4.1.3 (e,f,g,h,i) Addition reactions of Alkenes

Syllabus

(e) the reactivity of Alkenes in terms of the relatively low bond enthalpy of the π-bond

(f) addition reactions of Alkenes with:

(i) Hydrogen in the presence of a suitable catalyst, e.g. Ni, to form Alkanes

(ii) Halogens to form dihaloalkanes, including use of Bromine to detect the double C=C bond as a test for unsaturation in a Carbon chain

(iii) Hydrogen halides to form Haloalkanes

(iv) steam in the presence of an acid catalyst, e.g. H₃PO₄, to form Alcohols

(g) definition and use of the term electrophile (an electron pair acceptor)

(h) the mechanism of electrophilic addition in Alkenes by heterolytic fission (see also 4.1.1 h–i)

{For the reaction with halogens, either a carbocation or a halonium ion intermediate is acceptable.}

{Use of reaction mechanisms to explain organic reactions.}

(i) use of Markownikoff’s rule to predict formation of a major organic product in addition reactions of H–X to unsymmetrical alkenes, e.g. H–Br to Propene, in terms of the relative stabilities of carbocation intermediates in the mechanism

{Limited to stabilities of primary, secondary and tertiary carbocations.}

{Explanation for relative stabilities of carbocations not required.}

{Use of stability to explain products of organic reactions.}

What does this mean?

Reactivity of Alkenes

An addition reaction is one in which atoms or groups are added to a molecule without replacing anything.

Alkenes undergo electrophilic addition.

Electrophile literally means electron-loving - but don't write this in an exam.

It is obvious that positive or partially positive groups would be attracted towards the high electron density in the double bond.

It's less obvious that to form a new bond to the Carbon atom the electrophile will need to provide both electrons for the bond-pair.

But this is what an examiner will be interested in when asking for a definition of an electrophile as "an electron-pair donor".

Hydrogenation of Alkenes

You need to know that to add Hydrogen to Ethene (Ethylene is an industry name) an Nickel catalyst is needed.

But you won't be asked for a mechanism for this particular addition reaction.

Or this one, but you'll need to know that concentrated Phosphoric (or Sulphuric) acid is the catalyst.

And that the product is always an alcohol.

But you will need to know the mechanism for others.

Addition of HBr

To show the movements of electron pairs in mechanisms we use curly arrows.

Since these show movements of electron-pairs they either show bonds forming or breaking.

Learn the mechanism for electrophilic addition it will get you 3 or 4 marks in every exam.

This one is an examiner favourite.

Stage 1

Note that the first arrow starts on the C=C double bond where an electron pair can be found.

And that it moves towards the δ+ Hydrogen atom on the H-Br where it forms a new bond.

This is heterolytic fission because both electrons in the pair are moving towards one atom.

You must be able to draw the correctly δ+ / δ- arrangement the right way round so an understanding of electronegativity is required.

The second curly arrow must start in the H-Br bond and move to the Br atom yo show the H-Br bond breaking.

This makes logical sense since Hydrogen atoms can only form one bond at a time so, since one has formed, one must break.

Stage 2

When the H-Br bond broke both electrons in the bond-pair moved onto the Br.

But the Br had provided only one of these electrons and the H had provided the other.

So, the Br gained an electron that didn't "belong" to it.

That is why it has gained a negative charge to be Br^-.

But a negative charge cannot appear out of nowhere.

In stage one the overall charge was neutral, so the overall charge must remain neutral.

There must be a corresponding positive charge.

And it's logically got to be where the electrons have moved away from - which is why it is on the Carbon that only has 3 bonds.

A carbo-cation has formed.

Clearly one of the four lone pairs of electrons on the Br^- would move onto the positive carbon, forming a new bond.

This curly arrow should start on a lone pair (so draw one).

And this completes the molecule that we draw in Stage 3.

Addition of Halogens.

You'll notice that all the arrows and full charges on this mechanism are exactly the same as for the reaction with HBr.

The question is, why should one Br atom be δ+ and the other be δ- when both atoms have the same electronegativity.

But one Bromine atom is closer to the electron-density in the double bond.

This will repel electrons away from this Bromine (making it δ+), towards the other Bromine (making it δ-).

This is an induced dipole rather than the permanent one in HBr.

After which the reaction can proceed just as before.

This reaction could happen equally well with Chlorine in place of Bromine but Bromine is preferred as a test for unsaturation (presence of a C=C double bond) because the colour change from Orange to colourless is very obvious.

Markownikoff's rule.

If we added HBR to Propene we could make two products.

You would assume that they would be produced in a 50:50 ratio.

But you would be wrong.

Markownikoff's rule tells us that the δ+ Hydrogen atom will mostly join the Carbon atom that already has the most Hydrogen atoms.

But why?

Look at the two Carbocations that can form.

The top one has two alkyl groups (the CH₃ and the R group) that can push electrons towards it, stabilising the positive charge.

It will form quite easily.

The bottom carbocation only has one alkyl group to stabilise it so forms less well.

So, since the first carbocation forms in preference the Hydrogen atom on the H-Br will mostly end up on the lefthand Carbon atom in the alkene - and will only sometimes end up on the middle one.

And the Br atom is more likely to end up in the middle rather than on the lefthand Carbon atom.

Alkyl groups pushing electrons is known as the Alkyl Inductive Effect.

Video

Exam-Style Questions

1. The alkene CH₃CH=C(CH₃)CH₂CH₃ reacts with Hydrogen Bromide to form 3-bromo-3-methylpentane,

CH₃CH₂CBr(CH₃)CH₂CH₃, as the major product.

(a) Give the name of this alkene and state the type of stereoisomerism shown by this compound.

Name of alkene ..................................................................................................................................................

Type of stereoisomerism ....................................................................................................................................(2)

(b) Name and outline a mechanism for the reaction between the alkene and hydrogen bromide.

Name of mechanism ...............,...........................................................................................................................

Mechanism

(5)

(c) Give the structure of the isomeric product also formed in the above reaction and explain why the two isomers are obtained in unequal amounts.

Structure

Explanation.

(3)

2. Many naturally-occurring organic compounds can be converted into other useful products.

Oleic acid can be obtained from vegetable oils. Oleic acid is an example of an unsaturated compound.

CH₃(CH₂)₇CH=CH(CH₂)₇COOH

oleic acid

(i) Deduce the molecular formula and the empirical formula of Oleic acid.

Molecular formula ........................................................................................................................................

Empirical formula ........................................................................................................................................

(ii) State what is meant by the term unsaturated.

.............................................................................................................................................................................

(iii) Identify a reagent for a simple chemical test to show that Oleic acid is unsaturated. State what you would observe when Oleic acid reacts with this reagent.

Reagent ...........................................................................................................................................................

Observation with Oleic acid ........................................................................................................................(5)

(Total 5 marks)

3. Ethene is an important starting point for the manufacture of plastics and pharmaceutical chemicals. Most of the Ethene used by industry is produced by the thermal cracking of Ethane obtained from North Sea gas (Reaction 1). It is also possible to make Ethene either from Chloroethane (Reaction 2) or from ethanol (Reaction 3).

(a) Give essential conditions and reagents to reverse Reaction 3. (2)

(b) Suggest a reason why chloroethane is not chosen by industry as a starting material to make Ethene commercially.

Answers

1. (a) Name of alkene 3-methylpent-2-ene (1)

Type of stereoisomerism geometrical or cis-trans (1)

(b) Name of mechanism electrophilic addition (1)

Mechanism

(c) Structure

Explanation (CH₃CH₂)2C⁺CH₃ or tertiary Carbonium ion more stable (1)

than CH₃C⁺HCH(CH₃) CH₂CH₃ or secondary Carbonium ion (1)

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2. (i) C₁₈H₃₄O₂ Only; 1

C₉H₁₇O Only; 1

(empirical formula is not consequential on molecular formula)

(ii) (An unsaturated compound) contains (at least) one double bond

OR

Contains C=C; 1

(must be a positive statement)

(iii) M1: Bromine water

OR

Br_2(aq)

OR

Bromine

OR

Br₂; 1

(penalise “bromide water”, but mark on)

Ml: decolourised or goes colourless

OR

from brown/red/orange/yellow to colourless; 1

(Must be “colourless” not “clear” for M2)

(chemical error if no reagent or wrong reagent, loses both marks) (credit KMnO 4 for M1, (purple) to colourless for M2 (if acidified) OR (purple) to brown/brown precipitate (if alkaline or unspecified) (No credit for hydrogen or iodine as reagents)

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