Question

NUCLEOPHILIC DISPLACEMENT- FORMATION OF AN ETHER BY AN SN2 REACTION Bond formation by use of an SN2 reaction is very important for organic and biological synthesis. Many new bonds formed make use of this versatile reaction pathway. In this reaction, one partner is the nucleophile and the other partner is the electrophile. The electrophile will have a leaving group that will be lost in the reaction. In the SN2 reaction that you will be performing, the nucleophile is the phenoxide ion, which is formed from reaction of phenol with sodium hydroxide. The nucleophile will displace the iodide to form butyl naphthyl ether and sodium iodide. Can you think of any possible undesirable side reactions? MI NaOH + Nal + H2O EtOH, Heat phenol chlorbutane butoxybenzene 0.935g/mL 150.22 g/mol EXPERIMENTAL PROCEDURE To a 100 mL round bottom flask (rbf), add 0.56 g (14.0 mmol) of NaOH (about 5 or 6 pellets) and 1.0 g (10.62 mmol) of phenol. Add 20 mL ethanol and a few boiling chips to the flask. Swirl the flask. Reflux the solution for 15 minutes or until all of the solids have dissolved. After 15 minutes, remove the heat source and allow the solution to cool slightly.1 Remove the reflux condenser and add 2.0 mL 1-chlorobutane (17 mmol) to the round bottom flask. Replace the condenser and reflux for an additional hour. Before stopping the reaction, place about 50 mL of water in an ice bath for later use. After 1 hour of refluxing, remove the heat source and carefully pour the contents of the rbf over about 25 g of ice in a 250 ml beaker. Using the water that has been cooling, rinse the rbf and add this to the beaker containing the ice and your product. Swirl the contents of the beaker until all of the ice has melted. Extract the product using petroleum ether 10 mL three times. Separate the petroleum ether layers from the aqueous layers using a separatory funnel. Combine the petroleum ether layers and treat the solutions with calcium carbonate. Filter the solution and wash the filtrate with petroleum ether and combine the dried petroleum ether layers. Boil your petroleum ether off to leave behind butoxybenzene as a liquid. Using the volume or the weight, calculate the percent yield. Make sure the solution has stopped boiling before you remove the reflux condenser.

Please calculate the percent yield if my actual yield was 0.06g

Answer 1

1 millimoles quivalents density Compound Molar 11 Amount mass (ng/mol) 110719/ml Phenol . 94.11 NaOH 40 Chlorobutane 92.57 109. 10.621 0.56g 14.0 1.32 2mL 19 mmd 1.789 10.89 g/m OH OT+ Md Nach Cphenol) (Chlorobutane) ELOH Heat ne I trad+hol Etoh, Heat w butoxybenzene 0.935 g/mL 150. 22 g/mol Phenol Amount of phend = 1.09. (weight) Molecular weight of phend - 94.11 g/mol :: Number of moles - 19 ) 1 -0.01062 moes (94-11 g/mol = 10.62 millimoles Nuot; Weight of NaoH=0.56 g. Molecular. mass dy About - 40g/mol, Number of moles = (4:56) = 0.014 moles 4-0 = 14.0 mmot. chlorobutane ndarmass = 92.57 g/mol. volume of chlorobutano semi i density = 0.899/mo weight of chlorobutane - (volume (clensity) =(2) C0.89) = 1.789 . Number of moles = (1.78 g = 0.019 moly 9257g/ml ) 19 mmoles

Nach YOH. thachtho EtOH, A From the givers data, for 1 equivalent of phenol, 1,32 equivalents of Naotl, 1.79 equivalents of chloro butane are used. Phend is Limiting reagent. So, theoretically from 1 equivalent Cle 10.62mmoi) I equivalent of butoxybenzene should be formed. le Theoretical yield of butoxy benzene = 10.62 mmit. .: Amount of butoxybenzene = (10.62)x(Molarmare) Theoretical yield) -- (10.62)x (150.22) = 1595 milligrams H = 1.595 grams From data s' Actual Yield = 0.06g. X100 ** Percent yield - Actual yield Theoreticalyield X.100 1.5 - 37.617.%. .Percent yield = 37.6%