, ORGANIC REACTIONS (AS+A2) ALKANE REACTIONS NAME Cracking REACTANTS Alkanes only Chlorination or Bromination (Free radical substitution) Combustion Alkanes with chlorine or bromine CONDITIONS Heat with aluminum oxide catalyst Presence of UV light (sunlight) Alkanes with oxygen Burn PRODUCTS Alkanes and alkenes Alkyl halide Carbon dioxide and water ALKENE REACTIONS NAME REACT WITH Hydrogenation Hydrogen (Electrophilic addition) Electrophilic addition Steam Electrophilic addition Electrophilic addition Oxidation Hydrogen halide Halogen molecules Cold dilute acidified potassium manganate Oxidation Hot concentrated acidified potassium manganate with =CH2 Hot concentrated acidified potassium manganate with =CH Oxidation CONDITIONS PRODUCTS Heat with platinum or nickel catalyst Heat with phosphoric acid catalyst RTP RTP Alkane Alcohol Alkyl halide Alkyl halide Diol, purple solution goes colorless Carbon dioxide Aldehyde and then carboxylic acid Oxidation Hot concentrated acidified potassium manganate with =C Ketone Additional polymerization Other alkenes Long chain polymer ALKYL HALIDE REACTIONS NAME Nucleophilic substitution REACT WITH CONDITIONS Aqueous NaOH (or Heat aqueous solution other aqueous alkali) PRODUCTS Alcohol Nucleophilic substitution KCN Ethanolic solution heated under reflux Nitrile Nucleophilic substitution Ammonia Primary amine Hydrolysis Aqueous silver nitrate NaOH Ethanolic solution of aqueous ammonia heated under pressure In ethanol Elimination Ethanolic solution heated Alcohol and silver halide Alkene, salt, and water ALCOHOL REACTIONS NAME Combustion REACTANTS Oxygen CONDITIONS Ignited Nucleophilic substitution HX (KBr in sulfuric or phosphoric acid is used instead of HBr) PCl3 PRODUCTS Carbon dioxide and water Alkyl halide and water Heat Alkyl halide and H3PO3 PCl5 RTP Alkyl halide, hydrochloric acid, and POCl 3 SOCl2 Sodium Oxidation Alkyl halide, HCl, and SO2 Hydrogen and salt of alcohol Aldehyde and then carboxylic acid if not distilled Ketone Primary alcohol with potassium manganate or potassium dichromate Secondary alcohol with potassium manganate or potassium dichromate Acidic solution, warm alcohol Dehydration Acid, aluminum oxide, porous pot, or pumice Esterification Carboxylic acid Esterification Acyl chloride Concentrated if Alkene and water acid, heated if other Heated under Ester reflux with strong acid catalyst Ester Oxidation Acidic solution, warm alcohol CARBONYL REACTIONS NAME Reduction REACTANTS Aldehyde with NaBH4 or LiAlH4 Ketone with NaBH4 or LiAlH4 CONDITIONS Nucleophilic addition Condensation or Carbonyl group test Oxidation HCN KCN catalyst and heat Heat Oxidation Aldehyde with Tollen’s reagent Reduction 2,4-DNPH Aldehyde with Fehling’s solution PRODUCTS Primary alcohol Secondary alcohol Warmed in alkaline solution Warmed in alkaline solution Hydroxynitrile Water and deep orange precipitate Salt of acid and red copper oxide Salt of acid and silver mirror of Ag atoms ESTER REACTIONS NAME Hydrolysis (reversible) Hydrolysis (irreversible) REACTANTS Dilute acid CONDITIONS Heat under reflux Dilute alkali Heat under reflux PRODUCTS Carboxylic acid and alcohol Salt of acid and alcohol. Salt becomes carboxylic acid after acidification NITRILE REACTIONS NAME Hydrolysis REACTANTS Dilute acid CONDITIONS Heat Hydrolysis Dilute alkaline Heat PRODUCTS Carboxylic acid and ammonium salt Salt of acid and ammonia. Salt can be converted to acid through acidification. ARENE REACTIONS: NAME Halogenation Nitration REACTANTS X2 Concentrated HNO3 and concentrated H2SO4 Chloroalkane (CH3Cl) CONDITIONS Anhydrous AlX 3 catalyst Temperature between 25 o C and 60 o C under reflux AlCl 3 catalyst and heat PRODUCTS Aryl halide and HX Nitroarene Friedel-Crafts acylation Acyl chloride (CH3COCl) AlCl 3 catalyst and heat Acylbenzene and HCl Hydrogenation Hydrogen gas Heated with nickel or platinum catalyst Cyclohexane Friedel-Crafts alkylation Alkylbenzene and HCl ALKYLARENE REACTIONS (2-4-6 Directors): NAME Halogenation REACTANTS X2 CONDITIONS AlX3 Nitration Concentrated HNO3 and concentrated H2SO4 Temperature between 25 o C and 60 o C Complete oxidation Hot alkaline potassium manganate (VII) followed by dilute H2SO4 Hydrogen gas Hydrogenation Heated with nickel or platinum catalyst PRODUCTS Halogen is substituted on either the 2 nd or 4th carbon. If in excess, it goes in all positions in a varied fashion. Nitroarene where NO2 is attached to either the 2 nd or 4th carbon. Benzoic acid and water Alkyl cyclohexane PHENOL REACTIONS (2-4-6 Directors): NAME Acid-base reaction REACTANTS Base (NaOH) Acid-base reaction Reactive metals (Na) Diazonium ion Nitration HNO3 Bromination Bromine water CONDITIONS Alkaline solution PRODUCTS Soluble salt and water (sodium phenoxide if NaOH) Soluble salt (sodium phenoxide) and hydrogen Yellow-orange solution or ppt of azo compound Dissolve phenol in NaOH and cool it in ice before adding cold diazonium ion Room temperature Mixture of 2-nitrophenol and 4-nitrophenol if dilute HNO3 2,4,6-trinitrophenol if concentrated Room temperature 2,4,6-tribromophenol CARBOXYLIC ACID REACTIONS: NAME Redox CONDITIONS Neutralization Acid-base REACTANTS Reactive metals like sodium Alkali Carbonates Esterification Alcohols Concentrated sulfuric acid catalyst and heated under reflux Reduction LiAlH4 Solid PCl 5 Liquid PCl 3 Heat Liquid SOCl 2 HCOOH with any oxidizing agent HOOCCOOH with warm acidified KMnO4 Oxidation *Oxidation PRODUCTS Salt of acid and hydrogen Salt of acid and hydrogen Salt, water, and carbon dioxide Ester and water Primary alcohol and water Acyl chloride, POCl 3, HCl3 Acyl chloride and H 3PO3 Acyl chloride, SO2, HCl CO2 2CO2 + 2H2O ACYL CHLORIDE REACTIONS: NAME Hydrolysis Addition-elimination Addition-elimination Condensation Condensation REACTANTS H2 O Alcohol Phenol NH3 Amine CONDITIONS RTP RTP RTP RTP RTP PRODUCTS Carboxylic acid and HCl Ester and HCl Ester and HCl Amide and white fumes of HCl Amide and HCl which reacts with unreacted amine to form white organic ammonium salt PHENYLAMINE REACTIONS: NAME Bromination REACTANTS Aqueous bromine HNO2 and HCl CONDITIONS Room temperature Keep below 10 o C. Produce HNO2 by reacting NaNO2 and HCl PRODUCTS 2,4,6-tribromephenylamine Diazonium salt and water, which can be heated to give phenol. AMIDE REACTIONS: NAME Hydrolysis REACTANTS Aqueous acid or alkali Reduction LiAlH4 CONDITIONS Reflux PRODUCTS Carboxylic acid and: 1) NH3 if nonsubstituted amide 2) Primary amine if substituted amide In excess base, the acid will form CO2- ion In excess acid, NH3 will form a salt Nonsubstituted amide gives primary amine and substituted amide gives secondary amine. Both give water [ PREPARATION OF SOME ORGANIC COMPOUNDS Production of Amines: NAME Nucleophilic substitution REACTANTS CONDITIONS Halogenoalkanes with NH 3 Ethanol heated under pressure PRODUCTS Primary amine and NH4Cl Nucleophilic substitution Halogenoalkanes with primary amine Reduction Amides and LiAlH4 Reduction Nitriles with LiAlH 4 Heat in a sealed tube with ethanol under pressure Dry ether or H2 and Ni catalyst Secondary amine and HCl Primary amide forms primary amine, secondary amide forms secondary amine Primary amine Production of Benzoic Acid: 1) 2) 3) 4) Heat alkylbenzene under reflux with hot alkaline KMNO 4. The purple color of the Mn7+ ions disappears as they are reduced to Mn 4+ ions. Brown ppt of MnO2 forms. Add dilute acid to protonate the organic product formed. The result is benzoic acid. Production of Phenylamines: 1) Nitration of benzene to form nitrobenzene 2) Reduce with hot Sn/concentrated HCl 3) Add aqueous NaOH Production of Diazonium (Azo) Compounds: 1) First, a diazonium salt must be formed. For that, you need HNO2. This is produced in a test tube using NaNO2 and HCl. 2) Nitrous acid is reacted with a phenylamine to form a diazonium salt at temperatures below 10o C. 3) The diazonium salt reacts with phenol in NaOH in a coupling reaction and substi tutes in at the 4 th position of the benzene ring to form the azo compound. POLYMERIZATION: Production of Polyesters: 1) Can take place with the following reactants: diol and dicarboxylic acid, diol and dioyl chloride, hydroxycarboxylic acid. 2) Alcohols lose OH group and carboxylic acids lose H from COOH when reacting together. 3) Alcohols lose H from O-H and acyl chloride group loses Cl when reacting together. 4) Ester linkage is O-C=O. Production of Polyamides: 1) Can take place when diamines react with either dicarboxylic acids or with dioyl chlorides or when aminocarboxylic acids or amino acids react with each other. 2) Amine group loses H from N-H, carboxylic group loses OH from COOH, acyl chloride loses Cl from COCl. 3) Amide linkage is O=C-NH. Degradation of Polymers: 1) Polyalkenes are not biodegradable as they’re nonpolar and saturated. 2) Polyesters with a C=O group absorb UV light, weakening their bond and breaking down into smaller molecules. 3) Polyesters and polyamides can be hydrolyzed by acid, breaking down into their monomers. 4) Hydrolysis of polyesters and polyamides by alkalis produces salts of monomers. AMINO ACIDS 1) Amino acids contain both –COOH and –NH2 groups, so they are amphoteric. 2) Form zwitter ion with NH 3+ and COO- groups at isoelectric point. 3) Exist as soluble crystalline solids. 4) Act as buffer solutions. 5) When pH is lowered and acid is added, the COO- group accepts H+ to reform COOH group. 6) When pH is increased and alkali is added, NH3+ donates H+ to reform NH2 group. 7) Two amino acids react with each other to form a dipeptide with an amide linkage. 8) In electrophoresis, positively charged amino acids move towards the negative electrode and negatively charged amino acids move towards the positive electrode. 9) The greater the charge, the greater the movement. The greater the mass, the lesser the movement. MECHANISMS FOR ORGANIC REACTIONS Free Radical Substitution of Alkanes: INITIATION STEP: Homolytic fission of halogen molecule to form free radicals PROPAGATION STEP: These free radicals attack alkanes or other halogen molecules and turn them into free radicals. The cycle is then repeated. TERMINATION STEP: Two free radicals react with each other to form a single unreactive molecule. Electrophilic Addition of Alkenes: Step 1: The H-Br bond breaks heterolytically, forming a Br - ion and a H+ ion because bromine has a higher electronegativity. Step 2: The H+ ion attacks the organic compound and takes a pair of electrons from it, leaving a highly reactive carbocation which is the intermediate compound. Step 3: The bromide electrophile donates a pair of electrons and reacts with the carbocation to form an alkyl halide. The same process occurs when nonpolar halogen molecules react with alkenes, with one small difference: they must be polarized first before heterolytic fission occurs. When the halogen molecule comes close to the C=C bond, the halogen atom closer to the C=C bond gets a partial positive charge and the atom further away gets a partial negative charge, hence, heterolytic fission occurs. Nucleophilic Addition of HCN with Carbonyl Compounds: Step 1: Cyanide ion attacks carbonyl atom to form a negatively charged intermediate Step 2: The negatively charged oxygen atom in the reactive intermediate quickly reacts with aqueous H + (either from HCN, water or dilute acid) to form 2-hydroxynitrile SN1 MECHANISM (Tertiary Alkyl Halides): Step 1: C-X bond breaks heterolytically and the halogen leaves the halogenoalkane as an X - ion (this is the slow and rate-determining step) Step 2: Tertiary carbocation is attacked by the nucleophile Drawn twice second step should be the fast one SN2 MECHANISM (Primary Alkyl Halides): The nucleophile donates a pair of electrons to the δ+ carbon atom to form a new bond. At the same time, the C -X bond is breaking and the halogen (X) takes both electrons in the bond (heterolytic fission). The halogen leaves the halogenoalkane as an X - ion. Electrophilic Substitution of Arenes (Halogenation): 1) Halogen molecule forms a dative bond with AlX 3, producing X+ and AlX4-. 2) The electrophile attacks the benzene ring, receiving a pair of electrons from it. Since there are only four π electrons in the ring and the carbons have a net postive charge now, this disturbs the aromaticity of the molecule. 3) Heterolytic cleavage of C-H bond follows and the two electrons thus gained go into the benzene ring to restore its aromatic character. Electrophilic Substitution of Arenes (Nitration): 1) HNO3 reacts with H2SO4 to form an electrophile NO2+. 2) The electrophile attacks the benzene ring, receiving a pair of electrons from it. Since there are only four π electrons in the ring and the carbons have a net positive charge now, this disturbs the aromaticity of the molecule. 3) Heterolytic cleavage of C-H bond follows, and the two electrons thus gained go into the benzene ring to restore its aromatic character. Hydrolysis of Acyl Chlorides: 1) H2O acts as a nucleophile. The lone pair on the oxygen atom attacks the carbonyl carbon. 2) An HCl molecule is eliminated from the acyl chloride. Esterification of Acyl Chlorides with Alcohol: 1) Alcohol acts as nucleophile. The lone pair on the oxygen atom attacks the carbonyl carbon. 2) An HCl molecule is eliminated from the acyl chloride. Esterification of Acyl Chlorides with Phenol: 1) Phenol reacts with the base (NaOH) to form a phenoxide ion which is a better nucleophile than phenol. 2) The phenoxide ion attacks the carbonyl carbon of acyl chloride. 3) An NaCl molecule is eliminated from the acyl chloride. Formation of Amides From Acyl Chloride: 1) The nitrogen atom acts as a nucleophile and attacks the carbonyl carbon. 2) An HCl molecule is eliminated from the acyl chloride Reaction of Acyl Chloride with Ammonia: Reaction of Acyl Chloride with Primary Amines: Reaction of Acyl Chloride with Secondary Amines: v ALKYL HALIDES MISCELLANEOUS: SN1 Mechanism SN2 Mechanism Tertiary alkyl halides Two steps Rate of reaction depends on concentration of alkyl halide only Primary alkyl halides One step Rate of reaction depends on concentration of both alkyl halide and nucleophile Carbocation forms Polar solvent No carbocations formed Non polar solvent SHAPE OF BENZENE: 1) Regular planar compound with bond angles of 120o. 2) All bonds are the same length. 3) The carbons are all sp 2 hybridized. Two of their p orbitals mix with an s orbital, and one p orbital remains. 4) Neighboring p orbitals overlap sideways with each other to form pi bonds, leading to electrons being delocalized and able to spread over the entire ring. 5) Sigma bonds are formed by end to end overlap of sp orbitals. Each carbon forms 3 sigma bonds. REACTIVITY OF HALOGENOARENES: 1) Very unreactive compared to halogenoalkanes. 2) This is because of difference in C-X bond strength. 3) One lone pair of the halogen is delocalized over the benzene ring, strengthening the C-X bond and making it unreactive. PHENOLS MISCELLANEOUS: 1) There are different conditions and reagents for the nitration/bromination of phenol and benzene. 2) This is because phenols have a delocalized pair of electrons from the oxygen atom on the benzene ring, increasing reactivity. RELATIVE PROPERTIES OF VARIOUS COMPOUNDS: RELATIVE ACIDITIES OF WATER, ETHANOL, AND PHENOL: 1) Order of acidity can be explained by looking at relative stability of each compound’s conjugate base. 2) In a phenoxide ion, the charge density is evenly spread throughout the compound, making it less attractive to a proton. 3) In an ethoxide ion, there is an electron-donating ethyl group attached to oxygen, making the compound more attractive to a proton. 4) There are no electron-withdrawing or donating groups in a hydroxide ion. 5) Hence, phenol is the strongest acid, followed by water, followed by ethanol. RELATIVE ACIDITIES OF CARBOXYLIC ACIDS, PHENOLS, AND ALCOHOLS: 1) In carboxylic acids, the C=O group attracts the electron density in the O-H bond, weakening the O-H bond. 2) The charge on a carboxylate ion is spread out through the O=C-O group, making it less available for a proton. 3) In an alkoxide ion (conjugate base of alcohol), there is electron-donating alkyl group, making it more available for a proton. 4) In a phenoxide ion, the charge on the oxygen atom is spread throughout the ion, making it less available for a proton. 5) Hence, carboxylic acids are the most acidic, followed by phenols, followed by alcohols. RELATIVE ACIDITIES OF CHLORINE SUBSTITUTED CARBOXYLIC ACIDS: 1) Chlorine is an electronegative atom which attracts electron density in the O-H bond and stabilizes the carboxylate ion by delocalizing its negative charge. 2) Hence, the more the chlorine atoms in the compound, the stronger the acid. RELATIVE EASE OF HYDROLYSIS OF ACYL CHLORIDES, ALKYL CHLORIDES, AND HALOGENOARENES: 1) In acyl chlorides, the carbonyl carbon is attached to two electronegative atoms (O and Cl), making it electron deficient and weakening the C-Cl bond. 2) In alkyl chlorides, the carbonyl carbon is only attached to one electronegative atom (Cl), so it is less electron deficient and the C-Cl bond is a bit stronger. 3) In halogenoarenes, one of the lone pairs on Cl delocalizes over the benzene ring, making the CCl bond very strong. 4) Hence, acyl chlorides are the easiest to hydrolyze, followed by alkyl chlorides, followed by halogenoarenes. RELATIVE BASICITIES OF AMINES: 1) The nitrogen in alkylamines is connected to an electron donating group, making it more available for a proton. 2) The nitrogen in phenylamines is connected to an electron withdrawing group, making it less available for a proton. 3) Hence, alkylamines are more basic than phenylamines RELATIVE BASICITIES OF AMINES AND AMIDES: 1) The nitrogen in amides loses some electron density to the electronegative oxygen atom, making it less available for a proton. 2) This means they are much weaker bases than amines TESTS FOR ORGANIC COMPOUNDS TEST FOR Saturation (C=C bond) Halide ion TEST NAME TEST REAGENTS TEST PRODUCTS AND CONDITIONS Alkene with bromine water in absence of UV light Alkyl halide with aqueous silver nitrate in ethanol POSITIVE TEST OBSERVATIONS Bromine water decolorized OH group Sodium with organic compound 0.5 moles of hydrogen gas per OH group and salt of acid/alcohol White ppt if chlorine, offwhite/creamy ppt if bromine, yellow ppt if iodine Fizzing and bubbles which make pop sound with lighted splinter OH group PCl5 with compound Alkyl halide, HCl, and POCl 3 Misty white fumes of HCl Tertiary alcohols K2Cr2O7 with alcohol CH3CO group (methyl ketones, ethanal, some secondary alcohols) Carbonyl group Iodoform Aldehyde Tollen’s reagent Aldehyde Fehling’s solution Carboxylic acid Iodine with compound Yellow ppt of triin alkaline solution iodomethane and heated salt of acid 2,4-DNPH with compound Silver nitrate in excess ammonia with carbonyl compound Alkaline solution of copper ions with carbonyl compound Magnesium with compound No change, orange solution stays orange Yellow ppt Deep orange ppt Salt of carboxylic acid and silver atoms Salt of carboxylic acid and copper (I) oxide Salt of acid and hydrogen Silver mirror due to silver atoms Opaque red ppt of copper (I) oxide Fizzing and bubbles which make pop sound with lighted splinter Iodoform Test for Carbonyl Compounds: 1) Test for CH3CO group (methyl ketones and ethanal). 2) Compound is reacted with iodine in heated alkaline solution. 3) All three H-atoms in the -CH3 (methyl) group are replaced with iodine atoms, forming a -CI3 group. 4) The intermediate compound is hydrolyzed by the alkaline solution to form a sodium salt (RCO2- Na+) and a yellow precipitate of CHI3. OPTICAL ISOMERISM: Enantiomer Properties: 1) Compounds with chiral centers are called enantiomers. 2) They have identical physical and chemical properties. 3) They rotate plane polarized light in opposite directions: if one enantiomer rotates it clockwise, the other will rotate it anticlockwise. 4) Enzymes are chiral proteins that speed up chemical reactions. They have a specific binding/active site and only bind to molecules with the exact same shape. 5) Therefore, only one enantiomer of a compound can bind to an enzyme. Optically Active Mixtures: 1) When a solution has different amounts of each enantiomer, it will rotate the plane of polarized light. 2) Such mixtures are called optically active mixtures. 3) Mixtures with equal amounts of each enantiomer are called racemic mixtures and do not rotate the plane of polarized light as the enantiomers will cancel out each other’s effects. 4) The direction in which polarized light is rotated can be used to identify which enantiomer is present in a sample. Chirality and Drug Preparation: 1) As different enantiomers bind to different enzymes, it is possible that one enantiomer can act as a medicine and the other can harm health. 2) Hence, it is important to be able to separate both enantiomers. 3) Chiral catalysts can be used to separate enantiomers. Only small amounts are needed, and they can be reused. 4) Enzymes can be used as chiral catalysts. 5) Enzymes are placed on inert supports during the reaction so reactants have to pass over them without any byproducts being produced. 6) Due to the specific binding site of enzymes, only one enantiomer is produced in the reaction. CHEMICAL ENERGETICS: Effect of Ionic Charge and Radius on Magnitude of Lattice Energy: 1) As ionic radius increases, charge density decreases, and the ions become further apart. 2) Thus, the electrostatic forces of attraction are weaker and magnitude lattice energy decreases as it becomes less exothermic. 3) As ionic charge increases, charge density increases, and there is a stronger electrostatic force of attraction between the ions. 4) Hence, the magnitude of lattice energy increases and it becomes more exothermic. Effect of Ionic Charge and Radius on Magnitude of ΔHhydꝋ: 1) As ionic radius increases, charge density decreases, and the ion-dipole attractions between the ion and water molecules decrease. 2) Hence, ΔHhydꝋ becomes less exothermic and its magnitude decreases. 3) As ionic charge increases, charge density increases, and the ion-dipole attractions between the ion and water molecule increase. 4) Hence, ΔHhydꝋ becomes more exothermic and its magnitude increases. Factors Affecting Electron Affinities : 1) The greater the nuclear charge, the greater the attractive forces between nucleus and incoming electron. 2) The greater the atomic radius, the lesser the attractive forces between nucleus and incoming electrons. 3) The greater the number of electron shells, the greater the shielding effect, and the lesser the attractive forces. 4) The greater the attractive forces, the more exothermic the electron affinity. Electron Affinities of Group 16 and Group 17: 1) Generally become less exothermic down the group. 2) Fluorine is an exception. It is less exothermic than chlorine because it has a very small atomic radius, so its electron density is high, so there are some repulsive forces involved as well. GROUP II METALS THERMAL STABILITY OF GROUP II COMPOUNDS DOWN THE GROUP: 1) Thermal stability increases down the group. 2) This is because ionic radius increases down the group. 3) Hence, charge density decreases down the group. 4) Hence, polarizing effect of the cation decreases down the group. 5) Hence, anion is less polarized down the group and thermal stability increases. VARIATION IN SOLUBILITY OF HYDROXIDES DOWN THE GROUP: 1) Solubility increases down the group. 2) ΔH sol ꝋ = ΔH hydꝋ — ΔH lattꝋ 3) Down the group, ΔH hydꝋ and ΔH lattꝋ both decrease in magnitude. 4) However, ΔH lattꝋ decreases quicker than ΔH hydꝋ so ΔHsol ꝋ becomes exothermic down the group and the compounds become more soluble. VARIATION IN SOLUBILITY OF SULFATES DOWN THE GROUP: 1) Solubility decreases down the group. 2) ΔH sol ꝋ = ΔH hydꝋ — ΔH lattꝋ 3) Down the group, ΔH hydꝋ and ΔH lattꝋ both decrease in magnitude. 4) However, ΔH hydꝋ decreases quicker than ΔH lattꝋ so ΔHsol ꝋ becomes more endothermic down the group and the compounds become less soluble. 5) Key to remembering: ΔH hydꝋ is the negative term while – ΔHlattꝋ is the positive term. Also, solubility increases when ΔH sol ꝋ is more negative. So when solubility increases, ΔHsol ꝋ is getting more negative, which means negative term (ΔH hydꝋ) must be decreasing less and vice versa. TRANSITION METALS Orbital Shapes: Properties of Transition Metals: 1) Variable oxidation states. 2) Behave as catalysts. 3) Form complex ions. 4) Form colored compounds. Variable Oxidation States: 1) Transition metals lose electrons from both 3d and 4s subshells due to their similar energy levels. 4s electrons are lost first. 2) At the start of the period, transition elements can lose all the electrons in the 3d and 4s subshells. 3) Across the period, nuclear charge increases, and it becomes harder to remove electrons. 4) Thus, it becomes more likely that only 4s electrons are removed, making the 2+ oxidation state more stable. Ability to Act As Catalysts: 1) Having variable oxidation states means that transition metals can accept and lose electrons easily. 2) This is because transition metal ions have vacant d orbitals which are not of high energy and are thus available to make dative bonds with ligands. 3) Thus, they catalyze redox reactions by acting as both an oxidizing agent and a reducing agent. 4) The higher the oxidation state, the stronger the oxidizing agent. Ability To Form Complex Ions: 1) Transition metal ions have vacant d orbitals that do not have high energy levels. 2) These vacant orbitals can accept pairs of electrons from ligands and form dative bonds. Coordination Numbers and Structures: 1) Coordination number is the number of dative bonds formed by a metal ion. 2) Coordination number of 6 means the structure of the complex is octahedral with bond angles of 90o. 3) Coordination number of 4 means the structure of the complex is either tetrahedral with bond angles of 109.5 o or square planar with angles of 90 o. 4) Coordination number of 2 means the structure of the complex is linear with bond angles of 180o. 5) Chloride ions form tetrahedral structures with a charge of –2. 6) Cyanide ions form square planar structures. 7) Hydroxide ions and water molecules form octahedral structures with a charge of 0. Ligands: 1) A molecule or ion that has one or more lone pair s of electrons. 2) Monodentate ligands (such as H 2O, NH3, Cl-, and CN-) have one lone pair and form one dative bond. 3) Bidentate ligands (such as H2NCH2CH2NH2 and C2O42- ) have two lone pairs and form two dative bonds. 4) Polydentate ligands such as EDTA4- form more than three dative bonds. Copper Complex Ion Reactions: Initial Copper Compound [Cu(H2O)6]2+(aq) Final Products [Cu(H2O)4(OH)2](s) + 2H2O(l) [Cu(H2O)4(OH)2](s) Reactant/Condition NaOH drop by drop OR excess conc NH3 drop by drop Excess concentrated NH 3 [Cu(H2O)6]2+(aq) Excess concentrated HCl [Cu(Cl)4]2-(aq) + 6H2O [Cu(H2O)2(NH3)4]2+(aq) + 2H2O + 2OH- Cobalt Complex Ion Reactions: Initial Copper Compound Reactant/Condition Final Products [Co(H2O)6]2+(aq) [Co(H2O)4(OH)2](s) + 2H2O(l) [Co(H2O)6]2+(aq) NaOH drop by drop OR excess conc NH3 drop by drop Excess concentrated NH 3 [Cu(H2O)6]2+(aq) Excess concentrated HCl [Co(Cl)4]2-(aq) + 6H2O Orbital Splitting: 1) Initially, a transition metal ion has five degenerate d orbitals. [Co(NH3)6]2+(aq) + 6H2O oxidized in air to [Co(NH 3)6]3+ 2) These orbitals split into two energy levels when the ion bonds with a ligand. 3) This difference is caused by the repulsion from the ligand’s lone pair. 4) In an octahedral complex, the ligands approach the ion along the x, y, and z axes and the 3dx2-y2 and 3dz2 are located along these axes. Hence, there is more repulsion in these two orbitals, and they are at a higher energy level than the others. 5) In a tetrahedral complex, the ligands approach the ion such that they are lined up with the 3dyz, 3dxz, and 3dxy orbitals. Hence, there is more repulsion in these two orbitals, and they are at a higher energy level than the others. 6) The difference in energy is ΔE. Forming Colored Complex Compounds: 1) When light passes through a transition metal complex, an electron in the set of orbitals with less energy absorbs ΔE. 2) ΔE = hv, where h = Planck’s constant and v = frequency. 3) The electron uses this energy to jump into a higher energy orbital (electron promotion). 4) The unabsorbed frequencies of light combine to make the complementary color which is observed. Effects of Ligands on Complementary Colors: 1) ΔE depends on the repulsion from a ligand’s lone pairs, which will vary even if the geometry and charge of the ion are the same. 2) Frequency of light absorbed depends on ΔE, so this will also vary, and hence the complementary color varies as well. Copper Ion Colors as Examples: 1) [Cu(H2O)6]2+ and [Cu(H2O)4(OH)2] are light blue in color. 2) [Cu(NH3)4(H2O)2]2+ is deep blue in color. 3) [Cu(Cl)4 ]2- is yellow in color. Cobalt Ion Colors as Examples: 1) [Co(H2O)6]2+ is pink in color. 2) [Co(NH3)6]2+ is brown in color. 3) [Co(Cl)4]2- and [Co(H 2O)4(OH)2] are blue in color. Reaction of MnO4- & Fe2+ in acid: 1) The purple color of the manganate(VII) ions disappears during the reaction. 2) When the reaction is over, a permanent pink color appears. 3) Equation is MnO 4- (aq) + 8H + (aq) + 5Fe2+ → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l). Reaction of MnO4- & C2O42- in acid: 1) Ethanedioate ions are oxidized by manganate (VII) ions. 2) Endpoint is when a permanent pink color appears. 3) Equation is 2MnO4-(aq) + 16H +(aq) + 5C2O42-(aq) → 2Mn2+(aq) + 8H 2O(l) + 10CO 2(g) Reaction of Cu2+ & I-: 1) When an excess of copper II and iodide ions react, precipitate of copper (I) iodide forms. 2) Equation is 2Cu2+ (aq) + 4I-(aq) → I2(aq) + 2CuI (s). Stereoisomerism In Complex Compounds: 1) All complexes which have at least two ligands which are different from the rest display cistrans geometrical isomerism. 2) Octahedral complexes with bidentate ligands display optical isomerism. 3) Cis isomers display some polarity while trans isomers are nonpolar. Stability Constant: 1) Expression is similar to Kc, just ignore water as it’s in excess. 2) The larger the Kstab, the more stable the complex formed. ELECTROCHEMISTRY Determining Avogadro’s Constant Through Electrolysis: 1) Set up an electrolytic cell with an electrolyte copper II sulfate and copper electrodes. 2) Weigh both electrodes initially. 3) Keep a constant current of approximately 0.17A for some time. 4) Remove the electrodes, wash with distilled water, dry with propanone, and reweigh. 5) The cathode should have increased in mass while the anode should have decreased. 6) Find total charge that was passed through and divide it by moles of copper deposited on cathode. This gives charge needed to deposit 1 mole of copper. 7) Divide this by 2 to find charge on 1 mole of electrons and then divide it by the elementary charge to find L which is (charge of 1 mol electrons) / (charge of 1 electron). Standard Hydrogen Electrode: 1) A half cell used as a reference electrode. 2) Consists of hydrogen gas in equilibrium with H + ions of concentration 1 mol dm -3. 3) Inert platinum electrode in contact with both the gas and the ions. Metals In Contact With Their Aqueous Solution: Nonmetals In Contact With Their Aqueous Solution: Ions With Variable Oxidation States: Variation of E Under Non Standard Conditions: 1) If the concentration of the substance on the left increases, equilibrium shifts to the right, meaning reduction becomes easier and the E value becomes more positive/less negative. 2) If the concentration of the substance on the right increases, equilibrium shifts to the left, meaning reduction becomes harder and the E value becomes less positive/more negative. EQUILIBRIA Buffer Solution Working: 1) It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. 2) When acid is added, it reacts with and is neutralized by the base, producing more of the acid in the initial mixture. 3) As the concentration of the acid in the initial mixture is already large, this doesn’t affect concentration and hence pH much. 4) When base is added, it reacts with and is neutralized by the acid, producing water and more of the base in the initial mixture. 5) As the concentration of the base in the initial mixture is already large, this doesn’t affect concentration and hence pH much. Role of HCO3- in Controlling Blood pH: 1) Solution is set up by CO 2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq). 2) If there’s an increase in H + ions: H+ (aq) + HCO3- (aq) → CO2 (g) + H2O (l) 3) If there’s a decrease in H + ions: CO2 (g) + H2O (l) → H+ (aq) + HCO3- (aq) Common Ion Effect: 1) When you add more of an ion in a saturated solution, the concentration of that ion increases. 2) This means equilibrium moves towards the left, resulting in ppt of the solute. Factors Affecting Partition Coefficient: 1) If the solute is nonpolar, it will be more soluble in the organic solvent, resulting in larger Kpc. 2) If the solute is polar, it will be more soluble in water, resulting in smaller Kpc. REACTION KINETICS: Factors Affecting Rate Constant And Reaction Rate: 1) At a higher temperature, more molecules have the activation energy required to break bonds. 2) Thus, rate increases and so does rate constant when temperature increases. Heterogeneous Catalysts: Iron in the Haber Process: 1) The nitrogen and hydrogen gas are diffused to the iron surface. 2) The reactants are adsorbed on to the iron surface and bond with it. 3) These bonds weaken the covalent bonds in the gases, which then react with each other. 4) The bond between the produced ammonia and the iron surface then weakens, resulting in desorption and eventually diffusion of ammonia. Heterogeneous Catalysts: Catalytic Removal of Oxides of Nitrogen: 1) The catalyst can be palladium, platinum, or rhodium. 2) The reacting nitrogen oxides and carbon monoxide are adsorbed on to the catalyst surface. 3) The covalent bonds in the nitrogen oxide and carbon monoxide weaken. 4) Adjacent nitrogen atoms form nitrogen gas while carbon monoxide and oxygen atoms form carbon dioxide. 5) The nitrogen and carbon dioxide molecules desorb and eventually diffuse away from the surface. Homogeneous Catalysts: Oxidation of Sulfur Dioxide: 1) Sulfur dioxide in air oxidizes to sulfur trioxide, which eventually forms dilute sulfuric acid with the equation SO 3(g) + H2O(l) → H2SO4(aq). 2) Nitrogen helps in oxidation of sulfur dioxide: NO 2(g) + SO2(g) → SO3(g) + NO(g). 3) It is then regenerated: NO(g) + ½ O 2(g) → NO2(g). Homogeneous Catalysts: The Iodine-Peroxydisulfate Reaction: 1) The overall equation for this is S2O82- (aq) + 2I- (aq) → 2SO42- (aq) + I2 (aq). 2) Since both of the reacting ions are negatively charged, it would require a lot of energy to bring them close together to react with each other. 3) Therefore, iron III ions are used as a homogeneous catalyst. 4) Reduction of iron III: 2Fe3+ (aq) + 2I- (aq) → 2Fe2+ (aq) + I2 (aq) 5) Regeneration of iron III: 2Fe2+ (aq) + S2O82- (aq) → 2Fe3+ (aq) + 2SO42- (aq) ANALYTICAL CHEMISTRY: Thin Layer Chromatography: 1) The stationary phase is Al2O3 or SiO2 (coated on a thin metal sheet.) 2) The mobile phase is a polar or nonpolar solvent. 3) The greater the solubility of the compound in the mobile phase, the greater the Rf value. Gas Liquid Chromatography: 1) The stationary phase is a high boiling point nonpolar nonvolatile long chain hydrocarbon liquid. 2) The mobile phase is an inert gas like nitrogen. 3) Area under graph shows relative concentration of each peak. Use height if can’t calculate area. 4) The longer the retention time, the less polar the component. NMR Spectroscopy: 1) TMS is used as a reference compound as it is inert and volatile. It gives a single sharp peak at 0 ppm. 2) CDCl3 is used as a solvent in proton NMR to avoid any extra peaks. 3) D2O substitutes protons in O-H and N-H and removes their peaks. DEFINITIONS Enthalpy Change of Atomization: Enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions. Lattice Energy: Enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions (under standard conditions). First Electron Affinity: Enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms, to form 1 mole of gaseous ions each with a single negative charge under standard conditions. Enthalpy Change of Solution: Enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form a very dilute solution. Enthalpy Change of Hydration: Enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution. Entropy: Number of possible arrangements of the particles and their energy in a given system. Standard Electrode Potential: The voltage produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions. Standard Cell Potential: The voltage of an electrochemical cell made up of two half-cells. Conjugate Acid-Base Pairs: A pair of reactants and products that are linked to each other by the transfer of a proton. Buffer Solution: A solutions which resists change in pH upon addition of acid or alkali. Partition Coefficient: The ratio of the concentrations of a solute in two different immiscible solvents in contact with each other when equilibrium has been established (at a particular temperature). Order of Reaction: The power to which the concentration of that reactant is raised in the rate equation. Overall Order of Reaction: Sum of the powers of the reactants in a rate equation. Half Life: The time taken for the concentration of a limiting reactant to become half of its initial value. Rate Determining Step: Slowest step in a reaction. Intermediate: A substance derived from substances that react together to form it in the ratedetermining step. Transition Element: A d-block element which forms one or more stable ions with incomplete d orbitals. Ligand: A species that contains a lone pair of electrons that forms a dative covalent bond with a central metal ion. Coordinate Number: The number of dative bonds formed by a metal ion. Complex: A molecule or ion formed by a central metal ion surrounded by one or more ligands. Degenerate d orbitals: d orbitals at the same energy levels. Non-degenerate d orbitals: d orbitals at different energy levels. Stability Constant: Equilibrium constant for the formation of the complex ion in a solvent from its constituent ions or molecules. Racemic Mixture: A mixture with equal amounts of each enantiomer. Optically Active Mixture: A mixture with unequal amounts of each enantiomer which rotates the plane of polarized light. FORMULA SHEET Chemical Energetics: Born Haber Cycle: H fꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔH lattꝋ 2) ΔHsolꝋ = -ΔHlattꝋ + ΔHhydꝋ 1) Entropy: 1) 2) ΔSsystemꝋ = ΣSproductsꝋ - ΣSreactantsꝋ ΔGꝋ = ΔH reactionꝋ - TΔSsystemꝋ Electrochemistry: 1) F = Le 2) Ecellꝋ = Ereductionꝋ – Eoxidationꝋ 3) Nernst Equation: E = E ꝋ + (0.059/z) log 10 (oxidized species/reduced species) 4) Nernst Equation: E = E ꝋ + (RT/zF) ln (oxidized species/reduced species) 5) ΔGꝋ = - n x Ecellꝋ x F Equilibria: 1) pH = -log10 [H+] 2) Ka = [H+]2 / [HA] 3) pKa = -log10 Ka 4) Kw = [H+] [OH-] 5) Weak acid: [H+] = √(Ka*HA) 6) Buffer solution: pH = pKa + log10 ([salt]/[acid]) 7) Ksp = [Ax+ (aq)]a [By-(aq)]b Reaction Kinetics: 1) Rate = Δconcentration/ Δtime 2) Rate = k [A]m [B]n 3) T1/2 = 0.693/k Analytical Chemistry: 1) Rf value = distance travelled by component / distance travelled by solvent 2) %age composition of mixture in GL chromatography = area under substance / total area 3) Mass spectrometry: 100/1.1 * (M+1 peak)/(M peak)
