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Advanced Level Organic Chemistry: Alkenes - electrophilic addition of hydrogen bromide

Part 2.3 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

Part 2.3 Bonding in alkenes, reactivity of alkenes compared to alkanes, electrophilic addition reaction with hydrogen halides

Sub-index for this page

2.3.1 The covalent bonding in alkenes (σ and π bonds)

2.3.2 The reactivity of alkenes (compared to alkanes)

2.3.3 The addition of hydrogen halides to alkenes and using the Markownikoff rule

2.3.4 The electrophilic addition mechanism of HX addition to alkenes (non-aqueous conditions), carbocations and explaining and using Markownikoff's rule

2.3.5 The electrophilic addition mechanism of HX addition to alkene (aqueous conditions), and using Markownikoff's rule to predict and explain the likely products

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2.3.1 The covalent bonding in alkenes

From the quantum level rules, carbon's electron configuration is 1s22s22p2

In terms of the separate orbitals you can express it as 1s2, 2s2, 2px1, 2py1, 2pz0

This can be further expressed as an electron box diagram: 1s2s2p with two unpaired electrons.

However, as you should know by now, carbon usually forms four bonds (valency of 4) rather than two.

This is because it is energetically favourable to promote one electron from the 2s orbital into the third empty 2pz orbital, the energy required for this 'promotion' is far less than that released when the carbon atom forms four bonds rather than two..

This gives a theoretical electron configuration of 1s2, 2s1, 2px1, 2py1, 2pz1  or  1s2s2p

This gives four unpaired electrons, all of which can pair up with an electron from another atom to form four covalent bonds, but they may be of two varieties if a double or triple bond is involved in the carbon based molecule - read on ...

Reminders (if needed): The diagrams below show the bonding situation in alkenes, that conveniently involve two of the most important types of covalent bond between atoms, including carbon.

alkenes structure and naming (c) doc b      diagram of sigma and pi bonds in alkenes covalent bonding in alkenes explained advanced level chemistry  Sigma and pi covalent bonds in alkenes

In carbon based compounds a single bond (sigma bond, σ bond) is formed by the overlap of two orbitals which can be either an s orbital and p orbital, or a two orbitals (illustrated above).

Two electrons (an electron pair) are mutually attracted to the positive nuclei on either side.

A molecular orbital is formed and the axis of the bond lies on a central line between the two nuclei.

In this case. the pi bond (π bond) is formed by the overlap of two p orbitals, but, due to repulsion with the bonded pairs of the sigma bond electrons, they cannot form another sigma bond molecular orbital along the same central axis.

Instead, a pi orbital (containing one electron), is formed above and below the planar arrangement of the sigma bonds linking the two carbon, and other two atoms together (>C=C< planar bond arrangement).

Incidentally, the presence of the pi orbitals inhibits rotation around the double bond in unsaturated alkenes, because it requires a lot of energy to twist the orbitals around and break the pi bond.

(Incidentally, this accounts for the possibility of E/Z isomerism in alkenes)

However, in alkanes everything is free to rotate around the sigma bonds of the C-C bonds in saturated alkane molecules.

 


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2.3.2 The reactivity of alkenes (compared to alkanes)

Alkenes are reactive molecules, particularly when compared to alkanes.

They are reactive towards electron pair accepting electrophiles because of the high density of negative electron charge associated with the pi electrons of the double bond (diagram below)

The π and σ bonds of the C=C double bond are shown in the diagram

The sigma bond is present between all the atoms in organic molecules. However in alkenes, the double bond consists of a sigma bond (σ) and a pi bond (π). Two electrons are in the molecular orbital of the sigma bond which is directed linearly between the two carbon atoms.

BUT, the other two electrons of the C=C double bond are in the two pi orbitals which lie above and below the plane of the sigma bonds of the >C-C< system (1 electron per pi electron cloud).

It is the electrons of these pi orbitals that are susceptible to electrophilic attack by an electron pair acceptor - an electrophile.

The pi bond makes alkenes reactive in other ways, so most of alkene chemistry involves addition reactions.

It is the pi bond that breaks to form two new single sigma bonds with two other atoms - often (but not necessarily) both from the attacking reagent.

 Consider the bond enthalpies (bond energies)

Bond Bond enthalpy kJ/mol
σ bond of C-C (i) 348
σ + π bond in C=C (ii) 612
π bond only (ii) - (i) 264

The pi bond is considerable weaker than the sigma bond in a carbon - carbon double bond in alkenes.

This makes alkenes much more reactive than alkanes with their strong bond enthalpies - 348 kJ/mol for C-C and 412 kJ/mol for C-H (both sigma bonds).

A higher bond enthalpy means a higher activation energy inhibiting reactivity.

Irrespective of whether the reaction is electrophilic in nature, the overall addition reaction of alkenes is expressed as:

C=C  +  a-b  ===> a-C-C-b

e.g. alkenes structure and naming (c) doc b +  HBr  ===>     or    (c) doc b

The pi bond of the C=C bond is broken and two new sigma bonds (C-Br and C-H) are formed.

With some alkenes, there is the possibility of isomeric products depending on which way round the HX adds to an unsymmetrical alkene like propene.

Note that alkenes can also readily undergo free radical reactions  e.g. their peroxide catalysed polymerisation to form a poly(alkene) and these reactions also involve the interaction of free radicals with the π (pi) electrons.


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2.3.3 The addition of hydrogen halides to alkenes

Examples of the addition of hydrogen bromide to alkenes, which readily occurs under non-aqueous, gaseous conditions HBr(g) or aqueous conditions with concentrated hydrobromic acid HBr(aq), and all at room temperature and pressure.

(i) ethene + hydrogen bromide ===> bromoethane

alkenes structure and naming (c) doc b  +  HBr  ===>  (c) doc b  

This is a symmetrical alkene, so this reaction can only give one product.

Symmetrical here means the atoms/groups attached to each carbon of the double bond are identical.

But-2-ene, CH3-CH=CH-CH3 is another symmetrical alkene.

 

(ii) propene + hydrogen bromide ==> 2-bromopropane

alkenes structure and naming (c) doc b +  HBr  ===> (c) doc b

Propene is an unsymmetrical alkene - giving rise to two possible products depending on which way round the hydrogen bromide molecule adds to the carbon - carbon double bond..

Some of the other isomer of C3H7Br, 1-bromopropane, is also formed in this reaction.

The minority product (c) doc b is from a con-current reaction, when the H-Br molecule adds the 'other way round' - more on this later when discussing the mechanism.

This reaction involves addition of an unsymmetrical reagent (H-X) to an unsymmetrical alkene, where different atoms/groupings differ on either side of the double bond e.g. H2C=CH-CH3.

The major product is predicted from Markownikoff's Rule which can be stated in different ways e.g.

On the addition of HX to an alkene, the hydrogen atom attaches itself to the carbon that already has the most hydrogen atoms already directly attached.

(The rule is defined in other ways later in this section and only fully explained when the mechanism is considered in section 2.3.4)

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.

So in these reactions you think of it as a sort of  'H+Br-' addition.

 

(iii) The symmetrical E/Z isomers of but-2-ene can only give one product on addition of hydrogen chloride.

 alkenes structure and naming (c) doc b or  alkenes structure and naming (c) doc b  +  HCl  ===>  (c) doc b (2-chlorobutane)

 

(iv) Skeletal formulae to show  addition of hydrogen chloride to but-1-ene to give 1-chlorobutane or 2-chlorobutane.

alkene HCl  ===> (c) doc b  or  (c) doc b

 

(v) The addition of hydrogen iodide to cyclohexene to give iodocyclohexane.

alkenes structure and naming (c) doc b  +  HI  ===>  (c) doc b

 

Some general comments.

As you can see, hydrogen chloride (HCl), hydrogen bromide (HBr) and hydrogen iodide (HI) all add to alkenes.

The reactivity order is HI  >  HBr  >  HCl, because of the order of decreasing bond enthalpies (bond energy).

The smaller the bond energy of 'HX', the lower will be the activation energy for the reaction.

Bond enthalpies (kJ/mol): HI 299  <  HBr 366  <  HCl 431;  C-C (σ bond) 348;  C-C (π bond) 264;  C-H (412).

Apart from the HX bond enthalpy order, also note the low value to break the pi bond of the alkene functional group - the most reactive site on the molecule, it is also the weakest bond involved in the addition of a hydrogen halide to an alkene.

These are mechanistically described as electrophilic addition reactions (mechanism discussed in section next).

Hydrogen chloride (HCl) and hydrogen iodide (HI) also add via the same electrophilic addition mechanism, either mixing the gases, HX in a non-polar solvent or bubbling the alkene into a concentrated aqueous solution of HX.

The Markownikoff (Markovnikov) rule 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 but the IUPAC definition of 1997 states:

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

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

BUT the 'rule' only applies to the ionic mechanism described in the next sections 2.3.4 and 2.3.5.

(but you can get the opposite effect in free radical addition in the presence of peroxides!)

The Markownikoff rule is NOT based on the stability of the molecule, they are effectively ~ equally stable. the rule is based on the relative stability of the intermediate carbocations that can be formed - see later in section 2.3.4.

Examples of using the Markownikoff Rule

(i) methylpropene  +  hydrogen chloride  ===>  2-chloro-2-methylpropane (major product)

or 1-chloro-2-methylpropane (minor product)

alkenes structure and naming (c) doc b +  HCl  ===>  (c) doc b or (c) doc b  

  +  HCl  ===>  (c) doc b or (c) doc b  

(ii) pent-1-ene  + hydrogen bromide  ===> 2-bromopentane (major product) 

or  1-bromopentane (minor product)

CH3-CH2-CH2-CH=CH2  +  HBr  ===>  CH3-CH2-CH2-CHBr-CH3  

or   CH3-CH2-CH2-CH2-CH2Br


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2.3.4 Electrophilic addition mechanism of HX addition to alkenes (non-aqueous conditions) and explaining Markownikoff's rule

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

See Electrophilic addition of hydrogen bromide to form halogenoalkanes (aqueous and non-aqueous media)

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

The hydrogen bromide 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.

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

ionic mechanism of electrophilic addition of hydrogen bromide to alkenes advanced A level organic chemistry

Diagram mechanism 3 illustrates the general mechanism for adding an hydrogen halide electrophile HX (e.g. HBr) to an alkene R2C=CR2 (R = H, alkyl or aryl).

 

ionic mechanism of electrophilic addition of hydrogen bromide to ethene advanced A level organic chemistry

Mechanism diagram 53 shows the electrophilic addition of hydrogen bromide to the alkene ethene.

In this case, for step (1), the attacking electrophile is the already polarised hydrogen bromide molecule, Hδ+Brδ-, which splits heterolytically to protonate the alkene, forming the carbocation and leaving a free bromide ion.

The HBr molecule is an electrophile because it accepts a pair of electrons from the ethene π bond to form the new C-H bond.

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

Simultaneously the hydrogen - bromine bond of the hydrogen bromide 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 (H-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 (H 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 proton (C-H bond formed).

(ii) The H-Br bond pair shift completely onto the bromine atom to form the bromide ion.

In step (2) the negative bromide ion formed in step (1) rapidly combines with the positive carbocation to form the bromoalkane (bromoethane).

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.

Note that ethene is a symmetrical ethene, where the atoms/groups on either side of the double bond are identical, so only one product is possible (at least under non-aqueous conditions).

reaction progress profile for ionic mechanism of electrophilic addition of hydrogen halides to alkenes hydrogen bromide hydrogen chloride hydrogen iodide advanced A level organic chemistry

Mechanism diagram 64 for electrophilic addition:  >C=C<  +  H-X  ===> >CH-CX<

The reaction profile progress diagram for the addition of a hydrogen halide to an alkene is shown above.

Ea1 is the activation energy for step 1, the formation of the carbocation 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.

Ea2 is the activation energy for step 2, the formation of the final product when the carbocation 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 hydrogen bromide HBr(aq) is used, discussed further down.

The next reaction described is the addition of hydrogen bromide to propene.

ionic mechanism of electrophilic addition of hydrogen bromide to propene advanced A level organic chemistry

Diagram mechanism 54 shows the electrophilic addition of hydrogen bromide to the alkene propene.

The description of the steps and reaction profile are identical to that of ethene + HBr, so no repeats!

However, complications arise when an asymmetrical reagent (H-Br) adds to an asymmetrical alkene (where different atoms/groupings differ on either side of the double bond) e.g. H2C=CH-CH3.

The products can be 1-bromopropane or 2-bromopropane.

Note that bromine (Br-Br) and ethene (H2C=CH2) are both symmetrical molecules and give only one product under non-aqueous reaction conditions.

However, before we further discuss the possible products from the hydrogen bromide - propene reaction, the major product AND understand the Markownikoff Rule, you need to know more about the relative stability of carbocations - illustrated below.

inductive effect and the relative stability of carbocations advanced A level organic chemistry

A carbocation has three sigma bonds (via three single covalent bond pairs) to other atoms/groups e.g. H, alkyl etc.

A carbocation has empty 2p orbital, therefore it is electron deficient by one electron and so carries overall a single (but whole) positive charge.

The empty 2p orbital accepts the pair of electrons from the electron donating electrophile to form the new C-Br bond.

The three C-R bonds are actually in a trigonal planar arrangement, bond angles are all ~120o (R = H, alkyl).

Note on some structural terms used:

Tertiary means 3 alkyl groups (and no H's) attached to the relevant positive carbon atom (in this case the carbon atom carrying the positive charge) e.g. (CH3)3C+

Secondary means two alkyl groups (and 1 H) attached to the relevant positive carbon atom e.g. (CH3)2CH+

Primary means one alkyl group (and 2 H's e.g. propyl or ethyl carbocation) attached to the relevant carbon atom OR no alkyl group and 3 H's for the methyl carbocation e.g.  CH3CH2CH2+CH3CH2+  or  CH3+

The orientation of the products from non-symmetrical addition (HX or Br2(aq) see later) is governed by the stability of the carbocation intermediate formed by the protonation of the alkene by the attacking H-X electrophile, and explains the Markownikoff rule.

The order of carbocation stability is tertiary > secondary > primary, because alkyl groups give a slight electron donating inductive effect via the attraction of the positively charged carbon atom.

 This +I effect of the electron shift and its direction, is shown by the brown arrow heads in the diagram above and note from left to right the number of brown arrow heads goes 3, 2, 1 and then 0 as the inductive effect decreases, as does the stability of the carbocation.

The alkyl groups involve  bigger electron clouds than that of a hydrogen atoms and even lone pairs on bromine atom can have the same stabilising effect (see addition of bromine to alkenes).

This inductive electron cloud shift spreads the positive charge of the carbocation beyond the single carbon atom and gives the carbocation more stability by lowering its potential energy. It is possible that the vacant 2p orbital of the positive carbon accepts some of the electron cloud of the adjacent carbon atoms.

Whatever, the true full explanation, it is a general rule of physics, that spreading out electric charge lowers the potential energy and increases the stability of a situation - here the context is the stability of a carbocation involved in an organic chemistry reaction mechanism.

The most stable carbocation will be the one most likely to exist with a sufficient life-time to be hit by the electron pair donating ion (e.g. :X-, :OH-, :Br- or H2O:) or any other electron pair donor that acts as an electrophile.

relative stability of carbocations tertiary > secondary > primary > methyl advanced A level organic chemistry

The Markownikoff (Markovnikov) rule is NOT based on the stability of the molecule, the orientation of the major isomeric product is governed by the relative stabilities of the possible carbocations that could be formed.

e.g. for the following reaction between but-1ene and hydrogen chloride ...

alkeneHCl  ===> (c) doc b  or  (c) doc b

... you would expect 2-chlorobutane to be the major product, and 1-chlorobutane the minor product.

... and the mechanism possibilities are similar to reaction mechanism diagram 58 for addition of HBr.

 

ionic mechanism of electrophilic addition of hydrogen bromide to cyclohexene advanced A level organic chemistry

Mechanism diagram 56 shows the electrophilic addition of hydrogen bromide to cyclohexene to give bromocyclohexane, just one product (if non-aqueous), no need for Markownikoff Rule, its a symmetrical alkene.

 

ionic mechanism of electrophilic addition of hydrogen bromide to but-2-ene 2-butene advanced A level organic chemistry

Mechanism diagram 57 shows the electrophilic addition of hydrogen bromide to but-2-ene to give 2-bromobutane, just one product (if non-aqueous), no need for Markownikoff Rule, its a symmetrical alkene.

 

ionic mechanism of electrophilic addition of hydrogen bromide to but-1-ene 1-butene advanced A level organic chemistry

Mechanism diagram 58 shows the electrophilic addition of hydrogen bromide to but-1-ene to give 1-bromobutane or 2-bromobutane.

Here you need the Markownikoff Rule to predict the major product will be 2-bromobutane.

In equation 58b, the secondary carbocation is CH3CH2CH+CH3, and is more stable than the primary carbocation CH3CH2CH2CH2+ in equation 58a.

So 2-bromobutane will be the major product from the more stable secondary carbocation (pathway 58b),

and 1-bromobutane will be the minor product from the less stable primary carbocation (pathway 58a).

 


2.3.5 The electrophilic addition mechanism of HX addition to alkenes (aqueous conditions), and using Markownikoff's rule to predict and explain the likely products.

 

If a liquid/gas alkene is mixed with concentrated hydrobromic acid, HBr(aq) (hydrogen bromide dissolved in water) a bromoalkane is the expected product.

Expected overall reaction from section 2.3.4: R2C=CR2 + HBr ===> R2CH-CBrR2

but quite the full story by a long way!

In water, hydrogen bromide is a strong acid i.e. completely ionises to give the oxonium ion and bromide ion.

HBr(g) + H2O(l) ===> H3O+(aq) + Br-(aq) 

This means another attacking electrophile is more likely to be the proton donating oxonium ion H3O+. rather than a HBr molecule, which does NOT exist in aqueous solution.

reaction of hydrobromic acid with alkenes electrophilic addition mechanism to form bromoalkane advanced A level organic chemistry

Mechanism diagram 39 - electrophilic addition of hydrogen bromide to an alkene in aqueous media (R = H or alkyl)

In the acid solution via step (1) the H3O+ or oxonium ion (hydrated proton) is the 'attacking electrophile' and protonates the alkene to form the intermediate positive carbocation R2CHCR2+. The oxonium ion is an electrophile because it accepts a pair of electrons from the alkene π bond to form the new C-H bond.

In step (2) the (already present) negative bromide ion rapidly combines with the carbocation to form the bromoalkane product. The bromide ion donates a pair of electrons to form the new C-Br bond.

BUT .... with the high concentration of water present, a water molecule could also interact with the carbocation to eventually form a small amount of the alcohol R2CHCR2OH, this again provides evidence of an ionic mechanism with a carbocation intermediate.

reaction of hydrobromic acid with alkenes electrophilic addition mechanism to form an alcohol advanced A level organic chemistry

Mechanism diagram 39a - electrophilic addition of water to an alkene in acidic aqueous media  (R = H or alkyl)

Step2 in mechanism diagram 39a: R2CHCR2+ + H2O  === > R2CHCR2OH  + H+

Using abbreviated structural formula, I've set out the products that can be formed, indicating whether they are major or minor (if known?) for several alkenes.

(a) Ethene (symmetrical alkene)

(i)  H2C=CH2  +  HBr  ===>  CH3CH2Br    (bromoethane)

(ii) H2C=CH2  +  H2O  ===>  CH3CH2OH   (ethanol)

Not sure which is the major product?

A high concentration of bromide ion favours reaction (i), a low concentration (ii).

(b) Propene (unsymmetrical alkene)

(i)  CH3CH=CH2  +  HBr  ===>  CH3CH2CH2Br    (1-bromopropane)

(ii) CH3CH=CH2  +  H2O  ===>  CH3CH2CH2OH   (propan-1-ol)

(iii)  CH3CH=CH2  +  HBr  ===>  CH3CHBrCH3    (2-bromopropane)

(iv) CH3CH=CH2  +  H2O  ===>  CH3CH(OH)CH3   (propan-2-ol)

Not sure which is the major product?

A high concentration of bromide ion favours reaction (i) and (iii), but the Markownikov rule would make reaction (iii) give the major product - the secondary haloalkane, rather than the primary haloalkane.

A low concentration of bromide ion would favour reactions (ii) and (iv) (ii) but again, the Markownikov rule would make reaction (iv) give the major product - the secondary haloalkane, rather than the primary haloalkane.

 

Comment - evidence of the ionic mechanism in aqueous solution

 If other ions present from e.g. from dissolved salts like sodium chloride, NaCl, subsequence analysis shows as well as the products indicated by the equations above, you would also get chloroalkanes formed as well as bromoalkanes.

e.g. reaction (a) would give some chloroethane

and (b) would give 2-chloropropane >> 1-chloropropane (again using the Markownikov rule)


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