Edexcel A2 Chemistry 4 3 Notes

8/20/2019 Edexcel A2 Chemistry 4 3 Notes

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Unit 4: Chapter 1–

How Fast? - Rates

Rates – Definitions

Rate of reaction – is the change in the amount orconcentration of reactants or products per unittime (normally per second).

Rate equation – (rate = k[A]m[B]n). This is knownas a rate equation. [A] and [B] represent theconcentrations of A and B in mol dm-3, and k iscalled the rate constant. The indices m and n areusually whole numbers (1, 2 …) but they can befractional or zero.

Order of reaction – the order of reaction is thesum of the powers to which the concentrations ofthe reactants are raised in the experimentally determined rate equation (m +n).

Partial order – of one reactant is the power to which the concentration of thatreactant is raised in the rate equation.

Rate constant – the rate constant, k, is the constant of proportionality thatconnects the rate of the reaction with the concentration of the reactants. Itsvalue alters with temperature. A reaction with a large activation energy has alow value of k.

Half-life – is the time taken for any reactant concentration to fall to half of its

initial value. Rate-determining step – is the slowest step that controls how fast the overall

reaction occurs.

Activation energy – is the minimum energy needed by reactant particles(molecules or ions) before products can form.

Heterogeneous catalyst – these are catalysts that are in a different phase fromthe reactants.

Homogeneous catalyst – the catalyst and the reactants are in the same phase,usually the gas phase or in solution.

A Che

Specification Reference:

Demonstrate an understanding of the terms ‘rate of reaction’, ‘rate equation’, ‘order of

reaction’, ‘rate constant’, ‘half -life’, ‘rate-determining step’, ‘activation energy’,

‘heterogeneous and homogeneous catalyst’.


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Experimental Methods

1) Colorimetry

If a reactant or product is coloured, the concentration of the coloured species canbe measured using a spectrophotometer. The amount of light of a particularfrequency that is absorbed depends on the concentration of the colouredsubstance.

Colorimetry measures the intensity of a colour in thereaction mixture with time, such as the in the oxidationof iodide ions to give brown iodine. In clock reactionsthe reaction is timed until a sudden colour changehappens when a certain amount of product is formed.

For example, in the reaction between propanone andiodine, the brown colour fades.

CH3COCH3(aq) + I2(aq)→

CH3COCH2I(aq) + H

+

(aq) + I

+

(aq)Colourless Brown

Colourless

2) Mass Change

Mass change is used when a gas is produced. For example, when calcium carbonatereacts with acids to release carbon dioxide the mass of the flask decreases.


Select and describe a suitable experimental technique to obtain rate data for a given

reaction, e.g. colorimetry, mass change and volume of gas evolved.

Rate of Reaction (mol dm-3 s-1) =ℎ

ℎ


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CaCO3(s) + 2HCl (aq) → CaCl2(aq) + H2O(l) + CO2(g)

Recording the loss in mass over a certain period of time at regular intervals usingthe apparatus shown gives an indication of the rate of reaction.

3) Volume of Gas Evolved

If the reaction produces a gas, the volume of gas produced can be measured atregular time intervals. The volume of gas is proportional to the moles of gas andcan, therefore, be used to measure the concentration of the product.

The rate of the reaction of an acid with a solid carbonate can be studied in thisway. The acid is added to the carbonate and the volume of carbon dioxide notedevery 30 seconds.

CaCO3(s) + 2H+(aq)→

Ca2+(aq) + H2O(l) + CO2(g)

‘Clock’ Reactions

In a „clock‟ reaction, the reactants are mixed and the time taken to produce afixed amount of product is measured. The experiment is then repeated severaldifferent times using different starting concentrations.


Investigate reactions which produce data that can be used to calculate the rate of thereaction, its half-life from concentration or volume against time graphs, e.g. a clock

reaction.


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1) The Iodine ‘Clock’

The oxidation of iodide ions by hydrogen peroxide in acid solution can beconsidered a „clock‟ reaction:

H2O2(aq) + 2I-(aq) + 2H

+(aq)→

I2(s) + 2H2O(l)

25cm3 of hydrogen peroxide solution is mixed in a beaker with 25cm3 ofwater and a few drops of starch solution are added.

25cm3 of potassium iodide solution and 5cm3 of a dilute solution of sodiumthiosulfate are placed in a second beaker.

The contents of the two beakers are mixed and the time taken for thesolution to go blue is measured.

The experiment is repeated with the same volumes of potassium iodide andsodium thiosulfate but with 20cm3 of hydrogen peroxide and 30cm3 of

water, and then with other relative amounts of hydrogen peroxide andwater, totalling 50cm3.

The reaction produces iodine, which reacts with the sodium thiosulfate. When allthe sodium thiosulfate has been used up, the next iodine that is produced reactswith the starch to give an intense blue-black colour.

The amount of iodine produced in the measured time is proportional to the volumeof sodium thiosulfate solution taken. Therefore, the average rate of reaction foreach experiment is proportional to 1/time.


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2. The Sulf ur ‘Clock’

Sodium thiosulfate is decomposed by acid, producing a precipitate of sulfur.

S2O32-

(aq) + 2H+(aq) → S(s) + SO2(aq) + H2O(l)

A large X is drawn on a white tile with a marker pen.

2cm3 of sodium thiosulfate solution is mixed with 25cm3 of water in abeaker.

25cm3 of dilute nitric acid is placed in a second beaker.

The first beaker is placed on top of the X and the contents of the secondone are added.

The mixture is stirred and the time (t) taken for sufficient sulfur to beproduced to hide the X when looking down through the beaker is measured.

The experiment is repeated with different relative amounts of sodium

thiosulfate and water, totalling 50cm

3

.

The number of moles of sulfur produced is the same in all experiments. Therefore,the average rate of reaction for each experiment is proportional to 1/t.

Half-life

Half-life – is the time taken for any reactant concentration to fall to half of itsinitial value.

Half-life of First Order Reaction

The following equation and graph shows the decomposition of sulfur dichlorideoxide:

SO2Cl2(g) → SO2(g) + Cl2(g)


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The time it takes SO2Cl2 to halve is 2300 seconds.

Notice that this time is constant – it takes the same time for the concentrationto fall from 0.50 mol dm-3 to 0.25 mol dm-3 as it takes to fall from 0.25 mol dm-3

to 0.125 mol dm-3.

If the half-life is short, the reaction is rapid; if the half-life is long, the reaction

is slow. The half-life is independent of concentration.

All first order reactions have constant half-lives at a given temperature.

Half-life and Radioactive Decay

Radioactive decay is an example of a first order reaction because its rate isindependent of the concentration of the radioactive material.

Many radioactive elements have very long half-lives and scientists use these tohelp to work out the age of rocks in the Earth‟s crust.

Carbon-14 is radioactive.

Radioactivity of carbon-14 halves every 5730 years.

The difference between the historical ratio and that measured in the massspectrometer is the basis of carbon-14 i.e. the percentage abundance ofradioactive isotopes is measured.

Half-life in Second Order Reactions

The half-life of a first order reaction is independent of the initial concentration ofthe reactants. However, the half-life of a second order reaction does depend onthe initial concentrations of the reactants.

The graph above shows the change in concentration of HI with time for the

reaction:2HI(g) → H2(g) + I2(g) (@508⁰C)


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The initial HI concentration is 0.10 mol dm-3 and it takes 125 seconds for theconcentration to drop to 0.05 mol dm-3 – the half-life for the first part is 125seconds.

Think of 0.05 mol dm-3 as the new “initial concentration”, which takes 250seconds to fall to 0.0250 mol dm-3. So in second order reactions, halving the

initial concentration doubles the half-life.

Half-life is inversely proportional to the initial concentration of the reactants.

Concentration-Time Graphs

The simplest type of relationship between variables is the linear one of generalform y = mx + c, where m is the slope or gradient and c the intercept on the y -axis.

The gradient of a curve is found by drawing a tangent. The gradient of thetangent to the concentration–time graph at a particular time gives the rate

at that moment (concentration). The initial rate of reaction is found from the gradient of the tangent to the

concentration–time graph at t = 0.

It is often difficult to monitor reaction rates or concentration continuously.In practice you carry out a series of reactions where you vary the initialconcentration of each reactant in turn. You then plot a graph of initial rateagainst initial concentration for each reactant.

You can find the orders from the shape of each graph, and the value of kand units by substituting your measurements into the rate equation:

If it is a horizontal line, it means that the reaction is not taking place (rateof reaction = 0).

If a straight line sloping downwards is obtained, the slope is constant. Thismeans that the rate is constant. This only occurs when the reaction is zeroorder.

If a downward curve is obtained, with decreasing slope, the rate isdecreasing as [A] falls. Therefore, the reaction is first order or greater.

T1/2 =1

×


Present and interpret the results of kinetic measurements in graphical form, including

concentration-time and rate-concentration graphs.

k =


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Rate-Concentration Graphs

If the rate of a reaction, or some quantitythat is proportional to the rate, is plottedagainst the concentration of one reactant,the order with respect to that reactant canbe found.

1/time for the reaction to proceed to acertain point is often used as a measure ofthe rate.

If the graph of rate (or 1/time)against [reactant] is a straight line,the reaction is first order withrespect to that reactant.

If the graph of rate (or 1/time)against [reactant]2 is a straight line,the reaction is second order with

respect to that reactant.


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Deducing the Order of Reaction from Initial Rate Data

To do this, you need data from at least three experiments. For example, data fromthe reaction below could be used:

x A + y B → products

Experiment [A] /mol dm

-

[B] /mol dm

-

Initial Rate / mol dm

-

1 0.1 0.1 p

2 0.2 0.1 q

3 0.2 0.2 r

Comparing experiments 1 and 2 showed that [A] has doubled but [B] hasstayed the same.

If the rate is unaltered (q = p), the order with respect to A is 0.

If the rate doubles (q = 2p), the order with respect to A is 1.

If the rate quadruples (q = 4p), the order with respect to A is 2.

The order with respect to B can be determined in a similar way, byanalysing experiments 2 and 3.

Worked Example

In the table above, suppose p = 0.0024, q = 0.0096 and r = 0.0096.(1) Determine the partial orders of A and B.

From experiments 1 and 2: when [A] doubles, the rate quadruples (0/0096 = 4 x0.0024). Therefore the order with respect to A is 2.From experiments 2 and 3: when [B] doubles, the rate is unaltered. Therefore,

the order with respect to B is 0.

(2) Determine the total order.

Total Order is 2 + 0 = 2.

(3) Write the rate equation.

The rate equation is: Rate = k[A]2

(4) Calculate the value of the rate constant.

k = rate/[A] 2 = 0.0024 mol dm-3s-1/(0.1 mol dm-3)2 = 0.24 mol-1 dm3s-1.


Investigate the reaction of iodine with propanone in acid to obtain data for the order

with respect to the reactants and the hydrogen ion and make predictions about

molecules/ions involved in the rate-determining step and possible mechanism (details of

the actual mechanism can be discussed at a later stage in this topic).


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Iodine-Propanone Reaction

In acid solution iodine reacts with propanone to form iodopropanone and hydrogeniodide.

CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + HI(aq)

This reaction can be monitored by the following method:

Place 25cm3 of iodine solution of known concentration in a flask and add25cm3 of sulfuric acid solution (an excess) and 50cm3 of water. Add 5cm3 ofpure propanone from a burette, start the clock as you do so.

At noted intervals of time remove 10cm3 of the reaction mixture (aliquots)and quench it by adding it to a slight excess of aqueous sodiumhydrogencarbonate. (This reacts with the acid catalyst, stopping thereaction – ensuring no further change in iodine concentration occurs duringtitration.)

Quickly titrate the remaining iodine against standard sodium thiosulfatesolution, using starch as an indicator, near the end-point.

If the starch is added too soon, an insoluble starch-iodine complex is formedand the titre will be too low.

Repeat by removing samples at regular intervals.

The rate equation is of the form:

Rate = k[propanone] x [H+]y [I2]z

Because a large excess of propanone was taken, the concentration of propanone,written as [propanone], remains approximately constant. Even though acid isproduced in the reaction, the original amount of acid catalyst was large and so [H+]remains effectively constant. This means that the rate equation becomes:

Rate = k‟[I2]z where k’ = k[propanone] x [H+]y

The volume of sodium thiosulfate is proportional to the amount of iodine andhence the iodine concentration. So a graph of volume of sodium thiosulfate againsttime will have the same shape as the graph of [I2] against time.

The result of this experiment is shown in the graph below:

As this graph is a straight line, the slope andhence the rate is constant and so thereaction is zero order with respect to iodine.

A conclusion that can be drawn from this isthat iodine must enter the mechanism afterthe rate-determining step.

H+(aq)


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Experiment [CH3COCH3]/mol dm-3

[I2]/mol dm-3

[H+]/mol dm-3

Relative Rate

1 0.1 0.1 0.1 2

2 0.1 0.2 0.1 2

3 0.1 0.2 0.2 4

4 0.2 0.1 0.1 45 0.1 0.1 0.2 4

The table above is an example of some results obtained from this experiment. Wecan conclude that:

Doubling the concentration of iodine has no effect on the relative rate ofreaction (experiments 1 and 2)

Doubling the concentration of propanone doubles the relative rate ofreaction (experiments 1 and 4)

Doubling the concentration of hydrogen ions doubles the relative rate of the

reaction (experiments 1, 3 and 5)

Mechanism for the Reaction of Iodine and Propanone in Acid Solution

The reaction involves various steps and the slowest step, the rate-determiningstep, does not involve iodine. This is why iodine does not appear in the rateequation. Hydrogen ions act as a catalyst – they are regenerated during thereaction.

A possible mechanism for the reaction of iodine and propanone in acid solution is

given below.

Step 1An H+ ion protonates the oxygen atom in propanone:

(CH3)2C=O + H3O+⇌ (CH3)2C=O

+H + H2OThis is a reversible reaction involving proton transfer (acid-base reaction).Remember that a protonated water molecule, H3O

+, behaves as an H+ ion.

Step 2The electrons in the C=O bond partly shift to form a carbocation – i.e. the positivecharge is transferred from the oxygen to the carbon:

(CH3)2C=O+H ⇌ (CH3)2C

+-OH

Step 3This carbocation loses a proton and slowly changes into the enol form. The enolhas both alkene and alcohol functional groups and is isomeric with the originalketone:

(CH3)2C+-OH + H2O ⇌ CH3C(OH)=CH2 + H3O

+


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This involves breaking a strong C-H bond, hence this step has a high activationenergy and slow speed. The positive charge on the adjacent carbon of thecarbocation facilitates in „pulling‟ the C-H bond pair to form the C=C bond andreleases the proton to form an H3O

+ ion.

The rate of formation of the enol thus depends on the concentrations of theketone and the acid.

Step 4The iodine molecule acts as an electrophile and undergoes a quick electrophilicaddition reaction (like other alkenes). This produces a protonated iodoketone:

CH3C(OH)=CH2 + I2 → CH3C(=O+H)-CH2I + I

-

Step 5A water molecule then rapidly removes the proton in another acid-base reaction toform the iodoketone:

CH3C(=O+H)-CH2I + H2O → CH3COCH2I + H3O

+

In this example the low, rate-determining steps 1, 2 and 3 are the first stage in thereaction and need only CH3COCH3 and H

+. The remaining steps happen veryquickly. This is also the case when oxygen and hydrogen bromide react together at700K. However, the slowest reaction is not always the first.

In the reaction between bromide ions and bromate (V) ions in acid solution themost likely mechanism is that HBr and HBrO3 are made very rapidly before thethird, relatively slow, reaction between the two of them takes place. This is therate-determining step, and is followed by two more rapid steps to complete thereaction.


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Using a Colorimeter

A colorimeter measures the absorption of light during the progress of theexperiment. First, the colorimeter has to be calibrated using standard iodinesolutions. The reaction mixture containing iodine will be light brown in colour and

as the reaction proceeds the solution becomes paler and more light is transferred(when using a colorimeter there is no need for starch indicator).

Rate-Determining Steps

Reaction do not just happen whenever all the relevant molecules collide at once – they happen in steps. The slowest step controls how fast the overall reactionoccurs – it is called the rate-determining step.

Kinetic measurements establish the order of the reaction for each species. For areaction between A, B and C the rate equation could be:

Rate = k[A][C]2

Since substance B does not occur in the rate equation, any step involving moleculeB must be fast. This rate equation demonstrates that:

The reaction is first order with respect to A – so A is involved in the ratedetermining step.

The reaction is second order with respect to substance C – so two moles of Care involved in the rate-determining step.


Deduce from experimental data for reactions with zero, first and second order kinetics:

(i) Half-life (the relationship between half-life and rate constant will be given if

required)

(ii) Order of Reaction

(iii) Rate Equation(iv) Rate-determining step related to reaction mechanisms

(v) Activation energy (by graphical methods only; the Arrhenius equation will be given if

needed).

The chosen filter should let through

only the wavelength to be absorbed

by the coloured iodine solution.

Since iodine is brown-red, a blue-

green filter is used.


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Activation Energy and Catalysts

When reactant molecules collide, it may result in a chemical reaction. There is anenergy requirement before this can happen. The activation energy is the minimumenergy needed by reactant particles (molecules or ions) before products can form.

The diagram shows that the reaction is:• Exothermic – the products are at a lower energy level than the reactants• Subject to an energy barrier, the activation energy, in route 1• Able to follow an alternative pathway in route 2, with a lower barrier

A catalyst increases the reaction rate by providing an alternative reaction pathway

with a lower activation energy. Such catalysed reactions are faster.

Catalysts and reactants can be in the same physical state/phase, calledhomogeneous - for example, all liquids. Or they can be in different states/phases,when they are called heterogeneous.

Process Reactants Catalyst Type of Catalysis

Haber Synthesis Nitrogen,Hydrogen

Iron Heterogeneous

CatalyticConverter on car

Exhaust Gases Platinum Heterogeneous

Contact Process,Sulfuric Acidmanufacture

Sulfur Dioxide,Oxygen Gases

Vanadium (V)Oxide

Heterogeneous

Esterification Solutions of Acid,Alcohol

Hydrogen Ions Homogeneous


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Most catalysts are transition metals (or their compounds) because they havevariable oxidation states and can alter the number of bonds available to reactants.

Inorganic catalysts are used to catalyse a wide range of different reactions.Although catalysts are not permanently altered during the reactions that theycatalyse, they can be poisoned by some impurities and will not work again.

Catalytic Converters

A catalytic converter removes pollutant gases from the exhaust by oxidising orreducing them. The exhaust gases pass through a converter containing a preciousmetal catalyst, usually an alloy of platinum or rhodium. Several reaction may take

place. NOx and CO may take part in a redox reaction which removes both of themat the same time. NOx oxidises CO to CO2, and is itself reduced to harmless N2 gas.

2NO(g) + 2CO(g) → N2(g) + 2CO2(g)

CO and CxHy are oxidised by air:

2CO(g) + O2(g) → 2CO2(g)C7H16(g) + 11O2(g) → 7CO2(g) + 8H2O(g)


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For all three of these reaction to happen, you need a „three-way converter‟. Anoxygen monitor is fitted to the engine. This checks the quantity of oxygen goinginto the engine to make sure there is enough to carry out the oxidation reactions.

The overall result of passing exhaustgases through this kind of catalystsystem is to convert NOx, CO and CxHy to relatively harmless N2, CO2 and H2O.The catalytic reactions do not startworking until the catalyst has reached

a temperature of about 200⁰C, so theyare not effective until the engine haswarmed up.

Catalyst systems of this type cost several hundred dollars, mainly because of thehigh cost of the precious metals they contain. The catalyst is „poisoned‟ by lead,so unleaded fuel must always be used.

Unfortunately, the major environmental impact of motor vehicles cannot be solvedby catalytic converters – Climate Change.

Worked Example

1)

Natural gas (methane) and air react exothermically in a Bunsen flame.Explain why mixtures of these gases do not ignite spontaneously at roomtemperature.

The activation energy barrier for this combustion reaction is too high at roomtemperature. Colliding molecules have insufficient energy to react. Once ignited,the Bunsen flame heats arriving gases before they react so that enough moleculeshave energy equal to or greater than the minimum required for reaction.

Effect of Temperature on Rates

When the temperature increases, the rate of a reaction increases too because therate constant increases.

The rate constant k is only a constant for a particular temperature. Changing thetemperature changes the value of k because the proportion of molecules that havethe required energy (greater than the activation energy) is increased and thecolliding particles have a greater average energy.

The Arrhenius equation shows the relationship between the rate constant k andthe temperature T (in kelvin). The logarithmic (ln) form of this equation is:

lnk = −

+ a constant


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Remember – units for

activation energy are kJ mol-1

• E A is the activation energy for the reaction• R is the gas constant and has the value 8.31 J K-1 mol-1 • T is the kelvin temperature (absolute temperature)

• Gradient =−

Investigating the Activation Energy of a Reaction

If we know the rate equation for a reaction, it is easy to calculate the rateconstant k using the rate of reaction for a known concentration of reactants.Calculating the activation energy requires the results from experiments at a rangeof different temperatures, to give values of the rate constant k at eachtemperature. A suitable reaction to study is the oxidation of iodide ions by iodate(V) ions in acidic solution. The equation is:

IO3-(aq) + 5I

-(aq) + 6H

+(aq) → 3I2(aq) + 3H2O(l)

A small known amount of sodium thiosulfate is added at the start of the reactiontogether with starch as the indicator. The iodine released by the oxidation of theiodide ions first reacts with the thiosulfate, so the mixture remains colourlessinitially but then suddenly turns starch blue-black when all the thiosulfate hasreacted. This is an example of a clock reaction with a built-in time delay thatdepends only on the concentration of the iodide and iodate ions. The relativeinitial rate of production of iodine can be found from 1/t, where t is the timedelay.

The value for the activation energy can be found graphically as described above.

Examiner‟s Ti

Working out the value of anactivation energy. This showsthat a plot of lnk against 1/Tgave a straight line whose

gradient is−


Investigate the activation energy of a reaction, e.g. oxidation of iodide ions by iodate (V)


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Hydrolysis of Halogenoalkanes

There are three different types of halogenoalkane. They can all be hydrolysed

(split) by heating them with sodium hydroxide – but they react using differentmechanisms.

Primary Halogenoalkanes – halogen is joined to a carbonwith one alkyl groupattached.

Secondary Halogenoalkanes – halogen is joined to a carbonwith two alkyl groupsattached.

Tertiary Halogenoalkanes – halogen is joined to a carbonwith three alkyl groupsattached.

Two substitution mechanisms are possible when an iodoalkane reacts with aqueousalkali:

R-I + OH- → R-OH + I-

Only the experimental rate data can show which mechanism actually takes place.


Apply knowledge of the rate equations for the hydrolysis of halogenoalkanes to deduce

the mechanisms for primary and tertiary halogenoalkane hydrolysis and to deduce the

mechanism for the reaction between propanone and iodine.

Demonstrate that the mechanisms proposed for the hydrolysis of halogenoalkanes areconsistent with the experimentally determined orders of reactions, and that a proposed

mechanism for the reaction between propanone and iodine is consistent with the data

from the experiment in 4.3e.

Use kinetic data as evidence for SN1 or SN2 mechanisms in the nucleophilic substitution

reactions of the halogenoalkanes.


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The single step illustrated for the substitution of 1-iodopropane, a primaryhalogenoalkane, involves two different species – both the hydroxide ion andthe primary halogenoalkane.

The reaction will be second order – it depends on the concentration of boththe hydroxide ion and the primary halogenoalkane.

The hydroxide ion joins onto the central carbon atom at the same time asthe halogen atom was leaving.

Part of the energy required to break the C-Halogen bond was supplied bythe energy released on producing the C-OH bond.

Calculations show that the approach of the hydroxide ion along the line ofcentres of the carbon an bromine atoms is that of lowest energyrequirement.

So, rate = k[RI][OH-]We call this an SN2 mechanism, meaning substitution/nucleophilic/second order.

Tertiary halogenoalkanes hydrolyse by the alternative SN1 mechanism, meaningsubstitution/nucleophilic/first order.

The C-halogen bond heterolytically breaks first (slow step) forming atertiary carbocation – which are relatively stable.

Followed by attack by the hydroxide ion (fast step).

The slow step involves only one species and does not depend on theconcentration of hydroxide ions. Hence the reaction is first order overall.

The mechanism must be consistent with the evidence:

If the reaction is second order overall it must involve two different species

If the reaction is first order overall (only one species in the rate equation)then this is the rate-determining step and it must be a two-step reaction.

Nucleophile – a species that is attracted to an electron-deficient centre, ofan atom, molecule or ion that donates a pair of electrons to

form a new covalent bond.Electrophile – an atom, molecule or ion that accepts a pair of electrons to

form a new covalent bond. An electrophile is therefore aspecies that bonds to an electron-rich site in a molecule.

Substitution - one atom or group of atoms is replaced by another atom orgroup of atoms.

Hydrolysis - when a molecule is broken down by the addition of water. Thismay involve water, acid or alkali as the inorganic reagent.

Summary


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Primary Halogenoalkanes only react by SN2 mechanism.

Secondary Halogenoalkanes can react by both the SN1 and SN2 mechanisms – it depends what mood they‟re in.

Tertiary Halogenoalkanes only react by the SN1 mechanism.

It is an SN2 mechanism if the rate is dependent on the concentration of both

the reactants and the order with respect to each is 1. It is an SN1 mechanism if the rate is only dependent on the concentration of

the halogenoalkane.

Other Experimental Methods to Determine Rate of Reaction

Titration

If the concentration of a reactant or product can be estimated by a titration, the

reaction can be followed using this technique: Measure out samples of the reactants with known concentration.

Mix them together, start a clock and stir the mixture thoroughly.

At regular time intervals, withdraw samples using a pipette and quench(stop) the reaction. Quenching can usually be achieved either by adding thesolution from the pipette to ice-cold water or to a solution that reacts withone of the reactants, to prevent further reaction from taking place. Thetime at which half the contents of the pipette have been added to thequenching solution is noted.

The quenched solution is then titrated against a suitable standard solution.

The titre is proportional to the concentration of the reactant or product beingtitrated.

This method can be used when an acid, alkali or iodine is a reactant or product.Acids can be titrated with a standard alkali, alkalis with a standard acid and iodinewith a standard solution of sodium thiosulfate.

Infrared Spectroscopy

Infrared spectroscopy can be used in a similar way to colorimetry. Thespectrometer is set at a particular frequency and the amount of infrared radiationabsorbed at that frequency is measured at regular time intervals. The oxidation ofpropan-2-ol to propanone by acidified potassium dichromate (VI) can be followedby setting the spectrometer at 1700cm-1 (the absorption frequency due to thestretching of the C=O bond) and measuring the increase in absorption as the CHOHgroup is oxidised to the C=O group.

Conductimetric Analysis

If the number of ions changes, so will the electrical conductivity.

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Edexcel A2 Chemistry 4 3 Notes