Chalcone
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Chalcone

Course Number: CHEM 2311, Spring 2011

College/University: Central Ohio Technical...

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Green Synthesis and Hydrogenation of a Di-Substitution Chalcone Purpose: The purpose of this experiment was to synthesize Chalcone 1 with an aldol condensation reaction between para-chlorobenzaldehyde and para-methoxyacetophenone. The crude Chalcone 1 was purified by recrystallization, its identity and purity was confirmed by analyzing the melting point, mass spectrum, Infrared spectroscopy and 1H NMR...

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Synthesis Green and Hydrogenation of a Di-Substitution Chalcone Purpose: The purpose of this experiment was to synthesize Chalcone 1 with an aldol condensation reaction between para-chlorobenzaldehyde and para-methoxyacetophenone. The crude Chalcone 1 was purified by recrystallization, its identity and purity was confirmed by analyzing the melting point, mass spectrum, Infrared spectroscopy and 1H NMR spectroscopy. The purified Chalcone 1 was hydrogenated with methanol, ammonium formate and palladium to form a product of unknown identity. The structure of the product could be predicted by analyzing the melting point, IR, mass spectrum and 1H NMR of the unknown product. Reaction: Aldol Condensation Hydrogenation Mechanism: Procedure: The procedure followed was as is described in Laboratory Manual for Organic Chemistry 2311, Eighth Edition, Jane E. Wissinger, Cengage Learning, Mason, Ohio, 2010, pp. 105-110. Reagent/Product Table for Aldol Condensation: Sodium Hydroxid e 40 0.41 10.25 0.19 2.13 1390 318 2 p-chloro benzaldehyde 1 p-methoxy acetophenone 150.17 0.52 3.46 x x 152 36 Ethano l 46.07 x x x 0.789 78 -114 Chalcone 1 272.73 0.83 3.04 x x x 130-131 Molecular Mass (g/mol) 1 1 1 140.57 0.492 3.50 0.41 1.196 214 46 Mass (g) Milli Moles Volume (mL) Density (g/cm3) Boiling Point (C) Melting Point (C) (1) This information was collected during the experiment (2) Palleros, D.R. J. Chem. Educ. 2004, 81, 1345-1347 All other information was collected from www.chemblink.com Results and Observation for Experiment 1 (Synthesis): Para-chloro benzaldehyde and p-methoxyacetophenone are white crystals and dissolve in ethanol. 4 Sodium Hydroxide pellets were used, which were crushed and added to the solution. The mixture turned pale yellow initially then white precipitate forms. Few mls of water is added and light custard yellow froth forms. The mixture was heated by keeping temperature between 60-65C for 30 minutes, however for a few minutes the temperature did increase to 69C. The solution is filtered to give pale custard yellow powder, but after washing with water white powder is left. The crude mass was taken then the crystals were dissolved in ethanol to give yellow liquid. Recrystallized crystals were flaky, shiny, transparent and cream coloured. The mass was recorded and melting point taken. Yield and Properties of Chalcone 1 Theoretica l Yield Crude Pure 0.94 g 0.94 g Actua l Yield 1.35 g 0.83 g Corrected %Yield x 88.1% Melting (C) x 129.1 - 131.4 2 % Yield 143.7 % 88.3% Point Appearance white powder Flaky, shiny, transparent and cream coloured Literature Melting Point 130-131 (2) Palleros, D.R. J. Chem. Educ. 2004, 81, 1345-1347 All other information was collected collected during the experiment Chalcone 1 Yield Calculations Limiting Reagent The overall reaction had a 1:1 ratio for the reactants Para-chloro benzaldehyde, pmethoxyacetophenone and Sodium Hydroxide to produce 1 mole of Chalcone 1. Least moles of pmethoxyacetophenone were used 3.46 mmoles therefore it is the limiting reagent. Theoretical yield: N(Para-chloro benzaldehyde) = m/M = 0.492/140.57 = 3.50 mmol (2sf) N(p-methoxyacetophenone) = m/M = .52/150.17 = 3.46 mmol (2sf) N(Sodium Hydroxide) = m/M = 0.41/40 = 10.25 mmol (2sf) Overall reaction ratio: Para-chloro benzaldehyde : p-methoxyacetophenone : Sodium Hydroxide: Chalcone 1 1:1:1:1:1 Since p-methoxyacetophenone is present in least moles, it is limiting reagent Thus N(p-methoxyacetophenone ) = N(Chalcone 1) m(Chalcone 1) = nM = 3.46*10-3*272.73 = 0.94g Actual Yield: Crude = 1.35g %Yield = (1.35/.94) *100 = 143.7% Pure = 0.83g %Yield = (0.83/0.94) *100 = 88.3% Corrected % yield: The NMR spectrum shows .1 cm (1.632ppm) area for 2 hydrogens water, thus 1 hydrogen for water represents .05 cm. There is no ketol impurity. Total area for product 17.2 cm, area per hydrogen = 17.2/13 = 1.32. .05/1.32 = .037 moles of water per mole of chalcone 1. Moles of chalcone 1 = .00304 mol, which can be multiplied by water mole ratio and molar mass to give final product mass. Water present: .00304*.037*18 = .002g Chalcone 1 present =0.83 - .002 =0.828 (2dp) Corrected yield% =.828/.94*100 = 88.1% Infrared Spectroscopy of Chalcone 1 A NaCl salt disk was used to hold 2 drops of the product sample dissolved in CH2Cl2. A background reading was done before the spectroscopy using the same plates. IRs of only the pure product was taken. Vibration C-H stretch alkane C=O stretch ketone C=C stretch Aromatic Compounds C-O stretch ether =C-H bend Alkene C-H bend Aromatic Compounds Frequency (cm1 ) 2955.5 1652 1600.8 1249.8 981.89 815.77 Descripti on Weak Medium Strong Medium Strong Strong 1 H NMR Spectrum of Chalcone 1 (200 MHz, CDCl3) The attached spectrum has peak assignments with structure. The integration of proton Ha was defined as representing 3 hydrogen and was used to determine number of hydrogens. This works nicely for all hydrogens except Hb, which has a much lower integration than the rest, and by calculations computes to being near 0 hydrogens rather than 1, but it has been listed as 1 because that is what is expected. Compoun d Chalcone 1 Protons, Hx Ha Hb Hc Hd He Hf Hg Chlorofor m Water Hh Hi Chemical ppm Shift , Splitting Pattern 3.888 Singlet 7.468 Singlet 7.746 Doublet 8.032 Doublet 6.981 Doublet 7.568 Doublet 7.379 Doublet 7.258 Singlet 1.632 Singlet # J value, Integration Area / H Hz cm ' s 0 0 15.8 3 3 2.4 2.2 0 0 0.3 3.9 1.8 2.6 3.1 3 2.5 0.3 0.1 3 1 1 2 2 2 2 1 1 Mass Spectometry of Chalcone 1: Peak (m/z) Molecular ion (272) M+2 (274) Relative Abundance 160000 58000 Molecular Formula C16H13O2Cl35 C16H13O2Cl37 237 165 135 77 257 30000 47000 112000 35000 28000 C16H13O2 C10H9O2 C8H7O2 C6H5 C15H10O2Cl35 Results and Observation for Experiment 2 (Hydrogenation): Chalcone 1 from the previous experiment, ammonium formate, methanol and catalytic amount of palladium were added together, which resulted in the flask getting fogged up with white fog gas and makes the contents of the flask invisible. When methanol is added the solution turned black. The solution is refluxed for 20 minutes. Black solid is filtered out using a filter pipet with colorless liquid in the erlenmeyer flask. Methanol is rinsed added to the product mixture in flask through the filter. The methanol is evaporated by heating in a water bath, once bubbling stops the solution is cooled and crude crystals form. The crude crystals were light yellow in color, and were weighed. The crude crystals were recrystallized from 95% ethanol and water, and when vacuum filtered out. The pure crystals were white and the mass and melting point was recorded. Reagent/Product Table for Hydrogenation of Chalcone: 1-(4Methoxyphenyl)-2phenylethanone 242 0.13 0.54 x x x 1 Ammonium Formate 1 Methanol 32.04 x x x 0.791 -98 64.7 Chalcone 1 272.73 0.205 0.75 x x x 2 Pd/ C x 0.05 x x x x x Molecular Mass (g/mol) 1 1 1 63.06 0.35 5.55 0.28 1.26 115-120 x Mass (g) Milli Moles 3 Volume (mL) Density (g/cm ) Boiling Point (C) Melting Point (C) 130-131 76-79 (1) This information was collected during the experiment (2) Palleros, D.R. J. Chem. Educ. 2004, 81, 1345-1347 All other information was collected from www.chemblink.com Yield and Properties of Hydrogenated Chalcone 1 Theoretica l Yield Actua l % Yield Corrected %Yield Melting (C) Point Appearance Yield Crude Pure 0.18 g 0.18g 0.18 g 0.13 g 100 % 72.2 % x 2.78% x 76 -79 x Light yellow fine crystals White fine crystals Literature Melting Point All information was collected collected during the experiment Chalcone 1 Yield Calculations Limiting Reagent The overall reaction had a 1:1 ratio for the reactants Chalcone 1, ammonium formate, and methanol to produce 1 mole of 1-(4-Methoxyphenyl)-2-phenylethanone. Least moles of Chalcone 1 were used 0.75 mmoles. 10%Pd/C is not liminting reagent because it was a catalyst. T herefore Chalcone 1 it is the limiting reagent. Theoretical yield: N(Chalcone 1) = m/M = 0.205/272.23 = 0.75 mmol (2sf) N(ammonium formate) = m/M = .35/63.06 = 5.55 mmol (2sf) Overall reaction ratio: Chalcone 1 : ammonium formate : 1-(4-Methoxyphenyl)-2-phenylethanone 1:1:1 Since Chalcone 1 is present in least moles, it is limiting reagent Thus N(Chalcone 1) = N(1-(4-Methoxyphenyl)-2-phenylethanone) m(1-(4-Methoxyphenyl)-2-phenylethanone) = nM = 0.75*10-3*242 = 0.18g Actual Yield: Crude = 0.18g %Yield = (.18/.18) *100 = 100% Pure = 0.13g %Yield = (0.13/0.18) *100 = 72.2% Corrected % yield: The NMR spectrum shows 3 impurities: chalcone 1, water and ethanol. Peaks H a ,Hi and Hp will be used to correct the yield. The region between 7 8 ppm is too clustered and will make for difficult and inaccurate calculations. Moles of hydrogenated chalcone 1 = 0.00054 mol Ha Integration area = 2.9 and it represents 3 hydrogens so 1 hydrogen = 0.967 cm Chalcone correction: Hi Integration area = 1.9 and it represents 3 hydrogens so 1 hydrogen = 0.633 cm .633/.967 =0.654 moles of chalcone 1 per mole of hydrogenated chalcone 1 Chalcone 1 present = 0.654*0.00054*272.23 = 0.10 Water present: .00054*(4.5/2/.967)*18 = .02g Ethanol present =0.00054*(0.6/2/.967)*46.07 = 0.005 Corrected yield% =(.13-.1-.02-.005/.18)*100 = 2.78% Infrared Spectroscopy of Hydrogenated Product Vibration C-H stretch aromatic compounds C-H stretch alkane C=O stretch ketone C=C stretch Aromatic Compounds C-O stretch ether =C-H bend Alkene C-H bend Aromatic Compounds C-O stretch Alcohol Frequency (cm-1) 3021.4, 3057.9 and 3082.4 2933.6 and 2967.9 1666.1 1600 1258.7 974.65 814.71 1030.9 Descriptio n Weak Weak Strong Strong Medium Medium Strong Medium 1 H NMR Spectrum of Hydrogenated Chalcone 1 (200 MHz, CDCl3) The attached spectrum has peak assignments with structure. The integration of proton Ha was defined as representing 3 hydrogen and was used to determine number of hydrogens for hydrogenated chalcone 1. Hi was defined as representing 3 hydrogens and was used to calculate number of hydrogens for chalcone 1. The Integration lines were not clearly defined for some hydrogens which is why they are not listed. Compound Hydrogenate d Chalcone 1 Protons, Hx Ha Hb Hc Chemical ppm Shift , Splitting Pattern Singlet Doublet Doublet J Hz value, Integration cm 0 4.8 5 2.9 2 2 Area / #H' s 3 2 2 3.863 6.915 7.943 Hd He Hf Hg Hh Chalcone 1 Hi Hj Hk Hm Hn Ho Water Ethanol Hp Hq Hr 3.244 3.061 7.235 7.304 7.175 3.896 6.989 8.043 7.753 7.553 7.388 1.74 3.723 1.242 Multiplet Multiplet Doublet of Doublets Doublet of Doublets Triplet of triplets Singlet Doublet Doublet Doublet Doublet Doublet Singlet Quartet Triplet 2.8 2.8 4 1.4 2.4 0 2 2 15.6 6.6 4.6 0 7 7 2 2 2 2 2 2 1 1.9 1 1 3 2 2 1 2 2 4.5 0.6 0.6 2 2 3 Mass Spectometry of hydrogenated Chalcone 1: Peak (m/z) Molecular ion (240) Base (135) 77 Relative Abundance 670000 1630000 260000 Molecular Formula C16H16O2 C8H7O2 C6H5

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ECH 159, Fall 2003, Homework 10, November, 2003 Problem 1 The linearized Korteveg de Vries Equation reads3 t u = kx u, < x < ,u(x, 0) = f (x)(a) Fourier transform the equation from x to and solve for U (, t) = F (u(x, t) (b) Using the convolution theo
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HW7Q3 Summary of NLR by Mathematica and Polymath Mathematica (including diethyl ether data points) k KA Equation (mol/g-cat/ (atm-1) atm1 or 2) 0.000120 0.0038 A 0.000009 0.1 0.000821 1.65 C 0.00011 0.20 0.000120 0.00081 D 0.000008 0.04 k ' Pethanol 0.000
UC Davis - ECH - 146
ECH146 winter 2008HW7 Statistics% Score = Score 100% 115HW7 %Scores Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 68 3 72 83 19 376 0 0 91 17 109 3109 46HW7 %Score12 10 8 6 4 2 0
UC Davis - ECH - 146
ECH146 winter 2008HW8 Statistics% Score = Score 100% 80HW8 %Scores Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 75 3 81 100 23 507 -1 -1 75 25 100 3280 44HW8 %Score14 12 10 Frequ
UC Davis - ECH - 146
UC Davis - ECH - 146
UC Davis - ECH - 146
0.220.20.18Concentration of A, CA (mol/L)0.160.140.120.10.080.060.040.02050100 Time (min)150200250600050004000 1/-rA (L*min/mol)300020001000000.10.20.30.4 0.5 Conversion X0.60.70.80.918000160001400012000 1/-rA (L*min/m
UC Davis - ECH - 146
Fig 1.1: F vs VA21.81.6Molar flow rate of A, FA (mol/L hr)1.41.210.80.60.40.2005101520 25 30 Volume of PFR, V (L)35404550Fig 1.2: C vs A21.81.6Concentration of A, CA (mol/L)1.41.210.80.60.40.2005101520 25 30 Reside
UC Davis - ECH - 146
Fig 1.1: C vs tA504540Concentration of A, CA (mol/s)3530252015105050010001500 Time, t (s)200025003000Fig 1.2: -v/r vs CAA654 -v/rA (L), L2/mol3210 10152025 30 35 Concentration of A, CA (mol/s)404550
UC Davis - ECH - 146
ECH 146 Winter, 2008 Problem Set 4 Due 2/08/081. The data shown in the table below were obtained for the reaction A 2B + C taking place in a batch reactor isothermally and with no volume change. The feed to the batch reactor was pure A. CA/molcm-3 62.5 5
UC Davis - ECH - 146
ECH 146 Winter, 2008 Problem Set 5 Due 2/15/08 1. The liquid-phase reaction of shown below takes place in a batch reactor at 25o C A+B C + D For an equal molar feed the following concentration-time data was obtained for A: CA/(mol/dm3 ) t/h 1.0 0 0.95 0.2
UC Davis - ECH - 146
Fig 1.1: ln(r) vs ln(C )A98765 ln(r) 4 3 2 1 0 -1-0.500.51 ln(C ) A1.522.53Fig 1.2: ln(r/C2 ) vs ln(C )A B2.521.51 ln(r/C2 ) A 0.5 0 -0.5 -1 00.511.5 ln(C ) B22.53Fig 2.1 k /k =10ab1 CA/CA0 0.9 CI/CA0 CP/CA0 0.80.70.6 C/CA0
UC Davis - ECH - 146
ECH 146 Winter, 2008 Problem Set 7 Due 3/07/08This problem set gives students practice in applying the concepts of LangmuirHinshelwood kinetics and in analyzing kinetics of catalytic reactions and methods of nonlinear regression in data fitting. 1. Consi
UC Davis - ECH - 146
Fig 3.2: k 1obsvs d0.80.6 observed rate constant, kobs (min-1)0.40.20-0.2-0.4-0.600.20.40.60.8 1 1.2 diameter of particle, d (mm)1.41.61.82
UC Davis - ECH - 146
Q uiz 1 (1114/08) E CH 146, W inter 2008ICLOSED BOOK] TIME ALOTTED: 20 MINUTES1. A steady state reaction A ~ B was carried out at constant temperature in a p lug flow r eactor (PFR). The initial concentration o f A was 1.5 mol/L. There was no volume c
UC Davis - ECH - 146
UC Davis - ECH - 146
Grading Criteria for Quiz #1Criteria A1. Plot CA vs. residence time or FA vs. volume A2. Estimate slope (rA) B1. Integrate the design equation to get time; that is, show t = B2. Plot 1/-rA vs. time and indicate area for time B3. Estimate Area (time)CA 0
UC Davis - ECH - 146
ECH146 Winter 2008 Solution for HW3Q1 with Polymath Tutorial This tutorial is purely based on the personal experience of the grader, and thus, there should be better ways to use Polymath. The textbook uses Polymath and there are many good tutorials on the
UC Davis - ECH - 146
ECH146 Winter 2008 Solution for HW4Q1b with Polymath Non Linear Regression TutorialProblem Setting Characteristics of Reaction A 2B + C Isothermal no volume change rate equation known to follow rA = kC A Experimental condition: Bach reactor CA0 = 62.5 mo
Texas El Paso - ASE - 362
ASE 362K Problem Set #2 Clemens / Spring 2011 Due: Wednesday, February 9 Problem 1. Consider a turbojet engine whose exit Mach number is subsonic under all conditions. For a subsonic jet, the exit pressure is equal to the ambient pressure (i.e., Pe=P). (a
Texas El Paso - ASE - 453
ASE 362K Problem Set #2 Clemens / Spring 2011 Due: Wednesday, February 9 Problem 1. Consider a turbojet engine whose exit Mach number is subsonic under all conditions. For a subsonic jet, the exit pressure is equal to the ambient pressure (i.e., Pe=P). (a
Carl Sandburg College - ENG - 120
Khanda Sofy Professor Hayter English 120 February 8, 2011 The good and bad: Technology Do to all the accident caused by cell phones; technology happens to be the most danger when it comes to driving. According to researchers, Texting while driving takes a
Cornell - HD - 3620
Lecture 2 HD 3620 1/28/2010Nature and Function of Infant-Caregiver Bonds, Part 1 Background o Clinical observations of separation effects o Bowlby: A warm, continuous association with at least one person i.e. a relationship o Explanation? Because she fee
Cornell - HD - 3620
HD 3620Lecture 3 2/2/2010 Nature and Function of Infant-Caregiver Bonds, Part 2 Last time: the attachment behavioral system Today: more factors that promote bondingAttachment Behavioral System How do you choose who will be primary attachment figure? Sele
Cornell - HD - 3620
HD 3620 2/4/2010 Lecture 4 Oxytocin and Pair Bonding Research on prarievoles (CS Carter) o Monogamous species o Defining characteristics Vs. montanevoles o Promiscuous species o Differ in OT receptors (Later: vasopressin) o Effects of OT administration (h
Cornell - HD - 3620
H D 3620 2/9/2010 Lecture 5 Attachment Research : Normative Processes and Individual Differences Previously o Attachment system dynamics In theory (Bowlby) Attachment System Dynamics Is my attachment figure near? Yes Secure, content E xplore, play, social
Cornell - HD - 3620
HD 3620 2/11/2010 Lecture 6 Ainsworths Baltimore Study Method: o Every other week for first three months. o Observed and noted mother-infant interactions Finding: one reliable predictor of infants strange situation attachment pattern (from caregiver sensi
Cornell - HD - 3620
Article 1: Relationships, Human Behavior, and Psychological Science Psychological science rarely uses relationship context in its theoriesthis is because psychology usually focuses on the behavior of the individual. History of relationship events on human
Cornell - HD - 3620
Article 2: The origins of attachment theory The article discusses the negative effects on personality development of prolonged institutional care and/or frequency changes of mother-figure during the early years of life. In Maternal Care and Mental Health,
Cornell - HD - 3620
Article 3: The Nature of the Childs Ties John Bowlbys work on attachment theory o He found that major disruptions in the mother-child relationship are precursors of later psychopathology o The infants tie to his/her mother came from a biologically based d
Cornell - HD - 3620
Article 4 Why Johnny Cant Sleep Ferberizing putting a child to sleep and then letting it cry until it falls asleep According to Darwinism, the natural nighttime arrangement is for kids to sleep alongside their mothers Ferber says that learning to sleep al
Cornell - SOC 1101 - 101
Sociology 1101 Lecture 16: Social Networks and Social Capital11/17/2010The social network perspective Society is constituted through a set of actors and their connections to each other in larger networks, or webs o Social actors- entities that are capab
Cornell - SOC 1101 - 101
Sociology Lecture 17: Network Analysis and Dynamics of Diffusion Final Friday December 10, 2 pm, Barton 100 East Social network data: A simple example RB JT AF AP CM RB JT AF 1 1 1 1 1 1 1 1 1 1 1 1 Adjacency matrix (with k= 5 actors) AP 1 1 1 110/22/201
Cornell - SOC 1101 - 101
Introduction to Sociology Lecture 18: Diffusion (continued) and The Small World11/29/2010Important aspects of network structure Components: Sets of actors who are all connected to each other, directly or indirectly Distance: The average number of links
Cornell - SOC 1101 - 101
Lecture 19: The Small World (continued) 12/1/2010 Mathematical explanation -Diagram (bulls eye) -If everyone has a certain number of contacts (say, 100) in their social network, and their contacts do as well, then a given persons network will span outward
Cornell - HD - 1170
HD 1170 Lecture 1 1/26/2010Hills Model Changes in Adolescence Social Transitions Biological Transitions Cognitive Transitions Psychosocial development Identity Autonomy Intimacy Sexuality Achievement Problems Contexts Families Peer Groups Work Schools Le
Cornell - HD - 1170
HD 1170 Lecture 2 1/28/10 Adolescent Development in Context 3 Sources of Information Scientific Research Theories Case Studies Adolescence= anguish, turmoil Accurate view? 3 Views of Adolescence 1. Storm and Stress a. G. Stanley Hall, 1904 i. Adolescence