November 15, 2024, 03:48:28 AM
Forum Rules: Read This Before Posting


Topic: Isolation of Reaction Products by Extraction With a Solvent  (Read 8696 times)

0 Members and 2 Guests are viewing this topic.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Isolation of Reaction Products by Extraction With a Solvent
« on: November 08, 2013, 04:40:39 PM »
I was given an "unknown mixture" in a recent organic lab and asked to separate it into its aqueous and neutral fractions. I used 50ml Dichloromethane as a solvent with 10ml of NaOH being added to the sep funnel every time I extracted. At the end of the lab I was told that my Aqueous Fraction was Benzoic Acid and my Neutral Fraction was Acetanilide. I know I must get the % yield, but in order to do that I must first get balanced equations for both of these fractions...I was just wondering if it was the NaOH or the Dichloromethane I react each compound with? And what should the products be like? I know it is an Acid/Base extraction...but does that mean the balanced equations should be both acid/base and their associated products? I am fairly confused as I have only just finished stereochemistry and have not moved onto these topics yet! :( please assist!

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #1 on: November 08, 2013, 04:55:32 PM »
Have you provided us with the complete procedure that you followed?

Did you simply add dichloromethane and NaOH to your mixture or were there any more steps?

What was the weight of the material you started with? What was the weight of the product(s)?

“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #2 on: November 09, 2013, 07:39:20 AM »
1-Dissolved 2.003g of Mixture 9 un 50ml Dichloromethane.
2-10ml of 2N NaOH was added to the DCM soln while in the separating funnel (only the first time, all other additions are made directly to the separating funnel.)
3-The sep. funnel was then inverted vigourously & the pressure realeased frequently.
4-The tap was then opened to let the organic layer flow into a flask labellled Organic.
5-The tap was then closed and an Aqueous vessel placed beneath it, the contents were then allowed to flow into this vessel.
6-The organic layer was then added back into the sep. funnel.
7-Steps 2-7 are repeated four more times.
8-The aqueous fractions are all added to the same vessel.
9-The combined Aqueous fractions were poured into the sep funnel and 10ml DCM added to them. Following this, extraction was then carried out as previously described (step3-6)
10-Both organic layers were then combined
11-These organice fractions were washed with 10ml H20 three times.
12-The first washing fraction of Aqueous soln was added to the combined fractions of Aqueous Soln
13-The second and third washings were kept separate.

The initial weight of the mixture was 2.003g,
The weight of the crude neutral component was 0.969g.
The Final weight of the Neutral fraction was 0.827g and the weight of the Aqueous fraction was 0.811g.

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #3 on: November 09, 2013, 08:35:23 AM »
Was some acid added to the aqueous phase before step 9?
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #4 on: November 09, 2013, 01:43:36 PM »
No just the 5 x 10ml additions of NaOH during extraction and the 10ml DCM (step 9) were added.

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #5 on: November 10, 2013, 01:39:18 AM »
In essence what you have done in steps 1-8 is to take DCM soluble benzoic acid and acetanilide as a mixture and treated them with aqueous NaOH.

What happens to acids when they are treated with NaOH? What happens to acetanilide when it is treated with NaOH?
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #6 on: November 10, 2013, 05:11:43 PM »
Acid + Base= H20 + Salt so the first would be a Neutralisation reaction.

I know acetanilide is an Amide. But since we haven't gotten as far as these mechanisms in lectures I'm not sure what the step is.
My google search says Amide + Hydroxide ion = Amine + Carboxylate

So if this is true would be guess below be correct?

Acetanilide + NaOH = CH3CO2Na + C6H5NH2

Also, in this case the limiting factor would turn out to be Acetanilide...but I am not sure which product I would be calculating the % yield on?

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #7 on: November 10, 2013, 08:38:28 PM »
Amides are quite stable to alkaline hydrolysis, it can happen but requires some heat so in this case we can assume that acetanilide remains unchanged.

So if we focus on the acid, what is the product (salt) of benzoic acid and sodium hydroxide?

What are the physical properties of this salt? Is it hyrophillic or hydrophobic?
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #8 on: November 11, 2013, 04:41:43 AM »
NaOH + C6H5COOH = C6H5COONa + H2O

The salt would be Sodium Benzoate. This salt is hydophilic and it's solubility in water is 556 g/l in water at 20°C so therefore it is hygroscopic.

I forgot to mention that I recrystallised the Neutral Fraction only. This would have involved placing it on a water bath and evaporating it off until there was a sediment left on the bottom of the flask, it was then treated with a minimum amount of chloroform/light pertoleum, boiled, placed in an ice bath, filtered and air dried. This being the case, would alkaline hydrolysis then have occurred?

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #9 on: November 11, 2013, 05:47:38 AM »

I forgot to mention that I recrystallised the Neutral Fraction only. This would have involved placing it on a water bath and evaporating it off until there was a sediment left on the bottom of the flask, it was then treated with a minimum amount of chloroform/light pertoleum, boiled, placed in an ice bath, filtered and air dried. This being the case, would alkaline hydrolysis then have occurred?


No alkaline hydrolysis should have taken place under those conditions.

So you have made sodium benzoate, a very hydrophillic substance with a very low solubility in DCM.

What does not make sense to me is that you extracted your sodium benzoate into water from DCM but then you are extracting back into DCM from the aqueous phase.

I can't see how the salt would favour the organic phase when it is partitioned, I would expect the partition to be very much in favour of the aqueous phase.

This is why I am wondering whether there was an acidification step that you may have missed off your procedure.
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #10 on: November 11, 2013, 02:00:35 PM »
There was an addition of concentrated HCL after all of the washings of the Aqueous soln and after the Aqueous soln had been on ice for a few minutes, but straight after the addition of HCL we filtered it by suction at the pump? This happened long after the steps above...is this what I may have missed? Sorry it totally slipped my mind to put this bit in!

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #11 on: November 11, 2013, 04:58:09 PM »
It may be an important step in your procedure, can you provide a full description of the exact method that you followed so that we can provide a difinitive answer to your problem.

If you add HCl to Sodium benzoate, what would the product be? Would this be more soluble in DCM or water?
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #12 on: November 12, 2013, 04:32:13 AM »
After the steps I have above the manual says
Neutral Component:
The DCN solution is dried with anhydrous magnesium sulphate and filtered into a weighed conical flask. The solvent  is evaporated in the fume hood using a steam bath. The dry flask and residue is then weighed carefully. The residue is recrystallised from the minimum amound of chloroform/light petroleum. The pure material is collected by filtration and air dried. Record the weight and melting point of the compound before/after recrystallisation.

Acidic Component:
The combined sodium hydroxide extracts are cooled to about 0°C and acidified carefully with conc. HCL. Record your observations. isolate the solid by filtration, dry, weight and record the melting point. Recrystallise from the minimum amount of hot water. Dry and record the weight and melting point.

For the acidic component we hadn't enough time to recrystallise it so we were instructed to only go as far as isolating the solid by filtration, drying it, weighing it and recording the MP.

Offline Seons R

  • Regular Member
  • ***
  • Posts: 12
  • Mole Snacks: +0/-0
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #13 on: November 12, 2013, 04:36:46 AM »
C6H5COONa + HCl = NaCl + C6H5COOH.

The product would be Benzoic acid and would be more soluble in the DCM because it is denser than water?

Offline Archer

  • Chemist
  • Sr. Member
  • *
  • Posts: 1001
  • Mole Snacks: +85/-20
  • Gender: Male
Re: Isolation of Reaction Products by Extraction With a Solvent
« Reply #14 on: November 12, 2013, 08:56:33 AM »
C6H5COONa + HCl = NaCl + C6H5COOH.

The product would be Benzoic acid and would be more soluble in the DCM because it is denser than water?

This is correct but irrelavent in your case because you didn't extract the benzoic acid, you collected it by filtration. You acidified water soluble sodium benzoate and formed benzoic acid which is not very water soluble in cold water so it precipitated as a solid.

So you started with 2.003g and you ended up with two products, one weighing 0.969g and the other weighing 0.827g.

Your yeild is simply the percentage of the weight of each product recovered from the total weight of the mixture.
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Sponsored Links