[viet]

in my lab ill be adding 3eq. of acetyl chloride to anhydrous MeOH (to give a final concentration of Trp. of 0.75 M)

the procedure is not clear, i'm not sure how to determine how much of each amount i will need.

[Arkcon]

When you're told to "add equivalents" you're expected to use the reaction stoichiometry and compute the mole equivalents to mix.

[AC Prabakar]

in my lab ill be adding 3eq. of acetyl chloride to anhydrous MeOH (to give a final concentration of Trp. of 0.75 M)

the procedure is not clear, i'm not sure how to determine how much of each amount i will need.

Could you elaborate a bit?

[Arkcon]

Why not give it a try, post your attempt, and we'll see where we can help you.  What is the balanced chemical equation for the reaction you're trying to perform?  What is the molar mass of each reagent?  How will you measure them -- generally by mass, but you could convert the molar mass to a volume of liquid, if that's easier to dispense.

[viet]

CH3COCl + CH3OH   CH3COOCH3 + HCl

CH3COCl: 78.49 g/mol
CH3OH: 32.04 g/mol
CH3COOCH3: 74.08 g/mol

sorry, i'm pretty lost.

[fledarmus]

(to give a final concentration of Trp. of 0.75 M)

What is Trp.?

This is an odd reaction for you to be using 3 eq of acetyl chloride, unless this is only part of the sequence. When I've seen this reaction in use, it has been for preparing an anhydrous solution of HCl in methanol, usually for making an HCl salt. I don't know of any particular use for methyl acetate. The way you presented your question, it appears that you are supposed to be adding three equivalents of acetyl chloride to one equivalent of methanol, which would give a final solution of 1 equivalent of methyl acetate in two equivalents of acetyl chloride? Are you using any solvent, or is this the total reaction?

My guess, based on the ways I have used this reaction, is that you have already prepared some basic compound, such as an amine, and are now going to prepare an HCl salt. To do so, you are preparing three equivalents of HCl in a methanol solution, where the equivalents of HCl are measured against the basic compound rather than against the methanol (which being the solvent is present in large excess). After you have prepared your HCl solution, you will add it to a solution of the basic compound and either precipitate the salt or strip the solvents down. Excess HCl and the largely inert methyl acetate will disappear either way. Is this anywhere close to what you are doing, or am I completely off?

[viet]

trp is tryptophan. acetyl chloride and methanol is the first step in the tryptophan methyl ester preparation.

the procedures are:

acetyl chlorde (3eq) was added dropwide to anhydrous MeOH ( to give a final conc of Trp of 0.75) at 0C. After 5 min, L-tryptophan (1eq) was added and the soln was stirred for 2 hrs.After cooling, the solvent was evaporated and the residue was dissolved in water and saturated aq. NaHCO3 soln was added.

[Arkcon]

So maybe your reaction above is incorrect?  Since this is a reaction between acetyl chloride and the amino acid tryptophan, maybe that reaction is the one you should work with?  How much tryptophan do you have?  In moles.  What is 3 times that, in moles, of acetyl chloride, when measured in grams?

[viet]

lets say I use 1g of tryp.

1g / 204.3 g·mol^-1 = 0.00489 mol of tryp.

ratio of 1 tryp:3 acetyl chloride.

3(0.00489 mol) = 0.01467 mol of acetyl chloride

0.01467 mol (78.49 g·mol^-1 = 1.1514 g acetyl chloride

thats what i got so far. to convert to volume i would use the density of acetyl chloride (1.10g/cm³ which is 1.10g/1mL) right?

[Arkcon]

Yes, that's well worked out at each step.

[discodermolide]

trp is tryptophan. acetyl chloride and methanol is the first step in the tryptophan methyl ester preparation.

the procedures are:

acetyl chlorde (3eq) was added dropwide to anhydrous MeOH ( to give a final conc of Trp of 0.75) at 0C. After 5 min, L-tryptophan (1eq) was added and the soln was stirred for 2 hrs.After cooling, the solvent was evaporated and the residue was dissolved in water and saturated aq. NaHCO3 soln was added.

All you are doing is making a water free solution of HCl in excess methanol to esterify an amino acid. 0.75M Trp is quite a dilute solution, all you need to do is ensure that there is enough MeOH, a) to react with the acetyl chloride and b) to esterify the Trp.

[fledarmus]

What Discodermolide said. The acetyl chloride reaction with methanol is very fast, and essentially complete when you finish the addition, although you're stirring it for 5 minutes just to make sure. All you have in the flask now is methanol, methyl acetate, and HCl.

Your calculations are good so far. The only number you are still missing is the volume of methanol you would need to give a final concentration of tryptophan in the solution of 0.75 M.

[discodermolide]

What Discodermolide said. The acetyl chloride reaction with methanol is very fast, and essentially complete when you finish the addition, although you're stirring it for 5 minutes just to make sure. All you have in the flask now is methanol, methyl acetate, and HCl.

Your calculations are good so far. The only number you are still missing is the volume of methanol you would need to give a final concentration of tryptophan in the solution of 0.75 M.

I think the final concentration is irrelevant. Trp will esterify in any case, any 0.75M is very dilute. For a more efficient process you must be more concentrated. But I am not a research chemist (I was one but not anymore) 

[viet]

so i got the volume of acetyl chloride to be 1.0467 ml.

im lost at figuring out the volume of methanol.
i have 0.0489 mol of tryp (1000ml/0.75 mol) = 6.52ml of tryp

[discodermolide]

so i got the volume of acetyl chloride to be 1.0467 ml.

im lost at figuring out the volume of methanol.
i have 0.0489 mol of tryp (1000ml/0.75 mol) = 6.52ml of tryp

Trp is a solid so you can't measure it in mL.
As I said the methanol volume is irrelevant for this reaction, you only need enough to react with 3 x acetyl chloride + 1 for the esterification. Or say 1.1 to be safe.