How could you separate a mixture of the following compounds? T...

Question

How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH.

Chemical structures of compounds with their pKa values: carboxylic acid (4.17), ammonium chloride (4.60), alcohol (9.95), chloride (10.66).

Accepted Answer

Hello everyone. Let's do this problem. It says create a scheme to separate the following compound from a mixture. One molar hydro Bromma acid, one molar potassium hydroxide, water and Dietl ether are available for you to use. We are given five compounds, their structures and their respective P K A values. So let's read out those compounds. I'm going to label them A through D sorry A through E compound A is cyclo Heino acid which has a PKA of 4.8 compound B is the conjugate acid of iso quinoline with that nitrogen protend comp and that has a P K A value of 4.85 compound C is two chloro Phey and that has a P K A of 8.6 compound D is chloro cyclohexane. And that has no P K A given because there is no acidic proton. And we know that P K A comes from acid strength in the acid dissociation constant. So if it's not going to dissociate and donate a proton, it will not have a P K A value. And lastly compound E is the conjugate acid of cyclo Penil laine with that nitrogen protend. All right. So a couple things to note for this problem, we want to separate these from a mixture is that ionic compounds have a high solubility in aqueous solutions and poor solubility, an organic solvents such as ethyl ether, right? We are given ethyl ether as one of our solvents that we can use. On the other hand, the opposite is true four electrically neutral compounds. So electrically neutral compounds have poor solubility in aqueous solutions and good solubility in organic solvents. So this tool will help us separate these compounds because compounds B and E are ionic, right? They have a positive charge on those nitrogen atoms. So we can use that to our advantage in separating these. Also, we can use P K A to our advantage because remember that if the P H of the solution is less than the P K A, then our compound will be in its acidic form. So if we need to, let's say dep proin an acid like compound A, then that will give it a negative charge, right? We're losing that positive proton. So it will become a negative conjugate base which will make it ionic, which means we can extract it in an aqueous solution. OK. So let's start off by looking at all the P K A values. So notice how all of our P K A values are above four and all of our compounds are given to us in the pronated form. So at A P H of less than four in a solution with A P H, less than four, all the compounds are going to be protend except D, right. That's the exception here. So at A P H, less than four B and E are going to be positively charged while AC and D are remaining electrically neutral. Therefore, the first separation we can make is using that positive charge for B and E to separate them into the aqueous solution. Right? Because ionic compounds have good solubility and aqua solution. While Ac and D can go into an organic layer using diethyl ether. So let's create a solution that has a P H less than four, so that we can maintain these structures given to us and then we'll add an organic solution. So if we mix these compounds with water that has been adjusted to a PH of 3, and then we also add die ethyl ether, we can separate these two solutions into B and E which will be one layer, the water layer or the aqueous layer and in the organic layer, the die ethyl ether where we'll have our neutral compounds, our electrically neutral compounds which are AC and D. All right. So let's finish separating B and E before we tackle AC and D, since those are closer to being separated. So looking at compounds B and E compound B has A P K A of 4.85 compound E has A P K A of 10.7. So if we create a solution that has a P H between the P K A values of these two compounds, we can de proin compound B and then leave compound E proton it. And then once compound B is derotated, it will lose that charge, which means it will go into the organic layer, right? So let's do that. So first we can adjust the P H of the solution to, let's say between 7 and 8. And then we also want to add diethyl ether to create that organic layer for our electrically neutral compound. So in our organic layer, we'll have our derotated, I saw quinoline in the die ethyl ether layer. And in our water layer, we'll have our conjugate acid of cyclo penal amine, which still has a positive charge, which is why it will remain in the water layer. All right. So we've separated compounds B and E, let's go back to our organic mixture that has compounds AC and D. So let's look at the differences between AC and D. Well, similar to what we just did with B and E looking at the P K A s of A and C compound A has A P K A of 4.8 compound C has A P K A of 8.6. So if we adjust the P H of our solution, adding water that has a P H of, let's say around P H seven P H eight. Then we can dep proton the carboxylic acid group on compound A. And then that will extract the conjugate base of compound A into the water layer. Now that it's ionic and then compound D will remain in the or sorry. Yes, compound C N D will remain in the organic layer. And we can add some extra diethyl either to bulk up that organic layer. So first add water with P H 728 to de proin compound A which has a P K 4.8, then add diy ether to book up that organic layer. So now we'll have that conjugate base of compound. A coo minus attached to that cyclohexane. And that will be in the water layer. Now, since it is ionic and then remaining in the organic layer is compound C N D dot L ether layer. Now to separate compound C and D remember that D does not have an acidic proton or A P K A value. So we can't really keep adjusting the P H and compound C is still an acid. So we can use our basic solution to extract a proton from compound C and that will make it ionic and then we can separate it into the water layer. So let's add are one molar potassium hydroxide solution to the mixture of C N D and the diy layer. So then we'll get the conjugate base of two chloro in a water layer, I forgot to write water layer for compound A and then remaining in the ethyl ether layer is compound D. So this is a scheme you can use to separate all five of these compounds from a mixture which makes d the correct answer for this problem. That's it for this one. I hope this video was helpful and thank you for watching.

How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH.

Key Concepts

Acid-Base Properties

Solubility and Extraction

Reagent Selection

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