Calculation for organic synthesis and %Yield calculation and Limiting reagent

[@raiszeeshanali9565]

thank you!!!! very helpful vedio for organic synthesis students

[@Dr.Mosharraf]

Thank you for your support

[@sudhazala587]

Thank you sir for the detail explanation i want to know more that how we can fine out the amount or equivale for catalyst and CuI for sonographer reaction ?

[@Dr.Mosharraf]

Didn’t I explain how I took the catalyst?

[@johnjonesnyogbe5309]

I find this video really helpful! Thanks

[@Dr.Mosharraf]

Thank you. Please watch my other videos to learn more. If you learn that will be my success. Thank you for watching and your feedback

[@strife3084]

@ ~ 24 min. you start discussing mol weight of reactants, more specifically limiting reagent "A". how did you figure 6:34 5.9 mmol? I'm confused. I calculated 5g of reactant A (molar mas 334.59) = .0149 moles x 1000( for mmol) = 14.94 mmol of reactant A. Am I missing something? why wouldn't mmol be calculated by figuring mol of product x 1000 or gram weight x 1000 over mole? which you have written. but 5g x 1000 over 334.54 is 14.94 and not 5.98. or is this specific to balancing limiting reagents?

[@Dr.Mosharraf]

Good catch on 6:34 min. It was my mistake. You are correct. mmol = (amount in gram x 1000)/ Molecular weight

[@strife3084]

Thank you! thought I've lost my mind. @ 41 years old, been out of college for a long while now but made it to an Associates. Went on to use it in Metallurgical examination and failure analysis in areospace industry (welding) now. however OrgChem has ALWAYS been my true love! p.s. the rest makes perfect sense to me. however my issue is how does one calculate mol over limiting reagent? is it arbitrary? so say it takes 2 moles of reactant b to react stoichiometrically with a to be on the safe side you call it 2.3x? and what would one do in the case of potential side/competing reactions? would you consider stoichiometric weights at that point?

[@Dr.Mosharraf]

Theoretically you can do with stoichiometric weight. But to be simple and to keep remember quickly, or easy to do, I prefer to select one reagent which is less mole or mmol that will be LR.

[@reabetswemocheko963]

Hi sir, how do you calculate the amount of reactant needed when you’re given the amount of the product ?

[@Dr.Mosharraf]

By doing back calculation. First find out how much reactant will give you how much product if the reaction converts 100 % product. Then do a small scale reaction and purify it. See how much you get. Amount you get divided by amount you should get (if 100% conversion happen) times 100 = Will be yield percent. Suppose you want to use reagent A as 10 g and according to theoretical reaction you should get product B as 6 g (100 % converted) Now you did real reaction with 200 mg. If 100 % converted then you are suppose to get 120 mg from it. But you got 100 mg. Now calculate percent yield = 100 mg/120 mg X 100 = 83.3 %. Now do back calculation. If you use 10 g A . Theoretically you should get 6 g product (100% conversion) but from small scale reaction you got 83%. So find out 83% of 6 g = 6x.83 = 4.98 g. Now 4.98 g product will be found from 10 g A, 50 g product will be found = (10 x 50)/4.98 g = 100.4 g Reactant A I hope you will understand. If not clear then comment it.

[@reabetswemocheko963]

Thank you very much !

[@salehchem4376]

thank you doctor. if i synthesised product from two reactants, after reflux then i checked TLC , I found my product almost in same level and same Rf with one of reactant, i try changed solvent system (mobile phase) like ( ethyl acetate : hexane 1: 3 and 1:4 and 1: 9) or ( ethyl acetate : petrolium ether 1: 5), but still my product same level with one reactant; please could tell my another ratio or solvent to seperate it or get different level? thank you

[@Dr.Mosharraf]

Please watch my video on TLC kzbin.info/www/bejne/oKbceppnnrR9m68