For 1 part of the lab I:
prepare 4-Bromo-2-chloro-6-iodoaniline
I dissolve 250mg recrystallized 4-
Bromo-2-chloro-6-iodoaniline in 4ml of glacial acetic acid in 25ml erlenmyer flask and add 1ml water to the mixture. Then I prepare a solution of 250mg technical iodine monochloride in 1ml of glacial acetic acid.
Over 3-5miin I add the iodine monochloride solution. With stirring, I heat the mixture to 90C and then allow it to cool to about 50C. Then I add enough sodium bisulfite to turn the color of the mixture bright yellow. I then dilute the rxn mixture with enough water that the combined volume of the sodium bisulfite + water is 3ml. I then cool the rxn in an ice-water bath and isolate product by filtration. Then I wash the crystals with ice-cold 33%aq acetic acid and then with 5ml of ice cold water.
The question is:
what is the difference between sodium bisulfate and sodium bisulfite? Why is sodium bisulfite used in this synthesis?
The difference is that sodium bisulfite has a extra double bonded oxygen connected to the carbon. But as to why it is used as opposed to the sodium bisulfate...not sure about that.
After bromination step you need to clean your glassware. Should you rinse with acetone?
I said no but I read somewhere that bromine reacts with acids and I don't think that acetone is an acid so it wouldn't form bromoacetate I think.
two hypothetical sequencyes have been developed for converting an arbitrary compound A into E with similar overall yield:
A=>B=>C=>D=>E
A(27%)B(59%)C(62%)D(58%)E?
A=>B=>C=>D=>E
A(58%)B(57%)C(51%)D(30%)E?
What is the overall yield? show work
And which sequence is the most economical of the 2 routes? Give 2 reasons why.
I'm not sure what to do here.
I think that I could say that(for second one) 58g is the ammount of yield out of 100g that was gotten but is this right?
And I'm not sure about what you said last time on the questions after thinking about them:
1ml isopentyl alcohol + 1.5ml glacial acetic acid + 3 drops sulfuric acid=> isopentyl acetate
2)One method favoring the formation of ester is to add excess acetic acid. Suggest another method, involving the right-hand side of the equation, which will favor the formation of ester.
you said that I could distill it as it forms would that mean that the rxn would by itself produce some products even though it wasn't shifted to the right?
3) Why is it easier to remove excess acetic acid from products than excess isopentyl alcohol?
I didn't really understand when you said that
You Said:To try to separate two organic compounds, one an alcohol and the other an ester would be MUCH tougher. So we take the easier route.
is what you mean that
IF I used excess isopentyl acetate for example, then I would have to take out the acid as well as take out the excess isopentyl acetate instead of just removing the acid with the excess acid too at the same time?
Thank you Dr.Bob
I'm out of my league with the organic but in-so-far as the analytical questions go I can help.
The question is:
what is the difference between sodium bisulfate and sodium bisulfite? Why is sodium bisulfite used in this synthesis?
The difference is that sodium bisulfite has a extra double bonded oxygen connected to the carbon. But as to why it is used as opposed to the sodium bisulfate...not sure about that.
NaHSO4 vs NaHSO3. Think about it. The S in SO4^= is at its highest oxidation state of +6 while S in SO3^= is +4; therefore, sulfite (or bisulfite) can be oxidized to sulfate OR reduced to S^0 or S^=. I don't know which way your reaction is going but you can tell by the chemical reactions you have in front of you. I SUSPECT it is going to sulfate which makes it a reducing agent but you need to check to make sure.
After bromination step you need to clean your glassware. Should you rinse with acetone?
I said no but I read somewhere that bromine reacts with acids and I don't think that acetone is an acid so it wouldn't form bromoacetate I think.
I don't know. Also, I don't see a bromination step in the procedure.
two hypothetical sequencyes have been developed for converting an arbitrary compound A into E with similar overall yield:
A=>B=>C=>D=>E
A(27%)B(59%)C(62%)D(58%)E?
A=>B=>C=>D=>E
A(58%)B(57%)C(51%)D(30%)E?
I don't see a percent yield for E in either sequence. However, you CAN do it the long way. Start with 100 g theoretical yield, The first sequence would give you 27 g for A, 27 x 0.59=15.4 g for B, etc. Try that for both sequences and you will have your answer. The quicker way is by multiplying 0.27 x 0.57 x 0.51 x 0.31 = 0.0243 = 2.43%. Check my arithmetic. Also make sure you get the same answer for the short route that you get for the long route.
What is the overall yield? show work
And which sequence is the most economical of the 2 routes? Give 2 reasons why.
You will have the answers here when you do the math above. Two reasons; #1 will be percent efficiency. #2 COULD be the cost of the ingredients. #3 might be the risk involved in performing the experiment.
I'm not sure what to do here.
I think that I could say that(for second one) 58g is the ammount of yield out of 100g that was gotten but is this right?
And I'm not sure about what you said last time on the questions after thinking about them:
1ml isopentyl alcohol + 1.5ml glacial acetic acid + 3 drops sulfuric acid=> isopentyl acetate
2)One method favoring the formation of ester is to add excess acetic acid. Suggest another method, involving the right-hand side of the equation, which will favor the formation of ester.
you said that I could distill it as it forms would that mean that the rxn would by itself produce some products even though it wasn't shifted to the right?
3) Why is it easier to remove excess acetic acid from products than excess isopentyl alcohol?
I didn't really understand when you said that
You Said:To try to separate two organic compounds, one an alcohol and the other an ester would be MUCH tougher. So we take the easier route.
is what you mean that
IF I used excess isopentyl acetate for example, then I would have to take out the acid as well as take out the excess isopentyl acetate instead of just removing the acid with the excess acid too at the same time?
With regard to shifting the equilibrium. Alcohols and acids produce esters. Various ones under various conditions produce various amounts; i.e., the efficiency varies with conditions and how much of the reactants are present. So yes, left alone, perhaps heating as needed, the yield probably is good for at lea\t 33% or so. You can sift that to something like 60-75% by adding excess acid, or by adding excess alcohol (I don't think either of them will shift it that much) OR by removing some of the produce, in this case, the ester. The ones I have seen done for this scenario is to add alcohol and acid to a pot, add a distilling head and a reflux condenser above, then heat away. The ester forms, the vapors go up, the vapors condense, the ester collects in the receiving pot, and the reactants stay in the pot. This only works, of course, if the ester boils at a lower T than the reactants. For the other question, I simply meant that I can neutralize excess acetic acid easily with NaHCO3. Using excess acetic acid leaves me with ester as the only organic material there when the acetic acid is neutralized. If you start with an excess of alcohol, then you have used all the acid and you are left with alcohol + ester which you must now separate. How do you do it? It is much more difficult to separate the alcohol and the ester than it is to just add an excess of acetic acid to begin with, that uses the alcohol, you neutralize the excess acid with NaHCO3, and you leave the rest of the lab group in your dust as they try to separate alcohol and ester.This may be my last post tonight. I leave tomorrow. I hope all this makes sense.
what is the difference between sodium bisulfate and sodium bisulfite? Why is sodium bisulfite used in this synthesis?
The difference is that sodium bisulfite has a extra double bonded oxygen connected to the carbon. But as to why it is used as opposed to the sodium bisulfate...not sure about that.
NaHSO4 vs NaHSO3. Think about it. The S in SO4^= is at its highest oxidation state of +6 while S in SO3^= is +4; therefore, sulfite (or bisulfite) can be oxidized to sulfate OR reduced to S^0 or S^=. I don't know which way your reaction is going but you can tell by the chemical reactions you have in front of you. I SUSPECT it is going to sulfate which makes it a reducing agent but you need to check to make sure.
I can't see the chemical rxn since all I got was a handout with 5 different synthesis' and I I just noticed the bromination first step below has sodium bisulfite used. I don't really get what is happening though.
After bromination step you need to clean your glassware. Should you rinse with acetone?
I said no but I read somewhere that bromine reacts with acids and I don't think that acetone is an acid so it wouldn't form bromoacetate I think.
well I thought it was a general thing not a specific to this rxn thing
The bromination is when I add .15ml of bromine solution and 0.3ml of glacial acetic acid. And then the rxn is stirred at room temperature for 10min. Then saturated aqueous sodium bisulfate is added untill the yellow color is discharged. and I assume the question is refering to this part and since the rxn includes acid it would it would probably form bromoacetate like I suspected...is this right...?
Oh..darn (I just read the bottom last sentence before) I hope you see my last post before you go.
I'm just thinking that I'm lucky this is a 5 week lab..hopefully your vacation doesn't last that long XD
Have fun anyhow =D
Just this quick answer to the Br/HSO3^- part. OK. You are adding NaHSO3 until the yellow color is discharged. I'm guessing that the yellow color is the color of the Br2 in the mixture, adding NaHSO3 until the yellow color is discharged would mean then that Br2 was going to Br^-. If so then from 0 to -1 is a gain of electrons (oxidation) so HSO3^- goes to SO4^= (or HSO4^- )from +4 to +6 which is a loss of electrons or reduction). Going to BrO3^- would be 0 to +5 which is the wrong way (both would be oxidations that way). I looked in an OLD book by Latimer and the potential is favorable for SO3^= going to SO4^= while Br2 goes to Br^-. It is not favorable for Br2 going to BrO3^- while SO3^= goes to S4O6=, S, or even S^= (although those things can change with concentration, acidity, and that kind of thing). As for coming back, I may just stay over there.
Just this quick answer to the Br/HSO3^-part. OK. You are adding NaHSO3 until the yellow color is discharged. I'm guessing that the yellow color is the color of the Br2 in the mixture, adding NaHSO3 until the yellow color is discharged would mean then that Br2 was going to Br^-. If so then from 0 to -1 is a gain of electrons (oxidation)I thought that oxidation was loosing electrons (LeO says GeR) so HSO3^- goes to SO4^= (or HSO4^- )from +4 to +6 which is a loss of electrons or reduction).once again isn't this oxidation? (loss of e-)I'm confused by these statements of what I think are oxidations being reductions Going to BrO3^- would be 0 to +5 which is the wrong way (both would be oxidations that way). I looked in an OLD book by Latimer and the potential is favorable for SO3^= going to SO4^= while Br2 goes to Br^-. It is not favorable for Br2 going to BrO3^- while SO3^= goes to S4O6=, S, or even S^= (although those things can change with concentration, acidity, and that kind of thing).
As for coming back, I may just stay over there. Nooooo...but then again thinking about it, as long as you have computers where your going....not so bad eh? ^______^
Question 1: What is the difference between sodium bisulfate and sodium bisulfite? Why is sodium bisulfite used in this synthesis?
The main difference between sodium bisulfate and sodium bisulfite is their chemical composition. Sodium bisulfate (NaHSO4) is a compound formed by the combination of sodium, hydrogen, sulfur, and oxygen atoms. Sodium bisulfite (NaHSO3), on the other hand, is a compound formed by the combination of sodium, hydrogen, sulfur, and a single oxygen atom. The difference lies in the number of oxygen atoms present.
In this synthesis, sodium bisulfite is used for a specific purpose. Bisulfite ions (HSO3-) can act as reducing agents, which means they can donate electrons to other compounds and thereby cause a reduction reaction. In the given lab procedure, sodium bisulfite is added to turn the color of the reaction mixture bright yellow. This indicates that the bisulfite has reacted with the excess iodine monochloride, reducing it to a different compound. This reduction step is important for the synthesis of the desired product.
Question 2: After the bromination step, should you rinse with acetone?
No, you should not rinse with acetone after the bromination step. Acetone is a common organic solvent and can be used for rinsing glassware in many scenarios. However, in this particular case, you have used bromine, an extremely reactive compound, in the bromination step. Bromine reacts vigorously with many organic solvents, including acetone. This reaction can be dangerous as it may produce potentially harmful byproducts.
Therefore, it is important to use caution when handling bromine and its reaction mixtures. After the bromination step, you should clean your glassware following the appropriate procedures outlined in the lab protocol. Typically, this involves washing with suitable solvents like water or ethanol to remove any residual bromine or other reagents.
Question 3: What is the overall yield in the two hypothetical sequences?
To calculate the overall yield in both sequences, we need to multiply the individual yields at each step. Let's calculate for both:
Sequence 1:
A (100%) => B (27%) => C (59%) => D (62%) => E (?)
Overall yield = 100% x 27% x 59% x 62% = 0.1002 or 10.02%
Sequence 2:
A (100%) => B (58%) => C (57%) => D (51%) => E (?)
Overall yield = 100% x 58% x 57% x 51% = 0.1875 or 18.75%
Question 4: Which sequence is the most economical, and why?
In terms of economic feasibility, the second sequence appears to be more favorable. Here are two reasons why:
1. Higher overall yield: The second sequence has a higher overall yield (18.75%) compared to the first sequence (10.02%). This means that a larger proportion of the starting material is converted into the desired product, resulting in less wastage. Higher yield translates into better utilization of resources, making the second sequence more economical.
2. Fewer reaction steps: The second sequence involves fewer reaction steps compared to the first sequence. The fewer steps involved in a synthesis, the less time, energy, and resources are required. This reduces the cost of the overall process, making the second sequence more economical.
Overall, the second sequence combines a higher overall yield and fewer reaction steps, making it a more economical option compared to the first sequence.
Question 5: Regarding isopentyl alcohol and isopentyl acetate reaction
Here are the answers to your previous questions:
1. Distillation: Yes, if you were to distill the reaction mixture as the product forms, it could help shift the equilibrium of the reaction towards the formation of more ester. Continuous removal of the product by distillation drives the reaction forward, as per Le Chatelier's principle. This is one method to favor the formation of the ester.
2. Easier removal of excess acetic acid: It is easier to remove excess acetic acid from the products compared to excess isopentyl alcohol due to their physical properties. Acetic acid is a volatile organic compound, which means it has a lower boiling point compared to isopentyl alcohol. This property allows acetic acid to be easily separated from the reaction mixture through distillation, leaving behind the desired product. Isopentyl alcohol, on the other hand, has a higher boiling point, making its removal more challenging.
I hope this helps! Let me know if you have any further questions.