Orgo II Lab – Flashcards
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What is TMS? Why is it added to samples?
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Tetramethylsilane(TMS) a. Solvent used as the reference point for NMR b. Silicon is of low electron negativity (Lower than all atoms) i. Will not produce any deshielding effects on sample c. All the protons are equivalent (single peak at 0)
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What is one application of NMR?
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There are four basic pieces of information that can be obtained from standard proton NMR (pmr). These are 1) the number of non-equivalent protons 2) the chemical environment of each type of protons 3) the relative number of each proton present and 4) the number of adjacent protons.
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Types of solvents used in NMR.
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Solvent used cannot be NMR active, most are deuterated, we use deuterated chloroform (CDCl3) deuterated means H2 isotope. There is also a reference sample added, we use tetramethylsilane (TMS, (CH3)4Si)
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How does the methylene cyclohexane form? (Dehydration)
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Methylene cyclohexane forms after a carbocation rearrangement (hydride-shift) to a tertiary position. Elimination of the hydrogen on the proton make methylene cyclohexane.
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What is expected to be the major product of dehydration of 2-methylcyclohexanol reaction? Why?
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The major product under thermodynamic control is the most substituted alkene product (Zaitsev) a. The minor product under thermodynamic control is the least substitute alkene (Hoffman's) b. The major product forms because with the most substituted product forms in the lowest energy form (REFER to rxn coordinate)
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Why do 3 Products form in dehydration of 2-methylcyclohexanol reaction?
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The reason why 3 products might form is because of the 3 different ways the alkene can form. Without the hydride shift, the Hoffman product can happen. With the hydride shift, there's the possibility of two other products, the Zaitsev Product and an unstable alkene. Add it all together, that's three possible products.
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How does the IR change from the starting material to the product? (Dehydration of 2-methylcyclohexanol)
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The IR spectrum of 1-methylcyclohexane has a broad peak(3200-3400) for -OH a. The IR spectrum of the desired products contains a sp2 =C-H stretch around 3050cm-1. This peak appears as a shoulder on the normal sp3 C-H absorbance. A C=C stretch may also be observed around 1620cm-1.
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What complications can arise when interpreting the product IR? (Dehydration of 2-methylcyclohexanol)
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If water is present in the product, there will an absorbance peak at 3400.
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What is the purpose of the H3PO4? (Dehydration of 2-methylcyclohexanol)
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Phosphoric Acid is the catalyst and the reaction would occur slowly or not at all without it ( this is because the rxn is NOT stoichiometric). It protonates the alcohol and turns it into a good LG and the LG leaves (slow step).
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What are three purposes of using the Drierite? (Dehydration of 2-methylcyclohexanol)
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a. Absorb water from H3PO4 and water produced from the reaction i. LeChatlier's principle. Drive reaction forward b. Acts as a boiling stone c. Increases reflux by increasing surface area
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Why does product and not starting material collect in Hickman ring? (Dehydration of 2-methylcyclohexanol)
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The starting material has a high boiling temperature and will not collect in the Hickman ring.
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Give 2 reasons why percent yield may be less than 100%. (Dehydration of 2-methylcyclohexanol)
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Percent yield may be less than 100% because the starting material was not boiled long enough for all of the product reflux.
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Give one reason why percent yield may be over 100%. (Dehydration of 2-methylcyclohexanol)
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The product may have not been completely dried and water may be present in the product.
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What is one unique safety hazard associated with this reaction? (Dehydration of 2-methylcyclohexanol)
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H3PO4 is highly corrosive
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What are some benefits of using GC?
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Quick and accurate way of analyzing samples
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What type of compounds can (and can't) be analyzed by GC?
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Samples of GC must be able to be vaporized into a gas phase. (Volatile) a. Non-volatile liquids or solids cannot be analyzed.
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Why do compounds separate using GC?
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By their partinitoning in the stationary (cabowax) and mobile phase (H2 or N2).
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What main physical property is responsible? What are some secondary properties? (GC)
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4. Based on their boiling points. Their ability to be volatile.
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If 5 compounds were injected onto a GC, how could you predict the order of elution?
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The RT is used for identification. The area of the peak is used to determine the amount of each component present. a. Area = Height (cm) x Width at Half maximum peak height (cm) = cm2 b. RT =distance to A (cm) x (min/cm)
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Why did a solid form over the course of the reaction? (Diels-Alder and TLC)
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The beginning appearance of the reaction is yellow color solution. At the end of the reaction, the color of the reaction mixture should be a clear solution (which is an indication that all of the diene has been completely reacted with the dienophile). The product that is in this clear reaction mixture is an insoluble product, therefore, when the flask is left to cool, white ppt will form.
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How did we know the temperature of the reaction without using a thermometer? (Diels-Alder and TLC)
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The boiling point of xylenes are around 140 degree Celsius. The xylene solvent refluxes at that temperature.
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How is the purity of a reaction mixture determined using TLC?
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A pure compound on the TLC has one dot in the column
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How is the identity of a product in a reaction mixture determined using TLC?
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The identity of the compound has the same Rf value as a reference sample
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Why is anthracene bright purple under UV light? (Diels-Alder and TLC)
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Anthracene is bright purple under UV light because of the amount of conjugations in its structure. a. # of conjugated pi bonds increase = wavelength of maximum electromagnetic radiation absorption (λmax) increases i. UV range to visible range
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Explain the detailed mechanism of EAS.
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The first step is the rate-determining step. The first step is one of the i bonds of the aromatic acting as the nucleophile and forming a bond to the electrophile, generating a delocalized cation and simultaneously destroying the aromaticity of the ring. This cation intermediate can resonance delocalize around the ring. During the second step a proton leaves, aromaticity is regained, and the final product is formed. The presence of substituents on the aromatic ring, greatly influence the rate of the first step, and hence the overall rate of the reaction.
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What is an arenium ion? (EAS)
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Arenium ion is the resonance structures and delocalization of the carbocation.
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IF given a series of structures, predict which compound will react the fastest in an EAS using bromine. Slowest?(EAS)
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EDG> Benzene> EWG a. Electron donating groups stabilize the cation. b. Electron withdrawing groups destablizie the cation.
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What step is the RDS (rate determining step)? Why?(EAS)
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The i bonds of the aromatic acting as the nucleophile and forming a bond to the electrophile, generating a delocalized cation and simultaneously destroying the aromaticity of the ring.
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Why is phenol a strong EDG (electron donating group)?(EAS)
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Phenol has more than one lone pair electrons that can donate electrons to the resonance of the structure.
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Why is phenyl acetate a weaker EDG than phenol? (The answer is not because it goes faster.)(EAS)
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Phenyl acetate is a weaker EDG because there is also an EWG group attached to the compound.
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Why is Salicylic acid a weaker EDG than phenol?EAS)
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Although salicylic acid is connected to an O with several lone pairs attached, the lone pairs attached are pulled toward the EWG group (carbonyl).
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Why is acetanilide a stronger EDG than phenyl acetate?(EAS)
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The nitrogen on acetanilide is less electronegative than the oxygen on the phenyl acetate. Acetanilide reacts faster because the oxygen wants to keep the electrons more stronger than nitrogen.
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Why did the disappearance of bromine signal the end of the reaction?(EAS)
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because it is the limiting reagent
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How did we know the limiting reagent was the bromine without doing any calculation?(EAS)
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0.050 M of bromine is used in 0.20M of Beneze + Group
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The EAS experiment was not a synthesis. What was the purpose?(EAS)
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EAS Lab was not a synthesis. The lab was merely used to demonstrate the reaction rates of bromine being added to a benzene and how the existing groups affected the reaction time. A strong EDG produced the fastest reaction times and a weaker EDG (some with EWG) produced the slower reaction time. a. A more electronegative atom tends to react slower than a less electronegative atom.
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What was a unique safety hazard associated with the EAS experiment?(EAS)
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All of the organic molecules possess toxic characteristics (Benzene!)
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Phenyl acetate took a long time to react, does this mean it is a EWG (electron withdrawing group)?(EAS)
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Phenyl acetate is an EDG with an EWG group attached. The electron density being pulled from the Oxygen made this reaction take a longer time.
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Where does all the waste from this experiment go?(EAS)
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All the waste belongs in the halogenated liquid waste bin. Bromine, a halogen is added to the ring.
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What is the nucleophile in this reaction?(EAS)
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The nucleophile in this reaction is the benzene with the varying groups attached. The double bonds make this compound the more electron rich compound.
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What is the electrophile in this reaction?(EAS)
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The electrophile is bromine (Br-Br)
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What is the nucleophile? What is the electrophile? What is the leaving group?(Aspirin)
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The oxygen of the phenol in salicylic acid acts as the nucleophile. The electrophile is the carbonyl carbon of acetic anhydride (after made a good electrophile). The leaving group is acetic acid.
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Why did the test tubes need to be very dry for the reaction to proceed?(Aspirin)
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Water will destroy any acetic anhydride in the solution and turn it into acetic acid. The reaction will no occur in the presence of water.
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What is the role of each of the following compounds in the oxidation reaction; Na2SO4, Na2S2O3, Cyclohexanol, NaClO, NaCl, NaHCO3, and Cyclohexanone.(Oxidation of Alcohol)
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1st reaction - oxidation a. Sodium hypochlorite (NaClO) - oxidizing agent; bleach solution b. Acetic Acid (CH3CO2H) - further activates NaClO and forms Hypochlorous acid (HClO); a catalyst c. Cyclohexanol - what is being oxidized i. Donates H and gains electrons 2nd reaction - quenching and KI-Iodide starch test a. Sodium thiosulfate (Na2S2O3) - reducing agent that reacts with and quenches any remaining oxidizing agent. b. Potassium Iodide-starch paper test (KI) - indicates presence of oxidizing agent 3rd reaction - pH test a. Sodium bicarbonate (NaHCO3) - weak base that neutralizes any acetic acid 4th reaction - Reducing solubility of cyclohexanone a. a. Sodium Chloride (NaCl) - reduces the solubility of the product (cylcohexanone) in the aqueous phase and increases product yield. 5th Reaction - Drying agent to remove water b. Sodium Sulfate (Na2SO4) - Drying agent; removes residual water from the organic product Product a. Cyclohexanone - oxidized product; increased oxygen content
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What causes the cyclohexanone to separate into a new layer?(Oxidation of Alcohol)
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Cyclohexanone is slightly soluble in water and cyclohexanol is more soluble in water a. Cyclohexanone is only an hydrogen acceptor b. Cyclohexanol is a hydrogen acceptor and donor i. More soluble in water
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What causes the cyclohexanone layer to float on top of the solvent (instead of sinking)?(Oxidation of Alcohol)
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Cyclohexanone has a density less than water. a. This will float on top of the aqueous layer
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What material is being oxidized? (Oxidation of Alcohol)
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Cyclohexanol is oxidized
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What is the reducing agent in this reaction?(Oxidation of Alcohol)
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The reducing agent is Na2S2O3
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What was the solvent in the oxidation reaction?(Oxidation of Alcohol)
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The solvent is the water from the bleach solution
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Predict the IR of the starting material.(Oxidation of Alcohol)
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The IR of the starting material will have a broad peak at 3200-3400 (-OH), 2700-3000 (-CH2)
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Predict the IR of the desired product.(Oxidation of Alcohol)
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The IR of the product should be 1600-2500 (C=O) and from -CH2
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What would be the impact of not enough bleach being used?(Oxidation of Alcohol)
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Because the bleach used in this lab is highly concentrated with water, not using enough Bleach will further decrease the product yield of cyclohexanone. There will be more cyclohexanol left in the solution.
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What was the role of sodium hydroxide?(Synthesis of Oxime)
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Sodium hydroxide depronates hydroxylamine to make it a better nucleophile.
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Why couldn't hydroxyl amine hydrochloride have been used without the base?(Synthesis of Oxime)
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The quaternary nitrogen of hydroxylamine does not have any electron pairs to be used for nucleophilic attack.
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Why did solids form over the course of the reaction?(Synthesis of Oxime)
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The solid confirms the production of cyclohexanone oxime. a. Cyclohexanone oxime is a white solid at room temperature
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Many people obtained greater than 100% yield for this reaction. Why?(Synthesis of Oxime)
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A reason for a greater percent yield can be a result from forcing the reaction to be cooled too rapidly. This can trap impurities within the crystals.
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Why are many product solutions put into ice water before filtration?(Synthesis of Oxime)
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Putting the product into an ice water bath allows the product(solid at room temperature) to form as crystals and be separated from the impurities and solvent.
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What is the nucleophile?(Aldol)
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The nucleophile is the acetone because, after one of the hydrogens is removed by NaOH, acetone gets a minus charge (-) and becomes an excellent nucleophile.
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What is the electrophile?(Aldol)
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The electrophile is the carbonyl carbon on benzaldehyde.
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What does the NaOH do?(Aldol)
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NaOH removes the hydrogen on the alpha carbon from acetone. It is also a catalyst for the reaction.
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What is the role of the ethanol?(Aldol)
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The ethanol could be the base that is used during the elimination of the alcohol to form the double bond, hence aldol condensation.
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Why was the product highly colored?(Aldol)
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The product was highly colored because of the conjugated double bonds.
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Why did the product solidify?(Aldol)
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The product solidified because the product is highly insoluble (17 carbons!).
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When recrystallizing, why should solvent be added until all the solid dissolves and no more? What happens if additional solvent is added? What happens if not enough solvent was added?(Aldol)
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The solvent used in the recrystallization process was ethyl acetate. The solvent was used to dissolve all of the crude product. That was why you could see two layers, one layer was the product and the other was the impurities. If too much solvent was added, then the solvent would react with the product. If too little solvent was added, then there would be impurities in the product layer.
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Why should the product be allowed to crystallize slowly and not plunge directly into ice water?(Aldol)
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The reason why the product should be allowed to cool to room temperature first was because of impurities. Plunging the mixture directly into the ice bath, does not allow the impurities to evaporate off the product. If this were to happen, there would be impurities in the product crystals and a high percent yield.
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pH test(Qual 2)
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If a material is soluble in water, then test the pH of the liquid. Alcohols, ketones, esters, ethers, aromatics and hydrocarbons are neutral. Carboxylic Acids and to a lesser extent phenols are acidic. Amines are basic.
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Sobium Bicarbonate (Qual 2)
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The carboxylic acid protonates the sodium bicarbonate to form carbonic acid which decomposes to carbon dioxide and water. The formation of gas bubbles indicates the presence of a carboxylic acid
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Ferric Chloride(Qual 2)
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Ferric Chloride selectively reacts with phenols. Ferric Chloride forms a purple (dark gray-maroon) colored complex with most phenols. Enols can show red, violet or tan.
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Jones Test- Chrome VI) Oxide (Qual 2)
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When CrO3 oxidizes a material, the chrome is reduced to chrome +3 or +2. Chrome (VI) is normally a shade of orange, chrome (III) is normally a shade of green. If the color of the chrome changes, this indicates the presence of a functional group that is able to be oxidized. This test shows the presence of primary and secondary alcohols and aldehydes. Does not react with tertiary alcohols, ketones, esters, or carboxylic acids (Chrome is carcinogen)
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2,4-DNP (Qual 2)
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The terminal nitrogen of the hydrazine (the one one the very end) is nucleophilic. It will selectively bond to the carbon of a carbonyl group and form a highly colored conjugated hydrazone material in a reaction mechanism very similar to that of oxime formation. This test shows the presence of aldehydes and ketones. There should be no reaction with alcohols and acids. (2,4-DNP is mutagen)
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Tollen's test (Qual 2)
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is a way to distinguish between a ketone and an aldehyde. Only perform Tollen's test on materials which tested positive for carbonyls using the 2,4-DNP. Silver (I) is a very mild oxidizing agent. Diaminosilver selectively reacts with aldehydes (not ketones) to form acids. Formation of black grains or mirror is evidence of silver being reduced.
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Iodoform test (Qual 2)
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The iodoform test distinguishes between regular ketones and methyl ketones. Only perform the iodoform test on materials which tested positive to the 2,4-DNP test and negative to the Tollen's test. Hence, only perform iodoform test on ketones. Methylketones undergo multiple halogenations on the carbon of the methyl group. In the presence of hydroxide, the hydroxide attacks the carbonyl and the triiodomethyl group leaves. Triiodomethane is insoluble and precipitates as an opaque yellow suspension.
