Part 2. The chemistry of ALKENES - unsaturated hydrocarbons

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry

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My advanced A level organic chemistry index of notes on alkenes

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Part 2.4 Addition of halogens to alkenes

Sub-index for this page

2.4.1 Introduction to the addition of halogens to alkenes

2.4.2 A test to distinguish between alkenes and alkanes

2.4.3 Reaction and mechanism of bromine addition to alkenes in non-aqueous media

2.4.4 Reaction and mechanism of bromine addition to alkenes in aqueous media

2.4.5 Unsaturated triglyceride fats/oils iodination (and reaction with ICl (iodine(I) chloride)

2.4.1 Introduction to the addition of halogens to alkenes e.g. bromine

Alkenes are unsaturated molecules, atoms can add to them via the C=C double bond, so a reaction occurs.

Alkenes readily react with halogens e.g. bromine, at room temperature and pressure.

Bromine water is used in a simple test for unsaturated alkenes to distinguish them from saturated alkanes.

The pi bond of the double bond opens up and two new carbon – bromine bonds (C–Br) are formed.

This double bond makes alkenes much more reactive than alkanes, the bromine water test for alkenes is just one example.

The equations illustrate what happens if gaseous alkenes are bubbled into a solution of bromine.

Alkanes are saturated – no double bond – and atoms cannot add – so no reaction.

 

A few examples are set out below in various styles of formulae to give dibromo halogenoalkanes

(1) ethene + bromine ====> 1,2–dibromoethane

CH₂=CH₂ + Br₂ ====> Br–CH₂CH₂–Br

When you bubble an alkene gas or mix a liquid alkene with bromine solution (water or hexane) the colour of the mixture changes from red-brown-orange to colourless. The decolourisation clearly indicates a chemical reaction has take place and is a simple test for unsaturation. (For the moment I'm ignoring the complications of different products with bromine water)

 

(2) propene + bromine ====> 1,2–dibromopropane

....

CH₃CH=CH₂ + Br₂ ====> CH₃–CHBr–CH₂Br

The decolourisation of bromine is a simple and effective chemical test for an alkene – an unsaturated hydrocarbon. The same reaction happens with chlorine (just but Cl instead of Br)

This reaction is NOT given by alkanes because they do NOT have a carbon = carbon double bond.

 

(3) a butene + bromine ===> a dibromobutane

but-1-ene  + bromine  ===> 1,2-dibromobutane

+ Br₂ ===>

but-2-ene  + bromine  ===> 2,3-dibromobutane

+ Br₂ ===>

The addition of bromine to the two butenes giving two slightly different dibromobutanes.

Note again you have gone from an unsaturated alkene (can add atoms to it) to a saturated derivative of an alkane (cannot add atoms to it)

There are complications with aqueous bromine, the presence of water produces small amounts of isomeric halogenoalcohols (haloalcohols, halogen alcohols) - see section 2.4.4 for more details and explanation.

Appreciate, that in dilute bromine/chlorine water, there are far more water molecules than halogen molecules.

 e.g. but-2-ene can form 3-bromopropan-2-ol (3-bromo-2-propanol)

CH₃-CH=CH-CH₃  +  Br₂  +  H₂O  ===>  CH₃-CH(OH)-CHBr-CH₃  +  H⁺  +  Br^-

Here you have to think of it as a sort of 'Br⁺OH^-' addition, but only one possible alcohol product.

However, with but-1-ene, H₂C=CH-CH₂-CH₃ there are two possible products:

From the Markownikov rule (reminder below):

The major product will be 1-bromobutan-2-ol   H₂BrC-CH(OH)-CH₂-CH₃

The minor product will be 2-bromobutan-1-ol   HOCH₂-CHBr-CH₂-CH₃

_{You can write similar equations for chlorine water.}

Reminder of the Markownikoff (Markovnikov) rule which predicts which isomer is likely to predominate for adding a non-symmetrical reagent to a non-symmetrical alkene and the rule can be stated in various ways:

For the heterolytic addition of a polar molecule to an alkene (or alkyne), the more electronegative (the most nucleophilic like OH^-, H₂O or Br^- etc.) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms.

This is usually step 2 in the electrophilic addition mechanisms described below.

Or, you can say, the least electronegative (the most electrophilic like Br⁺ or H⁺ etc.) will attach to the carbon atom bonded with the most H atoms.

This is usually step 1 in the electrophilic addition mechanisms, noting which carbocation is the most stable in the mechanisms described below.

 

(4) Chlorine readily reacts in a similar way, but NOT iodine, e.g.

(i) propene  +  chlorine  ===>  1,2-dichloropropane

  +  Cl₂  ===>    

(ii) hex-1-ene  +  chlorine  ===>  1,2-dichlorohexane

+ Cl₂  ===>

(iii) The E and Z stereoisomers of hex-2-ene both produce 2,3-dichlorohexane on addition of chlorine

  or    +  Cl₂  ===>  

 

These are all mechanistically described as electrophilic addition reactions (discussed in section 2.4.3 non-aqueous conditions and section 2.4.4 aqueous conditions)

 

(5) Addition of chlorine or bromine to alkenes with multiple C=C double bonds

For every C=C double bond in the molecule, theoretically, one molecule of the halogen will be added.

e.g. addition of bromine to penta-1,3-diene gives 1,2,3,4-tetrabromopentane

TOP OF PAGE and sub-index

2.4.2 A test to distinguish between an alkane and an alkene

Hydrocarbons are colourless.

Bromine dissolved in water or trichloroethane solvent forms an orange (yellow/brown) solution.

When orange-brown bromine solution (bromine water) is added to both an alkane or an alkene the result is quite different.

The alkane solution remains orange-brown – no reaction, saturated, no C=C double bond to take up a bromine molecule.

However, the alkene rapidly decolourises the bromine as it forms a colourless dibromo–alkane compound (see section 2.4.1 for lots of equations)– see the word and balanced symbol equations below.

Test for unsaturation in fats and oils

If you shake a vegetable oil or saturated animal fat with bromine water, the unsaturated vegetable oil will decolourise the bromine water and a saturated fat will not.

TOP OF PAGE and sub-index

2.4.3 The reaction and mechanism of halogen addition to alkenes in non-aqueous media

I've already written a detailed 'theoretical' pages on this reaction, so I'm just repeating the essential points here.

see Bonding in alkenes, reactivity of alkenes compared to alkanes introducing electrophilic addition

and Electrophilic addition of halogens to alkenes to give dihalocompounds and haloalcohols

Alkenes readily react with liquid bromine or bromine dissolved in an organic solvent like tetrachloromethane or hexane.

The reaction is more complicated with bromine water, described in section 2.4.4

Remember: An electrophile is an electron pair acceptor and will attack the pi electron rich bond of an alkene.

The bromine molecule attacks the electron-rich pi bond of the alkene molecule (e.g. ethene shown below) and forms a C-Br sigma bond with the carbon atoms of the double bond of the alkene functional group.

The pi bond orbitals give the C=C bond a high electron density region for the electrophile to attack.

The bromine molecule is an electrophile, which is a molecule/ion that can bond to an electron rich site by accepting a pair of electrons to form a new covalent bond.

Bromine is a non-polar molecule, BUT, becomes in a polarised state on collision with the alkene molecule and a δ+ δ- state is induced in the bromine molecule i.e. Br^δ+Br^δ- (and similarly for chlorine addition Cl^δ+Cl^δ-)

The carbon carbon double bond is a region of high electron density due to the pi electron clouds above and below the plane of the >C=C< bond system

On collision of the alkene and halogen molecule, an induced dipole effect causes the bromine molecule to be polarised (so can act as an electrophile) and splits heterolytically so that that the equivalent of a Br⁺ bonds to one of the double bond carbons forming a C-Br covalent bond in a carbocation, simultaneously releasing a bromide ion.

The bromide ion than adds to the carbocation to give the final addition product.

Hence the reaction of bromine with an alkene is described as an electrophilic addition.

Mechanism diagram 4 illustrates the general mechanism for adding a halogen electrophile X₂ to an alkene R₂C=CR₂ (R = H, alkyl or aryl) - it should be stressed that this is the mechanism under non-aqueous conditions.

It is written in the same style as for the electrophilic addition of HBr to alkenes, that is via a carbocation.

BUT, this is not the true mechanism, a more correct mechanism is shown below .....

There is considerably evidence (beyond the academic scope of the page) to show that the 1st stage in the mechanism of bromine addition (non-aqueous or aqueous) actually goes via a triangular bromonium ion shown in mechanism diagram 43 above.

The general descriptions of steps 1 electrophilic attack and 2 halide ion addition, apply to both mechanistic descriptions, but you can think of the C-Br bonds in the triangle of the bromonium ion as partial bonds, but (where appropriate) the Markownikoff rule applies (see addition with aqueous bromine in section 2.4.4. for bromine water (aqueous bromine).

This bromonium ion mechanistic style of pathway must NOT be applied to the addition HBr to alkenes

I will now describe both 'styles' of mechanism, carbocation AND bromonium ion pathways and YOU must check with your TEACHER what you need to know for YOUR exam !!!

(1) The carbocation mechanism for the addition of non-aqueous bromine to ethene

Mechanism diagram 59a showing addition of bromine to ethene via a distinct carbocation (but this is not considered the correct mechanism).

In this case, for step (1), the attacking electrophile is the transitory polarised bromine molecule, Br^δ+Br^δ-, which splits heterolytically to brominate ethene, forming the carbocation and a bromide ion.

The pi bond orbitals give the C=C bond a high electron density region for the electrophile to attack.

The bromine molecule is an electrophile because it accepts a pair of electrons from the alkene π bond to form the new C-Br bond.

It is the positive end of the polarised bromine molecule that attracts the electrons of the ethene's pi bond.

The two pi electrons of the C=C bond move to form a covalent bond with the δ+ bromine atom of the polarised (on collision) electrophile Br^δ+Br^δ- which undergoes heterolytic bond fission.

Simultaneously the bromine - bromine bond of the bromine molecule is broken and a bromide ion formed - this is an example of heterolytic bond fission, where the bonding pair of electrons moves onto only one atom of the original bond (Br-Br).

The full curly arrow indicates the movement of a pair of electrons - it must start from the pi bond and go towards the atom forming the new bond (Br in this case).

Note there are two electron pair shifts in step (1).

(i) The pair of pi electrons of the alkene move onto the δ+ bromine atom to form C-Br sigma bond.

(ii) Simultaneously, the original Br-Br bonding pair move onto the bromine atom to form the bromide ion.

In step (2) the bromide ion formed in step (1) rapidly combines with the positive carbocation to form the dibromoalkane (1,2-dibromoethane).

The bromide ion donates a pair of electrons to form the new C-Br bond - shown by the full curly arrow indicating the shift of a pair of electrons from the bromide ion to the positive carbon atom.

(2) The bromonium ion mechanism for the addition of non-aqueous bromine to ethene

Mechanism diagram 59b shows the more correct mechanistic pathway for the electrophilic addition of bromine to ethene via a bromonium ion.

The pi bond orbitals give the C=C bond a high electron density region for the electrophile to attack.

The electrophilic attack by the polarised bromine molecule creates a bromonium ion, to which the bromide ion than adds to ethene, giving the expected product 1,2-dibromoethane.

It is the same reaction mechanisms with chlorine i.e. via a chloronium ion, with the same intermediate triangular arrangement between the two carbon atoms and the halogen atom.

 

The reaction profile for the addition of a halogen (Cl₂ or Br₂) to an alkene.

Mechanism diagram 65 for electrophilic addition:  >C=C<  +  X-X  ===> >CX-CX<

The reaction profile progress diagram for the addition of a bromine or chlorine to an alkene is shown above.

E_a1 is the activation energy for step 1, the formation of the chloro/bomocarbocation after the electrophile 'attack'.

This is the highest activation energy because it initially involves bond breaking processes.

This is also the reason why step 1 will be the slowest of the two steps.

E_a2 is the activation energy for step 2, the formation of the final product when the bromocarbocation combines with an anion (or any other electron pair donating species e.g. it can be a water molecule in aqueous media).

This is the lower activation energy because it only involves a bond making progress.

This will be the fastest of the two steps, especially as it involves two oppositely charged ions coming together.

ΔH is the overall enthalpy change for the reaction and not to be confused with either of the activation energies.

Note there are complications in this reaction if aqueous bromine Br₂(aq) or aqueous chlorine Cl₂(aq) are used, discussed further down in section 2.4.4.

 

(3) The carbocation mechanism for the addition of non-aqueous bromine to propene

The electrophilic addition of bromine to propene under non-aqueous conditions.

In step 1 (pathway 60a) the bromine atom attaches itself to the carbon atom of the double bond with the least hydrogen atoms giving a primary carbocation.

In step 1 (pathway 60b) the bromine atom attaches itself to the carbon atom of the double bond with the most hydrogen atoms - this gives the more stable secondary carbocation.

However, under non-aqueous conditions, It doesn't matter which carbocation is formed, the product is the same in the end.

Note in step 2, the bromide ion attacks from the other side away from the newly bonded bromine atom.

This has stereochemical implications because stereoisomers (due to R/S isomerism) are formed from RCH=CHR alkenes because four different atoms/groups maybe bonded to one of the carbon atoms of the original double bond.

You may not have done R/S isomerism yet. See Part 14.3 Stereoisomerism - R/S isomerism

 

(4) The bromonium ion mechanism for the addition of non-aqueous bromine to propene

Mechanism diagram 60c shows the more correct mechanistic pathway for the electrophilic addition of bromine to propene via a bromonium ion.

The electrophilic attack by the polarised (on collision) bromine molecule creates a bromonium ion, to which the bromide ion than adds to ethene, giving the expected product 1,2-dibromopropane.

It is the same reaction mechanism with chlorine i.e. via a chloronium ion, with the same intermediate triangular arrangement between the two carbon atoms and the halogen atom (just swap Cl for Br).

 

(5) Carbocation and bromonium ion mechanisms for the addition of non-aqueous bromine to cyclohexene

Diagram mechanism 61a shows the addition of bromine to cyclohexene to give 1,2-dibromohexane - carbocation mechanism.

Diagram mechanism 61b shows the addition of bromine to cyclohexene to give 1,2-dibromohexane - bromonium ion mechanism.

In diagrams 61a and 61b less of the non-bonding lone pairs of electrons are shown.

 

(6) Carbocation and bromonium ion mechanisms for the addition of non-aqueous bromine to but-2-ene

Diagram mechanism 62a shows the addition of bromine to but-2-ene to give 2,3-dibromobutane.

The E stereoisomer shown, but its a similar mechanism for Z-but-2-ene and gives the identical product.

Only one product is possible from this symmetrical alkene AND non-aqueous bromine.

The more correct bromonium ion mechanism is shown below in mechanism diagram 62b.

In diagrams 62a and 62b less of the non-bonding lone pairs of electrons are shown

E (trans) stereoisomer shown, but its a similar mechanism for Z-but-2-ene (cis isomer ) and gives the identical product.

 

(7) Carbocation and bromonium ion mechanisms for the addition of non-aqueous bromine to but-1-ene

Diagram mechanism 63 shows the addition of bromine to but-2-ene to give 1,2-dibromobutane.

The secondary carbocation formed in pathway 63b is more stable than the primary carbocation formed in pathway 63a, BUT, in non-aqueous media, the product is the same anyway.

In diagrams 63a-c less of the non-bonding lone pairs of electrons are shown in the formation of 1,2-dibromobutane.

Although not a symmetrical alkene, with pure bromine (symmetrical diatomic molecule), only one product is possible.

 

(8) Carbocation and bromonium ion mechanisms for the addition of chlorine

The same mechanisms apply in non-aqueous conditions, when chlorine adds to alkenes (swap Cl for Br), but iodine is not a sufficiently powerful electrophile to add to alkenes in this way, unless the carbon double bond (C=C) is 'activated' by the presence of an oxygen atom attached to one of the carbons.

TOP OF PAGE and sub-index

2.4.4 The reaction and mechanism of halogen addition to alkenes in aqueous media

I've already written a detailed 'theoretical' pages on this reaction, so I'm just repeating the essential points here.

see Bonding in alkenes, reactivity of alkenes compared to alkanes, introducing electrophilic addition

and Electrophilic addition of halogens to alkenes to give dihalocompounds AND haloalcohols

Again, I will describe both 'styles' of mechanism, carbocation AND bromonium ion pathways and YOU must check with your TEACHER what you need to know for YOUR exam !!!

 

Mechanism diagram 5 describes a generalised carbocation pathway to form an alcohol.

The generalised mechanism by a halogenated alcohol is formed when an alkene reacts with bromine water.

The reason form the formation of the halogenated alcohol is explained via the ionic electrophilic addition mechanism.

 

Again, I need to point out, the above carbocation mechanism is not strictly correct, but the bromonium ion mechanism shown below is accepted as the more accurate representation of the mechanistic pathway.

Mechanism diagram 43b describes a generalised bromonium ion pathway to form an alcohol.

Alternative Mechanism diagram 43b shows the addition of bromine water via the bromonium ion, leading to the formation of a bromoalcohol - this is the more correct representation of the mechanism.

When an alkene is mixed with bromine water, the major product is NOT the dibromoalkane, but a brominated alcohol.

This is because the intermediate carbocation ion is much more likely to interact and collide with a water molecule than a bromide ion.

For a '1-ene' alkene, 1,2-dibromoalkane is not the only product, and this result applies to symmetrical or unsymmetrical alkenes when dealing with aqueous bromine ('bromine water'), because the water, as well as the bromide ion,  can combine with the positive carbocation/bromonium ion.

e.g. for a symmetrical alkene like ethene

Step (i) H₂C=CH₂  +  Br₂  ===>  H₂C⁺CH₂Br  +   Br^-

Step (ii)  H₂C⁺CH₂Br  +   Br^-  ===>  H₂CBrCH₂Br

BUT, there is an alternative step (ii) to form 2-bromoethanol, which is actually the major product.

H₂C⁺CH₂Br  +   H₂O  ===>  HOH₂CCH₂Br  +  H⁺

or more correctly   H₂C⁺CH₂Br  +   2H₂O  ===>  HOH₂CCH₂Br  +  H₃O⁺

(1) The carbocation mechanism for the addition of aqueous bromine to ethene

The above mechanism diagram 66a/b shows how the two different products can be formed from the same intermediate carbocation when ethene is bubbled into bromine water.

However, because of the high concentration of water (the solvent) the main product is 2-bromoethanol via pathway 66b.

The proton released would immediately combine with a water molecule to give the aqueous cation H3O+ (hydronium ion, example of an oxonium ion).

(2) The bromonium ion mechanism for the addition of aqueous bromine to ethene

Using displayed formulae, compared to mechanism diagram 66a/b, the bromonium ion is the only difference in drawing the mechanism in the more correct style to form 2-bromoethanol, the major product - based on probability.

 

An unsymmetrical alkene like propene

Step (i) CH₃CH=CH₂  +  Br₂  ===>  CH₃CH⁺CHBr  or  CH₃CHBrCH₂⁺   +   Br^-

Step (ii)  CH₃CH⁺CHBr  or  CH₃CHBrCH₂⁺  +   Br^-  ===>  CH₃CHBrCH₂Br

(the minor product of this reaction 1,2-dibromopropane, irrespective of the carbocation/bromonium ion from which it is formed)

One version of the Markownikov Rule - the more electronegative (the most nucleophilic) like H₂O or Br^- atom is more likely to become attached to the carbon atom bearing the smaller number of hydrogen atoms AND you still need to be familiar with the carbocation stability trend to, even if the mechanism goes via a bromonium ion.

So, the major product, and governed by the Markownikoff rule, AND the greater probability of step 2  involving water, rather than the bromide ion. will be a secondary alcohol.

Step (i) CH₃CH=CH₂  +  Br₂  ===>  CH₃CH⁺CHBr  +   Br^-

Step (ii) CH₃CH⁺CHBr  +   H₂O  ===>  CH₃CH(OH)CHBr  +  H⁺

(major product is 1-bromopropan-2-ol because of the high concentration of water molecules compared to the concentration of bromide ions - probability of attack argument AND the Markownikoff Rule).

Another minor product, again governed by the Markownikoff rule, will be a primary alcohol.

Step (i) CH₃CH=CH₂  +  Br₂  ===>  CH₃CHBrCH₂⁺   +   Br^-

Step (ii)  CH₃CHBrCH₂⁺   +   H₂O  ===>  CH₃CHBrCH₂OH  +  H⁺

(so the other minor product is 2-bromopropan-1-ol as well as 1,2-dibromopropane)

Although I've simplified equations above, to help you use Markownikoff's Rule, the detailed diagrams are set out below.

(3) The carbocation mechanisms for the addition of aqueous bromine to propene

Mechanism diagram 67a shows the formation of a primary carbocation in step 1, after which two things can happen.

This primary carbocation is less stable than the secondary carbocation formed in pathway 67b below.

(i) It combines with the bromide ion to give 1,2-dibromopropane.

(ii) Due to the high concentration of water molecules, it can also combine with a water molecule to form the halogenated alcohol 2-bromopropan-1-ol

Simultaneously a proton is released, that would combine with a water molecule to form the oxonium ion.  H⁺  +  H₂O  ===>  H₃O⁺

 

Mechanism diagram 67b shows the formation of a secondary carbocation in step 1, after which two things can happen.

This secondary carbocation is more stable than the primary carbocation formed in pathway 67a above.

(i) It combines with the bromide ion to give 1,2-dibromopropane (same as in 67a).

(ii) Due to the high concentration of water molecules, it can also combine with a water molecule to form the halogenated alcohol 1-bromopropan--2-ol (a positional structural isomer of the alcohol formed in 67a).

Simultaneously a proton is released, that would combine with a water molecule to form the oxonium ion.  H⁺  +  H₂O  ===>  H₃O⁺

Comparing mechanistic pathways 67a and 67b for propene plus bromine water

The secondary carbocation BrCH₂CH⁺CH₃ in mechanistic pathway 67b, is more stable than the primary carbocation ⁺CH₂CHBrCH₃ formed in mechanism pathway 67a.

Therefore there will be much more of the 1-bromopropan-2-ol product than 2-bromopropan-1-ol.

The 1,2-dibromopropane might well be the other minority product, but I can't find any data on the relative amounts of the three possible products.

However, because of (i) the high concentration of water (the solvent), and (ii) application of the Markownikoff rule (based on carbocation stability), the main product is 1-bromopropan-2-ol.

Extra note: If carbocations containing a Br atom seem a bit strange, its worth pointing out that the lone pairs of electrons on the bromine atom allow a bigger electron shift (plus inductive effect) than the tightly bound electron pairs on the adjacent hydrogen or carbon atoms attached to the positive carbon atom i.e. a ⁺C-Br is more stabilised than a ⁺C-C situation.

(4) The bromonium ion mechanism for the addition of aqueous bromine to propene

The bromonium ion is the only difference in drawing the mechanism in the more correct style to form 1-bromopropan-2-ol, the major product - based on probability AND the Markownikov rule.

For mechanism diagram 67c, I've shown the three possible mechanistic pathways from the same bromonium ion to give the three possible products.

Major product: 1-bromopropan-2-ol  (1-bromo-2-propanol)

Minor products: 1,2-dibrompropane and 2-bromopropan-2-ol  (2-bromo-2-propanol)

 

(1) Evidence for the IONIC electrophilic addition mechanism in aqueous media

If ethene is bubbled through aqueous bromine solution containing sodium chloride, apart from 1,2-dibromoethane, you also get 1-bromo-2-chloroethane.

Proving that a positive ion (the carbocation) was formed and either the bromide ion or chloride ion could add to it, to give the final 1,2-dihaloalkane product.

Step 1.  H₂C=CH₂  +  Br^δ+Br^δ  ===>  BrCH₂CH₂⁺  +  Br^-

Step 2.  BrCH₂CH₂⁺  +  Br^- or Cl^-  ===>  BrCH₂CH₂Br  or  BrCH₂CH₂Cl

No 1,2-dichloroethane formed because there is no δ+ chlorine atom available, chlorine is only present as the chloride ion Cl^- so it can only add on in step 2 of the mechanism.

Think of the major product being formed by adding a theoretical Br⁺Cl^-.

This is extra evidence on top of the formation of 2-bromoethanol (already discussed above) where step 2. would be

Step 2.  BrCH₂CH₂⁺  +  H₂O  ===>  BrCH₂CH₂OH  +  H⁺

 

(2) Evidence for the IONIC electrophilic addition mechanism in an organic solvent (non-aqueous)

Similarly, if you bubble propene into a methanol solution of bromine and lithium chloride you find the products include 1-bromo-2-chloropropane (major product from the Markownikov rule) and 2-bromo-1-chloropropane (minor product).

Methanol is a polar organic solvent, in which lithium chloride will dissolve sufficiently to give a high concentration of 'non-aqueous' chloride ions!

CH₃CHClCH₂Br  (major product) and  CH₃CHBrCH₂Cl  (minor product).

You will also get smaller amounts of 1,2-dibromopropane, CH₃CHBrCH₂Br

AND molecules with a methoxy group (CH₃O) where the methanol solvent has reacted with e.g. the more stable secondary carbocation you get the sequence:

Step 1.  CH₃CH=CH₂  +  Br^δ+Br^δ  ===>  CH₃CH⁺CH₃  +  Br^-

Then the equations for the formation of the major isomeric products are ...

Step 2. CH₃CH⁺CH₂Br  + Cl^-  ===> CH₃CHClCH₂Br  +  H⁺

...  again think of the major product being formed by adding a theoretical Br⁺Cl^- from the presence of the chloride ion,

or alternatively, from the electron pair donating solvent (lone pair on oxygen atom)

Step 2. CH₃CH⁺CH₂Br  + CH₃OH  ===> CH₃CH(OCH₃)CH₂Br  +  H⁺

This compound would be called 1-bromo-2-methoxypropane.

Think of the major product being formed by adding Br⁺CH₃O^- and use Markownikov's Rule.

Future developments

Produce more skeletal formula mechanism diagrams

More bromonium triangular structures

More on chlorine/chlorine water and mixed ion medium

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2.4.5 Indirect addition of iodine to unsaturated compounds

Unsaturated fats or vegetable oils i.e. triglyceride esters of unsaturated carboxylic acids and glycerol (propane-1,2,3-triol), can be analysed by determining the iodine number - a measure of unsaturation via an iodination reaction.

Reagents containing iodine(I) bromide (iodine monobromide) or iodine(I) chloride (iodine monochloride will react with double bonds in unsaturated fats and oils.

Iodine is too weak an electrophile to engage in reaction with the pi electron cloud of the alkene functional group, BUT, these reagent contain stronger acting electrophiles, but still containing an iodine atom.

(1) The reagents provide the following electrophiles I^δ+Br^δ-  iodine(I) bromide or  I^δ+Cl^δ- iodine(I) chloride, which will then facilitate indirectly the addition of iodine to the C=C double bond.

(2) Excess unused reagent reacts with potassium iodide to release iodine.

(3) The iodine is measured with back titration with sodium thiosulfate/

(1)   R-CH-CH-R  + I₂  == via ICl/IBr ==> RCHI-CHI-R

(2)   Cl/IBr  +  I^-  ==> I₂  + Cl^-/Br^-

(3)   I₂  +  2Na₂S₂O₃ → 2NaI  +  Na₂S₄O₆

 

Test for unsaturation in fats and oils

If you shake a vegetable oil or saturated animal fat with bromine water, the unsaturated vegetable oil will decolourise the bromine water and a saturated fat will not.

 

See also sections

2.5 Uses of hydrogenation, structure and properties of oils and fats

2.9 The occurrence of the alkene functional group in biological molecules

 

Future extras

calculations of unsaturation, addition of bromine etc.

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