Question

can some one help me find a limiting reagent and percent tieldncalculation for acetanilide

i started with 0.090 of acetanilide and after recryatallization i ended up with the product weigh 0.2346g. p bromoacetanilide.

please explain and show each and every step

Dispose of the filtrate from the Hirsch funnel tiltration 8P Al container for halogenated organic solvents rude proauct into a this mixture. Place all other filtrates into the R E Running the Reaction To a tared 5-mL conical vial with a cap, add the given amount of one of the follow- ing compounds: 0.090g of acetanilide, 0.060 mL of aniline, or 0.070 mL of anisole Reweigh the conical vial to determine the actual weight of the aromatic co Add 0.5 mL of glacial acetic acid and a spin vane to the conical vial. Attach arn condenser and place the conical vial in a water bath at 23-27°C, as shown in Tech. mpound. air ic compound is completely nique 6, Figure 6.6. Stir the mixture until the aromat dissolved. While the compound is dissolving, pack a drying tube loosely with glass wool. Add about 0.5 mL of 1M sodium bisulfite dropwise to the glass wool it is moistened but not soaked. This apparatus will capture any bromine g during the following reaction. iven off Under the hood, obtain 1.0 mL of the bromine/hydrobromic acid m ixturel in a 3-mL conical vial. Place the cap on the vial before returning to your lab bench. e bromine/hydrobromic acid mixture through the top of While stirring, add all th the air condenser, using a Pasteur pipette. CAUTION Be careful not to spill any of this mixture. Attach the drying tube prepared above. Continue stirring the reaction mixture fo 20 minutes Note to the Instructor: The brominating mixture is prepared by adding 2.6 mL of bromine 7.4 mL of 48% hydrobromic acid. This will provide enough solution for 20 students, assum o waste of any type. This solution should be stored in the hood hts Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in

Answer 1

= 1 Amount of acetanilide started = 0.090g m.w. of acetanilide = = 135 g/mol. motes of acetanilice used = 0.020 = 6.667 x 10th moles = 0.6667 mmoles 135 (2 Amount of boomine equivalents used - 10 ml of boominating mixtual contains = 2.6 ml of Bo 1 ml " 11 " 11 = 0.26 ml of Br O Density of Br = 3.119g/ml Iml of boominating mixture contains = 0.81094 g Mw. of Br = 160g/mol moles of Brg in 1ml mix of boominating mix = 5.068375 X 10 roles. = 5.068 mmoles So, in reaction 5.0 68 mmoles of Boy are used which 18 equal to 10.13 mmoles equivalents of Bo-atom - (As only one Bratom is requided to deact) Total Br mmoles used = 10.13 mmoles + Br-atom mmoles from HB solution. From above two calculations it is clear that Br-atom equivalents are in excess than acetanilide. Hence, formation of product will depends on mmoles of acetanilide used as acetanilide should be consumed to its 100% ideal theasttically Hence, acetanilide is limiting / Reagent

> Theoretical yield of p-boomoacetanilide. God - (mg) mmoles of acetanilide used x M.W. of þ- Boomo aclteni lide - 0.6667 x 214 mg/mollsmoles = 142.674 mg obtained yield = 0.2346 g = 234.6 ang % age yield - The obtained field x 100 % Theoretical field = 234.6 mg x 100 % 142. 6 Ting = 164.43 %