Reactions And Conditions (Oragnic 1transition & Redox) Flashcards
Conditions and product for alkene oxidation
-KMnO4 in cold, aqueous, alkaline solution
-Diol
Conditions and product of Acid-catalysed hydration of an alkene
-use of steam
-300 degrees Celsius
-60atm
-phosphoric acid catalyst
-alcohol produced
Conditions and product of electrophilic addition of hydrogen halides to an alkene
-gaseous reagent bubbled into liquid alkene
-Halogenoalkane produced
Conditions and product of hydrogenation of an alkene
-Hydrogen gas as a reagent
-150 degrees Celsius
-nickel catalyst
-alkane produced
Conditions and product and nucleophilic substitution of Halogenoalkanes with cyanide
-KCN or NaCN as a reagent
-in ethanol
-reflux
-forms a cyanide/ nitrile
Conditions and product of nucleophilic substitution of Halogenoalkanes with ammonia
-concentrated ammonia in ethanol
-heat in a sealed tube
-excess ammonia to prevent further substitution
-forms an amine
Conditions and product of hydrolysis of Halogenoalkanes with aqueous hydroxide ions
-aqueous NaOH or KOH as reagent
-heat under reflux
-forms an alcohol
Half equations showing that reaction between ethandioate ions and manganate (VII) ions is autocatalytic
Mn2+ is catalytic because it reacts with MnO4- to make Mn3+.
4Mn2+(aq) + MnO4-(aq) + 8H+(aq) → 5Mn3+(aq) + 4H2O(l)
The Mn3+ then reacts with the C2O42-:
2Mn3+(aq) + C2O42-(aq) → 2Mn2+(aq) + 2CO2(g)
The Mn2+ is regenerated in this step
Show how iron can be used a catalyst in the reaction between iodide ions and peroxodisulphate ions
The uncatalysed reaction between iodide ions and persulfate ions is very slow because it involves the reaction between two negative ions. Repulsion between the ions causes a high activation energy.
The reaction is catalysed by Fe²⁺ ions.
In the first stage the Fe²⁺ ions are oxidised to Fe³⁺ ions as they react with persulfate ions, which are reduced to sulfate ions.
Fe²⁺ (aq) + S₂O₈²⁻(aq) → Fe³⁺ (aq) + SO₄²⁻(aq)
Fe³⁺ ions are reduced by iodide ions, to reform Fe²⁺ ions. Iodide ions are oxidised to iodine.
Fe³⁺(aq) + 2I⁻ (aq) → Fe²⁺(aq) + I₂
The overall reaction is:
S₂O₈²⁻(aq) + 2I⁻ (aq) → SO₄²⁻(aq) + I₂
The Fe²⁺ ions are reformed to be used again.
Bidentate ligands
-en (H2NCH2CH2NH2) ethylenediamine
-C2O4^2- ethanedioate
Multidentate ligand
-EDTA
- 4- charge
Oxidation of alcohols
-reagent= potassium dichromate (K2Cr2O7)
-reagent is acidified in sulphuric acid
-heat
-produces aldehydes (primary) or ketones (secondary)
Colour change for potassium dichromate
Things that can be oxidised will colour change from orange to green solution
Procedure to produce and extract a ketone
-produced with acidified potassium dichromate heated under REFLUX with alcohol
-separated with distillation
Procedure to produce an aldehyde
-acidified potassium dichromate heated under distillation with primary alcohol
Procedure to produce a carboxylic acid
-acidified potassium dichromate heated under reflux with alcohol
-separated via distillation
Conditions and product of elimination of a haloalkanes
-NaOH or KOH dissolved in ethanol (no water present)
-heated
Forms an alkene, water, and halogen ion
Colour of [Co(H2O)6]2+
Pink solution
Colour of [Co(OH)2(H2O)4]
Blue precipitate
Colour of [Co(NH3)6]2+
Orange solution.
Conditions and reactant to produce a chloroalkane from an alcohol
-Nucleophilic substitution mechanism (OH is replaced by a chlorine, forming a chloroalkane)
-Reagent=PCl5
-produces HCl (white steamy fumes) and a chloroalkane
Conditions and reactant to form a bromoalkane from an alcohol
A reaction mixture of 50% concentrated sulphuric acid and KBr (The KBr reacts with sulphuric to form HBr which reacts with the alcohol)
Forms a bromoalkane and water
Reactants and conditions to form an iodoalkane from an alcohol
A mixture of red phosphorus with iodine (the phosphorus reacts with the iodine to form PI3 which reacts with the alcohol)
Forms and iodoalkne and phosphoric acid
Half Equations for alkaline fuel cell
2H2 + 4OH- —> 4H2O + 4e-
O2 + 2H2O + 4e- —> 4OH-
