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by: Thurman Wilderman


Thurman Wilderman
GPA 3.66


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This 20 page Class Notes was uploaded by Thurman Wilderman on Saturday September 12, 2015. The Class Notes belongs to CHEM 2211L at University of Georgia taught by Hubbard in Fall. Since its upload, it has received 14 views. For similar materials see /class/202580/chem-2211l-university-of-georgia in Chemistry at University of Georgia.


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Date Created: 09/12/15
Organic Chemistry Lab Review 11112011 101000 PM 1 Recrystallization method of separating compounds Solvents have different solvation properties at higher temperatures Gravity and suction filtration used 5 steps of recrystallization O O O O 0 choose solvent for recrystallization dissolve impure mixture in the hot solvent remove insoluble impurities from hot solvent by filtration cool to crystallize the pure compound isolate the pure compound by suction filtration how to choose a solvent 0 O A good solvent dissolves your target compound at high temperatures but not at low ones and not the impurity Or dissolves your target at high temperatures but not the impurity Low BP or high volatility bumping expulsion of liquid 3 methods for inducing crystallization Scraping glass remove solvent addingseed crystal add water BEFORE hot plate add boiling water until solution turns clear 2 Distillation separation of two liquids based on separate boiling points boiling point 0 vapor pressure atmospheric pressure Know how to draw and label parts Fractional 0 better because it increases number of simple distillations distillated more pure Simple distillation Azeotrope constant boiling solutions of distinct proportions Theoretical plates minidistillation O 0 Measure efficiency of fractionating columns More plates efficient liquid separation Daltons Law o PPa Pb 0 Total pressure is sum of partial pressures Raoults Law o P an be 0 Total pressure is the sum of partial pressures times the mole fraction 3 Extraction of trimyristin solidliquid extraction liquidliquid extraction multiple extraction reflux apparatus solid liquid liquid liquid immiscible solvents Different solvent and temperature setting yield different extracted components simple distillation apparatus dichloromethane is more dense than water reflux how to separate liquid 4Extraction of a 3component mixture acidbase manipulation REVIEW FLOW CHART Acid has OH Base NH2 Fluorenone yellow 1 molar HCI 6 molar NaOH likes dissolve likes base and neutral are insoluble in the solvent of water 5 Thin Layer Chromatography how to calculate Rf values 0 retention factor used in 3 ways 0 compound purity determination o compoundmulticomponent mixture identification 0 monitoring the progress of a chemical reaction draw plates o solid liquid chromatography o mobilephase o the developing solvent o stationary phase 0 TLC plate o most polar lower 0 least polar higher 0 because TLC plate is made of silica gel o visualize spots under UV light 0 if they aren t there use iodine 6 transfer hydrogenation o hydrogen s came from cyclohexene o catalyst 10 palladium on carbon o molarity has to be matching o when boiling get rid of benzene by product o celite filtration o to filter out small impurities o oxidation 0 lose electrons o reduction 0 gain electrons o If the reaction between an organic compound and hydrogen is below 480 C it needs a catalyst o Raney Nickel 0 slow requires high temp and pressure o Transfer Hydrogenation 0 Transfer of Hydrogen from organic source rather than acquiring it from H2 gas Highly flammable o Cyclohexene is hydrogen Donor 0 Formation of Benzene drives reaction 7 Molecular modeling o draw Newman projections o rank acids and bases o module 2 acid rankings o usedSpartan o More electronegative atoms can stabilize a negative charge better o Inductive effect 0 Electronegative atoms can pull the atoms doing the acidbase interaction and delocalizes its negative charge o Resonance o Resonance delocalizes a negative charge and increases acidity 8 Preparation of Diphenylacetylene o addition of bromine to double bonds o look up mechanism o dehydrohalogenation of alkyl halides o alternate bromine source pyrindiniumhydrobromideperbromide 0 solid reagent KNOW MECHANISM 9Gas Chromatography o 4 factors that affect peak separation 0 column length 0 column temperature 0 carrier gas flow rate 0 stationary liquid phase o retention time 0 measure of amount compound has retained o Mobile gas phase o If chloride was higher then time threw off results 0 Bromide should have been more nucleophilic 0 Had excess chloride o Know mechanism o Goal is a carboncarbon triple bond o Stationary phase 0 liquid o Mobile phase 0 carrier gas He for us o GC is great for separating and analyzing components of a volatile mixture using gas chromatogram 10 Substitution Reaction o 1 AgNO3 in Ethanol Sn1 o 15 NaI in acetone Sn2 o Sn2 reactions are faster Sn2 0 One concerted step Bimolecular nucleophilic substitution Know mechanisms Carbocation reactivity stability 0 More substituents the more stable Stationary phase 0 liquid Mobile phase 0 carrier gas He for us GC is great for separating and analyzing components of a volatile mixture using gas chromatogram SN2 o prefers polar aprotic solvents due to their ability to solvate the carbocation intermediate SN1 o prefers polar protic solvents to because they help solvate the leaving group SN2 0 does not like protic solvents because it increases the basicity of the nucleophile and it won t want to attack the desired substrate 1A nucleophile selectively bonds with or attacks the electrophile which has a positive or partially positive charge 2The electrophile are usually connected to a leaving group 11112011 101000 PM Objective To design an experiment that determines the solubility of five compounds using only two grams of each compound and only one hundred grams of water Procedure 1 Weigh out 100 grams of water in a graduated cylinder Weigh out 2 grams of each compound in at least 100 mL beakers and label the 2 F 0quot 7 8 Repeat steps 3 through 6 to determine the solubility of each compound containers a Potassium Carbonate b Ammonium Chloride c Sodium Sulfate d Calcium Sulfate e Calcium Phosphate Beginning with the first compound listed add a little water at a time while constantly stirring to the solid until the entire amount solid is dissolved or until all of the 100 grams ofwater is used a If all of the solid is dissolved weigh the amount of water that is used If all of the water is used and the solid is not completely dissolved then use a filter funnel to filter the undissolved solid Once all of the water has filtered through the funnel heat the solid over a beaker ofboiling water on a watch glass to remove all of the water Solubility Once all of the solid is dry weight the solid and subtract the weight from the original two grams to determine the amount of solid that was actually dissolved Record all results wt solute X 100 g of compound100 g of water wtsolvent To determine the solubility of the compound use the following formula Results Compound Amount of Amount of water Solubility Rank of Solubility compound used used 1most soluble Potassium 200 grams 952 grams 210 g of comp g 2 Carbonate of water Ammonium 200 grams 947 grams 211 g of comp g 1 Chloride of water Sodium Sulfate 200 grams 973 grams 206 g of comp g 3 of water Calcium Sulfate 330 grams 100 grams 330 g of comp g 5 of water Calcium 112 grams 100 grams 112 g of comp g 4 Phosphate ofwater Experiment 1 Recrystallization Identi cation of an Impure Unknown Puri cation is an important part of any laboratory experiment Commonly the chemical processes by which we synthesize novel organic compounds result in impure mixtures of liquids and or solids In order for these compounds to be of any use commercially or educationally they must rst be isolated from one another and then puri ed In today s lab we will be focusing on the isolation and identi cation of an organic solid The method we will employ is known as recrystallization an essential laboratory technique used for the puri cation of solid compounds In this experiment you will be given an impure sample of an unknown organic solid which you will purify via recrystallization and identify by determining its melting point Additional background information concerning both recrystallization and melting point determination will be provided on your lab ELC website New Techniques Recrystallization Filtration Gravity Suction Melting Point Determination Table of Reagents Gather all relevant physical data structure MW MP for the following compounds acetanilide benzamide benzoic acid ethyl 35dinitrobenzoate mnitrobenzoic acid otoluic acid salicylic acid vanilin In today s experiment you will be using this data to determine the identity of your puri ed unknown Safety You will be working with boiling water and handling hot glassware Caution must be exercised at all times Be careful to never heat a closed vessel and always make certain to add a boiling chip to any liquid before heating it Safety glasses and lab gloves are to be worn at all times Procedure Obtain one gram of an impure unknown solid from your TA These unknowns are labeled with a speci c alphanumeric code identifying them to the instructors This code must be recorded in your data section along with all other important experimental information Make sure that you stir the mixture prior to measuring out your sample It is a heterogeneous blend of pure unknown target compound and inert sea sand impurity Add approximately 150 175 mL of water and several boiling chips to a 250 mL Erlenmeyer ask Heat the water to a gentle boil using a hotplate Be careful not to heat the ask too vigorously as this can evaporate a significant amount of the water prior to use On a second hotplate place a 125 mL Erlenmeyer ask into which has been added one gram of the solid unknown along with a boiling chip Using a ring clamp as a handle slowly pour approximately 510 mL of boiling water into the ask containing the impure solid This does not need to be an exact measure of liquid You should estimate the volume rather than measure it out in a graduated cylinder After the water has been added swirl the resulting mixture for one minute while maintaining the solution temperature on the hotplate You may need to continue adding small aliquots approximately 35 mL of boiling water to the mixture and swirling until the unknown solid is fully dissolved Be sure to swirl the mixture for approximately one minute between each water addition to aid the dissolution process If you notice clumps forming in the mixture during this time use your stirring rod or spatula to break them up Be aware that there will be insoluble impurities sea sand left in your ask alter the unknown solid has fully dissolved This is to be expected It is important to use only the minimal amount of solvent necessary to dissolve your unknown Adding too much solvent will result in a significantly decreased percent recovery at the end of the experiment Set up a gravity filtration system comprised of a short stem funnel glass lined with uted or folded filter paper covered with a watch glass and placed on top of a 125 mL Erlenmeyer ask see Figure l for filter paper folding instructions The ask should have a few milliliters of boiling water in it so that the apparatus will remain warm during the filtration process Maintain the water s temperature using a hotplate o Realize that the water you add to the ask will be added to the total amount of water your unknown solid must be recrystallized from later on It is vital that the solution remain hot during filtration If the temperature of the solution drops significantly some of your unknown may crystallize out in the filter paper rather than passing through into the ask below It is recommended that you moisten the filter paper with a small amount of hot water prior to filtration This will prevent cooling After filtration rinse the filter paper with 23 mL of boiling water to dissolve any crystals that may have formed despite these precautions Gravity ltration l f9 E Cvsasspaper 7 ngMy g Fulani Repeal qua srs pining mm m m slxleenms halt g Openoullmo Hilledoone Figurel Remove the ask from the hotplate and allow the filtrate to cool slowly to room temperature Once room temperature is reached check for crystal formation Lfthere is no evidence of crystals gently scratch the inner wall ofthe ask with a stirring rod to begin to form Once the initial crystal formation is finished cool the ask in an ice bath for W from solution Riichner mnel and a n I 39 39 L F t and nre them down with a spatula or stirring rodin order to expel any excess moisture Allow the r The m that is removed during this stage the better your results wi a a nu are water 11 be overall Next scrape the Ifyour sample is still somewhat damp continue to dry it under a heating lamp for several nee um point ofthe unknown and attempt to identify it from your table ofreagents Conduct a Variance 0f5 C or meanw mun u uui uuLu in Melting Point Determination You will find that most of the products generated in your organic chemistry lab this semester are either white or yellow crystalline solids at room temperature It can be difficult to tell them apart based solely on physical appearance One ofthe first tasks that you will be assigned in lab this semester is to purify and correctly identify an unknown organic compound There are many ways in which an organic unknown can be identified and we will touch on several methods during the course ofthe semester Some of the more advanced techniques involve the use of expensive instrumentation such as IR and NMR spectrometers These techniques offer us the most detailed information concerning the chemical makeup of a compound but they are also very expensive and complex to utilize Melting point comparison on the other hand offers a fine balance between ease of use and accuracy of information All it requires is a simple apparatus and an accurate thermometer The melting point of a chemical substance is the temperature at which its solid and liquid phases are in equilibrium This week you will be performing a melting point determination of an unknown solid and comparing it to known literature values in order to properly identify it The melting points of most readily available chemical compounds are easily obtainable from any number of library and online resources The CRC Handbook of Chemistry and Physics is an excellent source of physical data It is imperative that you have accurate melting point values to compare your laboratory data to otherwise your experiment will be an exercise in futility Make sure that your table of reagents is filled out in its entirety prior to beginning your lab A compound s melting point range is defined by the temperature at which the sample first shows signs of melting droplet formation within the pack of sample crystals and the temperature at which the sample becomes fully liquefied ie Benzoic acid 122 123 C A pure substance will exhibit a narrow melting point range 1 2 C As impurities are introduced into a sample the melting point range ofthe parent compound will begin to change Impurities have a twofold effect they both broaden and depress lower the observed melting point range A rough estimate of the purity of a mixed sample can be made via a comparison ofthe observed melting point and the literature value of the pure compound As a general rule a 1 impurity results in a 05 C depression In reality this effect works both ways Each compound in a two component mixture will end up depressing the melting point of the other This results in a melting point range that is typically both broader and lower than that of either of the pure compounds that make up the sample Figure 1 provides a graphical representation of this relationship The melting point of pure component B steadily declines as the mole percentage of component A slowly increases When we reach a mixture composition of 74 B and 26 A the melting point has decreased to below the pure value for either individual component TB and TA As we progress along this curve we eventually reach a minimum melting point value for the two component mixture At this specific composition of mixture AB both compounds Wm mekswukaneausw Tmsxszerwedzheeuzemt Dam vfzhe 95127quot maths speeme ampasmanatwmmxaccursvaneswzhthememwafzhemmuremmvanems Fartms mp e h remDamaccursaammvasmanmm eBandi 96A Dntewemwe haandzms Dam amvanem a snwmnmanmgasme WWHW and mmvanemA szhe Daemmmpaund Asmemmepereemsgemsteeauymereesessmaaaes hememng DmmrangeunmwereachzheNewquotmea ue vureAU Nwzhmwuundemmd hwwvurmestan affenyaur memngpmm daermmatmns m m r sumem eam hwm Prevare and evamate asamwe Temperaqu m Camposmon mom w gme 1 meme mm Dunn unwmmepueemeninzn simnln Make surezhawaur END E xsdw AWE samwewm he exeemmnaw mmmrzm mad mm a is stab mamaWm dry mare ef nemN m ma use aheatmg amvm a Dracesswfnetessaw DnceYaursaNWexs N EmshasmaHDamanmaa n vaw erarwaur wmgxess Maw Et s nad yuursamp e obtam a memng pmnt capwHarytubEfmm ynurTA mg 5 a m yuurquEand crysta s suthat a smaH pumun sfnrcgd ntamemba a2 carem m a avermad n Usmgtuu much samp E Wm th y bruadEn ynur memng pmnt rangE rEndEnng n nae ynur hgmytapme c nsEd End nnthe bEnchtnp m saneme samp emtnthEtube NExt set ynur Vuu are nuw rEadytu mavetnthe MELTEmp apparatus see ngura 2 thevmometer msen capmary ma wnn sample Figure 2 MeHemp Appzmlux m u u H m temperature at which the entire sample is fully liquefied These two temperatures will make up your melting point range Record this data in your lab notebook o Caution avoid touching the heating block while the Mel Temp is in use or in the process of cooling down Place your thermometer and sample capillary into their appropriate wells Make sure that the thermometer slides completely into its holder If it does not your results will likely be inaccurate The voltage control is used to determine how fast your sample will be heated For quick approximate melting point ranges a higher setting may be used This would correspond to a heating rate of roughly 6 8 C per minute This is a good rate to use when you are unsure ofthe exact melting range of your unknown compound and need to get a quick preliminary value Once you have obtained a rough estimate of the melting range you can set up a second measurement in order to obtain a more accurate reading For your second attempt heat the sample quickly to within 20 C of your original melting point range and then lower your heat setting to 1 2 C per minute for the remainder ofthe run This will give you a nice balance of speed and accuracy o In general heat too quickly and your accuracy suffers heat too slowly and you will run out oftime before you can collect your data 0 Each new run must be performed using a new sample Never reuse your samples 0 Allow the Mel Temp to cool to at least 20 C below your melting range prior to using it again If you place a new sample in a hot apparatus it will most likely melt immediately Once the melting point range of your unknown has been established you will need to verify its identity using mixed melting points Choose the two most likely candidates from your table of reagents we will call them compounds Aand B Typically these will be the compounds with melting points closest to that of your unknown Make sure that you are only choosing compounds with melting points equal to or higher than that of your sample Remember impurities cannot raise the melting point of a substance they can only lower it Prepare two samples the first containing a 5050 mixture of your unknown sample and compound A and the second containing a 5050 mixture of your unknown sample and compound B Place both capillaries in the Mel Temp and determine their melting point ranges If the unknown is indeed compound A there will be no noticeable change in the melting point range of your first mixed sample You will however see a marked change in the melting point range of your second sample In this case compound B will be acting as an impurity in the mixture and will depress the observed melting point range If your unknown happens to be Recrystallization mrymmmmn mm am an c um Figure 1 Recrystamzatmn Haw Chart Thefive steps nf recrynallizatinn 1 Choose an appropnate sowentfor the recrystamanon Dwsso ve the mpure rmxture m the sowent 0 wa n 5 i I u g 8 3 u o c 3 8 3 S o 3 so ate the pure compound we when Mtrauon Step 1 How to Choose a Solvent A solvent must have a number of specific characteristics in order for it to be deemed suitable for use in recrystallization The process of finding a suitable solvent can be quite difficult and time consuming depending on the physical properties of your impure mixture The following conditions must be met First it must fully solvate your compound at high temperature but not solvate it at room temperature Dissolution is necessary so that the crystalline lattice can be broken down and reformed in absence of the impurities However if the compound is soluble at room temperature it will be exceedingly difficult to retrieve it from the solution at the end of the experiment Second the impurities that you are removing from the mixture must be either soluble in the solvent at room temperature or insoluble at high temperature This temperature related contrast is essential for the proper separation of the impurities from the desired compound In either case the two components can be separated from one another via simple gravity filtration Third the solvent must be chemically inert with regards to both the desired compound and any impurities within the mixture If your compound reacts with the solvent a chemical transformation will occur rather than a purely physical dissolutionrecrystallization Fourth and final the solvent must be suitably volatile in nature so that it can be easily removed once the pure crystals have been isolated A solvent that lingers in your sample can adversely affect your percent recovery calculations as well as your melting point determinations On the other hand choose a solvent that is too volatile and it will evaporate before you can fully dissolve your impure solid in step two Step 2 Dissolving the Impure Solid This step is self explanatory The impure solid is placed in an Erlenmeyer flask and small volumes of hot solvent are added in order to dissolve it Stirswirl the flask after each addition to aid in the dissolution of your compound Be sure to break up any clumps that form in the flask as you swirl Care must be taken to use only the minimum amount of solvent necessary to dissolve the solid Using too much solvent at this step will make it far more difficult to recover your pure compound at the end ofthe experiment You will end up wasting time evaporating excess solvent at step four and your percent recovery will be adversely affected Once your compound has fully dissolved place the flask on a hotplate to keep it warm and move on to step three Step 3 Separation of Impurities via Hot filtration Set up a hot filtration apparatus in preparation for separating the insoluble impurities from your solution The set up includes the following an Erlenmeyer flask a short stem funnel glass a watch glass fluted or folded filter paper and a minimal amount of hot solvent 0 Place the funnel filter paper included on top of your flask 0 Add a few milliliters of solvent plus a boiling chip to the flask and place it on a hotplate 0 Cover the funnel with a watch glass to keep the system warm 0 Once the vapors of the boiling solvent have permeated the funnel remove the watch glass and quickly filter the hot solution 0 Pour a small amount of boiling solvent through the filter to dissolve any solid that may have crystallized during the filtration The pure compound is now isolated in the filtrate The insoluble impurities have been captured in the filter paper and are ready to be discarded Step 4 Cooling the Solution and Crystallizing the Solid Remove your flask from the hotplate and set it on the bench top to cool As the flask reaches room temperature crystals will begin to form in solution If crystals do not form even after the flask cools crystallization must be induced There are two primary methods for accomplishing this 0 Scratching the inner wall of the flask with a stirring rod 0 Introducing a seed crystal to the solution Be cautious not just any crystal will do Adding a crystal that does not match your pure compound will only introduce an impurity back into the system Each of these methods creates a point of nucleation on which the rest of your crystals can then begin to grow Watch closely as the crystals begin to form filling the bottom of your flask Place the flask in an ice bath to aid in the complete precipitation of your pure solid from solution 0 The most effective ice bath is made up ofa modest quantity ofice in a large beaker partially filled with water Use the ice water to submerge the bottom of your flask to make sure that the temperature drop is uniform through the filtrate Using too much ice and too little water can hamper the effectiveness of the ice bath In the event that your crystals are still not precipitating out of solution you may have used too much solvent during your dissolution step Place your flask back on a hotplate and reduce your solvent volume before attempting to cool and crystallize your compound again Once crystallization has been completed you are ready to isolate your pure solid step 5 isolating the Pure Compound via Suction Filtration The iihai step of the recrystaHi39zati39on processisthe coHecti39on ofyour pure crystais Thisis giassAare setup in detaii lter papei Simmer Mime Nenprene aaapiei m aspiraiuron water iauoei Hiiiaiian quotask Separation acnnlque Sumun quotmaniaquot Figure 5 Suction fiiuation apparatus up your giassAare ahd tiiterihg your soiutioh ake sure thatyou usea thickwaHed rubber hose to connectyour iter ask to the 39 apiratr 39 quot 39 h uhd in the common equipmehtdrawers at the frontof the iao room Latex tubing ithihiight tahi may hot be used for suctioh tiitratioh why do you think that this is t a i i p y iiu r hose Remember Buchhertuhheisare research ouaiitygiassware Do not forget to add the neoprene adaptor to your setup without it you wiH not be abie to maintain a seai and therefore wi39H have no suction 39 39 the r i39Li39iiyuiH39iLipdpi funnei Fai39iure to do sowi39H resuiti39n 05 of crystais Lab 8 Preparation of Diphenylacetylene Results Mass of TransStilbene 2003 g Mass of Stilbene Dibromide 1840 g Mass of Diphenylacetylene 0272 g Melting Point of Diphenylacetylene 71 C Limiting Reactant Calculations amp Percent Yield Step One 20 g of TransStilbene X 1 mol 18024 g of Trans Stilbene X 1 mol Stilbene Dibromide 1 mol TransStilbene X 34006 g Stilbene Dibromide 1 mol Stilbene Dibromide 40 g ofPMPB X 1 mol 31986 g of PMPB X 1 mol Stilbene Dibromide 1 mol PMPB X 34006 g Stilbene Dibromide 1 mol Stilbene Dibromide 425 g Stilbene Dibromide Percent Yield of Stilbene Dibromide 1840 g 3770 X 100 4881 Step Two 377 g of Stilbene Dibromide X 1 mol 34006 g Stilbene Dibromide X 1 mol Diphenylacetylene 1 mol Stilbene Dibromide X 17824 g Diphenylacetylene 1 mol Diphenylacetylene l gt 15 g of KOH X 1 mol 5612 g of KOH X 1 mol Diphenylacetylene 1 mol KOH X 17824 g Diphenylacetylene 1 mol Diphenylacetylene 476 g Diphenylacetylene Percent Yield of Diphenylacetylene 0272 g 1980 g X 100 1374 Discussion This experiment required two reaction mechanisms to form a carboncarbon triple bond The rst reaction produced the intermediate stilbene dibromide through the halogentaion of the alkane transstilbene using pyridinum hydrobromide perbromide as a bromine source The percent yield of stilbene dibromide was only 4881 A low percent yield could be caused by rushing through the experiment The transstilbene was not given enough time to dissolve completely in the glacial acetic acid thus the transstilbene may not have fully reacted with the pyridinium hydrobromide perbromide while being heated Another possible explanation for a lower percent yield could be that some of the crystals were either stuck to the pouring ask or went through the lter paper when conducting the suction ltration The second reaction for this experiment was the double dehydrohalogentation of the alkyl halide stilbene dibromide intermediate and potassium hydroxide The diphenylacetylene produced yielded a mere 1374 Seeing as how the intermediate stilbene dibromide had a low percent yield it is no surprise that the nal product had an even lower percent yield It is possible that the ethanol was not hot enough to fully dissolve the impure crystals Therefore when another suction ltration was conducted some of the precious diphenylacetylene crystals were left with the impure crystals instead of proceeding to the nal ltration This could explain why the diphenylacetylene percent yield was extremely low The melting range of the diphenylacetylene was abnormally high Usually if there are impurities in the crystals than the melting range will be wide and low However the melting point was high at 71 C compared to the literature value of 625 0C A possible explanation could simply be that there was some ethanol left in the diphenylacetylene crystals The crystals were not completely dried when conducting the melting point determination Ethanol has a higher melting point and it is likely the pure diphenylacetylene crystals had some traces of ethanol thereby skewing the melting point temperature results Conclusion Majority of the errors that occurred are due to time constraints The lab was already extremely long and results are not valid if things aren t given enough to react dissolve heat ltrate etc If this lab was repeated duting the rst mechamism the transstilbene would be given more time to dissolve in the glacial acetic acid For the second mechamism the ethanol should have either been more hot or given more time to dissolve the impure crystals And for the melting rate determination the crystals could have more dry before performing the melting point


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