contents structure of halogenoalkanes naming of halogenoalkanes preperation of haloalkanes by...

CONTENTS• Structure of halogenoalkanes

• Naming of Halogenoalkanes

• Preperation of haloalkanes by alcohols

• Stability

• Physical properties of halogenoalkanes

• SN1 with mechanism

• SN2 with mechanism

• Reactions : with aquous NaOH / KOH (nucleophillic substitution)

with alcoholic NaOH/KOH (elimination)

with ammonia (nucleophillic substitution)

with KCN (nucleophillic substitution)

with water (nucleophillic substitution)

• Tests for haloalkanes

• Uses of haloalkanes

• CFC’s

THE CHEMISTRY OF HALOGENOALKANESTHE CHEMISTRY OF HALOGENOALKANES

Before you start it would be helpful to…

• Recall the definition of a covalent bond

• Be able to balance simple equations

• Be able to write out structures for hydrocarbons and their derivatives

• Understand the different types of bond fission

• Recall the chemical properties of alkanes, alkenes and alcohols

• Intermolecular forces

THE CHEMISTRY OF HALOGENOALKANESTHE CHEMISTRY OF HALOGENOALKANES

STRUCTURE OF HALOGENOALKANESSTRUCTURE OF HALOGENOALKANES

Format Contain the functional group C-X where X is a halogen (F,Cl,Br or I)

Halogenoalkanes - halogen is attached to an aliphatic skeleton - alkyl group

Structuraldifference Halogenoalkanes are classified according to the environment of the halogen

Names Based on original alkane with a prefix indicating halogens and position.

CH3CH2CH2Cl 1-chloropropane CH3CHClCH3 2-chloropropane

CH2ClCHClCH3 1,2-dichloropropane CH3CBr(CH3)CH3 2-bromo-2-methylpropane

PRIMARY 1° SECONDARY 2° TERTIARY 3°

Reactivity of halogenoalkanes

Primary Secondary Tertiary

+ + +

Primary Secondary Tertiarycarbocation carbocation carbocation

•Thirtiary carbocation• Trigonal planar molecule to minimize repulsion between electron pairs• bond angles of 120º• Contains 3 bond pairs (no lone pairs) ofElectrons* Formed by hetrolytic fission

Naming haloalkanes are similar to those for naming alkanes

The halogens are written as prefixes: fluoro- (F), chloro- (Cl), bromo- (Br) and iodo- (I)

e.g.

When the parent chain has both a halogen and an alkyl substituent, the chain is

numbered from the end nearer the first substituent regardless of what substituents are

e.g.

The compound has the systematic name

A 2-chlorobutaneB 3-chlorobutaneC 1-chloro-1-methylpropaneD 1-chloro-2-methylbutane

Structural formula

Shows the atoms carbon by carbon, with the hydrogen and functional groups attached.

CH3CH2CH2CH2OH

Displayed formula

Shows how all atoms are arranged, and all the bonds between them.

H H H HH C C C C OH H H H H

Skeletal formula

Shows the bonds of the carbon skeleton only, with any functional groups. The hydrogen and carbon atoms aren’t shown. This is handy for drawing large complicated structures, like cyclic hydrocarbons.

OH

STRUCTURAL ISOMERISM IN HALOGENOALKANESSTRUCTURAL ISOMERISM IN HALOGENOALKANES

Different structures are possible due to...

Different positions for the halogen and branching of the carbon chain

2-chlorobutane

2-chloro-2-methylpropane

1-chlorobutane

1-chloro-2-methylpropane

PHYSICAL PROPERTIESPHYSICAL PROPERTIES

Boiling point Increases with molecular size due to increased van der Waals’ forces

Mr bp / °C

chloroethane 64.5 13

1- chloropropane 78.5 47

1-bromopropane 124 71

Boiling point also increases for “straight” chain isomers.Greater branching = less relative surface area = lower inter-molecular forces

bp / °C

1-bromobutane CH3CH2CH2CH2Br 101

2-bromobutane CH3CH2CHBrCH3 91

2-bromo -2-methylpropane (CH3)3CBr 73

Remember:the only methyl halide which is a liquid is iodomethane; chloroethane is a gas.

van der Waals dispersion forces (review)

These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up.

This is why the boiling points increase as the number of carbon atoms in the chains increases.

Dispersion forces get stronger as you go from 1 to 2 to 3 carbons in the chain.

It takes more energy to overcome them, and so the boiling points rise.

The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger van der Waals dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them!

Halogen atoms have grater number of elcetrons than hydrogen atom, increacing Van der Waals forces than alkenes

The carbon-halogen bonds (apart from the carbon-iodine bond) are polar, because the electron pair is pulled closer to the halogen atom than the carbon. This is because (apart from iodine) the halogens are more electronegative than carbon.The electronegativity values are:

C 2.5 F 4.0

Cl 3.0

Br 2.8

I 2.5

forces due to the attractions between the permanent dipoles (except in the iodide case). The size of those dipole-dipole attractions will fall as the bonds get less polar (as you go from chloride to bromide to iodide, for example). Nevertheless, the boiling points rise! This shows that the effect of the permanent dipole-dipole attractions is much less important than that of the temporary dipoles which cause the dispersion forces.The large increase in number of electrons by the time you get to the iodide completely outweighs (more significant than) the loss of any permanent dipoles in the molecules.

Boiling temperature comparision

Halagenoalkane C-F bonds stronger than C-C

Alkenes/alkanes only london forces

BoilingTemp.

No. of carbon atoms1 2 3

Solubility

The halogenoalkanes are at best only very slightly soluble in water. (As they have more hydrocarbon part than polar part)

In order for a halogenoalkane to dissolve in water you have to break attractions between the halogenoalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy.

Halogenoalkanes tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate halogenoalkane and solvent.

Preperation oh halo-alkane

PREPERATION OF HALAGENOALKANES FROM ALKENEPREPERATION OF HALAGENOALKANES FROM ALKENE

Reagent Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition Room temperature.

Equation C2H4(g) + HBr(g) ———> C2H5Br(l) bromoethane

Mechanism

Step 1 As the HBr nears the alkene, one of the carbon-carbon bonds breaksThe pair of electrons attaches to the slightly positive H end of H-Br.The HBr bond breaks to form a bromide ion.A carbocation (positively charged carbon species) is formed.

Step 2 The bromide ion behaves as a nucleophile and attacks the carbocation.Overall there has been addition of HBr across the double bond.

ELECTROPHILIC ADDITION OF HYDROGEN BROMIDE

ADDITION TO UNSYMMETRICAL ALKENESADDITION TO UNSYMMETRICAL ALKENES

Problem • addition of HBr to propene gives two isomeric brominated compounds

• HBr is unsymmetrical and can add in two ways

• products are not formed to the same extent

• the problem doesn't arise in ethene because it is symmetrical.

Mechanism

Two possibilities

ELECTROPHILIC ADDITION TO PROPENE

The general reaction looks like this:

Preperation oh halo-alkane from alcohols

ROH + HX RX + H2O

CHLORINATION OF ALCOHOLSCHLORINATION OF ALCOHOLS

When PCl5 is added to dry alcohol, clods of hydrogen cloride fumes are produced

CH3CH2OH + PCl5 CH3CH2Cl + POCl3 + HCL (g)

Hydrogen chloride testHydrogen chloride gas forms a white smoke with ammonia.

BROMINATION OF ALCOHOLSBROMINATION OF ALCOHOLS

C2H5OH + HBr C2H5Br + H2O

Dry conditions

Room temp

NaBr / KBr + 50% CONC. H2SO4

Heat under reflux

3C2H5OH + PBr3 3C2H5Br + H3PO3

Moist red Phosperous + Br2

KBr + H2SO4 ---> KHSO4+ HBr

Heat under refux

IODINATION OF ALCOHOLSIODINATION OF ALCOHOLS

3C2H5OH + PI3 3C2H5I + H3PO3

Moist red Phosperous + I2

Heat under refux

In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H3PO4, and the iodoalkane is distilled off.

The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol.

Phosphoric(V) acid is used instead of concentrated sulphuric acid because sulphuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide. A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, (but not enough to get in the way of the main reaction) . There is no reason why you couldn't use phosphoric(V) acid in the bromide case instead of sulphuric acid if you wanted to.

C2H5OH + HI C2H5I + H2O

NaI / KI + CONC. H3PO4

H3PO4 + KI ----> KH2PO4 + HI

Instead of using phosphorus(III) bromide or iodide, the alcohol is heated under reflux with a mixture of red phosphorus and either bromine or iodine.

SECONDARY ALCOHOLS WITH HALOGENSSECONDARY ALCOHOLS WITH HALOGENS

+ PCl5

Butane-2-ol + PCl5 2-cloro-butane

Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid,

but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance.A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature.

A tertiary halogenoalkane (haloalkane or alkyl halide) is formed

TERTIARY ALCOHOLS WITH HALOGENSTERTIARY ALCOHOLS WITH HALOGENS

Haloalkanes can be prepared from the vigorous reaction between cold alcohols and

phosphorus(III) halides

Preperation of halo-alkanes


H CH3 CH3

H3C C+ H C+ H3C C+

H CH3 CH3


Stability of carbocation increases

Which of these compounds is a secondary halogenoalkane?

A CH3CH(OH)CH3

B CH3CCl(CH3)CH3

C CH3CHClCH3

D CH3CH2CH2Cl

The formation of a carbocation from a halogenoalkane is an example ofA homolytic fission.B heterolytic fission.C an initiation reaction.D a propagation reaction.

When a chloroalkane is heated with aqueous sodium hydroxideA no reaction occurs with primary, secondary or tertiary chloroalkanes.B a reaction occurs with primary and secondary chloroalkanes but not with tertiary chloroalkanes.C a reaction occurs with tertiary chloroalkanes but not with primary and secondary chloroalkanes.D a reaction occurs with primary, secondary and tertiary chloroalkanes.

NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION

Theory • halogens have a greater electronegativity than carbon• electronegativity is the ability to attract the shared pair in a covalent

bond• a dipole is induced in the C-X bond and it becomes polar• the carbon is thus open to attack by nucleophiles• nucleophile means ‘liking positive’

the greater electronegativity of the halogen attracts the

shared pair of electrons so it becomes slightly negative;

the bond is now polar.

OH¯ CN¯ NH3 H2O

NUCLEOPHILES :

• ELECTRON PAIR DONORS

• possess at least one LONE PAIR of electrons

• don’t have to possess a negative charge

• are attracted to the slightly positive (electron deficient) carbon

• examples are OH¯, CN¯, NH3 and H2O (water is a poor

nucleophile)

NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - MECHANISMMECHANISM

the nucleophile uses its lone pair to provide the electrons for a new bond

the halogen is displaced - carbon can only have 8 electrons in its outer shell

the result is substitution following attack by a nucleophile

the mechanism is therefore known as - NUCLEOPHILIC SUBSTITUTION


Note

the nucleophile has a lone pair of electrons

the carbon-halogen bond is polar

a ‘curly arrow’ is drawn from the lone pair to the slightly positive carbon atom

a ‘curly arrow’ is used to show the movement of a pair of electrons

carbon is restricted to 8 electrons in its outer shell - a bond must be broken

the polar carbon-halogen bond breaks heterolytically (unevenly)

the second ‘curly arrow’ shows the shared pair moving onto the halogen

the halogen now has its own electron back plus that from the carbon atom

it now becomes a negatively charged halide ion

a halide ion (the leaving group) is displaced

The "electron pushing effect" of alkyl groups ??

You are probably familiar with the idea that bromine is more electronegative than hydrogen, so that in a H-Br bond the electrons are held closer to the bromine than the hydrogen. A bromine atom attached to a carbon atom would have precisely the same effect - the electrons being pulled towards the bromine end of the bond. The bromine has a negative inductive effect.

Alkyl groups do precisely the opposite and, rather than draw electrons towards themselves, tend to "push" electrons away.

The stability of the various carbocations (for understanding only )

This means that the alkyl group becomes slightly positive ( +) and the carbon they are attached to becomes slightly negative ( -)

The alkyl group has a positive inductive effect.

This is sometimes shown as, for example:

The arrow shows the electrons being "pushed" away from the CH3 group. The plus

sign on the left-hand end of it shows that the CH3 group is becoming positive. The

symbols + and - simply reinforce that idea.

(for understanding only )

Order of stability of carbocationsprimary < secondary < tertiary

The importance of spreading charge around in making ions stable

The general rule-of-thumb is that If a charge is very localised (all concentrated on one atom) the ion is much less stable than if the charge is spread out over several atoms.Applying that to carbocations of various sorts . . .

You will see that the electron pushing effect of the CH3 group is placing more and more negative

charge on the positive carbon as you go from primary to secondary to tertiary carbocations.

The effect of this, of course, is to cut down that positive charge.

At the same time, the region around the various CH3 groups is becoming somewhat positive. The

net effect, then, is that the positive charge is being spread out over more and more atoms as you go from primary to secondary to tertiary ions.

The more you can spread the charge around, the more stable the ion becomes.


When we talk about secondary carbocations being more stable than primary ones, what exactly do we mean?

This means that it is going to take more energy to make a primary carbocation than a secondary one.

If there is a choice between making a secondary ion or a primary one, it will be much easier to make the secondary one.

Similarly, if there is a choice between making a tertiary ion or a secondary one, it will be easier to make the tertiary one.

Steric hindrance

* It is simply as Prevention or retardation of reaction

* Since the SN2 proceeds through a backside attack, the reaction will only proceed if the empty orbital is accessible.The more groups that are present around the area surrounding the leaving group, the slower the reaction will be.

For the SN2, since steric hindrance increases as we go from primary to secondary to tertiary, the rate of reaction proceeds

From primary (fastest) > secondary >> tertiary (slowest).

* In another way , If the approach by the nucleophile to the carbon is made difficult by crowding by neighboring groups the transition state is more difficult to form, and the rate of the reaction slows. The blocking of access to a reactive site by nearby groups is referred to as steric hindrance.


Primary carbocation (SN2 takes place)

Primary and tertiary carbocation intermediates have different stabilities

Because of inductive effects of alkyl groups stabilize thirtiary carbocation

Steric hindrance differs for attack on primary and tertiary carbon (in the molecule) / less space available for attack by OH on tertiary carbon / more space for attack by OH on primary carbon

As bulky / three alkyl groups obstruct attack by nucleophile Inhibits formation of transition state

Tertiary carbocation (SN1 takes place)

Secondary carbocation ( SN1 or SN2 takes place )

Why the difference in mechanism??


SN1

Why SN1 for thirtiary halagenoalkane ??

Tertiary carbocation is more stable As inductive effects of alkyl groups stabilize tertiary carbocation

More Steric Hindrance / less space available for attack by OH¯on thirtiary carbocation As 3 alkyl groups obstruct attack by nucleophile with tertiary compound

Br (:)Br


SN2

OH¯

CH3

H

H

Br

CH3H

H

OH Br

-

CH3

H

H

Br (:)Br¯

Why SN2 for primary halagenoalkane ??

Primary carbocation is less stable As inductive effects of alkyl groups are less than that of secondary and tertiary carbocation

Less Steric Hindrance / more space available for attack by OH¯on primary carbocationAs central carbon atom is not surrounded by many alkyl / bulky groups which obstruct attack by the nucleophile


ANIMATION SHOWING THE SN2 MECHANISM

NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - RATE OF REACTIONRATE OF REACTION

Basics An important reaction step is the breaking of the carbon-halogen (C-X) bondThe rate of reaction depends on the strength of the C-X bond

C-I 238 kJmol-1 weakest - easiest to break

C-Br 276 kJmol-1

C-Cl 338 kJmol-1

C-F 484 kJmol-1 strongest - hardest to break

Experiment Water is a poor nucleophile but it can slowly displace halide ions

C2H5Br(l) + H2O(l) ——> C2H5OH(l) + H+ (aq) + Br¯(aq)

If aqueous silver nitrate is shaken with a halogenoalkane (they are immiscible)the displaced halide combines with a silver ion to form a precipitate of a silverhalide. The weaker the C-X bond the quicker the precipitate appears.

Ag+ (aq) + X¯(aq) ——> AgX(s)

AgCl white ppt AgBr cream ppt AgI yellow ppt

NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTIONReaction with Hydroxide ions

AQUEOUS SODIUM HYDROXIDE

Reagent Aqueous* sodium (or potassium) hydroxideConditions Reflux/Heat in aqueous solution (SOLVENT IS IMPORTANT)Product AlcoholNucleophile hydroxide ion (OH¯)

Equation e.g. C2H5Br(l) + NaOH(aq) ——> C2H5OH(l) + NaBr(aq)

Mechanism

* WARNING It is important to quote the solvent when answering questions. Elimination takes place when ethanol is the solvent - SEE LATER

The reaction (and the one with water) is known as HYDROLYSIS


OH¯

CH3

H

H

Br

CH3H

H

OH Br

-

CH3

H

H

Br (:)Br¯

ELIMINATIONELIMINATIONReagent Alcoholic sodium (or potassium) hydroxideConditions Reflux/Heat in alcoholic solutionProduct AlkeneMechanism EliminationEquation C3H7Br + NaOH(alc) ——> C3H6 + H2O + NaBr

Mechanism

the OH¯ ion acts as a base and picks up a protonthe proton comes from a carbon atom next to that bonded to the halogenthe electron pair left moves to form a second bond between the carbon atomsthe halogen is displacedoverall there is ELIMINATION of HBr.

Generally elimination happend more readily with secondary or thirtiary halagenoalkane

Complication With unsymmetrical halogenoalkanes, you can get mixture of products

ELIMINATIONELIMINATION

Complication

The OH¯ removes a proton from a carbon atom adjacent the C bearing the halogen. If there had been another carbon atom on the other side of the C-Halogen bond, its hydrogen(s) would also be open to attack. If the haloalkane is unsymmetrical (e.g. 2-bromobutane) a mixture of isomeric alkene products is obtained.

but-1-ene

but-2-enecan exist as cis and trans isomers

Also make sure you know how to name thirtiary halo-alkanesAnd the alkenes after elimination

ELIMINATIONELIMINATION

ANIMATED MECHANISM

With Alcoholic sodium (or potassium) hydroxide


AMMONIA

Reagent Aqueous, alcoholic ammonia (in EXCESS)Conditions Reflux in aqueous , alcoholic solution under pressureProduct AmineNucleophile Ammonia (NH3)

Equation e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

(i) C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr

(ii) HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes The equation shows two ammonia molecules.The second one ensures that a salt is not formed.


AMMONIA

Why excess ammonia?The second ammonia molecule ensures the removal of HBr which would lead to the formation of a salt. A large excess ammonia ensures that further substitution doesn’t take place - see below

ProblemAmines are also nucleophiles (lone pair on N) and can attack another molecule of halogenoalkane to produce a 2° amine. This too is a nucleophile and can react further producing a 3° amine and, eventually an ionic quarternary ammonium salt.

C2H5NH2 + C2H5Br ——> HBr + (C2H5)2NH diethylamine, a 2° amine

(C2H5)2NH + C2H5Br ——> HBr + (C2H5)3N triethylamine, a 3° amine

(C2H5)3N + C2H5Br ——> (C2H5)4N+ Br¯ tetraethylammonium bromide

a quaternary (4°) salt

POTASSIUM CYANIDE

Reagent Aqueous, alcoholic potassium (or sodium) cyanideConditions Reflux in aqueous , alcoholic solutionProduct Nitrile (cyanide)Nucleophile cyanide ion (CN¯)

Equation e.g. C2H5Br + KCN (aq/alc) ——> C2H5CN + KBr(aq)

Mechanism


POTASSIUM CYANIDE

Reagent Aqueous, alcoholic potassium (or sodium) cyanideConditions Reflux in aqueous , alcoholic solutionProduct Nitrile (cyanide)Nucleophile cyanide ion (CN¯)

Equation e.g. C2H5Br + KCN (aq/alc) ——> C2H5CN + KBr(aq)

Mechanism

Importance extends the carbon chain by one carbon atomthe CN group can be converted to carboxylic acids or amines.

Hydrolysis C2H5CN + 2H2O ———> C2H5COOH + NH3

Reduction C2H5CN + 4[H] ———> C2H5CH2NH2


POTASSIUM CYANIDE

ANIMATED MECHANISM


Amine

Lone pair


WATER

Details A similar reaction to that with OH¯ takes place with water.It is slower as water is a poor nucleophile.

Equation C2H5Br(l) + H2O(l) ——> C2H5OH(l) + HBr(aq)

Testing for halogenoalkanes

ion present observation

Cl- white precipitate

Br- very pale cream precipitate

I- very pale yellow precipitate

Testing for halogenoalkanesAdd aqueous acidified silver nitrate

Note :

It is reacted with NaOH to hydrolyse the halogenoalkane to an alcohol and release the halogen as a halide ion. It is heated to make the reaction faster.

The test for a halide ion is done using silver nitrate. The solution needs to be acidic to avoid interference by other ions, and nitric acid contains no ions (unlike hydrochloric acid) that would interfere

Halogenoalkanes and water and alakli

ELIMINATION v. SUBSTITUTIONELIMINATION v. SUBSTITUTION

The products of reactions between haloalkanes and OH¯ are influenced by the solvent

SOLVENT ROLE OF OH– MECHANISM PRODUCT

WATER NUCLEOPHILE SUBSTITUTION ALCOHOL

ALCOHOL BASE ELIMINATION ALKENE

Modes of attack

Aqueous soln OH¯ attacks the slightly positive carbon bonded to the halogen.OH¯ acts as a nucleophile

Alcoholic soln OH¯ attacks one of the hydrogen atoms on a carbon atom adjacentthe carbon bonded to the halogen.

OH¯ acts as a base (A BASE IS A PROTON ACCEPTOR)

Both reactions take place at the same time but by varying the solvent you can influence which mechanism dominates.

USES OF HALOGENOALKANESUSES OF HALOGENOALKANES

SyntheticThe reactivity of the C-X bond means that halogenoalkanes play animportant part in synthetic organic chemistry. The halogen can be replaced by a variety of groups via nucleophilic substitution.

PolymersMany useful polymers are formed from halogeno hydrocarbons

Monomer Polymer Repeating unit

chloroethene poly(chloroethene) PVC - (CH2 - CHCl)n –

USED FOR PACKAGING

tetrafluoroethene poly(tetrafluoroethene) PTFE - (CF2 - CF2)n -

USED FOR NON-STICK SURFACES

USES OF HALOGENOALKANESUSES OF HALOGENOALKANES

Chlorofluorocarbons - CFC’s – REFRIGERENT

dichlorofluoromethane CHFCl2 refrigerant

trichlorofluoromethane CF3Cl aerosol propellant, blowing agent

bromochlorodifluoromethane CBrClF2 fire extinguishers

CCl2FCClF2 dry cleaning solvent, degreasing agent

All are/were chosen because of their

LOW REACTIVITY

HAVE HIGH ENTHALPY OF VAPORIZATION NON-TOXICITYNON-FLAMMABLE NON-CORROSSIVE

HAS A MODERATE DENSITY IN LIQUID FORM AND RELATIVELY HIGH DENSITY IN VAPOUR FORM

CH3CH2CH2CH2OH + HBr CH3CH2CH2CH2 Br + H2O

NaBr + H2SO4 ---> NaHSO4+ HBr

Heat under reflux

Prepearing Halagenoalkanes in the lab

Sodium bromide, butane-1-ol and water are placed in the flask . The flask is put into a beaker of cold water and conc. Sulfuric acid is added slowly from the flask. This flask is cooled because the reaction at this stage is exothermic.

The dropping funnel is removed and the apparatus refluxed on water bath for 30 minutes

The apparatus shown below is set up , the mixture is distilled and is collected. The distillate will be collected as two layers - a lower organic layer upper aqueous layer

Continuation

1-bromobutane distilled into the organic layer

Organic layer Aqueous layer

Unreacted butane-1-ol Unreacted butane-1-ol

Bromine Oxides of sulfure

But-1-ene Hbr

Water

Impurities that maybe in the distillate

Continuation

Aqueous layer is discarded. The organic layer can be purified and finally redistilled to produce pure 1-bromobutane

PURIFYING BROMOALKANE BROM WATERPURIFYING BROMOALKANE BROM WATER

transfer the contents of the collection flask to a separating funnel.

To get rid of any remaining acidic impurities (including the bromine and sulphur dioxide), return the bromoethane to the separating funnel and shake it with either sodium carbonate or sodium hydrogencarbonate solution.

Add some anhydrous calcium chloride to the tube, shake well and leave to stand. The anhydrous calcium chloride is a drying agent and removes any remaining water. (It also absorbs ethanol, and so any remaining ethanol may be removed as well (depending on how much calcium chloride you use)

Transfer the dry bromoethane to a distillation flask and fractionally distil it, collecting what distils over at between 35 and 40°C.

1-bromo butane from butane-1 -ol

Reagent Alcoholic sodium (or potassium) hydroxideConditions Heat in alcoholic solutionProduct AlkeneNucleophile hydroxide ion (OH¯)

Equation e.g. C3H7Br + NaOH(alc) ——> C3H6 + H2O + NaBr

LABORATORY PREPARATION OF ALKENE FROM LABORATORY PREPARATION OF ALKENE FROM

HALAGENOALKANEHALAGENOALKANE

C3H7Br + NaOH(alc)

C3H6

NaOH/KOH inethanol/alcohol

NaOH/KOH in water/ aqueous

* NaBr/KBr & (50% or moderately conc) H2SO4 /* P & Br2 / PBr3 /PBr5 /* NaBr /KBr & H3PO4 /* HBr

NH3 (in alcohol /in a sealed tube /at high pressure)

REVISION CHECKREVISION CHECK

What should you be able to do?

Recall and explain the physical properties of halogenoalkanes

Recall and explain the chemical properties of halogenoalkanes based on their structure

Recall and explain the properties of nucleophiles

Write balanced equations for reactions involving substitution and elimination

Understand how the properties of a hydroxide ion are influenced by the choice of solvent

Recall the effect of CFC’s on the ozone layer

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

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contents structure of halogenoalkanes naming of halogenoalkanes preperation of haloalkanes by...

Documents

halogenoalkanes halogen

atoms carbon

carbon skeleton

kcn nucleophillic substitution

functional group cx

halogen f

carbon atoms arent

covalent bond