Complete Unit 5 Notes edexcel.doc

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A Level Chemistry

UNIT 5

GENERAL PRINCIPLES OFCHEMISTRY II

NOTES (2!"

#ritte$ %ySEL&AGAHNESH PRASA'

- 1 –



I$tr)*+ti$

This unit includes the following.• A continuation of the consideration of redox reactions from Unit 2, looking at the

more quantitative aspects with a study of the use of electrode potentials.•

Redox chemistry and the Transition etals.• !n the "rganic #hemistry section, arenas, amines, amides, amino acids andproteins will $e studied. %nowledge of the "rganic #hemistry from all the otherUnits will $e assumed.

• %nowledge from all the other Units will $e necessary for Unit &.

Assessme$t

The Unit examination will $e 'hour () minutes. !t will carry *) marks. !t willcontain three sections, A, + and #.

Se+ti$ A is an o$ective test

Se+ti$ , short-answer and extended answer questions.uestions on analysis and evaluation of practical work will also $e includedin this section.

Se+ti$ C will extended answer questions on contemporary contexts.

- 2 –



Re)- E.*ili%ri/

Revie0Unit & redox chemistry $uilds on the redox chemistry first encountered in Unit 2.!mportant terms introduced then were redox, oxidation num$er and half equations.

A redox redox reaction is an electron transfer reaction. Oxidation Is Loss of ele+tr$s

Reduction Is Gain of ele+tr$sThe oxidation num$er of an atom shows the num$er of electrons which it has lost or gained as aresult of forming a compound./alf equations involve looking at the electron gain and electron loss processes separately.

Sti+himetry0toichiometry is the ratio $etween su$stances in a chemical reaction. 0o in the reaction $etweensodium hydroxide and sulphuric acid1 2a"/ 3 /20"( 4 a20"( 3 2/2"The stoichiometry for this reaction is that 2 moles of a"/ react with ' mole of /20"(

!n a redox reaction, electrons are transferred from one material to another.The redox equation must $alance in terms of num$ers of electrons and oxidation num$er.

5or example in the reaction $etween silver6!7 and copper, metallic silver and copper6!!7 areformed.The silver half equation is Ag3

6aq7 3 e- 4 Ag6s7 The copper half equation is #u6s7 4 #u23

6aq7 3 2e- The electron loss and gain must $alance, so the silver half equation has to $e dou$led.

2Ag36aq7 3 #u6s7 4 #u23

6aq7 3 2Ag6s7

This stoichiometry can also $e deduced $y examining oxidation num$er change.• The oxidation num$er of copper increases $y 2.• The oxidation num$er of silver decreases $y '• This means that they react in the ratio 2 silver to ' copper.

Re)- Titr/ti$s!t is possi$le to use redox reactions in titrations.#ommon reagents used for these are manganate68!!7 and thiosulphate with iodine.

Pt/ssi*m m/$1/$/te(&II" titr/ti$s A known concentration of potassium manganate68!!7 can $e used to determine the quantity of are)*+i$1 /1e$t present in a sample.

!n titrations involving potassium manganate68!!7 the half reaction is9M$O

3 4 H4 4 5e3 M$24 4 H2O E.*/ti$ 6

This may react with ethanedioic ions following the half reaction9C2O23 2CO2 4 2e3 E.*/ti$ 2

To com$ine these half equations we must x :quation ' $y 2 and x :quation 2 $y & and add1

The full reaction is9 2M$O3 4 67H4 4 5C2O

23 2M$24 4 H2O 4 6CO2

The reaction requires excess dilute sulphuric acid and a temperature of a$out ;)o#.

- 3 –



anganate68!!7 is a deep purple colour in solution, $ut the manganese6!!7 ion, to which it isreduced, is almost colourless.

As the manganate68!!7 is added, from a $urette, it reacts turning colourless.<hen all the reducing agent has reacted, the manganate68!!7 no longer reacts and its colourremains in the flask.

5rom the concentration of the manganate68!!7 and the volume used, the num$er of moles can $e

determined. Using the chemical equation, the num$er of moles of reducing agent can $e found,and so its concentration.

E-/m8le An iron ta$let, containing an iron6!!7 compound was crushed. 2.&)) g of the ta$let were dissolved andmade up to 2&) cm= of solution. 2&.) cm= of this solution was transferred to a flask $y pipette, 2& cm= dilute sulphuric acid was added to acidify the solution and the flask contents titrated against ).))& moldm-= potassium manganate68!!7 solution. The average titre value was 2;.(& cm=.#alculate the percentage iron in the ta$let.

oles of potassium manganate68!!7 > 2;.(& ? '))) x ).))& > '.=22& x ')-( mol

:quation9 M$O3

4 5Fe24

4 H4

9 2M$24

4 H2O 4 5Fe:4

oles of iron6!!7 @in 2&.) cm= sample > '.=22& x ')-( x & > ;.;'2& x ')-( mol

oles of iron6!!7 @in 2&).) cm= sample > ;.;'2& x ')-( x 2&) ? 2& > ;.;'2& x ')-= mol

ass of iron6!!7 > ;.;'2& x ')-= x &; > ).=B)g

C iron > ).=B) ? 2.&)) x ')) > '(.D

- 4 –

A drop of potassium manganate68!!7 is added

The colour appears in the flask

"n swirling the potassium manganate68!!7reacts and the colour disappears

ore potassium manganate68!!7 isadded

Until the colourremains after swirling



This*l8h/te titr/ti$s

Thiosulphate and iodine titrations are used to determine the concentration of -i)isi$1 /1e$ts.

5irst of all the oxidising agent is added to a solution containing excess iodide ions.This oxidises the iodide ions to iodine giving a $rown colour.

2I3 9 I2 4 2e3

Thiosulphate is then added from a $urette1 this reacts with the iodine to form colourless products.2S2O:

23 4 I2 9 SO723 4 2I3

Euring the titration, the colour intensity decreases, eventually reaching a pale yellow colour. At this point, a few drops of starch solution are added to give the deep $lue complex showing thelast traces of iodine. Thiosulphate is then added dropwise, until the mixture $ecomes colourless.

5rom a known concentration of thiosulphate, it is possi$le to determine the num$er of moles ofchemical involved in the reaction.

E-/m8le2.)(*g of a copper alloy was dissolved in concentrated nitric acid and made up to 2&) cm= of solution.2&.) cm= of this solution was transferred to a flask $y pipette and an excess of potassium iodide addedto it. The resulting mixture was titrated against ).')) mol dm-= sodium thiosulphate solution. Theaverage titre value was 2(.;& cm=. #alculate the percentage copper in the alloy.

• oles of sodium thiosulphate > 2(.;& ? '))) x ).')) > 2.(;& x ')-= mol

• :quation9 2S2O:23 4 I2 9 SO7

23 4 2I3

• oles of !2 in flask > 2.(;& x ')-= x ).& > '.2=2& x ')-= mol

• :quation9 2C*24 4 I3 9 2C*I 4 4 !2 • oles of #u23@in 2&.) cm= sample > '.2=2& x ')-= x 2 > 2.(;& x ')-=

mol• oles of #u23 @in 2&).) cm= sample > 2.(;& x ')-=

x 2&) ? 2& > 2.(;& x ')-2 mol

• ass of #u23 > 2.(;& x ')-2 x ;=.& > '.&;&g• C copper > '.&;& ? 2.)(* x ')) > B;.(C

- 5 –

As thiosulphate is added thecolour of the iodine fades

0tarch is added to givea deep $lue colour

The titration is then continueduntil the mixture $ecomescolourless



St/$)/r) Ele+tr)e Pte$ti/l E;

!f a metal is placed in a solution of its ions at a concentration of '.) mol dm-= at 2&o#, the potentialo$tained 6the tendency to release electrons7 is the standard electrode potential, :o.

St/$)/r) Ele+tr)eThe standard electrode potential of a metal and its solution cannot $e measured directly."nly potential differences $etween a metal and a standard electrode can $e measured.

A standard cell, to measure them all against is required.This standard cell, taken as having a :F of ).)), is the hydrogen half cell1

To measure the :F of the Ginc half cell, the following set up is used

<hen the hydrogen half cell is connected to the negative side of a high resistance voltmeter, the e.m.f.of the cell gives the :F for that half cell.

The value o$tained from this is :F > -).B; 8The negative value indicates that the Ginc loses electrons more readily than the hydrogen, and that it isa more powerful reducing agent. The more negative a value is, the more powerful the reducing agent.

5or a cell of hydrogen and copper the value o$tained from this is :F > 3).=( 8The positive value indicates that copper is a weaker reducing agent than hydrogen, and the #u23 is amore powerful oxidising agent.

I$ 1e$er/l< the mre 8sitive the E;< the mre 80er=*l the -i)isi$1 /1e$t>

- 6 –

/ydrogen gas at ' atm pressure

0olution containing hydrogen ionsat '.) moldm-=

easured at 2*D%Hlatinised platinum electrode

Iinc electrode

0olution containing Ginc ions at '.) mol dm-=

0alt $ridge

V/igh resistance voltmeter



Use = E; v/l*es0ince the :F values can $e used to determine in which way the electrons will flow, it is possi$le to usethem to decide whether a reaction occurs.

#ells can $e represented using formulae. Using a and $ to represent the metals in two half cells,the formulaic representation would $e9

a6s7 J a236aq7 $236aq7 J $6s7

+y convention change in this representation takes place from left to right, so the reactions taking placein this cell would $e1

a6s7 4 a236aq7 3 2e- :lectrons move from left to right $23

6aq7 3 2e- 4 $6s7

The feasi$ility of a reaction can $e found directly from the :F values using the following equation9E;

CELL ? E;RHC 3 E;

LHC

A 8sitive v/l*e indicates that a reaction is =e/si%le.

otice that in the left hand cell the reaction is the reverse of the half equation as it is normally written.

For example, will a reaction take place when zinc is added to silver nitrate solution?Relevant half equations are In23 3 2e- In :F > -).B;8

Ag3 3 e- Ag :F > 3).D)8!f a reaction takes place the Ginc will $ecome Ginc ions.This involves losing electrons so it corresponds to the left hand cell.6This is the reverse of the usually written half equation, so this will $e the left hand cell.70o the Ag is the R/# and In is the K/#

:

F

#:KK > :

F

R/# - :

F

K/#:F#:KK > 3).D - -).B; > 46>57 The positive value indicates that the reaction is feasi$le9

2A14(/." 4 @$(s" 9 @$24

(/." 4 2A1(s"

Another example; will a reaction take place when acidified hydrogen peroxide is added to bromide ions?Relevant half equations are/alf equation ' +r 2 3 2e- 2+r - :F > 3'.)B/alf equation 2 "2 3 2/3 3 2e- /2"2 :F > 3).;D8/alf equation = /2"2 3 2/3 3 2e- 2/2" :F > 3'.BB8

The oxidation of $romide ions is 2,r 3 9 ,r 2 4 2e3 0ince this process involves losing electrons, it is the left hand cell

There are two half equation for hydrogen peroxide. Kooking at half equation 2 first1:F#:KK > :F

R/# - :FK/#

:F#:KK > 3).;D - 3'.)B > 3>:! The negative value tells us that this reaction is not feasi$le.

!nspection of the half equation would also tell us this of course since in a redox equation one halfequation always has to $e the reverse of the usual form.

Kooking now at the other hydrogen peroxide half equation1:F

#:KK > :FR/# - :F

K/#

:F#:KK > 3'.BB - 3'.)B > 4> The positive value indicates that the reaction is feasi$le9

H2O2(/." 4 2H4(/." 4 2,r 3(/." 9 2H2O(l" 4 ,r 2(/."

- 7 –

Keft hand cell 6electrode7.K/# K/:

Right hand cell 6electrode7,R/# 6R/:7



Limit/ti$s = E; v/l*esThe electrode potential values have limitations $ecause1

• they refer to standard conditions• they indicate the energetic feasi$ility of a reaction, not the kinetics.

!f the conditions are different from the standard, the emf can change. 5or example, the measurement

of :F is made at a concentration of 'moldm-=. !f the concentration of one of the solutions is changed,this will change the emf of the reaction.

For example 2A14(/." 4 C*(s" 9 C*24

(/." 4 2A1(s"

5or the #u and Ag cell, the standard value for the reaction is9:F

#:KK > :FR/# - :F

K/#

:F#:KK > 3).D - 3).=( > 3).(;8

8alues of emf for different silver ion concLentrations are shown in the ta$le $elow.#oncentration of #u23 ? moldm

=#oncentration of Ag3

? moldm-=emf

'.) '.) x ')-2 3).=(

'.) '.) x ')-= 3).2D'.) '.) x ')-( 3).22'.) '.) x ')-& 3).';'.) '.) x ')-; 3).')'.) '.) x ')-B 3).)('.) '.) x ')-D -).)2'.) '.) x ')-* -).)D

At low silver ion concentration, the reaction will tend to go in the opposite direction.

#hanges in temperature and pressure can also affect the emf for a particular reaction.

5or example the reaction9 M$O2(s" 4 H4(/." 4 2Cl3(/." 9 Cl2(1" 4 M$24(/." 4 2H2O(/.7 :F

#:KK > :FR/# - :F

K/#

:F#:KK > 3'.2= - 3'.=; > 3>6: The negative value tells us that this reaction is not feasi$le.

!f concentrated /#l and solid n"2 are mixed and then heated, a reaction readily takes place.

The em= v/l*es $ly i$)i+/te the e$er1eti+ =e/si%ility = / re/+ti$< it 1ives *s $ i$=rm/ti$/%*t the Bi$eti+s>

The :F of a cell gives no information a$out the kinetics of a reaction.

The com$ination of copper and hydrogen half cells can $e written9

HtM/26g7L J 2/36aq7 #u236aq7 J #u6s7

#alculating :F for this cell1 :F#:KK > :F

R/# - :FK/#

:F#:KK > 3).=( - ) > 4>:&

The positive value indicates that the reaction is feasi$le9 H2(1" 4 C*24(/." 9 2H4

(/." C*(s"

<hen hydrogen is $u$$led through copper sulphate solution no reaction can $e o$served. Thereaction has a high activation energy, so although the reaction is energetically favoura$le, the highactivation energy means that the reaction is so slow as to $e negligi$le.

- 8 –



Re)- Chemistry = &/$/)i*m

8anadium can form four oxidation states."xidation num$er 2 = ( &0pecies present 823 8=3 8"23 8"2

3

#olour Nellow +lue Oreen 8iolet

0tandard redox potentials for these conversions are shown $elow#onversion 0tandard redox potential ?88=3 3 e- 823 -).2;2/3 3 8"23 3 e- 8=3 3 /2" 3).=(2/3 3 8"2

3 3 e- 8"23 3 /2" 3'.))

8anadium can $e reduced from the 3& states right through to the 32 state $y Ginc.:ach step can $e predicted using :o values.

5or example for the first step 6reduction from 3& to 3(79

@$24

4 2e3

@$ E

? 3>7 &&O2

4 4 2H4 4 e3 &O24 4 H2O E ? 46> &

:Fcell > :oR - :o

K > 3'.)) 8 - 6-).B; 87 > 3'.B; 8The large positive value of :cell shows that the reaction to reduce vanadium is spontaneous.

2&O24 4 H4 4 @$ 2&O24 4 2H2O 4 @$24

Reaction of Ginc with vanadium6!87@$24(/." 4 2e3 @$(s" E ? 3>7 &

&O24(/. 4 2H4(/." 4 e3 &:4(/." 4 H2O(l" E ? 4>: &

:F

#:KK > ).=( - -).B; > 3'.')8 reaction is feasi$le

@$ 4 H4 4 2&O24 @$24 4 2&:4 4 2H2O

Reaction of Ginc with vanadium6!!!7@$24(/." 4 2e3 @$(s" E ? 3>7 &&:4(/." 4 e3 &24(/." E ? 3 >27 &

:F#:KK > -).2; - -).B; > 3).&)8 reaction is feasi$le

@$ 4 2&:4 9 @$24 4 2&24

/eating with Ginc and sulphuric acid will therefore reduce vanadate687 from yellow to $lue togreen, and finally to lavender 6violet7.

- 9 –



St/$)/r) Ele+tr)e Pte$ti/ls /$) E.*ili%ri*m

a6s7 J a236aq7 $23

6aq7 J $6s7

!t has $een esta$lished that the feasi$ility of a reaction can $e found directly from the :F values usingthe equation9 E;

CELL ? E;RHC 3 E;

LHC

!t has also $een noted that if the concentrations are changed the value of :F values will change.

Ket us apply some hypothetical values to the right and left hand cells of our hypothetical reaction.

K/# half equation a236aq7 3 2e- a6s7 3).;;8

R/# half equation $236aq7 3 2e- $6s7 3).;(8

:F

#:KK > :F

R/# - :F

K/# :F

#:KK > ).;( P ).;; > -).)28The negative value shows that the reaction $elow is not feasi$le.

a6s7 3 $236aq7 4 $6s7 3 a23

6aq7

/owever if the concentration of $23 is increased, the equili$rium$23

6aq7 3 2e- $6s7

will shift to the right and more $ is formed, so the :F of this half cell increases

!f the concentration of a23 is decreased, the equili$riuma23

6aq7 3 2e- a6s7

will shift to the left and more a23 is formed, so the :F of this half cell decreases

Applying new hypothetical values

K/# half equation a236aq7 3 2e- a6s7 3).;;8 3).;=8

R/# half equation $236aq7 3 2e- $6s7 3).;(8 3).;B8

:F#:KK > :F

R/# - :FK/# :F

#:KK > ).;B P ).;= > 3).);8

The positive value shows that the reaction $elow is now feasi$le.a6s7 3 $23

6aq7 4 $6s7 3 a236aq7

This demonstrates that these are equili$rium processes.

The relationship $etween :F

#:KK and the equili$rium constant, %, isE;

CELL l$D

<e know from the equili$rium section in Unit ( 6Topic (.&7 Q0tot > Rln%

Therefore it follows that1

E;CELL Stt

- 10 –

Left hand cell,

LHC

Right hand cell,

RHC



F*el Cells

:lectricity is generally produced $y $urning fuel and using the heat to generate electricity.

5uel cells are electrochemical cells which convert chemical energy in fuels directly to electrical energy

5uels cells can $e B)C or more efficient at converting the chemical energy in fuels to electrical energy,whereas a typical modern power plant is only capa$le of a$out ()C

5uel cells differ from other cells, such as the dry cell, in having a continuous supply of reactant togenerate the electrical current. 5uels include hydrogen, hydrocar$ons and alcohols.

The diagram $elow shows the $asic design of a fuel cell.

At the negative electrode the following reaction takes place /2 4 2/3 3 2e-

At the positive electrode the catalyst causes dissociation of the oxygen1 "2 4 "

And then the following reaction takes place " 3 2/3 3 2e- 4 /2"

The overall reaction is H2 4 O2 9 H2O

An ethanol fuel cell has a similar design, the reactions $eing9

At the negative electrode9 #2/&"/ 3 /2" 4 #/=#"2/ 3 (/3 3 (e-

At the positive electrode9 (/3 3 "2 3 (e- 4 2/2"

The overall reaction is9 #2/&"/ 3 "2 4 #/=#"2/ 3 /2"

- 11 –

Fel Heat !lect"icit#

Fel !lect"icit#

- $

Horous graphite electrodeimpregnated with nickel andnickel oxide

Horous graphite electrodeimpregnated with nickel

hydrogen oxygen

0olid polymer electrode



,re/th/lysers

+reathalysers measure the quantity of alcohol in the $lood $y determining the amount in a sample ofexpired air.

The original $reathalysers used potassium dichromate and sulphuric acid. The dichromate oxidiGed the

ethanol to ethanal and ethanoic acid. This caused a consequential reduction in the dichromate inwhich the orange crystals turned green. !t was possi$le to deduce the rough level of alcohol in 'dm= of$reath. Accurate measurements were difficult to o$tain. This com$ined with the toxicity of thedichromate led to dichromate $reathalysers $eing replaced $y an ethanol fuel cell which gave moreaccurate results. The ethanol fuel cell consists of two electrodes made of a material such as platinum with a permea$lemem$rane $etween containing sodium hydroxide.

Alcohol passing through the cell causes the reactions $elow. At the negative electrode9 #2/&"/ 3 /2" 4 #/=#"2/ 3 (/3 3 (e- At the positive electrode9 (/3 3 "2 3 (e- 4 2/2"

The cell voltage is directly proportional to the ethanol concentration.

The $reathalyGer is initially cali$rated with air containing a known ethanol concentration."nce cali$ated the cell can $e used to determine the ethanol concentration in a $reath sample, with thevalues $eing read directly from a scale.

This type of $reathalyGer does not give a printed read out so it is not usually given as evidence in court. !f the $reathalyGer gives a reading a$ove the legal limit, the driver has to give a second sample of$reath at the police station. At the police station, the ethanol concentration is determined $y usinginfra-red a$sorption. This measures the a$sorption of the #-/ stretching at 2*&)cm-'. The instrument

is again cali$rated to give an ethanol concentration measurement which can $e printed out and usedas evidence in court.

- 12 –



Tr/$siti$ Met/l Chemistry

Tr/$siti$ met/ls /s )3%l+B eleme$ts5or the elements up to #a the =d or$itals are higher in energy than the (s or$ital. Therefore, afterargon 6element 'D7, the (s or$ital is filled9 #a has electron configuration MArL (s2.

5rom scandium on, the =d or$itals are filled, until they have ten electrons at Ginc.The term Sd-$lock elements refers to those elements in which this d-su$shell is filling 60cPIn7,$ut the term Str/$siti$ eleme$ts is used for d-$lock elements which form one or more sta$leions with a 8/rti/lly3=ille) )3s*%shell.

This excludes 0c and In, since their only common oxidation states are 0c=3 6=d)7 and In23 6=d')7.

This distinction is made $ecause the main features of the chemistry of the transition elementsdepend largely on this partially filled d-su$shell.

Ele+tr$ +$=i1*r/ti$s = the )3%l+B eleme$ts /$) their sim8le i$s>Nou will $e expected to use your Heriodic Ta$le to deduce the electronic configurations of atomsand ions.

Remem$er that9%i& the sta$ility of the half-filled su$-shell means that d& and d') configurations are particularly

sta$le e.g. #r is 's2 2s2 2p; =s2 =p; s6 :)5 6not 4s !d 4 71#u is 's2 2s2 2p; =s2 =p; s6 :)6 6not 4s !d " 71

%ii& in ions the (s electrons are always lost $efore =d electrons9 5e23 is 's2 2s2 2p; =s2 =p; =d;.

#lement $ymbol #lectronic structure of atom %ommonion&s'

#lectronicstructure of ion

0candium 0c 6Ar7=d '(s2 0c=3 6Ar7

Titanium Ti 6Ar7=d 2(s2 Ti(3 Ti=3

6Ar76Ar7=d '

8anadium 8 6Ar7=d =(s2 8=3 6Ar7=d 2

#hromium #r 6Ar7=d &(s' #r =3 6Ar7=d =

anganese n 6Ar7=d &(s2 n23 6Ar7=d &

!ron 5e 6Ar7=d ;(s2 5e23 5e=3

6Ar7=d ; 6Ar7=d &

#o$alt #o 6Ar7=d B

(s2

#o23

6Ar7=d B

ickel i 6Ar7=d D(s2 i23 6Ar7=d D

#opper #u 6Ar7=d ')(s' #u3 #u23

6Ar7=d ') 6Ar7=d *

Iinc In 6Ar7=d ')(s2 In23 6Ar7=d ')

Nte 5e=3 is more sta$le than 5e23 as it has an electron configuration without electron repulsion inthe partially filled d su$-shell.

- 13 –

A tr/$siti$ eleme$t is / )3%l+B eleme$t th/t =rms $e r mre st/%le i$s th/th/ve i$+m8lete )3r%it/ls>



Ge$er/l Pr8erties = Tr/$siti$ Met/ls>Transition metals have higher melting points, higher $oiling points and higher densities than other metals. Transition metals also show the following characteristic properties9

6> &/ri/%le -i)/ti$ st/tes9- Transition metals have electrons of similar energy in $oth the =dand (s levels. This means that one particular element can form ions of roughly the same sta$ility$y losing different num$ers of electrons. Thus, all transition metals from titanium to copper canexhi$it two or more oxidation states in their compounds.

"xidation states of some Transition etals9Titanium- 32, 3=, 3( 8anadium- 32, 3=, 3(, 3&#hromium- 32, 3=, 3; anganese- 32, 3=, 3(, 3&, 3;, 3B!ron- 32, 3= #o$alt- 32, 3=ickel- 32, 3=, 3( #opper- 3', 32

2> Frm/ti$ = +m8le- i$s9- As a lot of the transition metals have some empty spaces intheir =d-or$itals, they can receive lone pairs of electrons and form dative covalent $onds thusproducing complex compounds.

:> Cl*re) +m8*$)s9- <hen electrons move from a d-or$ital 6with lower energy7 to anotherd-or$ital 6with higher energy7, energy is taken in. This energy is in the form of visi$le light.The transition metal appears the complementary colour to the ight a$sor$ed, thus producingcoloured compounds.

> C/t/lyti+ 8r8erties9- 5or any element its higher oxidation states give rise to covalentcompound formation. As Transition etals have varia$le oxidation states, they tend to havecatalytic properties.

Cm8le- i$s<ater molecules, hydroxide ions, ammonia molecules and cyanide ions can all link on totransition metal ions to form +m8le- i$s. They do so $y donating a lone pair to form a $ond Pthis is a dative covalent $ond. The ions or molecules that form these $onds are called ligands.

A +m8le- i$ is one in which a central positive ion is surrounded $y ligands, which are co-ordinately 6datively7 $onded to it1 e.g. #r6/2"7;

=3, 5e6#7;(P.

A li1/$) is a molecule or negative ion which has a lone pair of electrons, and can use its lonepair to form co-ordinate $onds to a metal ion1 e.g. /2", # P.

The transition metals are not unique in forming complexes 6there are small num$ers formed $ymetals in groups 2, = and (7, $ut they form a much wider range than other elements.This is $ecause the transition metal ions are small and polarising, since their nuclei are poorlyshielded, and so they attract ligands strongly.

N/mi$1 Cm8le- I$sThe names of complex ions contain four main components.

5irst part 0econd part Third part 5ourth partum$er of eachtype ligand

ame of ligand ame of transition metal 6ending in Pate if it is a negative ion7

#harge on transitionmetal

- 14 –



!on formula ame of ionM#u6/2"7;L23 hexaaquacopper6!!7M#r6/2"7;L=3 hexaaquachromium6!!7M#u#l(L2- Tetrachlorocuprate6!!7M#u6/=7(L23 tetraamminecopper6!!7

M5e6#7;L(-

hexacyanoferrate6!!7<hen writing formulae, the central atom is put first, then the negative ions and then follow anyneutral molecules. :verything is then put in square $rackets and the charge added. 5orexample, tetraaquachloro copper 6!!7 would $e written as M#u#l6/2"7(L3

Sh/8es = Cm8le- I$s#omplex ions can $e tetrahedral in shape, $ut the maority have an octahedral shape.

Li$e/r Eichlorocuprate6!7 M#u#l2L-

Tetr/he)r/l Tetrachlorochromate6!!!7, M#r#l(L-

Pl/$/r Eichlorodiamminoplatinum6!!7, MHt6/=72#l2L

O+t/he)r/l /exaaquacopper6!!7, M#u6/2"7;L23 /exaamminechromium6!!!7, M#r6/=7;L=3

- 15 –

C$ Cl-Cl-

C" 3

$

Cl-

Cl-Cl

-

Cl-

't2$

(H3

Cl-

Cl-

(H3

C" 3$

(H3

(H3

(H3

H3

(

H3 (

(H3

C2$

)H2

)H2

H2)

H2

)

H2)

H2)



,i)e$t/te /$) Ply)e$t/te li1/$)s All the a$ove complexes contain monodentate ligands 6ligands that have one set of teeth wihwhich they $ite onto the transition metal ion7 as they form one dative covalent $ond."ther ligands can form more than one dative $ond and are called polydentate.

An example of a polydentate ligand is ',2-diaminoethane1 this is can form two dative $onds socan also $e referred to $identate.

Another polydentate ligand is called $y letters which come from the old name of theethylenediaminetetraacetate, or :ETA.

This can form six dative $onds so is a hexadentate ligand.

Cl*r i$ Cm8le- I$s<hen ligands pack around a metal ion, the d-or$itals no longer have exactly the same energies.!f they are partially filled, it is possi$le for an electron to ump from a lower-energy d-or$ital to anunoccupied higher-energy d-or$ital.

These S)) tr/$siti$s are of an energy corresponding to a$sorption in the visi$le region, andso the compound appears to $e coloured. The colour is that of the light which is not a$sor$ed9

e.g. copper6!!7 ions look $lue $ecause they a$sor$ red light.

The energies of d-or$itals, and so the colour of the complex, are very sensitive to the ligandpresent.

8isi$le light stretches from purple?$lue at ())P&))nm, via yellow at around ;))nm to red at;&)nm. A complex a$sor$ing from ())P&&)nm will look red 6like Fe(H2O"5SCN24 $elow7 while onea$sor$ing from ;))-;&)nm will look $lue 6e.g. #u6/2"7;

237. "ne a$sor$ing in the middle, from&))-;))nm will $e a $lue?red mix i.e. purple.

#ompounds of d-$lock elements which are not transition elements cannot undergo these electrontransitions, so do not have coloured compounds.The compounds of Ginc and scandium are therefore white and colourless in solution.#opper6!7 which has a full d-shell also has white compounds.

- 16 –

(i2$

(H2

(H2

H2 (

H2 (

H2 (

H2 (

CH2

CH2

H2C

CH2

H2C

CH2

(-CH2-CH

2-(

CH2C

)

)-

CH2C

)

)-

CCH2

)

-)

CCH2

)

-)



Li1/$) E-+h/$1e Pr+esses!t is possi$le for one type of ligand to $e replaced $y another type.Li1/$) e-+h/$1e re/+ti$s =te$ h/ve / +l*r +h/$1e /ss+i/te) 0ith them>

Thiocyanate ions can replace one of the water ligands in M5e6/2"7;L=3.Fe(H2O"7

:4 4 SCN3 9 Fe (H2O"5SCN24 4 H2O

An aqueous #opper6!!7 sulphate solution is $lue in colour $ecause of the presence of M#u6/2"7;L2 ions. <hen concentrated hydrochloric acid is added to this solution the colour changes from $lueto green. This happens $ecause the M#u#l(L2- ion is produced. The #l- ions have replaced the/2" molecules in a ligand exchange.

C*(H2O"724 4 Cl3 9 C*Cl

23 4 7H2O

<hen ammonia is added a further change from green to deep $lue takes place as ammoniamolecules replace the chloride ions.

C*Cl23 4 NH: 4 2H2O 9 C*(NH:"(H2O"2

24 4 Cl3

!f :ETA is then added yet another ligand exchange takes place and the solution turns pale $lue.

C*(NH:"(H2O"2

24

4 e)t/ 9 C*(e)t/"

24

4 NH: 4 2H2O

These changes take place $ecause the complexes $ecome more sta$le.

The sta$ility of the M#u6edta7L23 complex can $e understood in terms of entropy. As the one edta molecule replaces the six smaller ligands, the small ligands are released intosolution and therefore have greater freedom of movement, so greater disorder and consequentlythe entropy increases.

- 17 –

*C%H2)&6+2$ *CCl4+

2- *C%(H3&4%H2)&2+2$ *C%edta&+2$

Increasing stability



O-i)/ti$ st/tes = tr/$siti$ eleme$ts C* /$) Cr

C88er Ar:)6s6

#opper, =d'), is the only mem$er of the transition series to have a significant 3' oxidation state,and even here the 3' state is only sta$le if in a complex ion, or in an insolu$le compound P insolution, it disproportionates.

The 3' state, with a full d su$-shell, is not coloured 6apart from #u2"7.The 32 state, with its familiar $lue and green complexes, is the normal sta$le state.

C*(I"• #u2", as made $y reduction of 5ehlings or +enedicts solution with a reducing sugar, is a

red insolu$le solid.• #u#l and #u20"( are white solids. +oth of these, when dissolved in water

disproportionate92#u36aq7 à #u6s7 3 #u236aq7

• This can $e understood in terms of the redox potentials9#u23 3 e- C*4 :o > 3).'&8C*4

3 e- #u :o

> 3).&28There is a reaction $etween the two underlined species, i.e. the disproportionation.

0o when a copper6!7 compound is dissolved in water a $lue solution 6#u236aq77 and a red-

$rown solid 6#u6s77 are formed.

C*(II"• ost copper6!!7 compounds are $lue, and in solution they give $lue #u6/2"7;

23 ions.• <hen +88er(II" s*l8h/te solution is treated with dilute aqueous /mm$i/, the solution

starts $lue $ecause of the #u6/2"7;23 ion. !t first forms a pale $lue precipitate of #u6"/72,

and then this dissolves to give a deep $lue coloured solution, containing the#u6/=7(6/2"72

23 ion.M.+. the hydroxide is formed first $ecause ammonia solution is alkaline, due to the

reaction/= 3 /2" /(3 3 "/ P. Then the high concentration of /= molecules displaces the

equili$rium in favour of forming the ammonia complex.L"verall9 #u6/2"7;

23 3 (/= #u6/=7(6/2"7223 3 2/2"

• <hen +88er(II" s*l8h/te solution is treated with +$+e$tr/te) hy)r+hlri+ /+i) 6or sodium chloride solution7, the solution starts $lue $ecause of the #u6/2"7;

23 ion. As #l P

ions are added they displace water molecules, forming the tetrahedral #u#l (2P, which is

yellow. The colour changes from $lue through lime-green to yellow-green, and $ecomesmore intensely coloured despite the dilution9

#u6/2"7;23 3 (#l P6aq7 à #u#l(2P 3 ;/2"

<ith s)i*m hy)r-i)e #u236aq7 turns from a $lue solution to give a mid?light-$lueprecipitate9 #u236aq7 3 2"/ P6aq7 à #u6"/726s7

Estim/ti$ = +88er(II"#u236aq7 ions will react quantitatively with iodide ions, oxidising the latter to iodine and $eingreduced themselves to a white precipitate of copper6!7 iodide9

2C*24(/." 4 I (/." 2C*I(s" 4 I2(s"

<hen excess potassium iodide solution is added, the $lue colour disappears and a $rownsolution with a white precipitate results. This can $e titrated with standard sodium thiosulphate,adding starch $efore the endpoint and continuing until the $lue colour disappears 6leaving thewhite precipitate7. I2(/." 4 2S2O:

2(/." SO72(/." 4 2I (/."

- 18 –



Chrmi*m Ar:)5s6

#hromium has common oxidation states of =3 and ;3, although 23 also exists.

#hromium 6!!7 #hromium 6!!!7 #hromium 68!7#r 23 #r =3 #hromate #r"(

2- Eichromate

#r 2"B

2-

+lue Oreen Nellow "range

Cr(II"• The #r6/2"7;

23 ion is readily oxidised to #r6/2"7;=3

Cr(III"• The #r6/2"7;

=3 ion is 8*r8le, as are crystals of chromium6!!!7 sulphate, #r 260"(7=.• #r 2"= is a 1ree$ solid, and #r6"/7= is o$tained as a green precipitate $y adding sodium

hydroxide to any solution of a chromium6!!!7 salt. !t is amphoteric and dissolves in excesssodium hydroxide to form a green solution of #r6"/7;

=P.• /ydrated chromium6!!!7 chloride is a 1ree$ solid, which gives a green solution with one or

more #l P ions in the aqua-complex e.g. #r6/2"7(#l 23.• #r =36aq7 can $e oxidised to chromium68!7 $y adding excess sodium hydroxide, then

hydrogen peroxide, and $oiling. The solution goes yellow as #r"(2P ions are formed.

Cr(&I"• Hotassium dichromate68!7, %2#r 2"B, is an orange solid which dissolves in water to give an

orange solution. !n alkali this changes to the yellow chromate68!7 ion, and $ack again toorange dichromate68!7 on acidification9#r 2"B

2P 3 2"/ P 2#r"(2P 3 /2" then 2#r"(

2P 3 2/3 à #r 2"B2P 3 /2"

orange yellow yellow orangeote that this is not a redox reaction.

• Hotassium dichromate is a good primary volumetric standard 6i.e. can $e o$tained pureand sta$le, so can $e weighed out to give a solution of relia$ly known concentration7, andis often used to titrate with iron6!!7 ions, using a redox indicator. #r 2"B

2P 3 '(/3 3 ;5e23 à 2#r =3 3 B/2" 3 ;5e=3

• The :" value of 3'.==8 for #r 2"B2P in acid 6to 2#r =37 shows that it is quite a strong

oxidising agent, and can $e reduced $y many moderate reducing agents 60"2, 0n23,ethanol on warming7, when it turns from orange to green.

Uses = Cr #hromium metal is used in making st/i$less steel is much more expensive than mild steel,resists corrosion effectively, $ut lacks some other useful properties 6e.g strength, hardness7 andso cannot always $e su$stituted for normal steel.

#hromium is added to iron in smaller amounts to make /lly steels which are very hard 6used for example in $all $earings7.

- 19 –



'e8rt$/ti$ re/+ti$sEeprotonation reactions involve water ligands losing hydrogen ions 6proton7 to a proton acceptorsuch as an hydroxide ion.

C*(H2O"724 4 OH3 C*(OH"(H2O"5

4 4 H2O

'e8rt$/ti$ re/+ti$s =te$ res*lt i$ the =rm/ti$ = / 8re+i8it/te>

Re/+ti$ = +m8le- i$s 0ith s)i*m hy)r-i)e /$) /mm$i/ sl*ti$s0odium hydroxide and ammonia solutions contain hydroxide ions. <hen these are added tosolutions containing transition metal ions a precipitate of the metal hydroxide is formed.

!f further quantities of these reagents are added to the mixture, the precipitate, in certain cases,dissolves.

!on insolution

Reaction with a few dropsof a"/6aq7 or /=6aq7

Reaction with excessa"/6aq7

Reaction with excess/=6aq7

#r =3 Hale green ppt Hpt dissolves to form a deepgreen solution

n23 +eige ppt o further reaction

5e23 Eirty green ppt o further reaction o further reaction5e=3 Red-$rown ppt o further reaction o further reactioni23 Oreen gelatinous ppt o further reaction Hpt dissolves to form a

$lue solution#u23 +lue ppt o further reaction Hpt dissolves to form a

deep $lue solutionIn23 <hite gelatinous ppt Hpt dissolves to form a

colourless solution

A simple way of looking at these is as hydroxide ions adding to the transition metal ion P thenum$er of hydroxide ions $eing equal to the charge on the ion.

M6/2"7xLy3 3 y"/- 4 M6/2"7x-y6"/7yL 3 y/2"

!n fact rather than a water molecule leaving and a hydroxide ion oining, the process actuallyconsists of a hydrogen ion moving.

This process is called a deprotonation reaction. All the reactions in which the transition ions form8re+i8it/tes are deprotonation reactions.<hen the precipitates dissolve in excess sodium hydroxide solution, it is $ecause a furtherdeprotonation reaction takes place. These reactions represent a metal hydroxide reacting with analkali, so it can $e regarded as $eing due to the amphoteric nature of the metal hydroxide andreflects a degree of non-metal character.

- 20 –

C2$)H2

)H2

H2)

H2)

H2)

H2)

)H-

)H-

H$

C2$

)H

)H

H2)

H2)

H2)

H2)

H2)

H2)H$



<hen the precipitate dissolves in excess ammonia solution, it is $ecause the ammonia moleculesreplace the hydroxide and water molecules around the transition metal ion, and so this is a ligandreplacement reaction.

i$ /.*e*s s)i*m hy)r-i)e /.*e*s /mm$i/ sl*ti$

Cr :4 #r =36aq7 3 ="/-

6aq7 #r6"/7=6s7

grey green

#r6"/7=6/2"7=6s7 3 ="/-6aq7 M#r6"/7;L=-

6aq7 3 =/2" deprotonation hexahydroxochromate6!!!7

#r =36aq7 3 ="/-

6aq7 #r6"/7=6s7

grey green

M$24

n

23

6aq7 3 2"/

-

6aq7 n6"/726s7 white?$rown n

23

6aq7 3 2"/

-

6aq7 n6"/726s7 white?$rown

Fe24 5e236aq7 3 2"/-6aq7 5e6"/726s7 muddy?green

5e236aq7 3 2"/-6aq7 5e6"/726s7 muddy?green

Fe:4 5e=36aq7 3 ="/-6aq7 5e6"/7=6s7 rust $rown

5e=36aq7 3 ="/-6aq7 5e6"/7=6s7 rust $rown

Ni24 i236aq7 3 2"/-

6aq7 i6"/726s7

lime greeni23

6aq7 3 2"/-6aq7 i6"/726s7

lime greenli1/$) e-+h/$1e - dissolves in excess togive $lue Mi6/=7;L236aq7

C*24 #u236aq7 3 2"/-

6aq7 #u6"/726s7

pale $lue

#u236aq7 3 2"/-

6aq7 #u6"/726/2"7(6s7

pale $lueli1/$) e-+h/$1e - dissolves in excess togive deep $lue M#u6/=7(6/2"72L236aq7

@$24 In236aq7 3 2"/-

6aq7 In6"/726s7

whiteIn6"/726s7 3 2"/-

6aq7 MIn6"/7(L2-6aq7deprotonation

In236aq7 3 2"/-

6aq7 In6"/726s7

li1/$) e-+h/$1e - dissolves in excess togive MIn6/=7(L236aq7

C/t/lyti+ Pr8erties = Tr/$siti$ Eleme$ts

- 21 –

Tr/$siti$ met/l sl*ti$

'"eciitate f"./

Addition of sodium hydroxideor ammonia solution Prt$ tr/$s=er

Addition of sodium hydroxide solution Addition of ammonia solution

'"eciitate di//le/ '"eciitate di//le/

Li1/$) e-+h/$1ePrt$ tr/$s=er



The a$ility of transition elements to change oxidation state allows them to $e used as catalysts.Transition elements or their compounds are used in a num$er of important industrial processes.

0u$stance Reaction catalysed!ron /a$er process to convert nitrogen and hydrogen to ammoniaickel argarine production to hydrogenate unsaturated hydrocar$ons

8anadium687 oxide #ontact process to convert oxygen and sulphur dioxide to sulphur trioxide There are two main types of catalysis. Heter1e$e*s +/t/lysis is where the catalyst and the reactants are in different states.Hm1e$e*s +/t/lysis is where the catalyst and the reactants are in the same state.

An example of heterogeneous catalysis is seen in the catalytic converters found in cars.Hlatinum is used to remove harmful gases such as car$on monoxide.

2CO 4 O2 9 2CO2

Hlatinum provides a surface with which the car$on monoxide and oxygen can forms weak $onds.The transition metal uses =d and (s electrons to form these $onds. The $onding of the moleculeson the surface $rings them close together and the formation of the weak $onds with the surface

weakens the $onds within the molecules, reducing the energy required to $reak them.The a$ility of the transition element to change its oxidation state is also important in catalysis.This is seen in the manufacture of sulphuric acid where the conversion of sulphur dioxide tosulphur trioxide is catalysed $y vanadium687 oxide.

2SO2 4 O2 2SO:

The sulphur dioxide first reacts with the vanadium687 oxide. SO2(1" 4 &2O5(s" 9 SO:(1" 4 &2O(s7

The vanadium6!87 formed in this reaction then reacts with the oxygen O2(1" 4 &2O(s" 9 &2O5(s7

An example of homogeneous catalysis is seen in the catalysis $y iron6!!7 of the reaction $etweenthe persulphate and iodide ions. S2O

23 4 2I3 9 2SO23 4 I2

Although the persulphate is a powerful oxidiGing agent, the reaction is slow $ecause it requiresnegative ions to come together and this repulsion gives the reaction a high activation energy.

The iron6!!7 reacts first wit the persulphate 2Fe24 4 S2O23 9 2SO

23 4 2Fe:4

The iron6!!7 formed in this step then reacts with the iodide ions 2Fe:4 4 2I3 9 2Fe24 4 I2

'evel8me$t = Ne0 C/t/lystsEevelopment of new catalysts is an important area for research:thanoic acid is a very important industrial chemical used in the manufacture of polymers,perfumes, flavourings and pharmaceuticals. Until '*B), ethanoic acid was manufactured $yoxidiGing naptha and $utane at a temperature of 2))o#, a pressure of &)atm and a catalyst ofco$alt ethanoate. The large num$er of $y-products gives the reaction a low atom economy.

!n the '*;)s a new process $egan to $e used starting with methanol and car$on monoxide.CO 4 CH:OH 9 CH:CO2H

A catalyst of co$alt and iodine is used for this reaction. !t had a theoretical atom economy of'))C, although in practice it did not reach this.

- 22 –



8arious improvements were made to the catalyst used, using rhodium and then iridium in place of co$alt. As improvements were made the conditions required $ecame milder, the reaction moreefficient and the atom economy improved.

Other Uses = Tr/$siti$ Eleme$ts

C/$+er tre/tme$t

#ancer involves cells dividing uncontrolla$ly forming tumours.#is Ht6/=72#l2 was found to $e a$le to inhi$it cell division and could therefore $e used as atreatment for cancer.#isplatin as the material was called is now one of the most widely used anti-cancer drugs. "ne of the pro$lems with it is its toxicity, so research continues and a new drug, car$oplatin, has $eendeveloped.

S*$1l/ssesHhotochromic sunglasses which $ecome darker as the light intensity increases use a redoxreaction. The lenses contain silver6!7 chloride and copper6!7 chloride. 0trong light causes thefollowing reactions to take place9

C*4 4 A14 9 A1 4 C*24

The silver produced in this reaction turns the glasses darker. <hen the light intensity decreases,the reverse reaction takes place and the glasses $ecome less dark.

- 23 –

't

(H3

(H3

Cl

Cl

C

)

)CH2

)

't

(H3

(H3

C

C )CH2

H2CCi/latin

Ca"latin



Or1/$i+ Chemistry Are$es /$) Phe$ls

Are$es 3 Str*+t*re = %e$Je$eThe term arene includes all compounds with a delocalised V-system9 these are also called aromatic compounds. e.g. $enGene, #;/;.

%ekulW suggested the following structure for this compound.

X-ray diffraction studies provide information a$out $ond lengths.+ond lengths #-# in cyclohexane ).'&( nm #># in cyclohexene ).'== nmThis would give $enGene a distorted hexagon. /owever, X-ray diffraction studies show that the #-#$ond lengths in $enGene are all ).'=* nm. This means that the %ekule structure is incorrect.

The $enGene ring is a flat, regular hexagon, with six electrons in the delocalised VPSsandwich a$oveand $elow the ring.The Y-$onds are $uilt up in a similar way to ethene, leaving an unused 2p or$ital on each of the sixcar$ons. These can overlap sideways in $oth directions, resulting in a delocalised V-electron cloud,containing six V-electrons and stretching over all six atoms, in a Ssandwich which lies a$ove and$elow the plane of the ring9

Thus the #P# $onds in $enGene are all equal, each of length $etween a single and a dou$le $ond.

The delocalised structure is much more sta$le 6lower in enthalpy content7 than one with three dou$le$onds - pro$a$ly $y more than '&) kZ?mol -' and so $enGene is much less reactive than ethene, sincereaction involves loss of at least part of this extra sta$ilisation. :thene also has two π-electrons$etween the # atoms, while $enGene has only one.

Therm+hemi+/l evi)e$+e:vidence for this extra sta$le structure for $enGene is provided $y thermochemical evidence.

!n the presence of a nickel catalyst hydrogen can $e added to a dou$le $ond. <hen this is carriedout with cyclohexene the following reaction takes place.

C7H6 4 H2 9 C7H62 H ? 362 BKml36

This would suggest that a similar reaction using $enGene would give us the following reactionC7H7 4 :H2 9 C7H62 H ? 3:7 BKml36

The measured value for the hydrogenation of $enGene is actually -2)D kZmol-'.

- 24 –

120

H

CC

C

CCCH

H

H

H

H

HH

C C

C CC C

HH

H H

HH

HH

C C

C CC C

H H



This indicates that $enGene is more sta$le than would $e expected from a structure with three #>#.

S8e+trs+8i+ evi)e$+eThe infra-red spectrum of a compound containing a dou$le $ond shows an a$sorption $etween ';')and ';D) cm-'. This a$sorption however is a$sent in $enGene compounds.

Re/+ti$s = ,e$Je$e+enGene requires much more forcing conditions to make it react than does ethene9

• $enGene does not decolorise $romine water.• $enGene will not react with hydrogen at an apprecia$le rate under conditions at which ethene

reacts 6e.g. '&)[# and normal pressures, over a nickel catalyst7.• $enGene will not react with oxidising agents like alkaline %n"(.

Re/+ti$s = Are$esThe main reactions which $enGene does undergo are with electrophiles, $ut since it has a lowerelectron density $etween car$on atoms than ethene, $enGene requires stronger electrophiles andmore forcing conditions."nce an electrophile has added on to the ring, the su$sequent step is likely to $e loss of a proton,leading to ele+tr8hili+ s*%stit*ti$ rather than addition, since the sta$le delocalised V-system isregained.

Cm%*sti$ Arenes have the equivalent of three dou$le $onds per molecule and so tends to produce incompletecom$ustion, so when it $urns it produces a smoky flame.

Nitr/ti$ = Are$es Arenes can $e nitrated if they are mixed with conc nitric and sulphuric acids at temps $elow &)o#.

C7H7 (l" 4 HNO: (l" C7H5NO2 (l" 4 H2O (l"

itro$enGene

The sulphuric acid protonates the nitric acid, which ionises, forming the nitronium ion, "23

HNO: 4 2H2SO NO24 4 H:O

4 4 2HSO

+ NO2

+

NO2

+ H+

!f the temperature is raised a$ove &)o# there is a chance of multiple nitrations occurring.

- 25 –

#yclohexane, #;/'2

#yclohexatriene, #;/;

-=;) kZmol-'

Hredicted value of $ased on astructure with = dou$le $onds enene C6H6

-2)D kZmol-' actualvalue of hydrogenation

E-tr/ st/%ility = %e$Je$e



,rmi$/ti$ = Are$es Arenes will react with halogens in the presence of a halogen carrier catalyst, 5e, 5e+r =, Al or Al#l=.

C7H7(l" 4 ,r 2(l" C7H5,r(l" 4 H,r

+ Br2

Br

+ HBr

S*l8h$/ti$ = Are$es Arenes will react with concentrated 6fuming7 sulphuric acid to form sulphonic acidsThe reaction is carried out at room temperature.

C7H7 4 H2SO 9 C7H5SO:H 4 H2O

3 /20"(

0"=/

3 /2"

0ulphonic acid groups are often used in organic synthesis to increase the water solu$ility of largeorganic drug molecules.

Fei)el3Cr/=t AlByl/ti$

AlByl/ti$9 Reaction with a halogenoalkane in the presence of the catalyst Al#l=.

This is an important reaction in organic synthesis as it is a #-# $ond forming reaction.

Fei)el3Cr/=t A+yl/ti$

A+yl/ti$9 Reaction with an acyl chloride in the presence of the catalyst Al#l=.#onditions9 Anhydrous Al#l= as a catalyst.

C7H7(l" 4 CH:COCl (l" C7H5COCH:(l" 4 HCl(1"

acid chloride

3 3 /#l

4lCl3

#l

"

"

This is an important reaction in organic synthesis as it is a #-# $ond forming reaction.

- 26 –

$ CH3Cl CH3

lCl3

$ HCl



A))iti$ re/+ti$ 0ith hy)r1e$This reaction is unlike all those a$ove in that whereas most reactions of $enGene are su$stitution thisis an /))iti$ 8r+ess. !n the presence of a nickel catalyst hydrogen can add on to a $enGene ringacross the dou$le $onds.This reaction requires Raney nickel, which is a particularly finely divided form of the metal, and the

temperature of 2))o

# is higher than that used for addition to an alkene.The more extreme conditions for this reaction compared with those needed for addition of hydrogen toan alkene $ecause of the extra sta$ility of the delocaliGed electron ring structure.

+ 3H2

Nickel

200oC

Me+h/$isms 3 Ele+tr8hili+ s*%stit*ti$>

Arenes commonly undergo ele+tr8hili+ s*%stit*ti$ reactions.!n this type of su$stitution two of the delocalised MπL electrons on the $enGene ring are donated to the

electrophile. An unsta$le π-complex containing $oth an electrophile and a leaving group is formed as anintermediate.

Nitr/ti$itration is carried out under reflux at &&-;)o# using a nitrating mixture.This contains equal amounts of concentrated nitric acid and sulphuric acid.The sulphuric acid protonates the nitric acid, which ionises, forming the nitronium ion, "2

3

6sometimes called the nitryl cation79HNO: 4 2H2SO NO2

4 4 H:O4 4 2HSO

The nitronium ion is a powerful electrophile, and this pulls out an electron pair from the VPsystem,adding on to the ring9

$ ()2

/

"2

$

ote a car$ocationic intermediate is formed. This first step, addition of an electrophile, is similar to theattack of $romine on ethene./owever, the positive $enGene intermediate can lose much more energy $y giving up a proton to a$ase than it could $y adding on a nucleophile.The loss of a proton 6to an /0"(

P7 ion restores the full delocalisation energy, and sulphuric acid isreformed, as a catalyst.

/

"2

$

H)4-

"2

$ H2)4

- 27 –



,rmi$/ti$The reaction with $romine also follows the same mechanism.

Aluminium chloride is used as a h/l1e$ +/rrier +/t/lyst which helps form the electrophile. ,r3,r 4 AlCl: ,r )4333,r333AlCl:

)3

The mechanism is then the same as for nitration.

$2"

/

+r

$

/

+r

$

4lCl32" -

+r

$ 4lCl3 $ H2"

The delocalised V-system requires a large activation energy to disrupt 6hence the need for a catalyst7.The positive $enGene intermediate once again restores the full delocalisation energy $y losing /3 6hence su$stitution rather than addition7.

#hlorine reacts in a similar way, giving #;/&#l and /#l, and also needing Al or 5e to $e present6forming Al#l= or 5e#l=7.

AlByl/ti$

Alkylation follows the same mechanism.

/alogenoalkanes are weak electrophiles $ecause their polar $ond. e.g.#/=d3-+r d- .

The catalyst Al#l= makes the halogenoalkane a $etter electrophile. CH:3,r 4 AlCl:

)4CH:333,r333AlCl:)3

A+yl/ti$

Acylation also follows the same mechanism.

The electrophile is again improved $y Al#l=. CH:COCl 4 AlCl: CH:C4?O 4 Cl3AlCl:

3

- 28 –



Phe$ls

Hhenols are compounds in which the -"/ group is directly attached to the $enGene ring.Hhenol itself is a white crystalline solid which is sparingly solu$le in water at room temperature.The $enGene ring helps to sta$ilise a negative charge on the phenoxide ion, #;/&" P, and thismakes phenol apprecia$ly acidic 6unlike ethanol, which is neutral, a solution of phenol in water

has a p/ of a$out &7.

Re/+ti$s = 8he$l

Phe$l 0ith s)i*m hy)r-i)eHhenol dissolves in aqueous sodium hydroxide $ecause phenol $ehaves as an acid and gives upits proton to the hydroxide ion which is a $ase. A solu$le ionic product is formed.

phenoxide ion

Phe$l 0ith %rmi$e 3 ele+tr8hili+ s*%stit*ti$>The hydroxyl group in phenol can donate electrons $ack to the delocalised V-system, helping tosta$ilise the intermediates of electrophilic su$stitution and so making phenol much more reactivethan $enGene. !t will react imme)i/tely with $romine water, )e+lrisi$1 it and forming a 0hite8re+i8it/te of 2,(,;-tri$romophenol.

Nitr/ti$ = Phe$l>

Hhenol can $e nitrated with )il*te $itri+ /+i). This once again shows that the delocalised V-system, makes phenol much more reactive than $enGene.

"/

dil H()3

"/

"2

$

"/

"2

2

Uses = 8he$lsHhenol, in dilute solution, was the first successful antiseptic used $y Kister 6called carbolic acid 7.ow su$stituted phenols are used $oth as /$tise8ti+s 6to keep surfaces free of pathogens7 andas )isi$=e+t/$ts 6to kill pathogens already present7.

- 29 –

)H $ )H – → ) – $ H2)

)H $ 3" 2 → )H"

"

"

2,4,6-t"i".henl

$ 3H" Mreaction with chlorine gives 2,(,;-trichlorophenol, or T#H, used as anantisepticL



Nitr1e$ Cm8*$)s /$) Plymeris/ti$

Ami$es A 8rim/ry /mi$e is one in which a single alkyl group is attached to the nitrogen, e.g. R/2.

A se+$)/ry /mi$e has two alkyl groups, and a terti/ry one three, directly attached to the nitrogen9C2H5 NH2 C2H5 NHC2H5 (C2H5":N

primary secondary tertiaryethylamine diethylamine triethylamine

Nte - This is a different usage from that employed for alcohols and halogenoalkanes, where it is thenum$er of alkyl groups directly attached to the carbon atom $onded to " or #l which determinesprimary, secondary or tertiary.

An /ryl /mi$e has the amine group directly attached to the $enGene ring.e.g. phenylamine, #;/&/2.

Ami$e Pre8/r/ti$

Aliphatic 6non aryl7 amines are produced $y heating the appropriate halogenoalkane in a sealed tu$ewith excess ammonia dissolved in ethanol. M0ee Unit 2L.This initially forms the salt of the amine and the amine can $e o$tained $y adding sodium hydroxide.

CH:CH2CH2CH2,r 4 NH: 9 CH:CH2CH2CH2NH:4,r 3

CH:CH2CH2CH2NH:4,r 3 4 N/OH 9 CH:CH2CH2CH2NH2 4 H2O 4 N/,r

Aromatic amines are produced $y the reduction of a nitro$enGene. This is carried out $y addingconcentrated hydrochloric acid to a mixture of nitro$enGene and tin and heating the mixture under reflux.

C7H5NO2 4 7H 9 C7H5NH2 4 2H2O Again in this reaction the amine is actually formed as the chloride salt, so sodium hydroxide hasto $e added to release the amine which is then extracted from the mixture $y steam distillation.

Pr8erties = /mi$es

0ince amines contain hydrogen atom directly $onded to a nitrogen atom, they can form hydrogen$onds. This ena$les amines to interact with water molecules and so amines with short car$on chainsare misci$le with water.<hen dissolved in water, like ammonia, they form an alkaline solution.

R3NH2 4 H2O R3NH:4 4 OH3

As the car$on chain length increases, the amines $ecome less solu$le in water.Hhenylamine is only slightly solu$le in water $ecause of the large car$on group.

Also like ammonia, they are $ases, and they react with acids to form salts. CH:CH2CH2CH2NH 2 4 HCl CH:CH2CH2CH2NH 3

+Cl -

+utyl ammonium chloride

- 30 –



This a$ility to react with acids to form salts ena$les phenylamine to dissolve in a concentratedhydrochloric acid. C7H5NH 2(l) 4 HCl (/." 9 C7H5NH 3

+Cl - (/."

Addition of sodium hydroxide to this solution causes the reverse reaction.C7H5NH:

4Cl3(/." 4 N/OH(/." 9 C7H5NH2(l" + H2O(l" + N/Cl(/."

Re/+ti$ 0ith h/l1e$/lB/$esThe lone pair on the nitrogen will attack areas of positive charge.

Amines are therefore nucleophiles and will attack the δ3 car$on in halogenoalkanes.Hrimary amines react with haolgenoalkanes to form secondary amines and then teriary amines.

CH:NH2 4 CH:CH2,r CH:NHCH2CH:

Hrimary amine /alogenoalkane 0econdary amine

The nitrogen atom in a secondary amine still has a lone pair that can attack the halogenoalkane.

CH:NHCH2CH: 4 CH:CH2,r CH:N(CH2CH:"2

0econdary amine /alogenoalkane Tertiary amine

Re/+ti$ 0ith /+i) +hlri)esThe lone pair on the nitrogen will attack areas of positive charge. Amines are therefore nucleophiles.

Amines react with acyl chlorides to form /+i) /mi)es.

CH:COCl 4 2RNH2 CH:CONHR2 4 RNH:Cl an amide

An example of this reaction is the formation of paracetamol.

3 #/=#"#l 4 3 /#l

(-aminophenol Haracetamol

Re/+ti$ 0ith li1/$)s t =rm +m8le- i$s

The lone pair of electrons on the nitrogen of the amine means that it can form complex ions. An example of the such a complex is the one formed $y the $identate ligand ',2-diaminoethane.

- 31 –

(H2

)H

(HC)CH3

)H



Frm/ti$ = AJ )yesitro$enGene is heated under reflux with tin in conc. /#l as a reducing agent.

amino$enGene 6phenylamine7

Re/+ti$s = 8he$yl/mi$e>itrous acid, /"2, is unsta$le, so hydrochloric acid and sodium nitrite are used to make it9

a"2 3 /#l → a#l 3 /"2

Hhenylamine is dissolved in moderately concentrated hydrochloric acid, and cooled to a$out &[#. A solution of sodium nitrite, a"2, is also cooled to &[#, and added slowly to the phenylamineand acid, cooling to keep the temperature $elow ')[#. The phenylamine reacts to form$enGenediaGonium chloride, #;/&2

3#l P9

+enGenediaGonium chloride is unsta$le, and is used, in solution, immediately after it has $eenprepared P for example, to make an aGo-dye.

<hen cold $enGenediaGonium chloride solution is added dropwise to a cold solution of phenol inaqueous sodium hydroxide, an yellow?orange precipitate of an aGo-dye is o$tained9

diaGonium ion phenoxide yellow?orange aGo-dye

Ami)es Amides are unreactive car$oxylic acid derivatives.

Amides have the general structure9 R-#"-/2 #/=#"/2 #/=#/2#/2#"/2

ethanamide $utanamide

Ami)e 8re8/r/ti$ Amides can $e prepared $y reacting an acyl chloride with concentrated ammonia.The reaction occurs vigorously at room temperature forming fumes of /#l and solid amide.

CH:COCl 4 NH: 9 CH:CONH2 4 HCl:thanoyl chloride ethanamide

- 32 –

()2 $ H$ $ 6*H+n $ cnc HCl (H3

$ $ 2H2)

(H3

$

$ )H

– net"ali/e ith

(a)H (H2

(H2 $ H()2 $ H$ $ Cl – (( $ Cl – $ 2H2)$

(( $$

) – (( )H



Plymeris/ti$A))iti$ 8lymeris/ti$The process of 8lymeris/ti$ is the $uilding up of long-chain molecules from smallmolecules, which are called the m$mers. !n /))iti$ 8lymeris/ti$ there are no otherproducts, while in +$)e$s/ti$ 8lymeris/ti$ a small molecule such as water or /#l iseected every time a link is made. Addition polymers are normally made from compounds

containing car$on-car$on dou$le $onds, so that the process is essentially similar to theformation of poly6ethene7.

Hhenylethene 6styrene7 can $e heated in an inert solvent 6paraffin7 under reflux, with a smallamount of a peroxide compound as initiator. !t gradually forms the polymer, which can $eextracted as a waxy solid9

C C

HH

H

n C

H

H

C

H

C

H

H

C

H

C

H

H

C

H

etc

PhPhPhPh

Ph C H6 5

phenylethene poly6phenylethene7 or polystyrene

C$)e$s/ti$ 8lymeris/ti$ As explained a$ove, +$)e$s/ti$ 8lymeris/ti$ is the process $y which long-chainmolecules are formed $y reaction $etween $ifunctional monomer molecules, with the loss of onesmall molecule 6such as water or /#l7 for each link which is formed. ormally there are twodifferent monomer molecules, as in the case of the diacid and diol in 6a7 $elow, $ut sometimes$oth the types of reactive group may $e contained within the same molecule.

(a) Polyesters!f molecule A contains two car$oxylic acid groups, and molecule + contains two alcohol groups,when they react to form an ester the product will have an acid group at one end and an alcohol atthe other9

.lecle .lecle $ H2)

2CH

2)H$ H)CH

2CH

2)H –C–)HH) –C–

) )

H)–C–

)

–C–)CH

)

enene-1,4-dica":#lic acid ethane-1,2-dil

The left hand car$oxyl can then react with the "/ of another molecule +, while the righthand "/can react with another molecule A. This process is repeated, $uilding up a long chain polymer9

ote that each time an ester group is formed a water molecule is lost. This type of polymer isknown as a polyester, and this particular example, made from $enGene-',(-dicar$oxylic acid 6A7

and ethane-',2-diol 6+7, is marketed $y !#! as Terylene\.

Holyesters are suita$le for fi$res, and are widely used for modern crease-resistant syntheticmaterials, for ropes and sails. 0ince they contain ester links, they will $e slowly hydrolysed $yacids, and more rapidly $y alkalis.

!f $oth the alcohol and the acid groups are in the same monomer, it may polymerise with itself.5or example, the molecule /"P6#/27; P#""/ might form9

P"P6#/27; P#"P"P6#/27; P#"P"P6#/27; P#"P"P6#/27; P#"P

- 33 –

)

–C– –C–)–CH2CH

2 –)

) )


2 –)

) )


2 –)–

)



(b) Polyamides<hen a compound containing two acid groups, e.g. /"#"6#/27(#""/, reacts with anothercompound containing two amine groups, e.g. /26#/27;/2, an amide is formed with loss of /2"9

HOCO(CH2"COOH 4 NH2(CH2"7NH2 HOCO(CH2"CONH(CH2"7NH2 4 H2O

The amide produced still has reactive groups at either end, and can react with another diamine onthe left, and a diacid on the right, to form a chain four units long. The process is repeated givingthe polymer called $yl$7<7 6$ecause there are ; car$on atoms in each monomer unit79

CO(CH2"CONH(CH2"7NHCO(CH2"CONH(CH2"7NHCO(CH2"CONH(CH2"7NH

The re8e/t *$it here is9 P#"6#/27(#"P/6#/27;/P

The product, a polyamide, is useful for making fi$res, since the long chains can $e drawn out intofilaments, which causes them to line up, and is also a hard-wearing solid polymer 6e.g. the casesof d-i-y tools, like electric drills, are made of nylon, as are curtain hooks7.

Devl/r is made from $enGene-',(-diamine and $enGene-',(-dicar$oxylic acid, and the repeatunit is9

%evlar is an extremely tough fi$re, and is used in e.g. $ullet-proof vests.5or a ca$le of a particular diameter %evlar has the same strength as steel, $ut is five times lighter.!t is used in $ullet-resistaent clothing and in aircraft wings. !t is used to some extent in tyres for/O8s where it can make the tyre *kg lighter. !ntroduction for this purpose has $een slow as tyremanufacturers have invested heavily in the use of steel for reinforcing tyres.

- 34 –

C

)

H

(C

)

H

(

n



'r/0i$1 8lymers /$) =i$)i$1 re8e/t *$itsNou need to $e a$le to predict the repeat unit of a polymer from a given set of monomers, or to $ea$le to identify the monomers from a given length of polymer.

e(g( )raw a section of polymer which might be obtained from the molecules

/"P#;/( P"/ and #/=#/6#"#l72, and identify the repeat unit(

Answer9

–)–C)–C–C)–)– –)–C)–C–C)–)–

"eeat nit

–) –

H

CH3

CH3

H

CH3

H

–)–C)–C–C)–)–

Eraw the two repeat unit lines in so that the same groups occur either side, in the same order.

Nte th/t, when an alkene is symmetrical, like #52>#52, the minimum needed to specify thepolymer chain is P6#527n P. Argua$ly, therefore, this is the repeat unit, though some examinersfavour showing the monomer unit in the $rackets, i.e. P6#52 P#527n P.!n the unlikely event of your getting such a simple question, explain the situation and put $othdown9e.g. *ive the repeat unit from polymerisation of tetrafluoroethene(

Answer9 the chain formed is P6#52 P#527n P, though the minimum repeat unit is P6#527n P

T 0rB *t the m$mer *$its =rm 8lymer +h/i$s

'es it +$t/i$ ester (CO2 < /ls 0ritte$ OCO< r COO " 1r*8s i$ the +h/i$ if so,it is a polyester, and each group splits into P#""/ and /"P when identifying the monomer units.

'es it +$t/i$ /mi)e (CONH" 1r*8s i$ the +h/i$ if so, it is a polyamide, and each groupsplits into P#""/ and /2 P when identifying the monomer units.

!f it contains neither, $ut has a continuous car$on chain, it is pro$a$ly an addition polymer, andeach pair of # atoms in the chain originally had a dou$le $ond $etween them.

5or a polyamide it is $est to draw lines $etween the #" and the /, and make the #" into#""/, and the / into /29

e(g( +hat are the monomer units of the following polymer?

(H C)CH CH C) (H (H C)CH CH C) (H2 2 2 2

5irst, sketch in dotted lines to identify the repeat unit6s79 make sure they go in sensi$le positions6i.e. at an ester or amide linkage79

(H C)CH CH C) (H (H C)CH CH C) (H2 2 2 2

- 35 –



!n this case, note that the repeat unit for the polymer contains two of these sections. Themonomer units are9

/"#"#/2#/2#""/ and /2 P P/2

e(g( +hat isare the monomer unit&s' of the following polymer?

]#]#]#]#]#]#]#]#]

Answer-

# #

#/ter Attr/+ti$1 A))iti$ Plymers Addition polymers can $e made from ethenol and propenamide.

#/2>#/"/

ethenol

#/2>#/#"/2

The alcohol group and the amide groups in these polymers allows them to form hydrogen $ondsand so interact with water.

Ply(ethe$l" dissolves in water. !t is used in dissolving laundry $ags. These are used inhospitals where soiled laundry can $e moved without $eing handled directly and when placed in awashing machine the plastic dissolves away. !t is also used for detergent capsules which are placed in washing machines, the plasticdissolves in the water and releases the detergent.

Ply(8r8e$/mi)e" does not dissolve in water, $ut can a$sor$ water molecules and so $ecomessofter in water, so it is used to make soft contact lenses.

Ami$ /+i)s

- 36 –

H

C

H

H

C

)Hn

'l#%ethenl&

H

C

H

H

C

C)

(H2

n

'l#%"ena.ide&

CH3 CH3 CH3CH3H H H H

H H H H " " " "

CH3

H

H

"



Amino acids contain $oth an amine 6/27 and a car$oxylic acid group 6#"2/7.Their general formula is R#/6/27#""/, where R represents a side-chain 6not ust an alkylgroup7.There are a$out twenty amino acids which are found in nature, and which com$ine to make upthe proteins found in living organisms.They are all ^-amino acids, i.e. the /2 and the #""/ groups are attached to the same car$onatom.Three examples are given $elow9

/2 ]#]#""/ /2 ]#]#""/ /2 ]#]#""/

glycine alanine cysteine

Re/+ti$ 0ith /+i)sKike amines they $ehave like $ases and react with acids.

CH:CH2 CH(NH2"COOH (/." 4 H4 (/." CH:CH2 CH(NH:"4COOH (/."

Re/+ti$ 0ith %/sesKike car$oxylic acids they react with $ases.

CH:CH2CH(NH2"COOH (/." 4 OH3 (/." CH:CH2 CH(NH2"COO3 (/." 4 H2O(l"

@0itteri$ i$ str*+t*re!n solution the acidic hydrogen of the amino acid is lost and can attach itself to the nitrogen atomin the same molecule.The result is called a Gwitterion. #/=#/2 #/#""-

_ /=

3

The species formed contains a cation and an anion group. The groups which are ionised will

depend upon the p/ of the solution in which they are dissolved.!n a solution of low p/, where there is a high concentration of hydrogen ions, the acid group willtend to accept a proton. !n a solution of high p/, where there is a low concentration of hydrogenions, the amine group will tend to release a proton. This means that a particular amino acid can$e found in three forms according to the p/.

- 37 –

R C

C)2

-

H

(H3

$R C

C)2H

H

(H3

$ R C

C)2

-

H

(H2

L H ;de"ate H High H

H

H H H

CH3 CH2

H



<hen two amino acid units oin together via and amide link, they form a dipeptide.This is a condensation reaction.

/2 P#/P#""/ 3 /2 P#/P#""/ → /2 P#/P#PP#/P#""/ 3 /2"

+y convention the amino acid with the un$onded /2 group is shown on the left.

<hen amino acids are oined in this way the amide link is called a peptide $ond.!t is possi$le for many amino acid to oin in this way to produce a polypeptide.

A protein is made up of one or more polypeptide chains.

The amino acid present in a protein can $e investigated $y• first hydrolysing the protein using concentrated hydrochloric acid• then $y separating the various amino acids and identifying them using chromatography.

0ince amino acids are colourless, they cannot $e seen on the paper or thin layer and so amaterial is used which reacts with them forming a coloured product1 such a material is known as alocating agent."nce the chromatography has $een carried out, the paper will $e sprayed with $i$hy)ri$ andplaced in an oven at a$out '))o#. The amino acids are shown up as purple spots 6although turn$rown with time7. The various amino acid can then $e found from their Rf value.

Rf >

Or1/$i+ Sy$thesis

"rganic synthesis is a$out making organic compounds.The synthesis of new materials is important in the production of new dyes, pharmaceuticals,polymers, catalysts and antiseptics.0ome of the medicines that have $een developed include aspirin, paracetamol 6analgesics7,sal$utamol 6asthma treatment7 and chloramphenicol 6typhoid treatment7.

- 38 –

ta"ting line

lent f"nt

Eistance moved $y compound

Eistance moved $y 0olvent front

#/= #/20/#/= #/20/

/

"

Amide &peptide' link



"rganic chemists design synthetic pathways to convert an availa$le starting material into adesired target molecule 6product7. The pathways may involve several steps.

!f the num$er of car$on atoms in the chain is9

` i$+re/se) $y one, consider9a7 halogenoalkane with %# 4 su$stitutes with halogen.$7 car$onyl with /# 4 forms hydroxynitrile.

` i$+re/se) $y mre than one, consider9 5riedel-#rafts reaction for aromatic su$stances.

` )e+re/se) $y one, consider the iodoform reaction. R3CO3 CH: R3CO23N/4

S/=ety i$ r1/$i+ sy$thesis>"rganic compounds may $e haGardous $ecause of9

• 5lamma$ility - Use in small amounts avoids the undue risk of fire. Avoid naked flames.Use electrical heaters and water $aths.

• Toxicity - The use of small amounts, fume cup$oards, gloves and normal la$oratory safety

procedures reduces the risk of harmful amounts of a chemical entering the $ody $yinhalation, ingestion or $y skin a$sorption.• on-$iodegrada$ility - 0ome su$stances do not decay naturally in the environment. The

haGard is reduced $y using small quantities, and pouring waste solvents in a suita$lecontainer rather than pouring it down the sink.

Or1/$i+ Pr/+ti+/l te+h$i.*es

He/ti$1 *$)er re=l*- is necessary when either the reactant has a low $oiling temperature orthe reaction is slow at room temperature. :na$les reactions to $e heated at their maximumtemperature without the loss of any volatile reagents or products.

Dey 8i$ts P #ondenser is vertical.Eo not put a $ung in the condenser. 6Hressure will $uild up7<ater enters the condenser at the $ottom./eat electrically P to avoid naked flames.

- 39 –



P*ri=i+/ti$ Te+h$i.*es/" Re+ryst/llis/ti$>

This is used to purify an impure organic solid.• #hoose a suita$le solvent. The solvent is suita$le if the product is insolu$le in the cold

solvent $ut solu$le in the hot solvent.• Eissolve the impure sample in the minimum volume of hot solvent.

• 5ilter the solution hot under reduced pressure and collect the filtrate. This removes solidimpurities which were insolu$le in the solvent.• Allow the filtrate to cool so that crystals of the product form.• Again filter the mixture under reduced pressure. 0olu$le impurities are now removed.• <ash the residue with a little cold solvent.

Ery the residue which should then $e the pure product.

$7 Fr/+ti$/l )istill/ti$>

This technique has a num$er of important applications9• used to separate the components of liquid air1 The air is compressed and cooled to liquefy

it. 5ractions are oxygen -'D=o#, argon -'D;o#, and nitrogen -'*;o#.• used to separate fractions from petroleum1 The fractions are $itumen b=&)o#, fuel oil =))

o#, diesel 2() o#, kerosene 2)) o#, naptha '2) o#, petrol () o#, KHO 2& o#.• used to produce whisky and other alcoholic drinks. 5ractions are ethanol BD o# and water

')) o#.

'etermi$/ti$ = melti$1 tem8er/t*re>• A pure solid has a sharp melting point which can $e found in a data $ook.• !f a solid product has $een purified it can $e identified from its melting point or if we know

what it is we can tell if it is pure.• !mpurities lower the melting point.

• The melting point apparatus is heated slowly until the solid is seen to melt.• The temperature is recorded.• After the sample is melted it is allowed to cool.• <hen the first crystals of solid appear in the sample the temperature is recorded again.

- 40 –



M)er$ Or1/$i+ sy$thesis 8r+esses

"ne approach is to consider the purpose of the molecule, determine the structure required andthen find a way of producing that molecule. +acteria $ecome resistant to anti$iotics, thedevelopment of new anti$iotics is important. The $asic structure for the active component of an

anti$iotic molecule for $locking certain receptor sites in the vital enGymes of $acteria are known,and so it is possi$le to find a way of introducing this component into a new molecule to which the$acteria will not have resistance.

"ften natural molecules perform a perfect function, yet cannot $e produced on sufficient scale. A synthetic chemist can carry out analysis to determine the structure of such a molecule anddetermine a way of producing the su$stance on a larger scale.

8ast num$ers of new molecules are produced $y +m%i$/tri/l +hemistry> +y this process, large num$ers of reactions are carried out on an almost haphaGard way.5or example it is known that amines react with acyl chlorides. ixing of various amines and acylchlorides can then $e carried out controlled $y computer

Amine ', 2, =, (, &, ;, B, D, *,'), '', '2 are mixed with acylchloride A

Amine ', 2, =, (, &, ;, B, D, *,'), '', '2 are mixed with acylchloride +

This process would $e carried out for D acyl chlorides. This process would yield *; newcompounds all made automatically. These would then $e tested to see if they are of medical orother use. +y this procedure, many thousands of new compounds can $e produced in a day.

Another way of carrying out com$inatorial chemistry is the use of polymers.<hen a given su$stance is treated in a num$er of steps, traditionally after each step the productwould need to $e separated and purified. +earing in mind that each step will involve loss ofmaterial, the final product will pro$a$ly have a low yield. !n a new technique, the molecules of thestarting material are attached to insolu$le polymer $eads. They can then $e treated with thevarious chemicals, after each stage impurities can ust $e washed away while the molecules stayfixed to the polymer. <hen all the processing has $een carried out, the new su$stance can $edetached from the polymer again giving a high yield.

- 41 –



Or1/$i+ A$/lysis Analysis is a key tool for organic synthesis. "n a $asic level this analysis has included9

5ormula mass determination uantitative analysis #om$ustion analysis #hemical tests to identify functional groups present

ore sophisticated methods, using sensitive instrumentation are vital to modern organicsynthesis. 0uch methods include9

• 8arious types of chromatography• ass spectrometry• !nfra red spectrometry• uclear magnetic resonance spectrometry

!t is important that analytical methods are very sensitive in order to• 0uccessfully analyse products that may $e in low concentration• Hrovide information on the purity of the synthetic product

• !dentification of impurities, especially those that may $e toxic at low concentration

!nformation a$out the presence of impurities is particularly important as any process can $ecustomised in terms of temperature, pressure, catalyst or solvent in order to maximise the yield of desired product and minimise the production of $y-products and impurities.

Sy$thesis 8r%lems!n a synthesis pro$lem it will $e necessary to descri$e how a particular compound can $e madefrom a given starting material.Reagents and reactions conditions will need to $e given, $ased on reactions looked at in thespecification.Hurification procedures will need to $e descri$ed, these include the following9

• <ashing with appropriate solvents.• 0olvent extraction 6in a separating funnel7.• Recrystallisation 6from a suita$le solvent7.• Erying 6with a suita$le drying agent e.g. anhydrous magnesium sulphate7• Eistillation.• 0team distillation 6if the product is temperature sensitive and water solu$le7.• Eetermination of melting point.• Eetermination of $oiling point.

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Complete Unit 5 Notes edexcel.doc

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