Cannizzaro Lab Reaction and Techniques

Okay, welcome to the Cannizzaro lab. Here's a quick outline of what I'll be going over in this video I'll look at the reaction itself for a little bit and then to show the procedure I'll be using a flow chart which is a nice and simple way of showing the steps to the procedure and how to separate two compounds and then I'll show the actual experiment in lab and finally, we'll take a look at the NMR because there's some techniques and strategies that will be helpful for this lab and Future labs as well. So this is the reaction itself We're taking 4-chlorobenzaldehyde, reacting it with potassium hydroxide to form two different products, one being 4-chlorobenzoic acid and the other 4-chlorobenzal alcohol. Here's a look at the mechanism. When the hydroxide reacts with the aldehyde, it forms a tetrahedral intermediate, and when that collapses, a hydride will attack another 4-chlorobenzaldehyde, and this forms the two products that we saw before. The deprotonated alcohol product will... pull a proton off of the acid because the acid has a lower pKa and that forms the alcohol product and the deprotonated acid product. This means it is a two to one ratio from the reagent to the products. So industrially if you're trying to make money off of this this wouldn't be a great reaction to do because the highest percent yield you could get for each product would be 50 percent. But in a lab setting this is a cool reaction and allows us the opportunity to learn some techniques in separating two products. both of which we would want to keep. To show that separation, I'll be using a flow chart which looks something like this. We have a conical vial with our two products formed after the reaction, and I'll use HA to represent the acid product and OH to represent the alcohol product. The reaction was done under basic conditions, so the acid product will be deprotonated and dissolved in that aqueous solution like we saw at the end of the mechanism, whereas the organic alcohol product won't be soluble in that solution so it'll precipitate out and form a solid on the spin vein. To separate the two compounds we can add an organic solvent like methylene chloride that will create a bilayer and start dissolving the alcohol product. Methylene chloride is more dense than water so it'll be on bottom with the dissolved alcohol product and the top aqueous layer will still contain the ionized acid product. Now we could get to the same place if initially instead of looking like this we had the two products mixed together as two solids. We would just add methane chloride to dissolve both of them and then add an aqueous base to deprotonate the acid and that would create the bilayer like we see here. Now we can separate the two phases and then to make sure that we get all of the organic product out of the aqueous layer, we'll go ahead and extract that aqueous layer two more times with methylene chloride and combine all the organic extracts together. and we'll save the aqueous phase for later because that contains our acid product. We don't want to toss that. Now we have our alcohol products separated, but just in case any acid product happened to move into the organic phase with the methylene chloride extracts, we'll add sodium bicarbonate that will deprotonate any remaining acid and move it into the aqueous layer. Then we can separate those two layers and dry the organic phase with sodium sulfate to remove any leftover water. Once we have our dry organic phase, we can evaporate the solvent leaving a crude alcohol product. And then for purification, we'll recrystallize that product using the groin and then remove the crystals. And we can then characterize those crystals by taking a melting point and running an IR spec. As for the acid product, we want to be able to get it out of solution. And the easiest way to do this is by adding a concentrated acid, which will put the proton back onto the carboxylic acid group. This will make the molecule neutral and it will no longer be soluble in the aqueous solution. So it'll crash out of solution and we'll see it precipitate as a white slurry. Then we can remove that crude product using vacuum filtration and then recrystallize it using ethanol. so we can characterize a pure product using melting points and IR. Okay, I'll start out by measuring the 150 milligrams of our reagent, the 4-chlorobenzyldehyde, and then I'll go ahead and take that and put it in a 5 milliliter conical vial with a large spin vane. Then I'll be adding 400 microliters of methanol. and 400 microliters of the potassium hydroxide. Now I can set up the apparatus and I'll let everything reflux for an hour. And you can actually start to see the solid alcohol product forming here. You can see in that kind of white pasty film on top of the liquid. Once the hour has passed, I'll lift everything off of the heat so the conical vial can cool. And here you can see the alcohol product gathering. on the spin vein and the acid product is going to be dissolved in the aqueous solution. Initially I'll dilute the basic solution by adding some distilled water and then I can go ahead and start the methylene chloride extractions. As you can see the alcohol product doesn't dissolve super well at the beginning so I'll be stirring it and trying to break it up to dissolve it better. So the solid alcohol still hasn't dissolved completely. but it should continue to dissolve as I keep doing further extractions. I ended up having too much methane chloride trying to dissolve the solid, so I transferred everything to a beaker. And now I can save the aqueous solution containing the acid product for later and move on to the sodium bicarbonate washes which I'll be doing two of. And then I'll rinse the organic extracts with some distilled water just to help remove any ions that might have been left over from the bicarb washes. Now I can transfer the methylene chloride solution to an earlier flask where I can dry it using sodium sulfate and I'll add that until it's free flowing. Once I've removed the dry methylene chloride solution I'll actually add a little bit more and rinse the sodium sulfate to reclaim any alcohol product that might have been left behind. Now I can just evaporate off the solvent and you can actually start to see the alcohol product gathering on the side of the beaker here. But this is what the crude product looks like once I've evaporated off all of the methylene chloride and now I'll be transferring that crude product to a Craig tube. so that I can purify it through recrystallization. I'll place it in a hot water bath so that it can be dissolved at high temperatures using LaGroin as the recrystallization solvent. And I'll place that in the hot water bath as well so that it's also at a higher temperature. Now I'll start working on dissolving the alcohol product by slowly adding LaGroin and stirring it. Just so that I add the minimal amount of solvent to dissolve it at higher temperatures. Which, it's looking like everything is dissolving pretty well now, so I'm going to put it into an ice bath to make sure that the solid can come out of solution, which it definitely can, so that's good. So I'll place it back in the hot water bath to dissolve the solid once again, and now leave the tube on a rack so that the crystals can form slowly at room temperature. Now this clip is really sped up. In reality it took about 8 minutes for the crystals to fully form. Once the recrystallization process is done, I'll just scrape the sides with a spatula so it'll be easier to remove the crystals and they don't stick to the sides as much. To remove the crystals from the solvent, I'll place this stopper in the top of the kreg tube and then I'll place both of them inside of a propylene test tube, flip it upside down, and then put it into a centrifuge. What that will do is pull all the liquid to the bottom of the... propylene test tube, but the stopper will keep the crystals inside the Craig tube and therefore removing them from any liquid. So here's the centrifuge. I'll put the test tube with the crystals on one side and then balance it with a test tube that just has water in it on the other side. And then I'll run the centrifuge for a couple minutes just to make sure that all of the solvent is pulled to the bottom of that test tube. Once that's done, I'll remove the test tubes from the centrifuge. And you can see pretty well the separation of the solvent from the crystals. Now I can just remove the Craig tube and we should have a purified alcohol product that we can characterize. First I'll want to weigh out the alcohol product on a watch glass so that we can calculate a percent yield for this experiment. Then I'm going to crush up the crystals just a little bit so it'll be easier to place them into an open closed capillary tube to run the melting point. Right now they're in the open side of the capillary tube so I'm just going to tap it against the counter to draw them to the bottom of the closed side and once there it'll be ready to run a melting point now. I placed it into the machine well below its melting point of 72 degrees Celsius just in case there were impurities and it started melting early. Crystals didn't start melting until about 71.3 degrees Celsius and didn't finish melting until 72 degrees Celsius. So that'll be the melting point range for the alcohol product. I'll run an IR spec using the neutral mole technique. So I'll add some of the crystals and then add some neutral and then I'm going to mole the salt plates together. Until more of a film is what's left in between them. Now I can run the IR spec and Here's what the absorption pattern ends up looking like which looks pretty good. We have the alcohol peak as expected We have some neutral which is okay. We see some aromatic overtones around 1800 1900 and a carbon oxygen peak So this one turned out pretty well. Let's move on to the acid product now Okay, I've got the conical vial with the acid product and I'm going to dilute the solution a little bit more with some additional distilled water. Now I can add the concentrated hydrochloric acid and you can see the product precipitating out of solution in that white slurry as it's neutralized by the added acid and I'll make sure that I add enough so that the solution itself is acidic, meaning that all of the acid product will be protonated. and no longer dissolved. To make sure I added enough acid, I'll test the pH of the solution, which should turn the pH paper pink, so it looks like we're good here. To remove the solid product from the aqueous solution, I'm going to use a hirsch funnel and the vacuum filtration technique. The hirsch funnel has a stopper, so when I pour the contents of the conical vial in, the solid product will gather on the stopper and the vacuum will pull the aqueous solution through into the flask. Turn on the lab vacuum valve now and you can see from the compression of this tube that the vacuum is working. So I'll pour the product into the hirsch funnel and start separating it from the aqueous solution. I'll rinse the conical vial with some distilled water to help remove remaining product that was left over. And then I'll dump it into the hirsch funnel. This will also help rinse off any ions or pull anything out that isn't the product. to help rinse it as you can see it's pretty sticky and clumpy right now but eventually that'll dry out i'll keep rinsing it with water just to make sure that i dissolve and remove anything that i don't want there and after a few minutes you can see that the crude product is a lot more dry now and is probably ready to be removed i'll remove the crude product from the hirsch funnel And follow the exact same process for recrystallizing this product as I did with the alcohol product, only this time I'll be using ethanol as the recrystallization solvent instead of lagroin. So like before, I'll be adding just enough ethanol to dissolve the crude product at high temperatures, which is looking pretty good right now. And then I'll place the Craig tube in an ice bath just to make sure that the solid can come out of solution. And I'll re-dissolve the solid in the hot water bath and leave the tube on a rack so that the product can recrystallize slowly at room temperature. I'll separate the crystals by using the stopper and propylene test tube again and then place it into the centrifuge and allow it to run for a few minutes. I'll weigh out the purified acid crystals now so that a percent yield can be calculated. Now I can characterize the product so I'll put some in an open closed capillary tube to run a melting point. And we see the crystals start melting around 239 degrees Celsius and finish around 240.1. So it's a little lower than literature value, so there's probably some contamination. Finally, I'll run an IR spec using the neutral mole technique. And it looks like the carboxylic acid product was definitely formed. We can see the carboxylic acid funnel ranging from like 3300 to 1800. with the carbon hydrogen peak from neutral sticking out in the middle there. And then we can see the carbonyl peak from the carboxylic acid group around 1682. So this looks pretty good. Okay let's take a quick look at the NMR. I'll just be focusing on the HNMR for this compound but the same concepts should help with the other two compounds as well. So let's take a look at the HNMR to figure out which molecule we're looking at. We see this peak around 13, which is indicative of a carboxylic acid, so this would be the HNMR for the carboxylic acid product. Now this also could have been confirmed by looking at the D2O exchange here, which is a tool to help identify protons on functional groups that can hydrogen bond, like carboxylic acids or alcohols, etc. Basically what happens is we introduce a bunch of D2O, which is water with deuterium instead of protons, and as they hydrogen bond with the compound that we're trying to analyze, the proton on the hydrogen bonding functional group will get switched out with a deuterium, and since deuterium is not a proton, it will no longer show up on a proton NMR. So the alcohol peak, or in our case, the carboxylic acid peak that we were looking at, will no longer show up on a D2O exchange. So this just helps confirm that we are looking at the carboxylic acid product, and we can go ahead and label all the protons on that molecule now. We already know that the peak above 13 is for the acid proton, so we can go ahead and label that as hydrogen A, and then we know the peak at zero is for TMS, so we can label that as well. From here it looks like we have four more peaks and only two hydrogens left to label, but if we look at these two peaks down here, They are not in the aromatic range, meaning they're not going to be for hydrogens B and C. They've got to be something else. Plus, if we look at their integration, both of them have an integration smaller than that of hydrogen A, meaning less than 1, which isn't really possible, so they have to be solvents. The one around 2.5 is for DMSO, dimethyl sulfoxide. And then when water is mixed with DMSO, it shows up around 3.4, so we can label that as water. Which actually makes sense why when we look at the... deuterium exchange the peak no longer shows up around 3.4 it shows up around 4 because as water interacts with d2o it changes into hdo which will have a different chemical shift than water now the remaining two peaks are in the aromatic range so that'll be for hydrogens b and c but both of them are doublets and both of them have an integration of two so it's going to be difficult distinguishing which one is which just by looking at the peaks so we're going to have to consider other things happening on the molecule to be able to label those. For that, let's look at benzoic acid and consider its resonance structures. The carboxylic acid is in conjugation with the aromatic ring and is an electron withdrawing group. So we can draw a resonance structure where electron density is being pulled out of the aromatic ring and a positive charge is placed on the ortho position. Then we can draw a further resonance structure where that positive charge is now placed on the para position, and you could even draw one more where the positive charge gets placed on the other ortho position. But you get the idea. Because of those resonance structures, if we looked at the hybrid structure for benzoic acid, we would see partial positives on the ortho and para positions of the aromatic ring, meaning that those carbons and the hydrogens attached to them will be more deshielded because they have less electron density around them. So if we look back at our product, we can do the same thing. We'll place partial positives on the ortho and para positions to the carboxylic acid, which would place a partial positive on the carbon that hydrogen B is attached to. So B is going to be more deshielded than hydrogen C and we can label it like this. And that is it. I recommend doing this for any future lab where we have an aromatic ring in the product. Look to see if there are electron drying or electron donating groups. Consider resonance structures and place the partial positive or partial negative charges appropriately and this will definitely help you analyze and label the peaks.

Here's a look at the mechanism. When the hydroxide reacts with the aldehyde, it forms a tetrahedral intermediate, and when that collapses, a hydride will attack another 4-chlorobenzaldehyde, and this forms the two products that we saw before. The deprotonated alcohol product will... pull a proton off of the acid because the acid has a lower pKa and that forms the alcohol product and the deprotonated acid product.

This means it is a two to one ratio from the reagent to the products. So industrially if you're trying to make money off of this this wouldn't be a great reaction to do because the highest percent yield you could get for each product would be 50 percent. But in a lab setting this is a cool reaction and allows us the opportunity to learn some techniques in separating two products.

both of which we would want to keep. To show that separation, I'll be using a flow chart which looks something like this. We have a conical vial with our two products formed after the reaction, and I'll use HA to represent the acid product and OH to represent the alcohol product. The reaction was done under basic conditions, so the acid product will be deprotonated and dissolved in that aqueous solution like we saw at the end of the mechanism, whereas the organic alcohol product won't be soluble in that solution so it'll precipitate out and form a solid on the spin vein. To separate the two compounds we can add an organic solvent like methylene chloride that will create a bilayer and start dissolving the alcohol product.

Methylene chloride is more dense than water so it'll be on bottom with the dissolved alcohol product and the top aqueous layer will still contain the ionized acid product. Now we could get to the same place if initially instead of looking like this we had the two products mixed together as two solids. We would just add methane chloride to dissolve both of them and then add an aqueous base to deprotonate the acid and that would create the bilayer like we see here.

Now we can separate the two phases and then to make sure that we get all of the organic product out of the aqueous layer, we'll go ahead and extract that aqueous layer two more times with methylene chloride and combine all the organic extracts together. and we'll save the aqueous phase for later because that contains our acid product. We don't want to toss that.

Now we have our alcohol products separated, but just in case any acid product happened to move into the organic phase with the methylene chloride extracts, we'll add sodium bicarbonate that will deprotonate any remaining acid and move it into the aqueous layer. Then we can separate those two layers and dry the organic phase with sodium sulfate to remove any leftover water. Once we have our dry organic phase, we can evaporate the solvent leaving a crude alcohol product. And then for purification, we'll recrystallize that product using the groin and then remove the crystals. And we can then characterize those crystals by taking a melting point and running an IR spec.

As for the acid product, we want to be able to get it out of solution. And the easiest way to do this is by adding a concentrated acid, which will put the proton back onto the carboxylic acid group. This will make the molecule neutral and it will no longer be soluble in the aqueous solution.

So it'll crash out of solution and we'll see it precipitate as a white slurry. Then we can remove that crude product using vacuum filtration and then recrystallize it using ethanol. so we can characterize a pure product using melting points and IR.

Okay, I'll start out by measuring the 150 milligrams of our reagent, the 4-chlorobenzyldehyde, and then I'll go ahead and take that and put it in a 5 milliliter conical vial with a large spin vane. Then I'll be adding 400 microliters of methanol. and 400 microliters of the potassium hydroxide. Now I can set up the apparatus and I'll let everything reflux for an hour.

And you can actually start to see the solid alcohol product forming here. You can see in that kind of white pasty film on top of the liquid. Once the hour has passed, I'll lift everything off of the heat so the conical vial can cool. And here you can see the alcohol product gathering.

on the spin vein and the acid product is going to be dissolved in the aqueous solution. Initially I'll dilute the basic solution by adding some distilled water and then I can go ahead and start the methylene chloride extractions. As you can see the alcohol product doesn't dissolve super well at the beginning so I'll be stirring it and trying to break it up to dissolve it better.

So the solid alcohol still hasn't dissolved completely. but it should continue to dissolve as I keep doing further extractions. I ended up having too much methane chloride trying to dissolve the solid, so I transferred everything to a beaker.

And now I can save the aqueous solution containing the acid product for later and move on to the sodium bicarbonate washes which I'll be doing two of. And then I'll rinse the organic extracts with some distilled water just to help remove any ions that might have been left over from the bicarb washes. Now I can transfer the methylene chloride solution to an earlier flask where I can dry it using sodium sulfate and I'll add that until it's free flowing. Once I've removed the dry methylene chloride solution I'll actually add a little bit more and rinse the sodium sulfate to reclaim any alcohol product that might have been left behind. Now I can just evaporate off the solvent and you can actually start to see the alcohol product gathering on the side of the beaker here.

But this is what the crude product looks like once I've evaporated off all of the methylene chloride and now I'll be transferring that crude product to a Craig tube. so that I can purify it through recrystallization. I'll place it in a hot water bath so that it can be dissolved at high temperatures using LaGroin as the recrystallization solvent. And I'll place that in the hot water bath as well so that it's also at a higher temperature.

Now I'll start working on dissolving the alcohol product by slowly adding LaGroin and stirring it. Just so that I add the minimal amount of solvent to dissolve it at higher temperatures. Which, it's looking like everything is dissolving pretty well now, so I'm going to put it into an ice bath to make sure that the solid can come out of solution, which it definitely can, so that's good. So I'll place it back in the hot water bath to dissolve the solid once again, and now leave the tube on a rack so that the crystals can form slowly at room temperature.

Now this clip is really sped up. In reality it took about 8 minutes for the crystals to fully form. Once the recrystallization process is done, I'll just scrape the sides with a spatula so it'll be easier to remove the crystals and they don't stick to the sides as much.

To remove the crystals from the solvent, I'll place this stopper in the top of the kreg tube and then I'll place both of them inside of a propylene test tube, flip it upside down, and then put it into a centrifuge. What that will do is pull all the liquid to the bottom of the... propylene test tube, but the stopper will keep the crystals inside the Craig tube and therefore removing them from any liquid. So here's the centrifuge.

I'll put the test tube with the crystals on one side and then balance it with a test tube that just has water in it on the other side. And then I'll run the centrifuge for a couple minutes just to make sure that all of the solvent is pulled to the bottom of that test tube. Once that's done, I'll remove the test tubes from the centrifuge.

And you can see pretty well the separation of the solvent from the crystals. Now I can just remove the Craig tube and we should have a purified alcohol product that we can characterize. First I'll want to weigh out the alcohol product on a watch glass so that we can calculate a percent yield for this experiment. Then I'm going to crush up the crystals just a little bit so it'll be easier to place them into an open closed capillary tube to run the melting point. Right now they're in the open side of the capillary tube so I'm just going to tap it against the counter to draw them to the bottom of the closed side and once there it'll be ready to run a melting point now.

I placed it into the machine well below its melting point of 72 degrees Celsius just in case there were impurities and it started melting early. Crystals didn't start melting until about 71.3 degrees Celsius and didn't finish melting until 72 degrees Celsius. So that'll be the melting point range for the alcohol product. I'll run an IR spec using the neutral mole technique. So I'll add some of the crystals and then add some neutral and then I'm going to mole the salt plates together.

Until more of a film is what's left in between them. Now I can run the IR spec and Here's what the absorption pattern ends up looking like which looks pretty good. We have the alcohol peak as expected We have some neutral which is okay. We see some aromatic overtones around 1800 1900 and a carbon oxygen peak So this one turned out pretty well. Let's move on to the acid product now Okay, I've got the conical vial with the acid product and I'm going to dilute the solution a little bit more with some additional distilled water.

Now I can add the concentrated hydrochloric acid and you can see the product precipitating out of solution in that white slurry as it's neutralized by the added acid and I'll make sure that I add enough so that the solution itself is acidic, meaning that all of the acid product will be protonated. and no longer dissolved. To make sure I added enough acid, I'll test the pH of the solution, which should turn the pH paper pink, so it looks like we're good here. To remove the solid product from the aqueous solution, I'm going to use a hirsch funnel and the vacuum filtration technique. The hirsch funnel has a stopper, so when I pour the contents of the conical vial in, the solid product will gather on the stopper and the vacuum will pull the aqueous solution through into the flask.

Turn on the lab vacuum valve now and you can see from the compression of this tube that the vacuum is working. So I'll pour the product into the hirsch funnel and start separating it from the aqueous solution. I'll rinse the conical vial with some distilled water to help remove remaining product that was left over. And then I'll dump it into the hirsch funnel. This will also help rinse off any ions or pull anything out that isn't the product.

to help rinse it as you can see it's pretty sticky and clumpy right now but eventually that'll dry out i'll keep rinsing it with water just to make sure that i dissolve and remove anything that i don't want there and after a few minutes you can see that the crude product is a lot more dry now and is probably ready to be removed i'll remove the crude product from the hirsch funnel And follow the exact same process for recrystallizing this product as I did with the alcohol product, only this time I'll be using ethanol as the recrystallization solvent instead of lagroin. So like before, I'll be adding just enough ethanol to dissolve the crude product at high temperatures, which is looking pretty good right now. And then I'll place the Craig tube in an ice bath just to make sure that the solid can come out of solution. And I'll re-dissolve the solid in the hot water bath and leave the tube on a rack so that the product can recrystallize slowly at room temperature. I'll separate the crystals by using the stopper and propylene test tube again and then place it into the centrifuge and allow it to run for a few minutes.

I'll weigh out the purified acid crystals now so that a percent yield can be calculated. Now I can characterize the product so I'll put some in an open closed capillary tube to run a melting point. And we see the crystals start melting around 239 degrees Celsius and finish around 240.1.

So it's a little lower than literature value, so there's probably some contamination. Finally, I'll run an IR spec using the neutral mole technique. And it looks like the carboxylic acid product was definitely formed.

We can see the carboxylic acid funnel ranging from like 3300 to 1800. with the carbon hydrogen peak from neutral sticking out in the middle there. And then we can see the carbonyl peak from the carboxylic acid group around 1682. So this looks pretty good. Okay let's take a quick look at the NMR. I'll just be focusing on the HNMR for this compound but the same concepts should help with the other two compounds as well. So let's take a look at the HNMR to figure out which molecule we're looking at.

We see this peak around 13, which is indicative of a carboxylic acid, so this would be the HNMR for the carboxylic acid product. Now this also could have been confirmed by looking at the D2O exchange here, which is a tool to help identify protons on functional groups that can hydrogen bond, like carboxylic acids or alcohols, etc. Basically what happens is we introduce a bunch of D2O, which is water with deuterium instead of protons, and as they hydrogen bond with the compound that we're trying to analyze, the proton on the hydrogen bonding functional group will get switched out with a deuterium, and since deuterium is not a proton, it will no longer show up on a proton NMR.

So the alcohol peak, or in our case, the carboxylic acid peak that we were looking at, will no longer show up on a D2O exchange. So this just helps confirm that we are looking at the carboxylic acid product, and we can go ahead and label all the protons on that molecule now. We already know that the peak above 13 is for the acid proton, so we can go ahead and label that as hydrogen A, and then we know the peak at zero is for TMS, so we can label that as well. From here it looks like we have four more peaks and only two hydrogens left to label, but if we look at these two peaks down here, They are not in the aromatic range, meaning they're not going to be for hydrogens B and C.

They've got to be something else. Plus, if we look at their integration, both of them have an integration smaller than that of hydrogen A, meaning less than 1, which isn't really possible, so they have to be solvents. The one around 2.5 is for DMSO, dimethyl sulfoxide.

And then when water is mixed with DMSO, it shows up around 3.4, so we can label that as water. Which actually makes sense why when we look at the... deuterium exchange the peak no longer shows up around 3.4 it shows up around 4 because as water interacts with d2o it changes into hdo which will have a different chemical shift than water now the remaining two peaks are in the aromatic range so that'll be for hydrogens b and c but both of them are doublets and both of them have an integration of two so it's going to be difficult distinguishing which one is which just by looking at the peaks so we're going to have to consider other things happening on the molecule to be able to label those.

For that, let's look at benzoic acid and consider its resonance structures. The carboxylic acid is in conjugation with the aromatic ring and is an electron withdrawing group. So we can draw a resonance structure where electron density is being pulled out of the aromatic ring and a positive charge is placed on the ortho position.

Then we can draw a further resonance structure where that positive charge is now placed on the para position, and you could even draw one more where the positive charge gets placed on the other ortho position. But you get the idea. Because of those resonance structures, if we looked at the hybrid structure for benzoic acid, we would see partial positives on the ortho and para positions of the aromatic ring, meaning that those carbons and the hydrogens attached to them will be more deshielded because they have less electron density around them.

So if we look back at our product, we can do the same thing. We'll place partial positives on the ortho and para positions to the carboxylic acid, which would place a partial positive on the carbon that hydrogen B is attached to. So B is going to be more deshielded than hydrogen C and we can label it like this. And that is it. I recommend doing this for any future lab where we have an aromatic ring in the product.

Look to see if there are electron drying or electron donating groups. Consider resonance structures and place the partial positive or partial negative charges appropriately and this will definitely help you analyze and label the peaks.

Transcript for:Cannizzaro Lab Reaction and Techniques

Transcript for:
Cannizzaro Lab Reaction and Techniques