amines amide - TangHua2012-2013
... is replaced by an alkyl or an aromatic group (contains at least one C6H6) • Important: Methylamine--CH3NH2 ...
... is replaced by an alkyl or an aromatic group (contains at least one C6H6) • Important: Methylamine--CH3NH2 ...
Protection (and Deprotection) of Functional Groups in Organic
... liquid phase and frequently involving protectiondeprotection steps. The use of blocking functions in organic synthesis, developed for nearly 100 years, makes more complex the entire synthetic plan since it requires at least two additional steps. At the same time, environmental and economic considera ...
... liquid phase and frequently involving protectiondeprotection steps. The use of blocking functions in organic synthesis, developed for nearly 100 years, makes more complex the entire synthetic plan since it requires at least two additional steps. At the same time, environmental and economic considera ...
Organic Chemistry Lecture Outline Chapter 20: Carboxylic Acids
... iii. The further away the EWG, the less effect it has on the pKa of the acid. iv. Electron donating groups destabilized the carboxylate anion and thus result in a weaker acid (higher pKa). b. BENZOIC ACIDS i. EWG on the benzene ring of benzoic acid stabilize the carboxylate anion. ii. EWG's on the b ...
... iii. The further away the EWG, the less effect it has on the pKa of the acid. iv. Electron donating groups destabilized the carboxylate anion and thus result in a weaker acid (higher pKa). b. BENZOIC ACIDS i. EWG on the benzene ring of benzoic acid stabilize the carboxylate anion. ii. EWG's on the b ...
document
... KOH is a strong base. H2S is a weak acid. CH3NH2 is a weak base. HClO4 is a strong acid. ...
... KOH is a strong base. H2S is a weak acid. CH3NH2 is a weak base. HClO4 is a strong acid. ...
101
... CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O() + CO2(g) For the net ionic equation, the oxidation number of each element in the reactants and products is as shown. CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O() + CO2(g) ...
... CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O() + CO2(g) For the net ionic equation, the oxidation number of each element in the reactants and products is as shown. CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O() + CO2(g) ...
Year 2 Chemistry Contents Guide
... Guide to selecting an appropriate indicator based on pH curves Animation illustrating approximations that can be made at the half-equivalence point, and how this can be used to find pH and Ka Identifying missing words relating to titrations Strong acid-base titrations and calculations with full work ...
... Guide to selecting an appropriate indicator based on pH curves Animation illustrating approximations that can be made at the half-equivalence point, and how this can be used to find pH and Ka Identifying missing words relating to titrations Strong acid-base titrations and calculations with full work ...
Experiments for an Introduction to Organic Chemistry
... Surprising results! Explain them by paying attention to the pH of the reaction medium. Test the pH value of the cerium nitrate reagent (E73) and of the DNPH reagent (E76) by means of indicator paper: It turns red. Note: the PHT test (E75) is performed in an alkaline medium, in contrast to the cerium ...
... Surprising results! Explain them by paying attention to the pH of the reaction medium. Test the pH value of the cerium nitrate reagent (E73) and of the DNPH reagent (E76) by means of indicator paper: It turns red. Note: the PHT test (E75) is performed in an alkaline medium, in contrast to the cerium ...
PREPARATION, STRUCTURAL STUDIES AND CHEMICAL
... compounds often with quantitative yields. The process includes two steps: the ligand exchange reaction with subsequent reductive elimination. This reaction is highly selective and primary alcohols are converted only into their respective aldehydes without any formation of carboxylic acids as byprodu ...
... compounds often with quantitative yields. The process includes two steps: the ligand exchange reaction with subsequent reductive elimination. This reaction is highly selective and primary alcohols are converted only into their respective aldehydes without any formation of carboxylic acids as byprodu ...
Basic definitions for organic chemistry
... involved. The characterisation takes place in a series of stages (see below). In earlier times relatively large amounts of substance were required to elucidate the structure but, with the advance in technology and the increased use of electronic instrumentation, only very small amounts are now requi ...
... involved. The characterisation takes place in a series of stages (see below). In earlier times relatively large amounts of substance were required to elucidate the structure but, with the advance in technology and the increased use of electronic instrumentation, only very small amounts are now requi ...
Answers - Pearson-Global
... The two liquids should be compared in identical apparatus, side by side so that the temperature is always identical for the two throughout the time needed to run the experiment. You would need equal volumes of liquids, and equal volumes of water. All this stresses the importance of a “fair test”. Li ...
... The two liquids should be compared in identical apparatus, side by side so that the temperature is always identical for the two throughout the time needed to run the experiment. You would need equal volumes of liquids, and equal volumes of water. All this stresses the importance of a “fair test”. Li ...
Alcohols, Phenols and Ethers
... Polarity: Water is a very polar molecule; alcohols and ethers also display polarity. Boiling point: Water has a higher boiling point than predicted on the basis of its molar mass. Alcohols also share this characteristic, but ethers do not. Solubility: Water is miscible with other polar liquids; alco ...
... Polarity: Water is a very polar molecule; alcohols and ethers also display polarity. Boiling point: Water has a higher boiling point than predicted on the basis of its molar mass. Alcohols also share this characteristic, but ethers do not. Solubility: Water is miscible with other polar liquids; alco ...
Equilibrium Notes - Chemistry Teaching Resources
... The colour change from 20°C to 0°C shows that more N 2 O 4 is formed at 0°C and therefore [N 2 O 4 ] increases and [NO 2 ] decreases. This leads to a fall in the value of K. Similarly, the colour change from 20°C to 80°C shows that more NO 2 is formed at 80°C and therefore [NO 2 ] increases and [N 2 ...
... The colour change from 20°C to 0°C shows that more N 2 O 4 is formed at 0°C and therefore [N 2 O 4 ] increases and [NO 2 ] decreases. This leads to a fall in the value of K. Similarly, the colour change from 20°C to 80°C shows that more NO 2 is formed at 80°C and therefore [NO 2 ] increases and [N 2 ...
Lecture 4 - Sugars, ring structures
... A hemiacetal is a compound having the C(OH)OR group that results from the reaction of an aldehyde with one mole of an alcohol. Sugar molecules contain an aldehyde group and lots of alcohol groups. PSE 406 - Lecture 4 ...
... A hemiacetal is a compound having the C(OH)OR group that results from the reaction of an aldehyde with one mole of an alcohol. Sugar molecules contain an aldehyde group and lots of alcohol groups. PSE 406 - Lecture 4 ...
Free Radical Chemistry and the Preparation of Alkyl
... The selectivity of Br . is much higher than Cl . but the reactivity is less Because the abstraction of H by Br . is endothermic, the transition states resemble the radical. So radical stability has a larger effect on reactivity. What's the percent distribution of products of bromination of 3-methylp ...
... The selectivity of Br . is much higher than Cl . but the reactivity is less Because the abstraction of H by Br . is endothermic, the transition states resemble the radical. So radical stability has a larger effect on reactivity. What's the percent distribution of products of bromination of 3-methylp ...
Click Chemistry in Peptide-Based Drug Design
... peptides. Exoproteases such as aminopeptidases or metallo-carboxypeptidases can recognize N- or Cterminal groups and hydrolyze peptides. As a result, head-to-tail cyclopeptides are relatively more resistant proteolytically than the linear counterparts. Meldal and coworkers [37] prepared head-to-tail ...
... peptides. Exoproteases such as aminopeptidases or metallo-carboxypeptidases can recognize N- or Cterminal groups and hydrolyze peptides. As a result, head-to-tail cyclopeptides are relatively more resistant proteolytically than the linear counterparts. Meldal and coworkers [37] prepared head-to-tail ...
Lecture - Ch 19
... • Cyanohydrins: Product of nucleophilic reaction between aldehydes and unhindered ketones with HCN – Addition of HCN is reversible and base-catalyzed, generating nucleophilic cyanide ion, CN– Addition of CN to C=O yields a tetrahedral intermediate, which is then protonated – Equilibrium favors cyan ...
... • Cyanohydrins: Product of nucleophilic reaction between aldehydes and unhindered ketones with HCN – Addition of HCN is reversible and base-catalyzed, generating nucleophilic cyanide ion, CN– Addition of CN to C=O yields a tetrahedral intermediate, which is then protonated – Equilibrium favors cyan ...
Synthetic Organic Chemistry - Name
... Due to the presence of bulky groups in both or OH either of the reactants in Grignard reagent the extent of addition is reduced or the reaction not take place or some abnormal product is formed. If Grignard reagent has β-hydrogen atom, then in some hindred ketone also show reduction of carbonyl grou ...
... Due to the presence of bulky groups in both or OH either of the reactants in Grignard reagent the extent of addition is reduced or the reaction not take place or some abnormal product is formed. If Grignard reagent has β-hydrogen atom, then in some hindred ketone also show reduction of carbonyl grou ...
Chapter #14 Newest CD
... CH3 CH3 H2 CH3 (d) CH3-CH2-CH-C-CH3 H2 H2 CH3 CH3 (e) CH3-CH2-CH=C-CH-CH3 CH2-CH3 CH3 Plan: For (a) to (c), we refer to Table 15.2. We first name the longest chain (root- + -ane). Then we find the lowest branch numbers by counting C atoms from the end closer to a branch. Finally, we name each branch ...
... CH3 CH3 H2 CH3 (d) CH3-CH2-CH-C-CH3 H2 H2 CH3 CH3 (e) CH3-CH2-CH=C-CH-CH3 CH2-CH3 CH3 Plan: For (a) to (c), we refer to Table 15.2. We first name the longest chain (root- + -ane). Then we find the lowest branch numbers by counting C atoms from the end closer to a branch. Finally, we name each branch ...
Line 4: Equation
... Since you start with a metal chlorate, you will end up with a metal chloride and oxygen. Line 1: metal chlorate metal chloride + oxygen Line 2: sodium chlorate sodium chloride + oxygen Line 3: 2 NaClO3 2 NaCl + 3 O2 Line 4: 2 NaClO3 2 NaCl + 3 O2 Example 8: Write the balanced equation to sh ...
... Since you start with a metal chlorate, you will end up with a metal chloride and oxygen. Line 1: metal chlorate metal chloride + oxygen Line 2: sodium chlorate sodium chloride + oxygen Line 3: 2 NaClO3 2 NaCl + 3 O2 Line 4: 2 NaClO3 2 NaCl + 3 O2 Example 8: Write the balanced equation to sh ...
Transformation of Carbon Dioxide
... carbonate synthesis. If Schemes 8 and 10 are combined, then water can be canceled out, which yields a new equation that is independent of alcohol (route c, Scheme 9). Indeed, the reaction of orthoesters proceeds in the presence of a metal complex such as Bu2Sn(OMe)2 in high yields without adding alc ...
... carbonate synthesis. If Schemes 8 and 10 are combined, then water can be canceled out, which yields a new equation that is independent of alcohol (route c, Scheme 9). Indeed, the reaction of orthoesters proceeds in the presence of a metal complex such as Bu2Sn(OMe)2 in high yields without adding alc ...
+ 2 H2O(l Ca(OH)2 aq)
... c) 2 KMnO4(aq) + 3 Na2SO3(aq) + H2O(l) 2 MnO2(s) + 3 Na2SO4(aq) + 2 KOH(aq) KMnO4 is the oxidizing agent (O.N.(Mn) goes from +7 to +4). Na2SO3 is the reducing agent (O.N.(S) goes from +4 to +6). d) 2 CrO42–(aq) + 3 HSnO2–(aq) + H2O(l) 2 CrO2–(aq) + 3 HSnO3–(aq) + 2 OH–(aq) CrO42– is the oxidizin ...
... c) 2 KMnO4(aq) + 3 Na2SO3(aq) + H2O(l) 2 MnO2(s) + 3 Na2SO4(aq) + 2 KOH(aq) KMnO4 is the oxidizing agent (O.N.(Mn) goes from +7 to +4). Na2SO3 is the reducing agent (O.N.(S) goes from +4 to +6). d) 2 CrO42–(aq) + 3 HSnO2–(aq) + H2O(l) 2 CrO2–(aq) + 3 HSnO3–(aq) + 2 OH–(aq) CrO42– is the oxidizin ...
1 - KFUPM Faculty List
... Al is a representative metallic element and its monoatomic ion is Al3+ (main group IIIa, 3 steps to the next smaller noble gas). Sulfite is SO32- (lower oxygen content), while SO42- is sulfate (higher oxygen content). Charge neutrality requires thus (Al3+)2(SO32-)3 = Al2(SO3)3 for aluminum sulfite g ...
... Al is a representative metallic element and its monoatomic ion is Al3+ (main group IIIa, 3 steps to the next smaller noble gas). Sulfite is SO32- (lower oxygen content), while SO42- is sulfate (higher oxygen content). Charge neutrality requires thus (Al3+)2(SO32-)3 = Al2(SO3)3 for aluminum sulfite g ...
Structure Investigations of Group 13 Derivatives of N
... iminate anion as the O,N-bidentate ligand. The influence of weak hydrogen bonds on the supramolecular structure of resulting complexes is also discussed. Results and Discussion Synthesis and Solution Structure. The reaction of Me3M with 1 equiv of N-phenylsalicylideneimine (saldPh-H) in toluene solu ...
... iminate anion as the O,N-bidentate ligand. The influence of weak hydrogen bonds on the supramolecular structure of resulting complexes is also discussed. Results and Discussion Synthesis and Solution Structure. The reaction of Me3M with 1 equiv of N-phenylsalicylideneimine (saldPh-H) in toluene solu ...
Curriculum Vitae
... 106. Deluca, R. J.; Stokes, B. J.; Sigman, M. S. “The strategic generation and interception of palladiumhydrides for use in alkene functionalization reactions,” Pure Appl. Chem. 2014, 86, 395-408. 105. Xu, L.; Hilton, M. J.; Zhang, X.; Norrby, P.-O.*; Wu, Y.-D.*; Sigman, M. S.*; Wiest, O.* “Mechanis ...
... 106. Deluca, R. J.; Stokes, B. J.; Sigman, M. S. “The strategic generation and interception of palladiumhydrides for use in alkene functionalization reactions,” Pure Appl. Chem. 2014, 86, 395-408. 105. Xu, L.; Hilton, M. J.; Zhang, X.; Norrby, P.-O.*; Wu, Y.-D.*; Sigman, M. S.*; Wiest, O.* “Mechanis ...
Strychnine total synthesis
Strychnine total synthesis in chemistry describes the total synthesis of the complex biomolecule strychnine. The first reported method by the group of Robert Burns Woodward in 1954 is considered a classic in this research field. At the time it formed the natural conclusion to an elaborate process of molecular structure elucidation that started with the isolation of strychnine from the beans of Strychnos ignatii by Pierre Joseph Pelletier and Joseph Bienaimé Caventou in 1818. Major contributors to the entire effort were Sir Robert Robinson with over 250 publications and Hermann Leuchs with another 125 papers in a time span of 40 years. Robinson was awarded the Nobel Prize in Chemistry in 1947 for his work on alkaloids, strychnine included. The process of chemical identification was completed with publications in 1946 by Robinson and later confirmed by Woodward in 1947. X-ray structures establishing the absolute configuration became available between 1947 and 1951 with publications from J. M. Bijvoet and J.H. Robertson .Woodward published a very brief account on the strychnine synthesis in 1954 (just 3 pages) and a lengthy one (42 pages) in 1963.Many more methods exist and reported by the research groups of Magnus, Overman, Kuehne, Rawal, Bosch, Vollhardt, Mori, Shibasaki, Li, Fukuyama Vanderwal and MacMillan. Synthetic (+)-strychnine is also known. Racemic synthesises were published by Padwa in 2007 and in 2010 by Andrade and by Reissig.In his 1963 publication Woodward quoted Sir Robert Robinson who said for its molecular size it is the most complex substance known.