LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
... 19. How will you synthesise the following compounds from benzaldehyde? a) Mandelic acid b) m-Benzaldehydesulphonic acid c) Toluene (2+2+1) 20. Write a note on acidity of substituted benzoic acid. 21. How will you prepare the following compounds from benzenediazonium chloride? a) p-aminoazobenzene b) ...
... 19. How will you synthesise the following compounds from benzaldehyde? a) Mandelic acid b) m-Benzaldehydesulphonic acid c) Toluene (2+2+1) 20. Write a note on acidity of substituted benzoic acid. 21. How will you prepare the following compounds from benzenediazonium chloride? a) p-aminoazobenzene b) ...
Ch. 1: Atoms: The Quantum World
... the pull of the nucleus in an electron is decreased, as if the charge of the nucleus were smaller than it actually is. En = − ...
... the pull of the nucleus in an electron is decreased, as if the charge of the nucleus were smaller than it actually is. En = − ...
PHYSICAL SCIENCE PAPER 2 QUESTIONS SECTION A
... coins contain silver sulphide due to the anaerobic bacteria found in the mud. Methods for converting silver corrosion products to metallic silver all employ reducing reactions for the silver compounds. Electrolysis is the most common method used. ...
... coins contain silver sulphide due to the anaerobic bacteria found in the mud. Methods for converting silver corrosion products to metallic silver all employ reducing reactions for the silver compounds. Electrolysis is the most common method used. ...
Iodine Clock Reaction and Photochemical Reduction
... the time at which blue colour appears. Then, again add immediately 1 mL of Na2 S2 O 3 and note the time of appearance of blue colour after each addition until the time of a appearance of blue colour becomes 5-6 times the initial time. Shake the reaction mixture throughout the addition of Na2 S2 O 3 ...
... the time at which blue colour appears. Then, again add immediately 1 mL of Na2 S2 O 3 and note the time of appearance of blue colour after each addition until the time of a appearance of blue colour becomes 5-6 times the initial time. Shake the reaction mixture throughout the addition of Na2 S2 O 3 ...
radicals
... In the Case of a (non-radical) Decarboxylation ,Recall that: Decarboxylation produce CO2, and are thus thermodynamically favorable. These reactions proceed most readily when the COOH moiety is attached to an electronegative atom, like oxygen or nitrogen. ...
... In the Case of a (non-radical) Decarboxylation ,Recall that: Decarboxylation produce CO2, and are thus thermodynamically favorable. These reactions proceed most readily when the COOH moiety is attached to an electronegative atom, like oxygen or nitrogen. ...
PRACTICE FINAL EXAM CHEMISTRY 152 This
... Consider the following three statements and choose the correct answer. (I) The copper metal is the negative electrode (anode). (II) A salt bridge will be unnecessary in this spontaneous cell. (III) The electrons flow from the Zn electrode to the Cu electrode [a] only I is true [c] II and III are tru ...
... Consider the following three statements and choose the correct answer. (I) The copper metal is the negative electrode (anode). (II) A salt bridge will be unnecessary in this spontaneous cell. (III) The electrons flow from the Zn electrode to the Cu electrode [a] only I is true [c] II and III are tru ...
Molecular Geometry and Hybrid Orbitals Molecular Geometry
... • Chemical properties: (CH3)3C-Br + HOH3C-Br + HO• Biological properties: Enzyme specificity ...
... • Chemical properties: (CH3)3C-Br + HOH3C-Br + HO• Biological properties: Enzyme specificity ...
10.2 Functional group chemistry Hydrocarbons
... Both of these are strong bonds with bond enthalpies of 348 kJ/mol and 412 kJ/mol, respectively. These molecules will only react in the presence of a source of energy strong enough to break these bonds. Most alkanes are stable under most conditions and can be stored, transported and compressed ...
... Both of these are strong bonds with bond enthalpies of 348 kJ/mol and 412 kJ/mol, respectively. These molecules will only react in the presence of a source of energy strong enough to break these bonds. Most alkanes are stable under most conditions and can be stored, transported and compressed ...
Table
... Alkanes and Aromatics Organic Halides 1. Substitution Reactions with halogens or hydrogen halides Pathway to other groups: Alkyl halides alkenes 1. Elimination Reactions, removing hydrogen and halide ions. ...
... Alkanes and Aromatics Organic Halides 1. Substitution Reactions with halogens or hydrogen halides Pathway to other groups: Alkyl halides alkenes 1. Elimination Reactions, removing hydrogen and halide ions. ...
enzymatic resolution of a racemic mixture by acylation in
... [5-8]. Of these, only a few refer to aliphatic alcohols, in particular of longer alkyl chain lengths [3], [5], [8]. The use of enol-esters as acylating agents has the advantage of being irreversible but the aldehyde formed can decrease enzyme activity. This problem has been circumvented by using suc ...
... [5-8]. Of these, only a few refer to aliphatic alcohols, in particular of longer alkyl chain lengths [3], [5], [8]. The use of enol-esters as acylating agents has the advantage of being irreversible but the aldehyde formed can decrease enzyme activity. This problem has been circumvented by using suc ...
File - Loreto Science
... • The following is a summary of some of the steps to make new substances out of others and back again! ...
... • The following is a summary of some of the steps to make new substances out of others and back again! ...
Ch 26 C-C bond formation
... • Nonhalogenated cyclopropanes can be prepared by the reaction of an alkene with diiodomethane, CH2I2, in the presence of a copper-activated zinc reagent called zinc– ...
... • Nonhalogenated cyclopropanes can be prepared by the reaction of an alkene with diiodomethane, CH2I2, in the presence of a copper-activated zinc reagent called zinc– ...
Molecular Orbitals and Hybridisation
... The sp3 orbitals formed are all half-filled, with the electron far more likely to be found in the larger lobe. Since electrons repel each other, the four sp3 hybridised orbitals surrounding a central carbon atom result in a familiar tetrahedral shape, with a maximum possible angle between each orbit ...
... The sp3 orbitals formed are all half-filled, with the electron far more likely to be found in the larger lobe. Since electrons repel each other, the four sp3 hybridised orbitals surrounding a central carbon atom result in a familiar tetrahedral shape, with a maximum possible angle between each orbit ...
This exam will consist of 30-35 multiple choice or short answer
... What is the product distribution? Why? How was this distribution determined? What is the electrophile? Reaction mechanism? Draw the intermediate, including all resonance structures. Is there a catalyst for this reaction? What is it? What did it do? How is GC used in this reaction? What did the chrom ...
... What is the product distribution? Why? How was this distribution determined? What is the electrophile? Reaction mechanism? Draw the intermediate, including all resonance structures. Is there a catalyst for this reaction? What is it? What did it do? How is GC used in this reaction? What did the chrom ...
Review Questions
... ∆Hf is equal to Zero (WHEN???) ∆ H = Σ n∆ Hof (products) - Σ n∆ Hof (reactants) Reaction Mechanism Problems: 1. The standard enthalpy of formation of sulfur dioxide is -296.8 kJ/mol. a. Write a thermochemical equation for the formation reaction b. Sketch a potential energy diagram for the reac ...
... ∆Hf is equal to Zero (WHEN???) ∆ H = Σ n∆ Hof (products) - Σ n∆ Hof (reactants) Reaction Mechanism Problems: 1. The standard enthalpy of formation of sulfur dioxide is -296.8 kJ/mol. a. Write a thermochemical equation for the formation reaction b. Sketch a potential energy diagram for the reac ...
Nucleophilic Substitution
... Let's look at how the various components of the reaction influence the reaction pathway: RReactivity order : CH3- > CH3CH2- > (CH3)2CH- > (CH3)3CIn an SN2 reaction, the transition state has 5 groups around the central C atom. As a consequence of the steric requirements at this center, less highly su ...
... Let's look at how the various components of the reaction influence the reaction pathway: RReactivity order : CH3- > CH3CH2- > (CH3)2CH- > (CH3)3CIn an SN2 reaction, the transition state has 5 groups around the central C atom. As a consequence of the steric requirements at this center, less highly su ...
Molecular Orbital Theory
... – combination of n atomic orbitals gives n MO – MOs are arranged in order of increasing energy – MOs fill by same rules as for atomic orbitals: • Aufbau principle: fill beginning with LUMO • Pauli exclusion principle: no more than 2 ein a MO • Hund’s rule: filling of degenerate orbitals ...
... – combination of n atomic orbitals gives n MO – MOs are arranged in order of increasing energy – MOs fill by same rules as for atomic orbitals: • Aufbau principle: fill beginning with LUMO • Pauli exclusion principle: no more than 2 ein a MO • Hund’s rule: filling of degenerate orbitals ...
SAMPLE PAPER -4 Time Allowed: 3 Hrs
... Q24 a) Account for the followings: (i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (ii) Grignard reagent should be prepared under anhydrous conditions. b) Which one of the following pairs of halogen compounds undergo S N2 faster and Why? ...
... Q24 a) Account for the followings: (i) Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (ii) Grignard reagent should be prepared under anhydrous conditions. b) Which one of the following pairs of halogen compounds undergo S N2 faster and Why? ...
PPT - Unit 5
... 2. Given the following data: -(C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300. kJ) 2( C(s) + O2(g) → CO2(g) ) 2(ΔH = -394 kJ) H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ Calculate ΔH for the following reaction: 2C(s) + H2(g) → C2H2(g) 2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ 2CO2(g) + H2O(l) → C2H2(g) + ...
... 2. Given the following data: -(C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300. kJ) 2( C(s) + O2(g) → CO2(g) ) 2(ΔH = -394 kJ) H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ Calculate ΔH for the following reaction: 2C(s) + H2(g) → C2H2(g) 2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ 2CO2(g) + H2O(l) → C2H2(g) + ...
Woodward–Hoffmann rules
The Woodward–Hoffmann rules, devised by Robert Burns Woodward and Roald Hoffmann, are a set of rules in organic chemistry predicting the barrier heights of pericyclic reactions based upon conservation of orbital symmetry. The Woodward–Hoffmann rules can be applied to understand electrocyclic reactions, cycloadditions (including cheletropic reactions), sigmatropic reactions, and group transfer reactions. Reactions are classified as allowed if the electronic barrier is low, and forbidden if the barrier is high. Forbidden reactions can still take place but require significantly more energy.The Woodward–Hoffmann rules were first formulated to explain the striking stereospecificity of electrocyclic reactions under thermal and photochemical control. Thermolysis of the substituted cyclobutene trans-1,2,3,4-tetramethylcyclobutene (1) gave only one diastereomer, the (E,E)-3,4-dimethyl-2,4-hexadiene (2) as shown below; the (Z,Z) and the (E,Z) diastereomers were not detected in the reaction. Similarly, thermolysis of cis-1,2,3,4-tetramethylcyclobutene (3) gave only the (E,Z) diastereomer (4).Due to their elegance and simplicity, the Woodward–Hoffmann rules are credited with first exemplifying the power of molecular orbital theory to experimental chemists. Hoffmann was awarded the 1981 Nobel Prize in Chemistry for this work, shared with Kenichi Fukui who developed a similar model using frontier molecular orbital (FMO) theory; because Woodward had died two years before, he was not eligible to win what would have been his second Nobel Prize for Chemistry.