A-LEVEL CHEMISTRY, ORGANIC CHEMISTRY III.

Question 1

WHAT IS BENZENE?

Answer 1

BENZENE IS A CYCLIC, PLANAR MOLECULE WITH THE FORMULA C6H6.

Question 2

BENZENE STRUCTURE.

Answer 2

IN BENZENE THE CARBON HAS FOUR,4, VALENT ELECTRONS.

EACH CARBON IS BONDED TO TWO,2, OTHER CARBONS AND ONE,1, HYDROGEN ATOM.

THE FINAL LONE ELECTRON IS IN A P-ORBITAL WHICH STICKS OUT ABOVE AND BELOW THE PLANAR RING.

Question 3

BENZENE LONE ELECTRON.

Answer 3

THE LONE ELECTROSN IN THE P-ORBITALS COMBINE TO FORM A DELOCALISED RING OF ELECTRONS.

DUE TO THE DELOCALISED ELECTRONS TRUCTURE ALL OF THE C-C BONDS IN THE MOLECULE ARE THE SAME.

THEY HAVE THE SAME BOND LEGNTH, 139pm.

Question 4

CARBON TO CARBON BOND LENGTH IN BENZENE.

Answer 4

THE C-C BOND LENGTH IN BENZENE LIES BETEWEEN 154pm, SINGLE BOND, AND 134pm, DOUBLE BOND.

Question 5

HOW IS BENZENE NORMALLY DRAWN?

Answer 5

BENZENE IS NORMALLY DRAWN IN THE SKELETAL FORMULA.

WE DO NOT SHOW THE HYDROGENS IN THE SKELETAL FORMULA.

Question 6

KEKULE’S STRUCTURE.

Answer 6

THIS REFERS TO THE STRUCTURE OF BENZENE.

THIS STRUCTURE SHOWS BENZENE WITH DOUBLE BONDS.

IT IS CALLED KEKULE’S STRCTURE NAMES AFTER AUGUST KEKULUE WHO DISCOVERED IT.

HE THOUGHT THERE WAS ALTERNATING DOUBLE AND SINGLE BONDS.

Question 7

DRAWING BENZENE TO SHOW DELOCALISED ELECTRONS.

Answer 7

THIS STRUCTURE IS THE HEXAGON WITH THE CIRCLE IN THE MIDDLE.

THE CIRCLE IS REFERD TO AS THE DELOCALISED RING.

THIS STRUCTURE SHOWS THE DELOCALISED ELECTRON SYSTEM AND YOU WILLL BE MORE LIKELY TO SEE THIS THAN THE KEKULE STRUCTURE, HOWEVER BOTH CAN BE USED.

THIS IS KNOWN AS THE DELOCALISED MODEL.

Question 8

BENZENE STABILITY.

Answer 8

BENZENE IS ACTUALLY MORE STABLE THAN THE THEORETICAL ALTERNATIVE CYCLOHEXA-1,3,5-TRIENE.

THE ALTERNATING SINGLE AND DOUBLE BOND MODEL.

Question 9

HOW DO WE MEASURE THE STABILITY OF BENZENE?

Answer 9

WE MEASURE THE STABILITY OF BENZENE BY COMPARING THE ENTHALPY CHANGE OF HYDROGENATION IN BENZENE AND CYCLOHEXA-1,3,5-TRIENE.

IF WE HYDROGENATE CYCLOHEXENE IT HAS THE ENTHALPY CHANGE OF -120kLmol^-1, CYCLOHEXENE HAS ONE,1, DOUBLE BOND.

IF BENZEE HAS THREE,3, DOUBLE BONDS WE WOULD EXPECT AN ENTHALPY CHNAGE OF HYDROGENATION OF 360kJmol^-1.

Question 10

BENZENE, REAL ENTHALPY CHANGE OF HYDROGENATION.

Answer 10

BENZENE HAS A PREDICTED ENTHALPY CHANGE OF HYDROGENATION OF -360kJmol^-1.

HOWEVER, WHEN WE MEASURE THE ENTHALPY CHANGE OF HYDROGENATION FOR BENZENE IT IS FAR LOWER AT -208kJmol^-1.

THIS IS THE EXPERIMENTAL VALUE.

ENERGY IS REQUIRED TO BREAK BONDS AND ENERGY IS RELEASED TO FORM BONDS.

THIS SUGGETS MORE ENERGY IS REQUIRED TO BREAK BONDS IN BENZENE THAN CYCLOHEXA-1,3,5-TRIENE.

LESS EXOTHEMIC, AS LESS HEAT IS GIVEN OUT FROM BOND FORMATION.
AS MORE ENERGY IS TAKEN IN FOR BOND BREAKING.

THIS SUGGESTS THAT BENZENE IS MORE STABLE THAN THE THEORETICLA CYCLOHEXA-13,5-TRIENE WITH THREE,3, DOUBLE BONDS.

THE STABILITY IS DUE TO THE DELOCALISED ELECTRON STRUCTURE.

Question 11

WHAT IS BENZENE?

Answer 11

BENZENE IS A HYDROCARBON.

Question 12

BENZENE BURNING WITH OXYGEN.

Answer 12

BENZENE BURNS READILY IN OXYGEN.

BENZENE BURNS IN OXYGEN TO PRODUCE CARBON DIOXIDE AND WATER IF BURNED COMPLETELY.

THIS IS NO DIFFERENT TO BURNING A STANDARD HYDROCARBON.

THE REACTION FOR THE COMPLETE COMBUSTION OF BENZENE IS:

2C6H6 + 15O2 -> 12CO2 + 6H2O.

Question 13

INCOMPLETE COMBUSTION.

Answer 13

IN REALITY CARBON DOES NOT BURN COMPLETELY AS THERE IS NOT ENOUGH OXYGEN IN THE AIR.

AS A RESULT WE GET A LOT OF UNREACTED CARBON ATOMS, SOOT, AND A BLACK SMOKY FLAME IS OBSERVED.

Question 14

ADDING BROMINE TO AN ALKENE.

Answer 14

ALKENES HAVE A DOUBLE BOND AND UNDERGO ELECTROPHILIC ADDITION.

ADDING BROMINE WATER TO AN ALKENE CAUSES A COLOUR CHANGE FROM BROWN-ORANGE TO COLOURLESS.

BROMINE, BROWN-ORANGE, IS THE ELECTROPHIE AND ADDS TO THE ALKENE FORMING A DIBROMOALKANE, COLOURLESS.

THIS IS BECAUSE BROMINE IS A DIATOMIC MOLECULE, Br2.

Question 15

ADDING BROMINE TO AN ALKENE THE MECHANISM.

Answer 15

Br2 IS POLARISED AS THE ELECTRONS IN THE DOUBLE BOND REPELS ELECTRONS IN Br2.

AN ELECTRON PAIR IN THE DOUBLE BOND IS ATRACTED TO THE DELTA POSITIVE BROMINE AND FORMS A BOND.

THIS BREAKS THE Br-Br BOND.

A CARBOCATION INTERMEDIATE IS FORMED AND Br- IS ATTRACTED TO C+.

COLOURLESS DIBROMOALKANE FORMED.

Question 16

ELECTROPHILE IN THE ADDITION OF BROMINE TO AN ALKENE.

Answer 16

BROMINE IS THE ELECTROPHILE AS IT ATTRACTED TO THE REGION OF HIGH ELECTRONDENSITY IN THE DOUBLE BOND.

Question 17

ARENES REACTION.

Answer 17

WHEN ARENES REACT THEY UNDERGO ELECTROPHILIC SUBSTITUTION REACTIONS.

Question 18

WHY IS BENZENE ATTRACTED TO ELECTROPHILES?

Answer 18

BENZENE HAS A HIGH ELECTRON DENSITY AS IT HAS A DELOCALISED RING OF ELECTRONS.

THIS IS ATTRACTIVE TO ELECTROPHILES, ELECTRON LOVING SUBSTANCES.

Question 19

BENZENE HIGH ELECTRON DENSITY.

Answer 19

BENZENE HAS A HIGH ELECTRON DENSITY AS IT HAS A DELOCALISED RING OF ELECTRONS.

Question 20

WHY CAN BENZENE NOT UNDERGO ELECTROPHILIC ADDITION REACTIONS?

Answer 20

AS WE HAVE SEEN BENZENE IS STABLE SO UNLIKE TRADITIONAL AKENES THEY DO NOT UNDERGO ELECTROPHILIC ADDITION REACTIOS, UNLIKE THE BROMINATION OF ALKENES, AS THIS WOULD DISRUPT THE STABLE RING OF ELECTRONS.

Question 21

ELECTROPHILIC SUBSTITUTION OF BENZENE.

Answer 21

INSTEAD OF UNDERGOING ELECTROPHILIC ADDITION THEY UNDERGO ELECTROPHILIC SUBSTITUTION REACTIONS, WHERE A HYDROGEN OR A FUNCTIONAL GROUP ON THE BENZENE RING IS SUBSTITUTED FOR THE ELECTROPHILE.

Question 22

HOW MANY MECHANISMS ARE THERE FOR THE ELECTROPHILIC SUBSTITUTION REACTION OF ARENES?

Answer 22

THERE ARE FOUR,4, MECHANISMS YOU NEED TO KNOW:

FRIEDEL-CRAFTS ACYLATION,
FRIEDEL-CRAFTS ALKYLATION,
HALOGENATION REACTION,
NITRATION REACTION.

Question 23

AROMATIC COMPOUNDS.

Answer 23

AROMATIC COMPOUNDS ARE MOLECULES THAT CONTAIN A BENZENE RING, THEY ARE ALSO KNOWN AS ARENES.

Question 24

NAMING AROMATIC COMPOUNDS.

Answer 24

THEY ARE NAMED IN TWO,2, WAYS.

WE CAN NAME THE BENZENE AT THE END.

FOR EXAMPLE, BROMOBENZENE.

WE CAN USE THE WORD PHENYL TO NAME THEM.
HERE WE USE PHENYL AS IF IT IS A FUNCTIONAL GROUP, C6H5.

FOR EXAMPLE.

C6H5OH,
PHENOL.

C6H5NH2.
PHENYLAMINE.

Question 25

ELECTROPHILIC SUBSTITUTION BENZENE, CONDITIONS.

Answer 25

THE DELOCALISED ELECTRONS IN THE BENZE RING ARE ATTRACTED TO THE CARBOCATION.

TWO,2, ELECTRONS MOVE TO FORM A BOND WHICH BREAKS THE RING AND A POSITIVE CHARGE DEVLOPS.

THE ELCTRONS IN THE C-H BOND THEN MOVE TO NEUTRALISE THE POSITIVE CHARGE AND RE-FORM THE RING.

HYDROGEN IS SUBSTITUTED.

Question 26

ELECTROPHILE TYPE, ELECTROPHILIC SUBSTITUTION BENZENE.

Answer 26

BENZENE RINGS ARE STABLE MOLECULES SO REACTIONS ARE DIFFICULT.

WE NEED A VERY STRONG ELECTROPHILE TO REACT.

THESE CAN BE CREATED BY USING A HALOGEN CARRIER CATALYST.

Question 27

HALOGEN CARRIERS, BASE.

Answer 27

HALOGEN CARRIERS ARE USUALLY ALUMINIUM HALIDES, IRON AND IRON HALIDES.

FOR EXAMPLE, AlCl3.

Question 28

BENZENE USE.

Answer 28

BENZENE IS USED WIDELY IN PHARMACEUTICALS AND DYE STUFFS HOWEVER DUE TO THE STABILITY OF BENZENE IT IS DIFFICULT TO REACT.

FRIEDEL-CRAFTS REACTIONS CAN HELP TO SOLVE THIS PROBLEM.

Question 29

FRIEDEL-CRAFTS DISCOVERIES.

Answer 29

CHARLES FRIEDEL AND JAMES CRAFTS CAME UP WITH A REACTION WHERE AN ACYL GROUP, RCO-, OR ALKYL GROUP, R-, IS ADDED ONTO A BENZENE MOELCULE.

AFTER THE ACYL OR ALKYL GROUP IS ADDED THE BENZENE STRUCTURE IS WEAKER AND IT MAKES IT EASIER TO MODIFY IT FUTHER TO MAKE USEFUL PRODUCTS.

Question 30

ADDING THE ACYL OR ALKYL GROUP ONTO THE BENZENE MOLECULE.

Answer 30

IN ORDER TO ADD ONTO THE BENZENE RING THE ELECTROPHILE MUST HAVE A VERY STRONG POSITIVE CHARGE.

ACYL GROUPS HAVE A POSITIVE CHARGE HOWEVER IT IS NOT POSITIVE ENOUGH.

Question 31

THE USE OF A HALOGEN CARRIER.

Answer 31

WE CAN USE A HALOGEN CARRIER TO ACT AS A CATALYST, FOR EXAMPLE AlCl3, WHICH WILL PRODUCE A MUCH STRONGER ELECTROPHILE WITH A STRONGER POSITIVE CHARGE.

Question 32

FRIEDEL-CRAFTS ACYLATION OR ALKYLATION.

Answer 32

IN THE FRIEDEL-CRAFTS ACYLATION OR ALKYLATION WE HAVE TO REACT AN ACYL CHLORIDE OR HALOGENOALKANE WITH THE HALOGEN CARRIER, AlCl3, TO CREATE A STRONGLY POSITIVE ELECTROPHILE.

Question 33

FRIEDEL-CRAFTS ACYLATION REACTION, MAKING THE ELECTROPHILE.

Answer 33

THE HALOGEN CARRIER, ALCl3, ACCEPTS A PAIR OF ELECTRONS AWAY FROM THE ACYL GROUP.

AS A RESULT, THE POLARISATION INCREASES AND A CARBOCATION IS FORMED AND SO IS AlCl4-.

A STRONGER ELECTROPHILE IS PRODUCED WHICH CAN NOW REACT WITH BENZENE.

Question 34

FRIEDEL-CRAFTS ACYLATION, REACTING THE ELECTROPHILE WITH THE BENZENE.

Answer 34

NOW WE HAVE MADE THE EELCTROPHILE WE NEED TO REACT IT WITH BENZENE TO MAKE A LESS STABLE PHENYLKETONE UNDER REFLUX AND A DRY ETHER SOLVENT.

THE ELECTROPHILE IS ADDED TO THE BENZENE RING.

THE DELOCALISED ELECTRONS ARE ATTRACTED TO THE CARBOCATION, OF THE ELECTROHPILE.

TWO,2, ELECTRONS MOVE TO FORM A BOND WHICH BREAKS THE RING AND A POSITIVE CHARGE DEVELOPS.

THE NEGATIVE, AlCl4-, IS THEN ATTRACTED TO THE POSITIVELY CHARGED RING AND ONE OF THE CHLORINE ATOMS BREAKS AWAY TO FORM A BOND WITH THE HYDROGEN.

THE ELECTRONS IN THE C-H BOND MOVE TO NEUTRALISE THE POSITIVE CHARGE AND RE-FORM THE RING.

WE ALSO FORM HCl AND AlCl3 CATALYST IS REFORMED.

Question 35

MAKING THE ELECTOPHILE, FOR FRIEDELS-CRAFTS ALKYLATION REACTION.

Answer 35

TO MAKE THE POWERFUL ELECTROHPILE WE USE AlCl3 AS THE HALOGEN CARRIER JUST LIKE IN THE ACYLATION REACTION.

AlCl3 ACCPETS A PAIR OF ELECTRONS AWAY FROM THE HALOGENOALKANE.

AS A RESUST, A CARBOCATION IS FORMED.

A STRONGER ELECTROPHILE IS PRODUCED WHICH NOW REACT WITH BENZENE.

WE ALSO FORM AlCl4- AS A PRODUCT.

Question 36

THE ELECTROPHILES REACTION WITH BENZENE, FRIEDELS-CRAFTS REACTION.

Answer 36

NOW WE HAVE MADE THE ELECTROPHILE WE NEED TO REACT IT WITH BENZENE TO MAKE A LESS STABLE ALKYLBENZENE UNDER REFLUX AND DRY ETHER SOLVENT.

THE DELOCALISED ELECTROSN ARE ATTRACTED TO THE CARBOCATION.

TWO,2, ELECTRONS MOVE TO FORM A BOND WHICH BREAKS THE RING AND A POSITIVE CHARGE DEVELOPS.

THE NEGATIVE HALOGEN CARRIER, AlCl4- IS THEN ATTRACTED TO THE POSITIVELY CHARGED RING AND ONE THE CHLORINE ATOMS BREAKS AWAY TO FORM A BOND WITH THE HYDROGEN.

THE ELECTRONS IN THE C-H BOND MOVE TO NEUTRALISE THE POSITIVE CHARGE AND REFORM THE RING.

WE ALSO FORM HCl AS A PRODUCT AND AlCl3 CATALYST REFORMED.

Question 37

ALCOHOL ADDED TO A BENZENE RING.

Answer 37

ALCOHOL BASED GROUPS CAN ALSO BE ADDED TO A BENZENE RING TOO.

IF WE USE AN ELECTROPHILE THAT CONTAINS AN ALKYL CHAIN WITH OAlCl3- THEN THIS CAN ADD AN ALCOHOL BASED GROUP TO THE BENZENE RING.

THIS WORKS IN A SIMILAR WAY TO FRIEDEL-CRAFTS REACTIONS AS THE OXYGEN GROUP HAS A LONE PAIR OF ELECTRONS WHICH IT TO ACT AS A NUCLEOPHILE.

WE ALSO FORM AlCl3.

Question 38

NITRATING BENZENE USE.

Answer 38

NITRATING BENZENE IS USEFUL AS IT ALLOWS US TO MAKE DYES FOR CLOTHING AND EXPLOSIVES.

Question 39

HOW DO WE NITRATE BENZENE?

Answer 39

IF WE HEAT BENZENE WITH CONCENTRATED NITRIC ACID, HNO3, AND SULFURIC ACID, H2SO4, WE FORM NITROBENZENE.

HOWEVER LIKE WE HAVE SEEN BEFORE WE HAVE TO MAKE A POWERFUL ELECTROPHILE FIRST.

Question 40

ELECTROPHILE, NITRATING BENZENE.

Answer 40

TO MAKE THE ELECTROPHILE WE REACT SULPHURIC ACID WITH NITRIC ACID.

HNO3 + H2SO4 -> H2NO3 + + HSO4-.

THE H2NO3 + DECOMPOSES TO FORM THE ELECTROPHILE, NITRONIUM ION NO2 +.

H2NO3 + -> NO2+ + H2O.

Question 41

USING THE ELECTROPHILE TO NITRATE BENZENE.

Answer 41

THE NITRONIUM ION IS ATTACKED BY THE BENZENE RING FORMING AN UNSTABLE, POSITIVELY CHARGED RING.

THE ELECTRONS IN THE C-H BOND MOVE TO REFORM THE DELOCALISED ELECTRON RING.

NITROBENZENE IS FORMED AND A H+ IS FORMED WHICH REACTS WITH HSO4- TO MAKE H2SO4 AGAIM.

H2SO4 IS A CATALYST.

Question 42

NITRATING BENZENE, TEMPERATURE.

Answer 42

A TEMPERATURE BELOW 55 DEGREES CELSIUS WILL ENSURE A SINGLE NO2 SUBSTITUTION.

ABOVE THIS WILL RESULT IN MULTIPLE SUBSTITUTIONS.

Question 43

PHENOLS.

Answer 43

PHENOLS HAVE A HYDROXYL GROUP, -OH, ATTACHED TO A BENZENE RING.

THE CARBON WITH THE, -OH, GROUP IS THE CARBON ONE,1.

WE NUMBER OTHER GROUPS FROM THIS.

Question 44

PHENOLS REACTIVITY.

Answer 44

PHENOLS ARE REACTIVE THAN BENZENE DUE TO THE ELECTRON DENSITY IN THE RING BEING HIGHER.

ELECTROPHILIC SUBSTITUTION REACTIONS ARE MOR ELIKELY TO OCCUR WITH PHENOL THAN WITH BENZENE DUE TO THE -OH GROUP AND ORBITAL OVERLAP.

THE ELECTRONS IN THE P-ORBITAL OF THE OXYGEN OVERLAP WITHT HE DELOCALISED RING STRUCTURE AND SO THEY ARE PARTIALLY DELOCALISED INTO THE PI-SYSTEM.

THE ELECTRON DENSITY INCREASES WITHIN THE RING STRUCTURE AND SO IS MORE SUSCEPTIBLE TO ATTACK FROM ELECTROPHILES.

Question 45

ASPIRIN.

Answer 45

ASPIRIN IS AN ESTER AND IS MADE BY REACTING ETHANOIC ANHYDRIDE OR ETHANOYL CHLORIDE AND SALICYLIC ACID.

Question 46

MAKING ASPIRIN USING ETHANOIC ANHYDRIDE.

Answer 46

ETHANOIC ANHYDRIDE + SALICYLIC ACID -> ASPIRIN + ETHANOIC ACID.

Question 47

THE USE OF ETHANOIC ANHYDRIDE COMPARED TO THE USE OF ETHANOYL CHLORIDE.

Answer 47

ETHANOIC ANYHDRIDE IS USED INSTEAD OF ETHANOYL CHLORIDE IN INDUSTRY BECAUSE:

IT IS SAFER AS IT IS LESS CORROSIVE.

IT IS CHEAPER.

IT IS SAFER AS IT DOES NOT PRODUCE HARMFUL HCl GAS.

IT IS SAFER AS IT DOES NOT REACT VIGOROUSLY WITH WATER.

Question 48

WHAT IS ASPIRIN?

Answer 48

ASPIRIN IS AN ESTER.

Question 49

PHENOLS DISSOCIATE.

Answer 49

PHENOLS PARTIALLY DISSOCIATE WHICH MEANS THEY ARE WEAK ACIDS.

Question 50

PHENOLS DISSOCIATE, PRODUCTS.

Answer 50

PHENOLS DISSOCIATE WEAKLY TO FORM A PHENOXIDE ION AND H+ ION.

Question 51

PHENOLS REACTION WITH ALKALI.

Answer 51

PHENOLS REACT WITH ALKALIS TO FORM A SALT AND WATER.

FOR EXAMPLE SODIUM POHENOXIDE AND WATER ARE MADE WHEN PHENOL WITH SODIUM HYDROXIDE.

Question 52

PHENOLS, BROMINE WATER.

Answer 52

PHENOLS CAN REACT WITH BROMINE WATER, 3Br2.

PHENOLS REACT WITH BROMINE WATER AS PHENOLS ARE MORE REACTIVE THAN BENZENE.

WE OBSERVE THE BROWN BROMINE WATER DECOLOURISING.

AS OH IS AN ELECTRON DONATING GROUP, SUBSTITUTION OCCURS AT CARBON 2,4 AND 6.

THE PRODUCT IS THEREFORE 2,4,6-TRIBROMOPHENOL.

Question 53

2,4,6-TRIBROMOPHENOL.

Answer 53

IT SMELLS OF ANTISEPTIC AND IS INSOLUBLE IN WATER.

Question 54

PHENOLS, DILUTE NITRIC ACID.

Answer 54

PHENOLS CAN REACT WITH NITRIC ACID.

PHENOLS REACT WITH NITRIC ACID TO PRODUCE NIITROPHENOLS AS PHENOLS ARE MORE REACTIVE THAN BENZENE.

REMEMBER WITH BENZENE WE NEED CONCENTRATED NITRIC ACID AND CONCENTRATED SULFURIC ACID AS A CATALYST.

AS OH IS DONATIING AN ELECTRON GROUP, WE SEE SUBSTITUTION AT CARBON TWO,2, AND FOR,4,

TWO,2, ISOMERS ARE PRODUCED.

2-NITROPHENOL.
4-NITROPHENOL.

Question 55

AMINE.

Answer 55

AN AMINE IS DERIVED FROM AMMONIA MOLECULES AND ALL CONTAIN A NITROGEN ATOM WHERE HYDROGENS ARE REPLACED WITH AN ORGANIC GROUP, FOR EXAMPLE AN ALKYL GROUP.

WE GET DIFFERENT TYPES OF AMINES, PRIMARY, SECONDARY, TERTIARY AND QUATERNARY.

Question 56

PRIMARY AMINES.

Answer 56

A CENTRAL NITROGEN ATOM,
WITH ONE,1, ORGANIC GROUP GROUP AND TWO,2, HYDROGEN GROUPS.

FOR EXAMPLE, METHYLAMINE.

Question 57

SECONDARY AMINES.

Answer 57

THEY HAVE TWO,2, ORGANIC GROUPS AND ONE,1, HYDROGEN ATOM AROUND THE CENTRAL NITROGEN.

FOR EXAMPLE,
DIMETHYLAMINE.

Question 58

TERTIARY AMINES.

Answer 58

A CENTRAL NITROGEN ATOM WITH THREE,3, ORGANIC GROUPS.

FOR EXAMPLE,
TRIMETHYLAMINE.

Question 59

QUATERNARY AMINES.

Answer 59

A CENTRAL NITROGEN ATOM, WHICH HAS A POSITIVE CHARGE, N+.
THIS IS BECAUSE NITROGEN CAN ONLY FORM THREE,3, BONDS.
THIS IS WHY IT IS A SALT.

WITH FOUR,4, ORGANIC GROUPS AROUND IT.

FOR EXAMPLE,
TETRAMETHYLAMINE ION.

Question 60

AROMATIC AMINE.

Answer 60

PHENYLAMINE.

AROMATIC AMINE IS PRIMARY.

A NITROGEN MOLECULE WITH TWO,2, HYDROGEN MOLECULES ATACHED TO IT AND A BENZENE RING.

THIS IS STILL A PRIMARY AMINE, AS THE NITROGEN STILL ONLY HAS ONE,1, ORGANIC GROUP ATTACHED TO IT.

Question 61

NON AROMATIC AMINES.

Answer 61

NON AROMATIC AMINES ARE KNOWN AS ALIPHATIC AMINES.

Question 62

MAKING ALIPHATIC AMINES.

Answer 62

ALIPHATIC AMINES ARE MADE IN TWO,2, WAYS EITHER BY REACTING A HALOGENOALKANE WITH EXCESS AMMONIA OR BY REDUCING A NITRIL.

Question 63

MAKING ALIPHATIC AMMINES, EXCESS AMMONIA.

Answer 63

REACTING HALOGENOALKANE AND EXCESS AMMONIA.

THE MECHANISM FOR MAKING EACH AMINE IS SIMILAR,

INSTEAD OF TWO,2, AMMONIA MOLECULES WE USE TWO,2, AMINES INSTEAD.

FOR EXAMPLE, THE PRIMARY AMINE METHYL AMINE REACTS WITH CHLOROETHANE TO FORM A SECONDARY AMINE.

THE REACTION:

CHLOROETHANE AND OUR AMMONIA.

AMMONIA IS A NUCLEOPHILE THAT ATTACKS THE DELTA POSITIVE ON THE CARBON OF THE CHLOROETHANE.

AN INTERMEDIATE IS FORMED, ALKYLAMMONIUM SALT, WITH A POSITIVE NITROGEN AND A Cl- ION.

A SECOND AMMONIA GIVES UP A LONE PAIR OF ELECTRONS TO HYDROGEN WHICH BREAKS AWAY FROM THE SALT.

A PRIMARY AMINE AND AMMONIUM CHLORIDE SALT IS PRODUCED.
NH4 +Cl-.

Question 64

PROBLEMS WITH USING AMMONIA TO MAKE ALIPHATIC AMINES.

Answer 64

IN THIS REACTION WE SAW THE PRODUCTION OF A PRIMARY AMINE HOWEVER THIS REACTION PRODUCES SECONDARY, TERTIARY AND QUATERNARY SALTS TOO SO WE HAVE AN IMPURE PRODUCT.

THIS OCCURS AS PRIMARY AMINES STILL HAVE A LONE PAIR OF ELECTRONS ON THE NITROGEN SO ALSO ACTS AS A NUCLEPHILE.

THE AMINE CAN REACT WITH ANY REMAINING HALOGENOALKANES TO PRODUCE A SECONDARY AMINE, THEN REACT FURTHER TO MAKE TERTIARY AND QUATERNERY SALTS.

Question 65

MAKING ALIPHATIC AMINES, REDUCING NITRILES.

Answer 65

THE CHEAPEST WAY TO MAKE PRIMARY AMINES IN INDUSTRY IS TO REDUCE NITRILES USING HYDROGEN GAS AND A NICKEL OR PLATNIUM CATALYST.

THIS REACTION IS CALLED CATALYTIC HYDROGENATION AND UNLIKE USING HALOGENOALKANES, THIS REACTION PRODUCES PRIMARY AMINES ONLY SO A PURE PRODUCT IS MADE.

CONDITIONS:
HIGH TEMPERATURE AND HIGH PRESSURE.

WE CAN REDUCE NITRILES USING A STRONG REDUCING AGENT, LiAlH4, AND DILUTES ACID.

THIS METHOD IS MORE EXPENSIVE THAN USING HYDROGEN GAS AND A NICKEL OR PLATINUM CATALYST AS LiAlH4, IS EXPENSIVE AND SO IS NOT USED IN INDUSTRY.

THE REACTION IS CALLED REDUCTION AND WE USE [H] TO SYMBOLISE A REDUCING AGENT.

THIS IS DISSOLVED IN A NON-AQEOUS SOLVENT SUCH AS DRY ETHER.

USE LiAlH4 AND A DILUTE ACID.

Question 66

MAKING AROMATIC AMINES.

Answer 66

AROMATIC AMINES ARE MADE BY REDUCING NITRO COMPOUNDS SUCH AS NITROBENZNE.

WE HEAT, UNDER REFLUX, NITROBENZENE WITH CONCENTRATED HCl AND A TIN CATALYST TO FORM A SALT SUCH C6H5NH3 +Cl-.

THE SALT PRODUCED IS REACTED WITH AN ALKALI SUCH AS NaOH TO PRODUCE AN AROMATIC AMINE SUCH AS PHENYLAMINE AND WATER.

Question 67

AROMATIC AMINES USE.

Answer 67

AROMATIC AMINES ARE USED TO MAKE DYE STUFFS AND PHARMACEUTICALS.

Question 68

AMINES AS A BASE.

Answer 68

AMINES HAVE A LONE PAIR ELECTRONS THAT ALLOWS THEM TO ACCEPT A PROTON AND HENCE ACT AS A BASE.

A PROTON BONDS TO AN AMINE VIA A DIATIVE COVALENT, COORDINATE, BOND.

BOTH ELECTRONS IN THE BOND ORIGINATE FROM THE LONE PAIR ON THE NITROGEN.

THE STRENGTH OF THE BASE IS DEPENDANT ON THE AVAILABILITY OF THE LONE PAIR OF ELECTRONS ON THE NITROGEN.

THE HIGHER THE ELCTRON DENSITY THE MORE READILY AVAILIBLE THE ELCTRONS ARE.

Question 69

BASE STRENGTH ORDER.

Answer 69

THE ORDER OF BASE STRENGTH IS AS FOLLOWS:

AROMATIC AMINES -> AMMONIA -> PRIMARY ALIPHATIC AMINES.

THE WEAKEST BASE TO THE STRONGEST BASE.

BENZENE IS AN ELECTRON WITHDRAWING GROUP SO IT PUTS ELECTRONS AWAY FORM THE NITROGEN INTO THE RING STRUCTURE.

ELECTRON DENSITY AT NITROGEN REDUCES SO THE LONE PAIR AVAILABILITY IS REDUCED AND AROMATIC AMINES ARE LESS BASIC,

ALKYLN GROUPS ARE ELECTRON PUSHING GROUPS SO THEY PUSH ELECTRON TOWARD THE NITROGEN.

ELECTRON DENSITY INCREASES SO LONE PAIR OF ELECTRON AVAILABILITY IS INCREASED AND PRIMARY ALIPHATIC AMINES ARE MORE BASIC.

Question 70

AMINE SOLUBILITY.

Answer 70

AMINES CAN HYDROGEN BOND WITH WATER AND SO SOME HAVE THE ABILITY TO DISSOLVE IN WATER TO FORM ALKALI SOLUTIONS.

THE LONE PAIR OF ELECTRONS ON THE NITROGEN CAN FORM HYDROGEN BONDS WITH THE HYDROGEN ATOMS ON WATER MOLECULES.

THE LONE PAIR ON OXYGEN CAN ALSO FORM HYDROGEN WITH HYDROGEN ON THE AMINE.

HOWEVER ONLY SMALLER AMINES WILL DISSOLVE.

LARGER ONES HAVE LONGER HYDROCARBON COMPONENTS AND ARE NON-POLAR AND THESE CAN DISRUPT THE HYDROGEN WITH WATER MOLECULES.

IF THE AMINE IS LARGE ENOUGH THEN THE LONDN FORCES BETWEEN THE NON-POLAR HYDROCARBON CHAIN WILL BE STRONGER THAN THE HYDROGEN BONDING BETWEEN THE NITROGE AND H ON WATER MOLECULES.

THIS MEANS LARGER AMINES WILL NOT DISSOLVE.

Question 71

AMINES REACTION WITH COPPER.

Answer 71

AMINES REACT WITH COPPER COMPLEX IONS TO FORM A DEEP BLUE SOLUTION.

IF WE ADD A SMALL AMOUNT OF BUTYLAMINE TO COPPER SULFATE SOLUTION A PALE BLUE PRECIPITATE IS FORMED, [Cu(OH)2(H2O], AS THE AMINE REMOVES TWO,2, H+ ACTING AS A BASE.

IF WE ADD MORE BUTYLAMINE THEN FOUR,4, OF THE LIGANDS WILL BE EXCHNAGED WITH THE AMINE FORMING A COMPLEX WHICH IS DEEP BLUE SOLUTION.

Question 72

FORMING COPPER COMPLEX.

Answer 72

IT CAN BE FORMED BY DISSOLVING COPPER (II) SULFATE IN WATER.

Question 73

ACYL CHLORIDES REACT WITH AMINES.

Answer 73

ACYL CHLORIDES REACT WITH BUTYLAMINES.

THE Cl IS SUBSTITUTED FOR A NITROGEN.

REACTION WITH PRIMARY AMINES PRODUCES N-SUBSTITUTED AMIDES.

WE ALSO FORM HCl.

THIS IS A VIGOROUS REACTION THAT PRODUCES A SOLID WHITE PRODUCT.

FOR EXAMPLE,

ETHANOYL CHLORIDE + BUTYLAMINE ->

N-BUTYL ETHANAMIDE + HYDROCHLORIDE.

THE BUTYL AMINE CAN ALSO REACT WIT HCl TO PRODUCE BUTYLAMMONIUM CHLORIDE, [C4H9NH3]+ Cl-.

Question 74

AMINES REACTION WITH ACID.

Answer 74

AMINES ALSO REACT WITH ACIDS AND FORM ALKALINE SOLUTIONS.

AMINES ARE BASES SO REACT WITH ACIDS TO FORM SALTS HOWEVER UNLIKE TRADITIONAL ACID BASE REACTION THEY DO NOT FORM WATER.

CH3CH2CH2CH2NH2 + HCl -> CH3CH2CH2CH2NH3+ Cl-.

SMALLER AMINES ARE SOLUBLE IN WATER AND DISSOLVE TO FORM ALKALINE SOLUTIONS.

UNLIKE SOME BASES WHERE THERE IS AN OH GROUP, FOR EXAMPLE NaOH, AMINES REACT WITH WATER TO PRDUCE THE OH- ION WHICH MAKES THE SOLUTION BASIC.

FOR EXAMPLE:

CH3CH2CH2CH2NH2 (aq) + H2O (l) -> CH3CH2CH2CH2NH3+ (aq) + OH- (aq).

Question 75

AMIDES.

Answer 75

AMIDES ARE DERIVATIVES OF CARBOXYLIC ACIDS AND HAVE THE FUNCTIONAL GROUP OF -CONH2.

INSTEAD OF HAVING AN -OH LIKE IN CARBOXYLIC ACIDS WE HAVE -NH2.

Question 76

N-SUBSTITUTED AMIDE.

Answer 76

IN AN N-SUNSTITUTED ONE OF THE HYDROGENS IS REPLACED WITH AN ALKYL GROUP.

FOR EXAMPLE,
NH2, TO, NHCH3.

Question 77

MAKING AMIDES.

Answer 77

ACYL CHLORIDES REACT WITH AMMONIA AND PRIMARY AMINES.

REACTION WITH AMMONIA PRODUCES AMIDES.

ETHANOYL CHLORIDE + NH3 -> ETHANAMIDE + HCl.

THIS IS A VIGOROUS REACTION AND WHITE MISTY FUMES OF HYDROGEN CHLORIDE GAS IS PRODUCED.

REACTION WITH PRIMARY AMINES PRODUCES N-SUBSTITUTED AMIDES.

ETHANOYL CHLORIDE + METHYL AMINE -> N-METHYL ETHANAMIDE + HCl.

THIS IS A VIGOROUS REACTION AND WHITE MISTY FUMES OF HYDROGEN CHLORIDE GAS IS PRODUCED.

Question 78

CONDENSATION POLYMERS.

Answer 78

CONDENSATION POLYMERS ARE COMPRIMISED OF THREE,3, MAIN TYPES:

POLYPEPTIDES, POLYAMIDES AND POLYESTERS.

CONDENSATION POLYMERISATION IS WHERE TWO,2, DIFFERENT MONOMERS WITH AT LEAST TWO,2, FUNCTIONAL GROUPS REACT TOGETHER.

WHEN THEY REACT A LNIK IS MADE, AND WATER IS ELIMINATED, HENCE CONDENSATION POLYMERISATION.

THE LINK DETERMINES THE TYPE OF POLYMER PRODUCED.

Question 79

POLYPEPTIDES.

Answer 79

POLYPEPTIDES, FOUND IN PROTIENS.

Question 80

POLYAMIDES.

Answer 80

FORMED BY REACTING DIAMINES AND DICARBOXYLIC ACIDS.

Question 81

POLYESTERS.

Answer 81

FORMED BY REACTING A DIOL AND A DICARBOXYLIC ACID TOGETHER.

Question 82

POLYAMIDES, FORMING.

Answer 82

POLYAMIDES ARE FORMED BY REACTING DICARBOXYLIC ACIDS AND DIAMINES TOGETHER.

AMIDE LINKS ARE FORMED WHEN DICARBOXYLIC ACIDS REACT WITH DIAMINES.

WE HAVE TO USE DICARBOXYLIC ACIDS AND DIAMINES AS THEY HAVE FUNCTIONAL GROUPS EITHER SIDE WHICH ALLOWS CHAINS TO BE FORMED.

Question 83

POLYAMIDE, KEVLAR.

Answer 83

KEVLAR IS A POLYAMIDE THAT IS USED IN BULLETPROOF VESTS, CAR TYRES, AND SOME SPORTS EQUIPMENT AS IT IS LIGHT WEIGHT BUT STRONG.

KEVLAR IS MADE FROM BENZENE-1,4-DICARBOXYLIC ACID AND 1,4-DIAMINOBENZENE.

Question 84

NYLON.

Answer 84

NYLON 6,6 IS A POLYAMIDE THAT IS USED IN ROPES, CARPETS, CLOTHING AND PARACHUTE FABRIC.

NYLON 6,6 IS MADE FROM HEXANEDIOIC ACID ADN 1,6-DIAMINOHEXANE.

Question 85

POLYESTERS, FORMING.

Answer 85

POLYESTERS ARE FORMED BY REACTING DICARBOXYLIC ACIDS AND DIOLS TOGETHER.

ESTER LINKS ARE FORMED WHE DICARBOXYLIC ACIDS REACT WITH DIOLS.

Question 86

TERYLENE.

Answer 86

TERYLENE IS A POLYESTER THAT IS USED IN PLASTIC DRINK BOTTLES, SHEETING AND CLOTHES.

IT ALSO HAS THE ACRONYM PET.

TERYLENE IS MADE FROM BENZENE-1,4-DICARBOXYLIC ACID AND ETHANE-1,2-DIOL.

Question 87

CONDENSATION POLYMERS.

Answer 87

WE CAN WORK OUT THE MONOMER FROM THE POLYMER CHAIN.

THE MONOMER CAN BE DETERMINED BY FINDING THE REPEAT UNIT.

WE LOOK FOR EITHER AN AMIDE LINK, HN-CO, OR AN ESTER LINK, CO-O.

THE MONOMER CAN BE FOUND BY BREAKING THE BONDS IN THESE LINKS AND ADD H OR OH TO EITHER END OF BOTH MOLECULES.

Question 88

CONDENSATION POLYMERS, HYDROLYSIS.

Answer 88

CONDENSATION POLYMERS CAN BE HYDROLYSED, SPLIT USING WATER, TO PRODUCE THE ORIGINAL MONOMERS.

IT IS JUST THE REVERSE OF POLYMERISATION SEEN BEFORE.

Question 89

AMINO ACIDS.

Answer 89

AMINO ACIDS HAVE AN AMINO GROUP, -NH2, AND A CARBOXYL GROUP, -COOH.

AMINO ACIDS ARE AMPHOTERIC WHICH MEANS THEY ACIDIC AND BASIC PROPERTIES.

AMINO ACIDS HAVE AN ORGANC SIDE CHAIN, REPRESENTED BY R, WITH THE EXCEPTION OF GLYCINE WHERE R IS HYDROGEN.

AMINO ACIDS, EXCEPT GLYCINE, ARE CHIRAL MOLECULES AS THEY HAVE FOUR,4, DIFFERENT GROUPS AROUND A CENTRAL CARBON ATOM.

THEY ROTATE PLANE POLARISED LIGHT.

Question 90

NAMING AMINO ACIDS.

Answer 90

AMINO ACIDS ARE NAMED IN TWO,2, DIFFERENT WAYS; THEY HAVE A COMMON NAME AND A SYSTEMIC NAME.

FIND THE LONGEST CARBON CHAIN, NAME IT AS AN ACID BECAUSE OF THE CARBOXYL GROUP.

NUMBER THE CARBONS, CARBON ONE,1, IS IN THE CARBOXYL GROUP.

NOTE THE NUMBER WHERE THE NH2 GROUP SITS, AND WE CALL IT AMINO.

NAME ANY OTHER GROUPS THAT ARE NOT NH2 IN THE STANDARD WAY, FOR EXAMPLE -OH IS HYDROXY.

Question 91

AMINO ACIDS, EXIST.

Answer 91

AMINO ACIDS SOMETIMES EXIST AS ZWITTERIONS.

A ZWITTERION IS A MOELCULE WITH BOTH POSITIVE AND NEGATIVE IONS.

ZWITTERIONS ONLY EXIST AT THE AMNIO ACID’S ISOLELECTRIC POINT.

IF THE pH IS LOWER THAN THE ISOELECTRIC POINT THE COO- IS LIKELY TO ACCEPT A H+.

A ZWITTERION IS LIKELY TO BE FORMED WHEN AT A pH AT THE ISOELECTRIC POINT.

BOTH THE CARBOXYL AND AMINO GROUP ARE IONISED.

IF THE pH IS HIGHER THAN THE ISOELECTRIC POINT THEN THE NH3+ IS LIKELY TO LOSE AN H+.

Question 92

ISOELECTRIC POINT.

Answer 92

THE ISOELECTRIC POINT IS THE pH AT WHICH THE AVERAGE OVERALL CHARGE IS ZERO.

THIS IS DEPENDANT ON THE ‘R” GROUP.

Question 93

THIN LAYER CHROMATOGRAPHY.

Answer 93

THIN LAYER CHROMATOGRAPHY, TLC, ALLOWS US TO SEPARATE AND IDENTIFY AMINO ACIDS AS THEY HAVE DIFFERENT SOLUBILITIES.

TLC USES A STAITIONARY PHASE OF SILICA OR ALUMINA MOUNTED ON A GLASS./.= METAL PLATE.

A PENCIL BASE LINE IS DRAWN AND DROPS OF AMINO ACID MIXTURES ADDED.

PLACE THE PLATE IN IN A SOLVENT, THE BASE LINE MUST BE ABOVE THE SOLVENT.

LEAVE UNTIL SOLVENT HAS MOVED UP NEAR THE TOP OF THE PLATE.

REMOVE, MARK THE SOLVENT FRONT AND ALLOW TO DRY.

IT WORKS BY THE AMINO ACID MIXTURE SPOTS DISSOLVING IN THE SOLVENT.

SOME CHEMICALS IN THE MIXTURE MAY NOT DISSOLVE AS MUCH AND STICK TO THE STAITIONARY PHASE QUICKLY.

WHAT WE ARE LEFT WITH IS A CHROMATOGRAM.

Question 94

CALCULATING Rf VALUES.

Answer 94

AMINO ACIDS CAN BE IDENTIFIED BY CALCULATING THE Rf VALUE FROM A CHROMATOGRAM.

THE NUMBER OF SPOTS ON THE PLATE TELLS YOU HOW MANY AMINO ACIDS MAKE UP THE MIXTURE.

THE AMINO ACID CAN BE IDENTIFIED BY CALCULATING THE Rf VA;UE AND COMPARING THESE TO A LIBRARY OF KNOWN Rf VALUES.

Rf VALUE OF AMINO ACID.= DISTANCE TRAVELLED BY SPOT. / DISTANCE TRAVELLED BY SOLVENT.

WE MEASURE BOTH OF THESE VALUES FROM THE BASE LINE.

Rf VALUES ARE FIXED FOR EACH AMINO ACID.

HOWEVER THIS CHANGES IF THE TEMPERATURE, SOLVENT OR MAKE-UP OF THE TLC PLATE CHANGES.

Question 95

GRINGARD REAGENTS, USE.

Answer 95

GRINGARD REAGENTS ARE USED TO HELP CARBON-CARBON BOND FOMRATION, WHICH WITHOUT GRINGARD REAGENT, WOULD BE VERY DIFFICULT.

Question 96

GRINGARD REAGENTS, COMPOSITION.

Answer 96

GRINGARD REAGENTS ARE AN ORGANOMAGNESIUM COMPOUND THAT ARE MADE BY REACTING A HALOGENOALKANE WITH MAGNESIUM IN DRY ETHER.

Question 97

GRINGARD REAGENT, CARBOXYLIC ACIDS.

Answer 97

CARBOXYLIC ACIDS CAN BE MADE BY REACTING A GRINGARD REAGENT WITH CARBON DIOXIDE.

IN DRY ETHER, WE BUBBLE CARBON DIOXIDE IN GRINGARD REAGENT.

WE ADD A DILUTE ACID TO THE SOLUTION.

A NEW C-C BOND IS FORMED WHEN THE ‘R” BREAKS OF THE GRINGARD REAGENT AND BONDS WITH THE C IN CO2.

THIS IN TURN BREAKS THE CO BOND TO FORM R-COO-.

Question 98

GRINGARD REAGENT, ALCOHOL.

Answer 98

ALCOHOLS CAN BE MADE BY REACTING A GRINGARD REAGENT WITH ALDEHYDES AND KETONES.

IN DRY ETHER, WE BUBBLE CARBON DIOXIDE IN GRINGARD REAGENT.

WE ADD DILUTE ACID TO THE SOLUTION.

A NEW C-C BOND IS FORMED WHEN THE ‘R” BREAKS OFF THE GRINGARD REAGENT AND BONDS WITH THE C IN THE CARBONYL GROUP.

THIS IN TURN BREAKS THE C=O BOND.

FINALLY, THE HCl PROTONATES TO FORM THE ALCOHOL.

Question 99

ALKENE FUNCTIONAL GROUP.

Answer 99

C-CC.

PROPERTIES:
UNREACTIVE, NON-POLAR BOND.

TYPICAL REACTION:
RADICAL SUBSTITUTION.

Question 100

AROMATIC COMPOUNDS FUNCTIONAL GROUP.

Answer 100

C6H5-.

PROPERTIES:
DELOCALISED ELECTRON RING, STABLE.

TYPICAL REACIONS:
ELECTROPHILIC SUBSTITUTION.

Question 101

ALKENE FUNCTIONAL GROUP.

Answer 101

C=C.

PROPERTIES:
ELECTRON RICH DOUBLE, NON-POLAR BOND.

TYPICAL REACTIONS:
ELECTROPHILIC ADDITION.

Question 102

ALCOHOL FUNCTIONAL GROUP.

Answer 102

C-OH.

PROPERTIES:
LONE PAIR ON OXYGEN CAN ACT AS A NUCLEOPHILE.

POLAR C-OH BOND.

TYPICAL REACTIONS:
ESTERFICATION AND NUCLEOPHILIC SUBSTITUTION.

DEHYDRATION./.= ELIMINATION AND NUCLEOPHILIC SUBSTITUTION.

Question 103

HALOALKANE FUNTIONAL GROUP.

Answer 103

C-X.

PROPERTIES:
POLAR C-X BOND.

TYPICAL REACTIONS:
NUCLEOPHILIC CSUBSTITUITION AND ELIMINATION.

Question 104

AMINE FUNCIIONAL GROUP.

Answer 104

C-NR2.

PROPERTIES:
LONE PAIR ON N IS BASIC, CAN ACT AS A NUCLEOPHILE.

TYPICAL REACTIONS:
NUCLEOPHILIC SUBTITUTIONS AND NEUTRALISATION.

Question 105

NITRILE FUNCTIONAL GROUP.

Answer 105

C-C=_N.

PROPERTIES:
ELECTRON DEFICIENT CARBON CENTRE.

TYPICAL REACTIONS:
REDUCTION AND HYDROLYSIS.

Question 106

ALDEHYDES AND KETONES FUNTIONAL GROUPS.

Answer 106

C=O.

PROPERTIES:
POLAR C=O BOND.

TYPICAL REACTIONS:
NUCLEOPHILIC ADDITION, OXIDATION OF ALDEHYDE,S REDUCTION.

Question 107

CARBOXYLIC ACID FUNCTIONAL GROUP.

Answer 107

-COOH.

PROPERTIES:
ELECTRON DEFICIENT CARBON CENTRE.

TYPICAL REACTIONS:
ESTERFICATION AND NEUTRALISATION.

Question 108

ESTER FUNCTIONAL GROUP.

Answer 108

RCOOR’.

PROPERTIES:
ELECTRON DEFICEITN CARBON CENTRE.

TYPICAL REACTIONS:
HYDROLYSIS.

Question 109

ACYL CHLORIDE FUNCTIONAL GROUP.

Answer 109

-COCl.
PROPERTIES:
ELECTRON DEFICIENT CARBON CENTRE.

TYPICAL REACTIONS:
NUCLEOPHILIC ADDITION-ELIMINATION, CONDENSATION AND FRIEDEL-CRAFTS ACYLATION.

Question 110

ACID ANYHDRIDE.

Answer 110

RCOOOCOR’.

PROPERTIES:
ELECTRON DEFICIENT CARBON CENTRE.

TYPICAL REACTIONS:
ESTERFICATION.

Question 111

EMPIRICAL FORMULA.

Answer 111

THE SIMPLEST WHOLE NUMBER RATION OF ELEMENTS IN A COMPOUND.

WRITE OPUT THE ELEMENTS INVOLVED.

WRITE THE PERCENTAGES AS MASSES.

DIVIDE THESE BY THE RELATIVE ATOMIC MASS TO GET NUMBER OF MOLE.

DIVIDE ALL TE NUMBERS BY THE SMALLEST NUMBER OF MOLES.

Question 112

MOLECULAR FORMULA.

Answer 112

COMBUSTION ANALYSIS CAN ALSO BE GIVEN AS VOLUMES BY USING MOLAR RATIONS OF GAS VOLUMES WE CAN WORK OUT THE WORK THE MOLECULAR MASS OF INKOWN SUBSTANCES.

Question 113

REFLUX.

Answer 113

REFLUX IS A TECHHNIQUE USED WHEN YOU WANT TO HEAT VOLATILE LIQUIDS.

REFLUC ALLOWS STRONG HEATING WITHOUT LOSING VOLATILE REACTANTS AND PRODUCTS.

VOLATILE COMPOUND EVAPORATE AND CONDENSE AND FALL BACK INOT THE FLASK.

THE LIEBIG CONDENSORT HAS COLD WATER RUNNING THROUGH THE WALL.

WHEN HOT EVAPORARTING SUBSTANCES HIT THE COLDER CONDENSER THEY TURN BAKC INTO LIQUID AND RETURN BACK THE ROUNF BOTTOMED FLASK TO REACT FURTHER.

AS WE ARE USING FLAMMABLE LIQUIDS HETAING IS DONE VIA A WATER BATH OR ELECTRIC HEATER CALLED A MANTLE.

THIS IS SAFER THAN USING A NAKED FLAME.

Question 114

DISTILLATION.

Answer 114

DISTILLATION IS USED WHEN WE WANT TO SEPARATE SUBSTANCES WITH DIFFERENTBOILING POINTS.

GENTLY HEATING A MIXTURE WILL RESULT IN COMPOUNDS SEPARATING OUT IN ORDER OF BOILING POINT.

Question 115

STEAM DISTILLATION.

Answer 115

STEAM DISTILLATION IS USED WHEN WE WANT TO SEPARATE SUBSTANCES WITH HIGH BOILING POINTS OR DECOMPOSE WHEN HEATED.

IF THE PRODUCT IS IMMISCIBLE THEN STEAM DISTILLATION IS USES TO SEPARATE COMPOUND THAT COULD NOT BE DONE UNDER STANDARD DISTILLATION.

THE STEAM IS PRODUCED AND PUSHED THROUGH THE IMPURE SAMPLE.

THE STEAM LOWERS THE BOILING POINT OF THE IMMISCIBLE PRODUCT AND ALLOWS FOR IT TO BE DISTILLED OUT OF THE MIXTURE BEFORE IT DECOMPOSES.

THIS METHOD IS ALSO USEFUL IF THE SUBSTANCES WE WANT TO SEPARATE HAS A HIGH BOILING POINT AS THE STEAM REDUCED THIS.

IF THE SUBSTANCE WE ARE TRYING TO COLLECT IS LESS VOLATILE THAN THE CONSITUENT SUBSTANCES WE ARE TRYING TO SEPARATE IT FROM, THEN THE DEIRED PRODUCT WILL EVAPORATE OUT OF THE FLASK WITH THE STEAM.

THIS IS THEN CONDENSED AND COLLECTED IN A SEPARATE FLASK.

Question 116

SEPARATION AND PURIFICATION.

Answer 116

SEPARATION TECHNIQUES ARE USED TO REMOVE IMPURITIES THAT ARE DISSOLVED IN WATER.

SEPARATION;

ADD THE PRODUCTS FROM DISTILLATION INTO A SEPARATING FUNNEL.

ADD WATER TO DISSOLVE SOLUBLE IMPURITIES AN CREATE AN AQEOUS SOLUTION.

AFTER ALLOWING THE SOLUTION TO SETTLE TWO,2, LAYERS WILL FORM:

TOP LAYER, IMPURE PRODUCT.

BOTTOM LAYER, AQUEOUS LAYER CONTAINING WATER SOLUBLE IMPURITIES.

DRAIN THE AQUEOUS LAYER.

WE NOW NEED TO PURIFY OUR SAMPLE THROUGH TWO,2, FURTHER STEPS:
WASHING
AND DRYING.

Question 117

FILTRATION.

Answer 117

GRAVITY FILTRATION IS USED TO SEPARATE SOLIDS FROM LIQUIDS.

THIS METHOD ISUSED IF WE WISH TO KEEP THE LIQUID PART AND DISPOSE OF THE SOLID.

PLACE A FLUETED FILTED PAPER, FOLDED INTO A CONCERTINA TO INCREASE SURFACE AREA, IN THE FUNNEL AND DAMPEN SLIGHTLY TO MAKE A SEAL.

POUR THE REACTION MIXTURE INTO THE FUNNEL SLOWLY TO STOP OVERFLOWING.

GRAVITY WILL PULL THROUGH THE LIQUID COMPONENT INTO THE VESSEL BELOW WHCIH CAN THEN BE PURIFIED IF REQUIRED.

THE SOLID LEFT IN THE FILTER PAPER CAN BE DISPOSED OF SAFELY.

Question 118

RECRYSTALLISATION.

Answer 118

RECRYSTALLISATION IS A METHOD TO PURIFY SOLIDS AND THE SOLVENT CHOSEN IS VERY IMPORTANT.

ADD JUST ENOUGH HOT SOLVENT TO ALLOW THE IMPURE SOLID TO DISSOLVE.

THIS WILL MEAN YOU HAVE A SATURATED SOLUTION OF YOUR IMPURE PRODUCT.

ALLOW SOLUTINO TO COOL DOWN SLOWLY, CRYSTALS WILL START TO FORM.

YOUT IMPURITIES WILL REMAIN DISSOLVE DIN SOLUTION AS THERE IS A SMALLER QUANTITY OF THEM, IT TAKES A LOT LONGER FOR THEM TO CRYSTALLISE.

FILTER TO GET YOUR SOLID PURIFIED CRYSTALS.

WASH WITH VERY COLD SOLVENT AND DRY THEM OFF.