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CHEM 281 – Virtual Labs Virtual Lab 281.1: Recrystallization - Recrystallization is a purification technique that is used in organic chemistry. - To read: o Technique 1 – Crystallization o Chapter 2 of textbook Recrystallization Technique: - How does it work? o Compounds dissolve better in hot solvent than in cold. o The idea is to dissolve your impure compound in hot solvent. o Then, with an ideal recrystallization, as the solution cools, your compound will crystallize out, but the impurities will stay in the solution. o You can then collect your pure crystals by filtration. - How much solvent? o Use as little hot solvent as possible. o If you use too much, then your compound will just remain dissolved as the solvent cools. Lab: Preparation of An Organic Acid: - Information: o Benzoic acid is a solid organic acid. o The preparation and isolation of benzoic acid from its salt, sodium benzoate, relies on the difference in solubility between sodium benzoate (very soluble in cold water) and benzoic acid (less soluble in cold water). o Sodium benzoate is about 200 times more soluble in water than benzoic acid. o The reaction is an example of a strong acid (HCl) displacing a weak acid (benzoic acid) from its salt (sodium benzoate). o Sodium benzoate + hydrochloric acid  benzoic acid + sodium chloride -

Experimental Procedure: o 1. Boil a kettle or beaker of water. o 2. Weigh out 5-10g of sodium benzoate into a pre-weighed (or tared) 250cm3 beaker. o 3. Carefully measure out 80cm3 of boiling water using a measuring cylinder. Add this water to the sodium benzoate in the beaker. Stir with a stirring rod to dissolve. o 4. Using a measuring cylinder, measure out 50cm3 of 2.0 mol dm-3 hydrochloric acid and add this to the hot sodium benzoate solution. o 5. Set up an ice bath by putting some crushed ice into a tray. Put the beaker into the ice bath to cool down the solution. As the solution cools, crystals of benzoic

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acid will precipitate out ‘like snow’. Also put a distilled or deionized water bottle in your ice bath for later.  At this stage you have made your product and now you need to separate it from the reaction mixture. 6. Set up the Buchner filtration apparatus. Don’t forget to add the filter paper and wet it with some distilled water. 7. Check the temperature of the solution. When it has cooled to room temperature take the beaker out of the ice bath and pour through the Buchner filtration apparatus. 8. Wash your crystals of benzoic acid with 50cm3 of ice-cold distilled water to remove any impurities in the moisture trapped between the crystals. 9. Observation – Pre-weigh a large watch glass or petri dish. Scrape out your crystals and put onto the watch glass. Put them in a drying cabinet for at least 15 minutes, preferably longer. Take them out and weigh the, calculate the mass of the product you have made. 10. Observation – Extension: You can purify the benzoic acid by recrystallization and check its purity by determining the melting point of the final solid.

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Exam Practice: o Q1: Calculate the % yield  Mass of sodium benzoate = 8.02g  Mass of benzoic acid = 4.95g  1. Moles of sodium benzoate (mol) = m/M  = 8.02g/144g mol-1  =0.0557 mol  2. Maximum moles of benzoic acid produced = 0.0557, as 1:1 ratio in equation  3. Actual moles of benzoic acid produced = m/M  = 4.95g/122g mol-1  = 0.0406 mol  4. Percentage yield of benzoic acid = moles reactant/moles product x100  = 0.0406/0.0557  = 72.9 %

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Practice Questions: o What is the bonding between sodium and the benzoate structure in sodium benzoate?  Ionic. o This experiment could be described as ‘one acid displacing a weaker acid from its salt.’ Which is which? Match the reactants and products correctly.  Benzoic acid – weak acid  Hydrochloric acid – strong acid  Sodium benzoate – salt

o What is present in the filtrate when the crude benzoic acid is filtered?  Benzoic acid  Hydrochloric acid  Sodium chloride  Water o Why in this procedure should a small volume of ice-cold distilled water be used to wash the crude crystals held in the Buchner funnel.  Lots of ice-cold water would dissolve some of the benzoic acid.  Room temperature or hot distilled water would dissolve some of the benzoic acid.  Washing removes the solution (similar in chemical composition to the filtrate) that is present between the crystals.  Using tap water would introduce additional impurities into the benzoic acid. o How could the percentage purity of the benzoic acid produced be determined?  By titration with a known concentration of sodium hydroxide. Virtual Lab 281.2: Infrared Spectroscopy - Infrared spectroscopy is an instrumental method for detecting functional groups. - The technique relies on the fact that different molecules (i.e. functional groups) vibrate at different frequencies as they absorb infrared radiation. - This absorbance can be measured and used to elucidate certain structural information. - To read: o Chapter 2 of textbook Lab: Infrared Spectroscopy: Theory: - Interpretation of an infrared spectrum, (commonly called IR) provides information about the functional groups present in a molecule: specific absorption bands indicate the presence of a particular functional group. o Knowing that a particular group is NOT present is just as important: if there is no characteristic absorption, then that functional group is NOT present. - By itself, an IR spectrum does not provide enough information to derive the complete chemical structure of a compound, but it is very helpful when combined with other spectroscopic techniques. -

How IR works: o Molecules are always in motion.  The bonds between atoms stretch, bend, rock, wag, twist – so called “molecular vibration” o Organic compounds absorb infrared radiation.

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Different types of bonds (in different functional groups) absorb infrared radiation at different characteristic frequencies, causing an increase in amplitude of that bond vibration. o An infrared spectrum is a plot of wavelength (frequency) vs. absorption.  By scanning through varying infrared frequencies and nothing where absorptions occur, we detect the presence (or absence) of different functional groups, based on their bonds’ IR absorption. -

A typical IR spectrum: o A infrared spectrum has many peaks (bands), but only a few correlate to structure. o The wavelength of the absorption maximum (%transmission minimum) is used to identify the band (peak)

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Describing IR bands: o An IR band can be completely described by its location, intensity and shape. o Location – reported as the wavenumber value of the absorption minimum. Very broad peaks (i.e. an acid -OH) should be reported as an absorption range. o Intensity – describes the % transmittance axis, i.e. the size (strong, medium, weak) of the absorption band (peak) relative to other peaks in the spectrum. o Shape – describes the width of a band – is it broad, narrow, parabolic or “W” shaped?

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Interpreting an IR spectrum: o When learning to “read” an infrared spectrum, be systematic. First locate and identify the C-H stretching region, then look to the LEFT of this region, then to the RIGHT of it, to determine the presence (or absence) of the various functional groups.  Completing a table will help you to organize this data. o Step 1:  Locate the C-H stretch region of your spectrum ~3000 cm-1  A strong absorption band just to the right (2900-3000 cm-1) suggests alkyl sp3 C-H bonds.  Weak bands between 3000 and 3200 cm-1 suggests aromatic or alkene sp2 C-H bonds.

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A sharp, strong peak at ~3300 cm-1 represents the sp C-H stretch of a terminal alkyne.

o Step 2:  Look to the left of the C-H stretch region for OH (medium strong) and NH (medium weak) stretches. Each group gives rise to an absorption band with distinctive shape and intensity at ~3350 cm-1.  If no absorption band is present, then there is no OH or NH in the molecule.  Note the presence or absence in the appropriate area of the chart.  

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Alcohol: a broad, parabola shaped peak around 3400-3200 Acid OH: this peak is extremely broad and more triangular than the alcohol – OH stretch. It is often shifted significantly to the right compared to the alcohol peak. Around 3300-2500. Primary amine: shows two medium-weak peaks that correspond to the symmetric and asymmetric stretching of the two N-H bonds. Around 3450-3400.  Secondary amine: produces a single medium peak in the same region.

o Step 3:  Look to the right of the C-H stretch region. There are three separate areas that are important to investigate:  A) around 2200-2300 cm-1 the triple bond stretches of alkynes (weak, sharp) and nitrile (cyano, strong sharp) compounds appear.  B) the very important region between 1650 and 1800 cm-1 shows the presence of a carbonyl group, C=O.  C) between 1500 and 1600 cm-1 shows C=C of alkenes and aromatic compounds.      

Alkyne: sharp peak with medium intensity at ~2200 cm-1 typically weaker than a nitrile peak. Terminal Alkyne: shows the characteristic alkyne stretch at ~2200 as well as an sp C-H stretch at 3300. Nitrile: The CN stretch occurs in the same region as the CC but can often be distinguished by its stronger intensity. Around 2260-2400. Ketone: shows a carbonyl stretch at 1700-1720 Ester: the carbonyl stretch shows up at 1730-1750. A distinguishable C-O band is usually seen in the fingerprint region around 1200 Aldehyde: carbonyl band at 1725. A weak “W” shaped peak appears at 2800 from the CH stretching of the aldehyde hydrogen.

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Acid: carbonyl stretch at 1700-1725. A broad OH band is seen between 3000-2500. Primary amide: a strong carbonyl peak at 1650 with 2 close, rounded NH peaks at 3500. Secondary amide: carbonyl peal at 1680 with 2 sharp bands at 3300 and 3070 representing the NH vibrations. Anhydride: two carbonyl stretches at 1820 and 1760. A band at ~1200 in the fingerprint region can usually be seen for the C-O bond.

 Alkene: a sharp, medium-weak peak always found at 1650.  Aromatic: up to 3 sharp peaks in the region between 1600 and 1450. Step 4:

Sample Preparation: o Salt plate care:  The sample is prepared and placed on potassium bromide salt plates. These are NEVER EXPOSED TO WATER or other polar solvents, since these expensive plates dissolve in water. Prevent moisture from your fingers from ruining the plate by handling the plates by their edges.  After recording a spectrum, clean the salt plates by rinsing with CH2Cl2 or any other non-hydroxylic solvent. o For liquid samples: liquid film:  Simply place a single drop of the liquid on one salt plate, then carefully place the second plate on top. Insert the plates into the cell holder in the infrared spectrometer, then follow the directions on the screen of the spectrometer to obtain the spectrum (you may need to adjust the amount of sample applied until you obtain a suitable spectrum)

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Clean the salt plates by rinsing with CH2Cl2 or any other nonhydroxylic solvent.

o For solid samples: film cast:  Find an organic solvent that completely dissolves the unknown. Place a drop of this sample solution on a single salt plate and wait for the solvent to evaporate, leaving behind crystals of the sample. Place the salt plate with your sample into the cell holder and continue as above.  Clean the salt plates by rinsing with CH2Cl2 or any other nonhydroxylic solvent. o For solid samples: nujol mull  Occasionally your sample will not dissolve in a suitable solvent. Grind your sample with the mineral oil “nujol” using a mortar and pestle and apply the resulting paste to a single salt plate. Place the salt plate with your sample into the cell holder and continue as above.  Clean the salt plates by rinsing with CH2Cl2 or any other nonhydroxylic solvent....