Posted by ~christina~ on Tuesday, March 11, 2008 at 5:27pm.
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 yuo 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? ^______^
HI
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