Alcohols, Aldehydes, and Ketones, Exercises of Organic Chemistry

Download Alcohols, Aldehydes, and Ketones and more Exercises Organic Chemistry in PDF only on Docsity! 1 Identification of an Unknown: Alcohols, Aldehydes, and Ketones How does one determine the identity and structure of an unknown compound? This is not a trivial task. Modern x-ray and spectroscopic techniques have made the job much easier, but for some very complex molecules, identification and structure determination remains a challenge. In addition to spectroscopic information and information obtained from other instrumental methods, chemical reactions can provide useful structural information, and physical properties can contribute significantly to confirming the identity of a compound. In this experiment, you will be asked to identify an unknown liquid, which will be either an alcohol, aldehyde, or ketone. Identification will be accomplished by carrying out chemical tests, called classification tests, preparing a solid derivative of the unknown and determining its melting point (MP), making careful observations, and analyzing the NMR spectrum of the unknown. R OH alcohol R H O aldehyde R R O ketone the carbonyl group A list of alcohols, aldehydes, and ketones, along with the MP of a solid derivative of each compound, is posted on the website. The unknown will be one of these listed compounds. If one can determine to which functional group class (alcohol, aldehyde, or ketone) the unknown belongs, two of the three lists need not be considered and the task will be greatly simplified. To accomplish this, classification tests will be carried out. First, consider some general ways in which alcohols, aldehydes, and ketones react. 2 CLASSIFICATION TESTS, which are simple chemical reactions that produce color changes or form precipitates, can be used to differentiate alcohols, aldehydes, and ketones, and also to provide further structural information. Because color plays such an important role in this experiment, a separate handout on this topic is available on the course website. 2,4-Dinitrophenylhydrazine: Aldehydes and ketones react with 2,4-dinitrophenylhydrazine reagent to form yellow, orange, or reddish-orange precipitates, whereas alcohols do not react. Formation of a precipitate therefore indicates the presence of an aldehyde or ketone. The precipitate from this test also serves as a solid derivative. A discussion on derivatives will be given later in this handout. The mechanism of this reaction is that of imine formation and can be found in any organic lecture text. R O NO2 NO2 H N H2N+ a series of steps NO2 NO2 H N NR a ketone if . . . R = carbon an aldehyde if . . . R = H 2,4-DNP a 2,4-DNP hydrazone (generally a solid) Ceric Ammonium Nitrate (CAN): Alcohols react with this yellow reagent to produce a color change (from yellow to red), but the carbonyl group is unreactive. This is a good experiment to test for the presence of an alcohol or to prove the absence thereof. Note that changing the groups attached to certain inorganic ions such as Ce4+ results in a change to the electronic structure, which results in a color change. Production of a magenta color, therefore, indicates the presence of an alcohol group. The 2 ammonium cations are present as spectators and do not participate. R OH + (NH4)2[Ce(NO3)6]2- R O Ce(NO3)5 an alcohol CAN, a yellow solid an alkoxy cerium(IV) derivative 2- Schiff’s Reagent: Before looking at the reaction of Schiff’s reagent, consider a much simpler system. The sulfur in the bisulfite ion acts as a nucleophile and adds to the carbonyl carbon. Because this is such a bulky nucleophile, it will add only to a relatively sterically unhindered carbonyl. This requires the carbonyl to be part of an aldehyde in which one of the R groups is the very small hydrogen, or a ketone having small ‘R’ groups. A ketone having large groups attached to the carbonyl will not react with bisulfite. O S OH O + R O R aldehyde or sterically unhindered ketone S O O O R R OH bisulfite bisulfite addition complex Aldehydes react with Schiff's reagent to produce a color change (magenta-colored addition product). In the same way, the Schiff reagent acts as a nucleophile that adds to the carbonyl group of an aldehyde. 5 O O conjugated carbonyls O O non-conjugated carbonyls 3,5-Dinitrophenylbenzoates (3,5-DNB): Alcohols react with 3,5-dinitrobenzoyl chloride to produce solid 3,5-DNB esters that follows the mechanism outlined below. R O H + NO2 NO2 Cl O R O H O Cl R' R' O O R H + Cl R'O O R a 3,5-DNB Ester where R' = NO2 NO2 3,5-dinitrobenzoylchloride SUMMARY: The results of the classification tests enable one to limit the search to one of three lists of possible compounds. The results of these tests will provide information on whether the unknown is an alcohol, aldehyde, or a ketone, and if it is an aldehyde or ketone, whether it is a methyl aldehyde or ketone, and possibly whether the carbonyl group is conjugated or not. Narrowing the possibilities further requires carefully obtaining the melting point of the purified solid derivative. Once this has been determined, the list of possible compounds, along with the MPs of the derivatives, can be consulted. For many unknowns, the MP of the derivative, together with the results of the classification tests will provide sufficient information to make a final conclusion as to the identity of the compound. Often, however, two or even three possibilities may have very similar test results and derivative MPs. In such a case, the NMR spectrum can be used to make a final determination. A SAMPLE ANALYSIS: Unknown X produces a red/orange DNP, MP 159-161° and gives a neg Schiff and neg iodoform test. Pos DNP -> ald or ket. Red/orange DNP -> probably conjugated carbonyl. Neg Schiff -> not ald, therefore ketone. Neg iodoform -> not methyl ketone. Look up MP of derivative in table of ketones. Three compounds fall within likely range: cyclohexanone, isobutyrophenone, and 1- methoxy-2-propanone. O O cyclohexanone, a non-conjugated carbonyl isobutyrophenone, a conjugated carbonyl O O 1-methoxy-2-propanone, a methyl ketone 6 The results point towards isobutyrophenone as being the unknown. 1H-NMR would readily confirm this by indicating the presence of aromatic hydrogens and the common splitting pattern of an isopropyl group. Use the following flow diagram to help carry out the experiment. Prelab: You may either print out your prelab and bring it with you to lab, or bring your computer. Your TA will grade it on the spot before you begin the experiment. For the in lab observations, you may use scratch paper and record later in your ELN, or bring your computer and record directly in your ELN. Postlab Report: Make sure to use the non-formal postlab report template on the course website! THE EXPERIMENT: (revised 4/20). Note that an incorrect identification of unknown will result in an automatic deduction of 5 points (10%). WARNING: acetone is a methyl ketone. If it is used to clean glassware, the glassware must be completely dried or else the acetone will interfere with your results. Be careful to not accidentally contaminate reagent bottles by using pipets contaminated with acetone, known compounds or unknowns. Many of the unknowns have a very disagreeable odor. To minimize this odor in the lab, be sure to rinse used pipets with a LITTLE acetone in the hood before disposing of them in the boxes in the waste hood labeled “contaminated pipets”. Do not dispose of them in the "Glass Only" waste boxes. The yellow pipet bulbs can be used indefinitely and should not be thrown out. It is not necessary to use a new pipet each time you measure out your unknown. Use the same one for the whole experiment. Conserve whenever possible. Unknown 2,4-DNPH ppt no ppt aldehyde or ketonealcohol Ce+4 yellow to red no color change NOT alcoholalcohol Schiff magenta aldehyde no color change ketone I2 /OH- no ppt not methyl ald/ket yellow ppt methyl ald/ket - recyst ppt - dry - mp 3,5-DNBC ppt - recyst ppt - dry - mp 7 You will be assigned one unknown to identify. The unknown will be a liquid alcohol, aldehyde, or ketone (taken from the list of possible compounds posted on the Chem 269 website). Record the number of the unknown in your ELN. Each unknown is unique and will require a slightly different approach. The identification will be made by doing chemical classification tests, by determining the MP of a derivative, and by interpreting the 1H-NMR spectrum. From the results of the tests and the MP of the derivative, you will narrow the identity of the unknown to a few possible compounds. Once this is done, interpreting the 1H-NMR spectrum of the unknown will allow you to make a final conclusion. The amount of unknown that you are given is more than enough to do each test several times. Conserve it. If you need more, it will cost you 1 point. (Aldehydes oxidize to carboxylic acids in the presence of oxygen and light. In the short amount of time that you will work in the lab, away from direct sunlight, this will not be a problem. If your unknown is an aldehyde and if for some reason you need to do additional tests on another day, ask your TA to store the sample in the refrigerator until you need it again.) Also note that some unknowns have low BPs and will evaporate unless kept tightly stoppered. Prelab exercise: as part of the prelab outline, summarize a logical approach to identifying your unknown by preparing a flow chart (use the chart above as a template but provide more detail). In heating reaction mixtures NEVER use a wooden boiling stick. Boiling sticks can be used only in nonreactive solutions such as in recrystallizations. Carry out the procedure in the order given below. Success depends upon very careful work. (1) Reaction with 2,4-dinitrophenylhydrazine: Using the reagent pipet attached to the bottle, measure 2 mL of the DNP reagent solution (2,4-dinitrophenylhydrazine dissolved in phosphoric acid and ethanol) into a reaction tube, and then, using your own "unknown" pipet, add 2 drops of liquid unknown. The amounts need only to be approximate. Do not contaminate the reagent or its pipet with your unknown. Mix the solution thoroughly and allow it to stand at room temperature for a few minutes. During this time, if the test is positive, a precipitate will form. If no precipitate forms, go to step (4). Collect the solid by suction filtration, and rinse it with several portions of water to wash off most of the unreacted DNP reagent. (Remember the correct way to rinse crystals in a suction filtration: break the vacuum by lifting the funnel slightly off of the flask, cover the crystals with water, then reattach the funnel. This helps to ensure that all of the crystals are rinsed. The product has a low solubility in water so there is little danger of redissolving them.) Press a piece of indicator paper onto the crystals to be sure that most of the acidic reagent has been removed. If it is still acidic, rinse it further. Note though that even after rinsing the crystals well, a small of amount of acid may still be present. If so, just proceed. Recrystallize the product using ethanol as solvent. Set the 2,4-dinitrophenylhydrazone (DNP) derivative aside to dry and go on to the next step. Some DNPs will not be very soluble in hot ethanol and will therefore not completely dissolve. If the derivative does not dissolve in about 5 mL of hot ethanol, just heat the suspension for a few minutes, allow it to cool and crystallize, collect it, rinse it, and allow it to dry as in a normal recrystallization. Even though it is not a normal recrystallization, this treatment will remove most impurities. Waste: place all filtrates and rinses into the Organic Liquid Waste container. (2) Schiff's test: For a meaningful interpretation of the results, run the test on a control (known compound which gives a + test: use benzaldehyde) and a blank (known compound which gives a - test: use acetone), right alongside the test for the unknown. In other words, run three reactions at the same time, one on benzaldehyde, one on acetone, and one on the unknown. To prevent contamination of the reagent, first add the reagent to the tubes using the reagent pipet and then add the knowns and unknown to the tubes using separate pipets. To 0.7 mL of Schiff's reagent in a reaction tube, add 1 drop of 10 unknown. Note that your MP will be close to the MP of more than one compound on the list. Even by considering the results of chemical tests, in some cases you may only be able to narrow the possibilities down to a few compounds. In such a case, the NMR spectrum may be especially important in making a final determination. To help you interpret the spectrum, read the notes on NMR at the end of this handout. DO NOT ASK YOUR TA IF YOUR IDENTIFICATION IS CORRECT. THIS EXPERIMENT SHOULD BE CONSIDERED TO BE A LAB PRACTICAL. JUST REPORT WHAT YOU FIND IN YOUR POSTLAB WRITE-UP. SAFETY: As with any laboratory chemicals, assume that those used in this experiment, including the unknowns, are toxic. Keep them off of your skin. If you become contaminated, wash thoroughly with soap and water. DISPOSING OF UNUSED UNKNOWN: place any unused unknown into the Organic Liquid Waste container, rinse the vial with a little acetone and add this to the waste container, and leave the unknown vial in the fume hood with the cap off. BEFORE LEAVING THE LAB: be sure to turn off Mel-Temps, hot plates, vacuum and air valves, clean and put away your equipment, clean up your work areas, close the fume hood sash all the way if you are the last person working in that hood, and get a signature from your TA. -------------------------------------------------------------------------- Postlab Questions 1.) An unknown sample does not react with 2,4-dinitrophenylhydrazine reagent, but a color change is observed on reaction with ceric nitrate reagent. Draw the structure of a compound that would give this result. 2.) An unknown sample produces a precipitate upon reaction with 2,4-dinitrophenylhydrazine reagent. Draw the structure of a compound that would give this result. 3.) The unknown in question 4 causes Schiff’s reagent to turn a magenta color. Draw the structure of a compound that would give this result. 4.) An unknown sample produces a reddish precipitate upon reaction with 2,4-dinitrophenylhydrazine reagent, no color change with Schiff’s reagent, and a yellow precipitate when mixed with iodine and base. Draw the structure of a compound that would give this result. 5.) Using NMR only, how would you distinguish among 2-pentanone and 3-pentanone? 11 Proton NMR of Alcohols, Aldehydes and Ketones (revised 7/20) These notes are designed only to help you gather a little extra structural information about your unknown. The notes are not meant as a stand-alone lesson in nmr interpretation. If you wish to obtain more structural information from your NMR spectrum, read the references given in your "Schedule of Experiments." Four characteristics of a proton NMR (1H) spectrum can be used to gain structural information: number of signals, positions of absorption, area of signals, and multiplicity of signals. In a proton nmr spectrum, we observe the absorption of energy by hydrogens in a molecule. Look at a 1H NMR spectrum (e.g., find one in your lecture text). It consists of peaks rising from a baseline, some appearing towards the right end of the spectrum and some towards the left (in a very simple case, only one peak would appear). The peaks may be sharper single lines (singlets) or sets of multiple lines (multiplets). Some peaks have a larger area than others. These are the characteristics that you should consider when interpreting a 1H nmr spectrum. Number of signals: Sets of equivalent hydrogens. Symmetry of the molecule is the key. Hydrogens in different environments in a molecule may give rise to different signals. Beware: sometimes the environments are not different enough and the signals will overlap or coincide (accidental equivalence). Consider benzene. The molecule is very symmetrical. All 6 H's are in equivalent environments. Substitution of any one H by Cl for example would result in the same molecule: chlorobenzene. Therefore, benzene gives rise to one peak. Consider 1,4-dimethoxybenzene. All 4 ring H's are in equivalent environments and give rise to one signal. All 6 methyl H's are in equivalent environments (but different from the ring H's) and give rise to a second signal. This molecule therefore gives rise to 2 signals. Consider the spectrum of chloroethane. 3.5 ppm 1.5 The three methyl H's are equivalent to one another and both methylene H's are equivalent to one another, so we see 2 signals in the spectrum. However, the signals are not sharp single peaks. They are split into relatively symmetrical, closely-grouped subsets of peaks. The CH3 signal is split into 3 peaks and the CH2 is split into 4 peaks. Counting up the number of signals in a spectrum to determine the number of sets of equivalent H's is complicated by this splitting. For the purposes of counting up numbers of signals, count a multiplet as one signal. So, although chloroethane has 7 peaks in the spectrum, it only has 2 sets of peaks, or 2 signals, meaning it has 2 sets of equivalent H's. In 1H NMR spectra, because of splitting and accidental equivalence, counting numbers of signals to determine numbers of sets of equivalent H's is often not very useful. However, if your spectrum has few signals, you can assume that the H's in your unknown exist in few different environments. Symmetry in the molecule tends to decrease the complexity of the spectrum. For example, 2-pentanone has 4 sets of equivalent H's and 12 would therefore have four signals (+ splitting). 3-Pentanone has only 2 sets of equivalent H's and 2 signals (+ splitting). Position of absorption or chemical shift: H's in different environments in a molecule will absorb energy at different magnetic field strengths in a 1H NMR spectrum. The x-axis of the spectrum represents the magnetic field strength and is shown in units of parts per million (ppm), with 0 ppm at the right and 10 ppm at the left. The position of absorption in the spectrum is known as the chemical shift and is helpful in recognizing the identity of functional groups (FGs). The alkyl portion of a molecule away from a FG absorbs towards the right, at lower values of ppm. A CH3 group away from a FG absorbs around 0.9 ppm, while a CH2 absorbs around 1.2 ppm and a CH around 1.5 ppm. Most nearby FGs will cause the absorption to move towards the left (higher ppm) so a CH3 attached to an oxygen will absorb around 3.3 ppm. FUNCTIONAL GROUPS CAUSE THE POSITION OF ABSORPTION OF HYDROGENS IN A MOLECULE TO OCCUR AT CHARACTERISTIC CHEMICAL SHIFTS (PPM). These can be very helpful in interpreting a spectrum. For typical unknown aldehydes and ketones, the following absorptions are important. The H attached to the carbonyl group in an aldehyde absorbs at about 10 ppm. This peak is not present in the spectrum of a ketone. H's on the carbon alpha to a carbonyl absorb at about 2-2.5 ppm. Other chemical shifts that may be important in your spectrum are as follows: H's attached to a benzene ring absorb at about 7 ppm. If your spectrum has a peak in this region, then your unknown has aromatic H's. Alkene H's absorb at about 5-6 ppm. H's in an environment such as H-C- X, where X= Cl, O, N absorb around 3-4.5 ppm. Area of the signals: The size of a signal, as measured by its area, is proportional to the number of H's giving rise to that signal. For example, for chloroethane, there are 2 CH2 H's and 3 CH3 H's. Therefore the relative areas of the signals would be 2:3. The areas are measured electronically and are usually shown as a stepwise curve superimposed on the NMR spectrum. The height of the step for a given signal is proportional to the area of that signal. You may or may not see an integral on the spectrum of your unknown, but you may be able to estimate areas. Multiplicity: This is more complicated but can provide the most structural information. Nonequivalent neighboring H's cause the signal of a given H to be split into a multiplet of n+1 peaks, where n=number of neighboring H's. This is only true when the chemical shift of the neighbors is considerably different than the chemical shift of the given H. Look again at chloroethane. The CH3 hydrogens have 2 neighbors (CH2) and will therefore be split into n+1 or 3 peaks (triplet). The CH2 hydrogens have 3 neighbors (CH3) and will therefore be split into n+1 or 4 peaks (quartet). Spectrum of chloroethane: A triplet of relative area 3 at about 1 ppm and a quartet of relative area 2 at about 3 ppm. In general, an upfield triplet of relative area 3 and a lower field quartet of relative area 2 indicates the presence of an ethyl group attached to an electron withdrawing group. For (CH3)2CH-X where X=withdrawing group: higher-field doublet of relative area 6 and a lower-field septet (7) of relative area 1. Interpreting the spectrum of your unknown: Do not expect to do a complete interpretation unless you have a simple spectrum. However, try to gather as much information as possible from your spectrum. Narrow down the number of possible compounds to 2 or 3 using the results of your chemical tests and the MP of the derivative. Draw the structure of these possibilities and predict how the nmr spectrum of these should appear. First, determine the number of sets of equivalent H's in each structure. The number of signals in the spectrum (excluding multiplicity) should be less than or equal to this number (less than