breaking and bond formation.
Comparing the three reactions pictured in Scheme 1 draws attention to the effect of radical philicity on hydrogen-atom abstraction from carbohydrates. In each of these reactions the oxygen-centered radical 22 either abstracts a deuterium atom from Bu3SnD to give the deuterated alcohol 24, or it reacts internally with H-3 to generate the carbon-centered radical 23.36 After the radical 23 forms, it then abstracts a deuterium atom for Bu3SnD to give the second reaction product (25). The relative amounts of products 24 and 25 provide a measure of external (deuterium) versus internal (hydrogen) abstraction. As H‑3 becomes less electron rich, internal reaction (22$\rightarrow$23) becomes less competitive. External abstraction (22$\rightarrow$24), on the other hand, should not be noticeably affected by changes in substituents at C-3. If, as expected, the transition states for the internal hydrogen-abstraction reactions shown in Scheme 1 are early, these reactions support the idea that polarity matching has a critical role in determining the favored reaction pathway for the radical 22.
B. Radical Addition
Pyranos-1-yl radicals add readily to electron-deficient, carbon-carbon double bonds but are much less reactive toward double bonds lacking electron-withdrawing substituents.37,38 A group of reactions that illustrates this difference in reactivity is found in Scheme 2.37 The ability of the pyranos-1-yl radical 26 to add to the unsaturated compounds shown in Scheme 2 correlates with the reduction potentials of these compounds; that is, addition to compounds with less negative reduction potentials occurs more rapidly than addition to compounds with more negative reduction potentials.37
Since the reduction potential in a substituted alkene is a measure of the ease of introducing an electron into a π* orbital, this potential becomes an indicator of energy-level positioning. When comparing two reduction potentials, the less negative one has a lower energy level for the π* orbital. Because this lower energy level causes the π* orbital (LUMO) to interact more effectively with the SOMO of the adding radical, transition-state stabilization due to frontier-orbital interaction increases as the reduction potential for the substituted alkene becomes less negative (Figure 11).
Although the addition of a nucleophilic radical to a π bond that is not electron-deficient is too slow to be observed in the reactions shown in Scheme 2, the situation changes when reactions become intramolecular. For π bonds that are 1,5- or 1,6-related to a radical center, intramolecular addition can take place even if the π bond is not decidedly electron-deficient (Scheme 3).39 As far as overall reaction rate is concerned, forced, close proximity of the radical center to the π bond can compensate for a small transition-state stabilization caused by a large separation in energy levels of interacting, frontier orbitals.
C. Polarity-Reversal Catalysis
The philicity of radicals involved in hydrogen-abstraction reactions provides the basis for a phenomenon known as polarity-reversal catalysis.4,33–35,40 This type of catalysis, which is responsible for the effect that thiols have on the reactions of carbohydrate acetals and ethers, is discussed in Section III of Chapter 5 in Volume II. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/07%3A_Radical_Philicity/V._Examples_of_Radical_Philicity_in_Reactions.txt |
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