CHEM219/ CHEM 219 Module 6 – Principles of Organic Chemistry with Lab | Portage Learning | Updated 2026–2027 | Complete Questions & Verified Answers | Grade A

CHEM219/ CHEM 219 Module 6 – Principles of Organic Chemistry with Lab | Portage Learning | Updated 2026–2027 | Complete Questions & Verified Answers | Grade A 2026 / 2027 Academic Year Q: kekulē model Answer suggested that atoms joined by alternate single and double bonds Q: evidence against kekule Answer 1. lack of reactivity of benzene 2. length of bonds 3. hydrogenation enthalpies Q: lack of reactivity Answer does not undergo electrophilic addition reactions does not decolourise bromine Q: delocalised model of benzene Answer 1. planar, cyclic, hexagonal hydrocarbon 2. overlapping of p-orbitals creates a system of pi-bonds 3. electrons are delocalised Q: phenol acts as... Answer a weak acid Q: phenol vs benzene Answer phenol is more reactive due to the lone pair of electrons from the oxygen p-orbital being donated into the pi-system the electron density of the ring is increased Q: oxidation of aldehydes Answer qto carboxylic acid Q: ketones and aldehydes undergo Answer nucleophilic addition Q: reducing agent Answer NaBH4 sodium tetrahydridoborate(iii) Q: ketone reduced to Answer secondary alcohol Q: aldehyde reduced to Answer primary alcohol Q: reactions of carbonyls with HCN Answer forms hydroxynitriles Q: testing for carbonyl group use 2,4-DNP Answer 1. ad 5cm depth of 2,4-DNP into a test tube 2. add three drops of unknown compound 3. no crystals formed, add a few drops of sulfuric acid 4. yellow/orange ppt indicated presence of a carbonyl Q: reaction with 2.4-DNP Answer condensation reaction Q: tollen's reagent Answer silver nitrate in aqueous ammonia, in presence of aldehyde a silver mirror is produced Q: testing for aldehyde group with Tollen's Answer 1. add 3cm of aq silver nitrate 2. add aq sodium hydroxide until a brown ppt is formed 3. add dilute ammonia until brown ppt dissolves 4. add unknown solution to clean tube 5. add equal vol of tollens Q: silver ions Answer act as an oxidising agent in presence of ammonia, reduced to silver Q: carboxylic acids are classified as... Answer weak acids Q: redox of carboxylic acid and metals Answer form hydrogen gas and carboxylate salt, metal dissolves and effervescence Q: neutralisation of COOH and metal oxides Answer salt and water Q: tests for carboxyl group Answer neutralisation with carbonates Q: naming esters Answer alcohol then carboxylic acid Q: acid anhydride Answer formed from removal of water from two carboxyls Q: esterification Answer alcohol and carboxylic acid to form an ester Q: acid hydrolysis of esters Answer heated under reflux with dilute aq acid produce carboxylic acid and alcohol Q: alkaline hydrolysis of esters Answer reflux with aq hydroxide ions form carboxyl ion and alcohol Q: acyl chloride prep Answer thionyl chloride Q: acyl chloride and water Answer makes carboxylic acid Q: ammonia and amines act as Answer nucleophiles by donating the lone pair of electrons on the nitrogen to an electron-deficient species Q: ammonia and acyl chloride Answer primary amide Q: primary amine and acyl chloride form secondary amide Q: amines behave as bases as the lone pair of electrons can accept a proton Q: amines and salt formation they are bases and they neutralise acids to make salts Q: ammonia in substitution reactions lone pair on the nitrogen atom allowing ammonia to act as a nucleophile Q: amine formation salt and sodium hydroxide Q: for amine formation, essential conditions; 1. ethanol as the solvent preventing further substitution to produce alcohols 2. excess ammonia reduces further sub of amine to form secondary and tertiary amines phenylamine produced reduction of nitrobenzene, heated under reflux with tin and hydrochloric acid esterification of amino acids heating with an alcohol in presence of conc sulfuric acid zwitterions dipolar ion exists at certain pH melting and boiling point will be higher isoelectric point the pH at which the zwitterion is formed amides produced Answer reactions of acyl chlorides with ammonia and amines chiral centre Answer carbon atom that is attached to four different atoms or groups of atoms enantiomers Answer non-superimposable mirror image structures (optical isomers) condensation polymerisation Answer joining of monomers with the loss of a small molecule, usually water polyester Answer monomers are joined together by ester linkages in a long chain to form a polymer -carboxylic acid and alcohol example polyesters Answer terylene poly(lactic acid) terylene Answer monomers: benzene-1,4-carboxylic acid and ethane-1,2-diol uses: clothing and ropes alkali hydrolysis produces: alcohol and acid salt poly(lactic acid) Answer monomer: amino acid benefit: much more biodegradable uses: waste produce bags, internal stitches, disposable utensils polyamindes Answer condensation polymers formed when monomers are joined together by amide linkages -carboyxlic acid/acyl chloride and an amine polyamide examples Answer 1. nylon-6,6 2. Kevlar nylon-6,6 Answer monomers: hexandioic acid and hexan-1,6-diamine uses: clothing and fibres Kevlar Answer monomers:benzene-1,4-diamine and benzene-1,4-dicarboxylic acid uses: bulletproof technique for separating a solid product from a solvent or liquid reaction mixture Answer filtration under pressure filtration under pressure Answer 1. connect one end of the pressure tubing to the filter pump whilst attaching the other end to the Buchner flask 2. fit the buchner funnel to the buchner flask ensuring that there is a good tight fit 3. switch on the tap 4. check for good suction by placing your hand across the top of the funnel recrystallisation Answer the solid product obtained after filtration will contain impurities which can be removed by carrying out recrystallisation. how to recrystallise Answer 1. pour a quantity of the solvent into a conical flask, place over a bunsen burner 2. tip the impure sample into a second conical flask or beaker 3. slowly add the solvent to the impure sample until it dissolves in the solvent 4. allow solution to cool melting point determination Answer determines whether a solid compound is pure melting range Answer difference between the temp at which the sample starts to melt and the temp at which melting is complete impure sample melting point Answer melts over a wide range of temperatures using electrically heated melting point apparatus Answer 1. place the capillary tube containing the sample into a sample hole and thermometer in its hole 2. using rapid heating setting start to heat whilst observing 3. once the solid is seen to melt, record the melting point 4. prep a second sample and place in the melting point apparatus and heat again 5. at melting p, set to low and observe slowly capillary tube setting up Answer 1. sample completely dry and free flowing 2. take glass capillary tube, rotate in bunsen flame 3. fill with crystals melting point using oil bath method Answer 1. set up oil bath 2. attach capillary tube containing sample to thermometer 3. clamp thermometer, end of thermometer and tube should dip into the oil 4. use a micro-burner and record melting point chromatography Answer used to separate individual components from a mixture of substances stationary phase Answer does not move and is normally a solid or liquid supported on a solid mobile phase Answer does move, normally a liquid or gas applications of chromatography Answer analysis of drugs, plastics and forensic science TLC Answer Thin layer chromatography is a quick and inexpensive analytical technique that indicates how many components are in a mixture adsorbent Answer solid substance usually silica, the stationary phase adsorption Answer process by which the solid silica holds the different substances in the mixture to its surface retention factor Answer distance moved by component/ distance moved by solvent retention time Answer in gas chromatography, the time for a component to pass from the column inlet to the detector use of gas chromatography Answer separating and identifying volatile organic compounds present in a mixture gas chromatography stationary phase Answer high boiling liquid adsorbed onto an inert solid support gas chromatography mobile phase Answer inert carrier gas gas chromatography Answer 1. volatile mixture injected into apparatus 2. carrier gas caries components through capillary column 3. components slow down as they interact with the liquid stationary phase 4. the more soluble the component the slower it moved through the capillary column information obtained from a gas chromatogram Answer 1. retention times 2. peak integrations used to determine the concs of components in the sample test for alkene Answer add bromine water, decolourised from orange to colourless test for haloalkane Answer add silver nitrate and ethanol and warm to 50 test for carbonyl Answer add 2,4-dinitrophenylhydrazine forming an orange ppt carbon-13 NMR info Answer types of chemical environments present from the chemical shift the number of carbon environments from the peaks benzene into chlorobenzene Answer chlorine and aluminium chloride benzene to nitrobenzene conc nitric acid conc sulphuric acid reflux at 55 electrophilic substitution benzene with alkyl group alkylation haloalkane anydrous aluminium chloride RTP dry inert solvent benzene with acyl group acylation acyl chloride anyhrous AlCl3 reflux 50 dry inert solvent nitrobenzene to phenylamine tin conc HCl reflux reduction phenylamine to benzene diazonium nitrous acid HCl keep below 10 diazotisation benzene diazonium to phenol water warm above 10 nitrile to carboxylic acid hydrolysis dilute nitric acid amide to carboxylic acid hydrolysis acyl chloride to amide excess ammonia low temp addition-elimination alcohol to alkene conc sulphuric acid heat elimination nitrile to amine reduction LiAlH4 haloalkane to amine excess alcoholic ammonia reflux amide to amine reduction carboxylic acid to acyl chloride 1 PCl5 dry conditions acyl chloride to carboxylic acid cold water addition-elimination acyl chloride to ester reflux in dry anhydrous conditions addition-elimination with alcohol acyl chloride to N-substituted amide anhydrous amine addition-elimination alkane to haloalkane free radical sub UV light radicals alkene to haloalkane hydrogen halide room temp electrophilic sub alcohol to aldehyde sulphuric acid and potassium dichromate distill oxidation alkene to alkane hydrogen nickel catalyst hydrogenation alkene to alcohol steam 300c 60atm H3PO4 catalyst hydration aldehyde to alcohol NaBH4/ sodium borohydride aq/alcoholic reduction nucelophilic addition alcohol to carboxylic acid sulphuric acid/ potassium dichromate refluc excess oxidising agent oxidation haloalkane to alkene KOH in ethanol heat elimination ketone/aldehyde to hydroxynitrile potassium cyanide dilute acid reflux nucleophilic addition carboxylic acid to acyl chloride SOCl2 dry condition carboxylic acid to ester alcohol and acid catalyst reflux pKb strength of base smaller=stronger Kb strength of base larger=stronger electron releasing more basic CH3 groups electron withdrawing less basic calculate isoelectric point pI given as the average of the pKa's that involve the zwitterion