CONTENTS Structure of halogenoalkanes Naming of Halogenoalkanes Preperation of haloalkanes by...
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Transcript of 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
Primary Secondary Tertiary
H CH3 CH3
H3C C+ H C+ H3C C+
H CH3 CH3
Primary Secondary Tertiary
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
NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - MECHANISMMECHANISM
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.
(for understanding only )
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.
(for understanding only )
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??
NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - MECHANISMMECHANISM
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
NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - MECHANISMMECHANISM
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
NUCLEOPHILIC SUBSTITUTION - NUCLEOPHILIC SUBSTITUTION - MECHANISMMECHANISM
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
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
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
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
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.
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
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
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
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
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
POTASSIUM CYANIDE
ANIMATED MECHANISM
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
Amine
Lone pair
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
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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