...Test For Carbonyls: Aldehydes and Keytones Abstract Qualitative analysis of aldehydes and ketones was conducted in the experiment. Twelve test samples namely, acetophenone,2-butanone, 2-heptanone, benzaldehyde, 3-pentanone, hexanal, butyraldehyde, octanal, 4-heptanone, propiophenone, cyclohexanone, propionaldehyde were used in the experiment. Various parameters regarding the physical properties of aldehydes and ketones were determined using tests such as oxidation with sodium nitrate and sodium hydroxide, 2,4-dinitrophenylhydrazones test, Semicarbones test, Tollens’ test, Fehling’s test, Iodoform test, and the chromic acid test. The objectives of the experiment were met since the physical properties of the aldehydes and ketones were observed using these tests. Give an Unknown Letter D sample was tested with 2,4-dinitrophenylhydrazones test, Semicarbones test, Tollens’ test, Fehling’s test, Iodoform test, and the chromic acid test. Unknown D presumed miscible with a color change of Tang orange precipitate in 2,4-Dinitrophenyldrazones test. Semicarbones Test Unknown D had a melting of point 123C. Tollen’s test for unknown D tested negative with no formation of black precipitate of reduced silver. In Fehling’s test for unknown D tested negative with no formation of reddish precipitate of copper(I) oxide metallic deposits. Iodoform test of unknown D tested positive and Chromic Acid test for unknown D tested negative. Once conducting all test unknown D tested positive for a Keytone...
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...Carbonyl Compounds Carbonyl Compounds Aldehydes Name Methanal (Formaldehyde) Ethanal (Acetaldehyde) Propanal Ketones Name Structural Formula Structural Formula HCHO CH3CHO CH3CH2CHO CH3CH2CH2CHO Propanone (Acetone) Butanone (CH3)2CO CH3COCH2CH3 Butanal Glucose & Fructose (Simplest sugars) Glucose (Chain Form) Glucose (Ring Form) Fructose (Chain Form) Fructose (Ring Form) Naturally Occurring Carbonyls with Odours Carvone Typical odour of Heptan – 2- one Almond Flavours Spearmint Caraway Seeds Gen Properties • General Formula: CnH2nO • Polarity: • Miscibility: The polarity is sufficient to enable the lower members of the homologous series of Aldehydes and Ketones to be completely miscible with water. Water will form H- bonds to the carbonyl group. Redox Reactions 1. Reduction: Aldehydes and ketones may be reduced to 10 alcohols and 20 alcohols, respectively using the reducing agent sodium tetrahydridoborate, NaBH4. (Water or ethanol is used as a solvent.) (a) Reduction of Ethanal CH3CHO + 2[H] NaBH4 in water CH3CH2OH (b) Reduction of Propanone CH3COCH3 + 2[H] NaBH4 in ethanol CH3CH(OH)CH3 Regarded basically as the addition of H2 across the C=O double bond. SAQ 19.4 • Draw the structural formulae for products obtained when the ff are treated with NaBH4: (i) Butanone (ii) Pentanal (iii) 2-Methylhexan-3-one Redox Reactions 2. Oxidation: Aldehydes require only mild conditions to oxidize to carboxylic acids CH3CHO + [O] K2Cr2O7/H+ CH3COOH (reflux) This reaction...
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...Tetraphenylcyclopentadienone—An Example of a Double-Crossed Aldol Addition Reactions of Carbonyl Groups The chemical reactivity of aldehydes and ketones is closely associated with the presence of the carbonyl group in their structures. For example, both aldehydes and ketones undergo addition reactions such as the addition of a Grignard reagent to the carbonyl group as shown in Figure 1. Figure 1. Addition reactions. The reactions in Figure 1 differ only because the pink H of an aldehyde is replaced by R′′ in the ketone. The addition reaction occurs at the carbonyl group. The carbonyl group is polarized so that the carbon atom bears a partial positive charge and the oxygen atom bears a partial negative charge. The R′ group of the Grignard reagent is negative relative to the positive Mg atom. Thus, the negative R′ group bonds to the positive carbon atom, and the negative oxygen and metallic magnesium form an ionic bond, yielding a salt in each reaction. The addition product is acidified in each case to make a covalent alcohol. The aldehyde produces a IIo alcohol; whereas, the ketone produces a IIIo alcohol owing to the R′′ group. The two equations for addition reactions in Figure 1 are summarized in Figure 2. A nucleophile (negative species) bonds to the carbonyl carbon (positive), breaking the π bond of the carbonyl group. Figure 2. Addition of a nucleophile to a carbonyl group. Figure 2 focuses our attention on the salient part of an addition reaction that...
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...NAME: AKINWANDE OLUFISAYO PARNTNER: COURTNEY DATE: JULY 8TH 2014. ALDOL CONDENSATION TO MAKE VANILIDENEACCETONE. DATA: REACTION: TABLE 1: DATA FROM EXPERIMENT | Vanillin | Acetone | Vanillindeneacetone | Chemical formula | C8H803 | C3H60 | C11H1203 | Molecular weight | 152.25 g/mol | 58.08 g/mol | 192.21g/mol | Density (g/cm3) | 1.06 | 0.788 | | Amount used | | | | Total moles | | | | Theoretical yield | | | | Percentage yield | | | | TABLE 2: DATA FROM EXPERIMENT: REAGENTS | Sodium Hydroxide | Hydrochloric Acid | Chemical formula | NaOH | HCl | Molecular weight | 39.997 g/mol | 36.46 | Denisty | 2.13 | 1.18 | Amount used | 3mL | 7mL | Total moles | 0.i598 | 0.227 | TABLE 3 : IR SPECTRUM Peaks (cm-1) | Functional group | 3699.8 | OH group | 3267.6 | Alkene | 1634.79 | Aromatic | CALCULATIONS: Total moles of Vanillin = 0.5g/152.15 = 0.0033 moles of vanillin. Total moles of acetone = (4mL * 0.788)/ 158.08 = 0.0199 moles of Acetone. Theoretical yields: Vanillin = 0.5g * 1mole * 1mole * 192.21 = 0.632g 152.15 1mole 1mole Acetone = 4ml * 0.788 * 1mole * 1mole * 192.21 = 3.832g 1mole 158.08 1mole 1mole Therefore, vanillin is the limiting...
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...Hydrocarbons O Fats O Fossil fuels O C-H bonds store Nucleus Fat droplets Carbon skeletons vary in length Double bond Carbon skeletons may have double bonds, which can vary in location ENERGY! Carbon skeletons may be unbranched or branched Carbon skeletons may be arranged in rings 10 m (a) Part of a human adipose cell (b) A fat molecule O Insoluble (C-H =nonpolar) Functional Groups are attached to Carbon O The carbon skeleton has “functional groups” How much difference can functional groups make? Estradiol Testosterone attached. O Functional groups are directly involved in chemical reactions and/or modify the overall shape of the molecule. 2 2/7/2014 Common Functional Groups O Hydroxyl group O Carbonyl group O Carboxyl group O Amino...
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...some time now. More specifically, the combination of dienes and carbonyl containing compounds have been studied primarily for their ability to combine together to form six membered oxygen heterocycles (1). The synthesis of oxygen heterocycles is a valuable tool, as these molecules are precursors to some natural products and they are also components in the synthesis of various carbohydrates (4). The general method utilized to form oxygen heterocycles was to combine an aldehyde as the heterodienophile and a diene in a hetero-Diels-Alder reaction. Extensive studies on this reaction have been completed. Huang and Rawal chose to study this reaction using hydrogen bond promoted ketones as the heterodienophile rather than aldehydes. This is the first time that unactivated ketones have been successfully utilized in a hetero-Diels-Alder reaction. This research is extremely valuable as it expands the extent by which oxygen heterocycles can be formed. This is extremely important to the biological chemists who study and utilize oxygen heterocycles frequently (1). Huang and Rawal began their study by observing the hetero-Diels Alder reaction rates between various aldehydes and the diene 1-amino-3-siloxybuadiene. They found that the reaction rate was siginificantly higher in polar protic solvents rather than polar aprotic solvents (2). They attributed this change in the reaction rate to the fact that the carbonyl oxygen in the aldehyde could readily hydrogen bond to the hydrogen...
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...Date: March 9, 2015 Introduction: Aldehydes and ketones are two of several types of compounds that contain the carbonyl group. Reactions that occur because of the presence of the carbonyl group include nucleophilic addition reactions and base catalysed condensations. Aldehydes are also easily oxidized, which provides a convenient means to distinguish them from ketones. The carbonyl group in aldehydes and ketones is highly polarized; the carbonyl carbon bears a substantial partial positive charge and is susceptible to nucleophilic attack. Further, since it is sp2 hybridized it is relatively open to attack. Because the carbonyl contains no good leaving group, addition occurs rather than substitution. Aldehydes and ketones are polar compounds; however, the pure compounds do not undergo hydrogen bonding as the alcohols do. Thus the boiling points of aledehydes and ketones are lower than alcohols, but higher than alkanes or ethers. Low molecular weight carbonyl compounds are water soluble. Tests used to classify aldehydes and ketones Question 1 Identify the tests used to classify aldehydes and ketones. For each test identify the reagent(s) used and tell what observations/ results are expected. Write equations to show reactions (where appropriate). There are various tests that can be used to identify aldehydes and ketones. Some tests can be used to identify the carbonyl group in compounds while others can be used to distinguish between and aldehyde and a ketone. 1)2,4-Dinitrophenylhydrazone...
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...6. OBJECTIVES Atomic Structure Periodic Classification p-Block elements d-Block elements f-Block elements Co-ordination and Bio-coordination compounds - KNOWLEDGE E/LA SA VSA 1(3) UNDERSTANDING O E / L A SA VSA 1(1) 1(5) 1(3) 1(3) 1(1) - APPLICATION SKILL O E/LA SA VSA O TOTAL 10 09 O E/LA SA VSA 1(5) 1(3) - 1(1) - - 1(5) - - 1(5) 1(5) 1(5) - 1(1) 12 18 07 11 1(5) 1(3) - 1(3) 1(1) 1(1) 1(1) 1(1) 1(5) 1(1) - 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. Nuclear Chemistry Solid state Thermodynamics-II Chemical equilibrium-II Chemical Kinetics-II Surface Chemistry Electrochemistry-I Electrochemistry-II Isomerism in Organic Chemistry Hydroxy Derivatives Ethers Carbonyl Compounds Carboxylic Acids Organic Nitrogen Compounds Bio molecules Chemistry in Action Problems in Chemistry TOTAL 1(5) 1(5) - - 1(3) 1(3) 1(3) 1(3) 1(3) 1(3) - 1(1) 1(1) 1(5) - 1(5) 1(5) 1(5) - 1(3) - 1(1) 1(1) 1(5) - 1(5) 1(5) 1(5) 1(5) 1(3) - 1(1) 1(1) 1(1) - 1(5) - 1(5) - 1(3) 1(3) 1(3) - - 09 1(1) 09 10 1(1) 10 12 11 14 10 08 12 1(3) 1(1) 1(3) 1(3) 1(3) 2(1) 1(1) 1(5) 1(1) 1(1) 1(5) 1(5) 1(5) - 1(1) 1(5) - 1(1) 07 14 14 11 07 08 1(1) 1(5) 1(1) 1(1) - 1(1) 1(5) - 20 5 24 6 25 20 24 10 15 25 6 10 20 10 9 4 233 No. of Questions Summary / E / LA 8 Short Answers (SA) No. 12 Very Short Answers (VSA) No. 21 Objective...
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...In the second part of the reaction excess reagents are used and the synthesized product is in very small quantities. The product synthesized is diastereomers of 1-(4-nitrophenyl)-3-oxobutyl 3,3,3-trifluoro-2-methoxy-2-phenylpropanoate. The identity of this compound is confirmed by the following peaks seen in 1H NMR: 8.19 and 7.62 ppm, 5.47 ppm, 2.90, 2.13 ppm, 3.30 ppm, 7.36 – 7.38 ppm. The melting point, IR spectrum, theoretical yield or percentage yield is not found since all of it used in 1H NMR analysis. The product 1H NMR shows a mix of both the diastereomers, but it is difficult to ascertain which one is in excess. Introduction The aldol reaction that was experienced in this lab is the nucleophillic addition of an enolate to a carbonyl group to form a B-hydroxycarbonyl. This reaction is a very powerful method to construct of carbon-carbon bonds. Nature has developed the aldolase enzymes that catalyze biological aldol reactions. Aldolase is the enzyme that produces only one enantiomeric product that is hard to achieve. A simple amino acid L-proline can be used as the aldolase catalyst in asymmetric aldol reaction of acetone with a variety of aldehydes. This method produced 68% and 76% of...
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...Prelab UNKNOWN Number 1 Element % Carbon 67.01744 Hydrogen 7.311602 Nitrogen 7.817121 Oxygen 17.853837 mmass 179.2 g/mol --------------------------------------------------------------------------------------------------------------------- Carbon 67.01744% x 179.2 = 120.095 / 12.011 = 9.999 = 10 Hydrogen 7.311602% x 179.2 = 13.102 / 1.008 = 12.998 =13 Nitrogen 7.817121% x 179.2 = 14.008 / 14.008 = 1.004 = 1 Oxygen 17.853837% x 179.2 = 31.994 / 15.999 = 1.999 = 2 C10H13NO2 [pic]--------------------------------------------------------------------------------------------------------------------- Degree of Unsaturation 2C + 2 + N - X - H = 5 2 --------------------------------------------------------------------------------------------------------------------- Functional Groups amide, aromatic, ether Tests melting point, solubility Physical Properties white crystalline powder, odorless Introduction Qualitative analysis in organic chemistry is used in the identification of unknown substances. Although there is not a rule of order when performing these tests it is more important to have a systemic approach. If the same general order is always...
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...Purpose: The purpose of this experiment was to synthesize 4-chloro-4'-methoxychalcone from para-chlorobenzaldehyde and para-methoxyacetophenone by using an aldol condensation reaction. The compound was purified by recrystallization. The product was characterized by analyzing melting point, TLC, IR spectroscopy, 1H NMR spectroscopy, and mass spectroscopy. Mechanism: Key features of the aldol condensation reaction mechanism: (1) the formation of an enolate by an acid-base reaction; (2) nucleophilic attack of the aldehyde by the enolate in a nucleophilic addition type reaction; (3) deprotonation of water by the alkoxide creating an aldol product. The rest of the reaction undergoes an E1cB mechanism: (4) aldol is converted to an enolate ion by hydroxide; (5) the OH on the enolate ion is lost and the electrons on the carbon move to form a double bond, giving a disubstituted chalcone. Procedure: The procedure in the Wissinger Manual was followed, with some minor modifications. 0.530 grams of para-methoxyacetophenone and 0.495 grams of para-chlorobenaldehyde were used. This, however, did not change the mol ratios of the reaction. The product was washed with excess water during vacuum filtration, instead of 5-10 mL of water, to remove residual NaOH. The product was rinsed with ethanol during vacuum filtration. Results and Observations: P-methoxyacetophenone, p-chlorobenzaldehyde, and sodium hydroxide pellets were all white solids. When mixed with ethanol, the solution turned...
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...CHEM 232 1. Organic Chemistry II Problem Set 12 Identify each of the following compounds as reducing or nonreducing sugars: H HO HO H HO CH2 O H OH H CH2OH OH B Nonreducing sugar (acetal) CHO H H E OH OH CH2OH CHO OH OH OH CH2OH H H G OH HO CH2OH O H H H OCH2CH3 H HO H OH OH CH2OH O H H H OH CH2 O C OH H CH2OH H OH H H O H HO A Reducing sugar (hemiketal) Nonreducing sugar (acetal and ketal) CH2OH C O OH OH CH2OH CH2OH HO OH H O H CH2OH H H H H F OH D Reducing sugar (hemiketal) Reducing sugar (aldose) Reducing sugar (aldose) Reducing sugar (ketose) 2. Which of the structures shown above are: a. Pyranose monosaccharides A and B b. Furanose monosaccharides D c. Tetroses E d. Aldopentoses F e. Ketopentoses G f. Disaccharides C 3. Give the Haworth formula for α-D-Glucopyranose. H HO H OH CH2OH O H H OH H OH 4. Complete the following reactions: a. CH3CH2CH(OH)CH2OH b. O HOCH2 C CH2OH HIO4 CH3CH2CHO + CH2=O OH O O AgNO3-NaOH NH3 HOCH2 CH C Ag0 OH HOCH2 C c. H H CHO OH OH CH2OH HIO4 3 HCO2H OH CHOH HOCH2 CH C d. O H CHO H H OH OH CH2OH HCN CN H H H OH OH OH CH2OH HO CN H OH OH CH2OH + CH2=O + H H e. H HO H CHO OH H OH CH2OH NaBH4 H2O H HO H CH2OH OH H OH CH2OH h. H HO f. HO H H CHO H OH OH CH2OH CHO 1. Br2/H2O 2. H2O2 Fe2(SO4)3 H H OH OH CH2OH g. CHO H OH H H OH OH CH2OH 3 PhNHNH2 CH NNHPh C H H NNHPh OH CH2OH O H H H OH OH H OH CH3OH H+ ...
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...CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON 15.15 a) Octane denotes an eight carbon alkane chain. A methyl group (–CH3) is located at the second and third carbon position from the left. CH3 CH3 CH CH CH2 CH2 CH2 CH2 CH3 CH3 b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically, numbering starts at the top carbon atom of the ring for convenience. The ethyl group (–CH2CH3) is located at position 1 and the methyl group is located at position 3. CH2 CH3 1 6 2 3 5 CH3 4 c) The longest continuous chain contains seven carbon atoms, so the root name is “hept-.” The molecule contains only single bonds, so the suffix is “-ane.” Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other will result in side groups at positions 4 and 5. Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3,4– dimethylheptane. Note that the prefix “di-” is used to denote that two methyl side groups are present in this molecule. d) At first glance, this molecule looks like a 4 carbon ring, but the two –CH3 groups mean that they cannot be bonded to each other. Instead, this molecule is a 4–carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is 2,2–dimethylbutane...
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...Organic Chemistry Laboratory CLASSIFICATION TESTS FOR HYDROXYL AND CARBONYL CONTAINING COMPOUNDS ABSTRACT Hydroxyl- and Carbonyl- containing samples were tasked to be distinguished and classified respectively; the three types of alcohols were to be differentiated along with aldehydes and ketones. Hydroxyl groups are functional groups containing OH- atoms acting as substituents in an organic compound, carbonyl- groups refer to a divalent group consisting of a carbon atom double-bonded to an oxygen atom. Hydroxyl groups are characteristic functional groups of alcohols and phenols, while carbonyl groups are characteristic functional groups of aldehydes and ketones. The samples include ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzyl alcohol, acetaldehyde, n-butyraldehyde, benzaldehyde, acetone, acetophenone, and isopropyl alcohol. The samples were analyzed through different tests including the solubility of alcohols in water, Lucas Tests, Chromic Acid Tests (Jones Oxidation), 2-4 Dinitrophenylhydrazone (2,4-DNP) Tests, Fehling’s Tests, Tollens’ Silver Mirror Tests, and Iodoform Tests. Lucas tests differentiate primary, secondary and tertiary alcohols. Chromic tests distinguish a sample’s capability to oxidize. 2,4-DNP tests differentiates aldehydes and ketones from other functional groups. Fehling’s tests and Tollens’ Silver Mirror tests distinguish aldehydes from other functional groups. Iodoform tests distinguish methyl carbinol and methylcarbonyl...
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...Classification Tests for Hydroxyl- and Carbonyl-containing Compounds Miguel, K.D., Moron, R.S.S., Pazon, A.D., Ramirez, C.V., Raquepo, T.M.R., Razon, D.N.A.Jr. 2B-PH, Group No. 6, Department of Pharmacy, Faculty of Pharmacy, University of Santo Tomas, España Boulevard, 1015 Manila, Philippines Abstract In organic chemistry it is very common to see molecules comprised mainly of a carbon backbone with functional groups attached to the chain. The functional group gives the molecule its properties, regardless of what molecule contains it.[1] Examples of functional groups are that of hydroxyl (-OH) which is usually seen in alcohols, and carbonyl (C=0) which is seen in aldehydes and ketones. In this experiment, several tests were conducted to distinguish and differentiate various sample compounds such as ethanol, n-butyl alcohol, sec-butyl alcohol, tert-butyl alcohol, benzyl alcohol, n-butyraldehyde, benzaldehyde, acetone, acetophenone, isopropyl alcohol, and acetaldehyde. The tests are the following: solubility test of alcohols in water, which gave a soluble result in ethanol, sec-butyl alcohol and tert-butyl alcohol. Next is the Lucas test, which is used to differentiate 1°, 2° & 3° alcohols. In Lucas test, tert-butyl alcohol gave an immediate turbid result; the rate of reaction was noted. Chromic Acid test (Jones Oxidation) which gave a positive result by producing a blue-green solution with the sample n-butyl alcohol, acetaldehyde, benzaldehyde, and isopropyl alcohol. 2...
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