14. Hydrocarbons Flashcards
(22 cards)
What are the two ways alkanes can be produced?
- Hydrogenation of alkenes
- Cracking
How are alkanes produced via the hydrogenation of alkenes?
- Alkenes are heated with hydrogen gas and a nickel or platinum catalyst
- The hydrogen is added across the double bond, so this is an electrophilic addition reaction
- An alkane with the same number of carbons as the original alkene will be produced
This is an exothermic reaction
How are alkanes produced via cracking?
- Long chain alkanes are broken down to form smaller alkanes and alkenes
- Crude oil is heated to around 650 °C and passed over a zeolite catalyst (silica or alumina) in anoxic conditions
- An example reaction is C₁₀H₂₂ –> C₈H₁₈ + C₂H₄
This is an endothermic reaction
Why is cracking useful?
- Lighter fractions of crude oil, like gasoline, are in higher demand than heavier fractions like fuel oil
- Gasoline is used to fuel cars
- Small alkenes are very reactive and can be used for polymerisation and the formation of alcohol
- Therefore, cracking allows more useful hydrocarbons with a lower Mᵣ to be obtained at the expense of less useful hydrocarbons
What is an equation for the complete combustion of propane in oxygen?
C₃H₈ + 5O₂ –> 3CO₂ + 4H₂O
This occurs when oxygen is abundant
What are two equations for the incomplete combustion of propane in oxygen?
- C₃H₈ + 3.5O₂ –> 3CO + 4H₂O
or - C₃H₈ + 2O₂ –> 3C + 4H₂O
- This partial or absent oxidation occurs when oxygen is limited
- Carbon monoxide binds preferentially to haemoglobin, preventing oxygen and carbon dioxide transport
What does free radical substitution of alkanes produce and what are the three steps called?
- Free radical substitution is the substitution of a hydrogen atom on an alkene for a halogen
- It produces a halogenoalkane
- The three steps are initiation, propagation and termination
What happens in the initiation step of free radical substitution?
Use the example of Cl₂ with ethane, but remember that it can happen with other alkanes and other halogens like Br₂
- UV light (a necessary condition for free radical substitution) induces homolytic fission in a Cl₂ molecule
- Two Cl* free radicals are formed
What happens in the propagation step of free radical substitution?
- The Cl* free radicals are very reactive, so attack ethane, taking a hydrogen that breaks off homolytically to form an alkyl free radical and hydrogen chloride
- C₂H₆ + Cl* → *C₂H₅ + HCl
- The alkyl radical then attacks a stable Cl₂ molecule, regenerating the Cl* free radical and forming a halogenoalkane
- *C₂H₅ + Cl₂ –> C₂H₅Cl + *Cl
- The cycle repeats, with the *Cl attacking the newly formed halogenoalkane, substituting another hydrogen to form HCl and a radical alkyl
- Eventually, if there is enough chlorine, all the hydrogens will be substituted, forming C₂Cl₆, though usually a mixture of halogenalkanes will be formed
What happens in the termination step of free radical substitution?
- Two free radicals react to form a stable compound, with examples shown below for the reaction of chlorine and ethane
- If this happens until no more free radicals remain, all free radical substitution will stop
- *C₂H₅ + *C₂H₅ –> C₄H₁₀
- *C₂H₅ + Cl * –> C₂H₅Cl
- (Cl* + Cl* –> Cl₂)
Why are alkanes unreactive?
- C-C and C-H bonds are strong and require a lot of energy to overcome
- They are non-polar as C and H have similar electronegativity values
- Therefore, as they have no regions of partial charge, they do not react with polar reagents
- This also means they do not attract nucleophiles and electrophiles
What are the three ways alkenes can be produced?
- Elimination of HX from a halogenoalkane
- Dehydration of an alcohol
- Cracking of a long chain alkane
How are alkenes produced via the elimination of HX from a halogenoalkane?
- A halogenoalkane is heated with ethanolic sodium hydroxide under reflux
- Hydrogen and a halogen are heterolytically eliminated from the halogenoalkane, forming a H⁺ and X⁻
- The H⁺ reacts with the OH⁻ from NaOH to form water and the X⁻ reacts with Na to form NaX
It is essential that the NaOH is ethanolic; otherwise, nucleophilic substitution will take place
How are alkenes produced via dehydration of alcohol?
- In an elimination reaction, an alcohol loses H₂O to become an alkene
- It loses its OH group and another hydrogen
- Alcohol vapour is passed over a catalyst, which can be heated solid Al₂O₃ or hot concentrated H₂SO₄
- The smaller alkenes are collected as gases, while the larger alkenes must be distilled from the water produced
What are the conditions and reagents required for the hydration of alkenes and what does this reaction form?
- An alkene is reacted with H₂O (g), which is added across the double bond
- Heat, high pressure and a H₃PO₄ catalyst adsorbed on silica are required
- The reaction forms an alcohol in an electrophilic addition reaction
What is the reaction between an alkene and cold dilute acidified KMnO₄?
- Cold dilute potassium permanganate (KMnO₄) is reacted with an alkene
- A diol (compound with two hydroxyl groups) with the same number of carbon atoms as the original alkene is formed
- This is an oxidation reaction
- The purple colour caused by the KMnO₄ will be replaced by the colourless solution of the diol
What is the reaction between an alkene and hot concentrated acidified KMnO₄?
Include the identities of each product depending on the alkene’s structure
- Hot concentrated potassium permanganate (KMnO₄) is reacted with an alkene
- The double bond cleaves and oxidation of the fragments occur
- The products of this oxidation depend on the structure of the original alkene
- In the case of RHC=CHR, each fragment is oxidised to form an aldehyde then a carboxylic acid
- In the case of H₂C=CH₂, each fragment forms carbon dioxide (and water), with aldehyde and carboxylic acid intermediaries
- In the case of R₁R₂C=CR₁R₂, each fragment forms a ketone
- Alkenes with two different structures on either side of the double bond form two appropriate products
The identity of the final products of oxidation provides an indication of the position of the double bond in the alkene
How can aqueous bromine be used to test for the presence of an alkene?
- Add the sample to bromine water and shake
- If it remains orange/brown, the compound is an alkane or another unreactive compound
- If a colourless solution is formed, the compound is an alkene or another unsaturated compound as electrophilic addition has occurred
What is the mechanism of the reaction between C₂H₄ and Br₂?
This is one example of X₂ being added across a double bond
- The high electron density of the double bond creates a temporary dipole in a Br₂ molecule
- The double bond cleaves heterolytically
- The closest Br atom with a partial positive charge acts as an electrophile, simultaneously accepting a pair of electrons from the breaking double bond to form a C-Br bond
- The Br₂ splits heterolytically as this happens, forming a Br⁻ ion and leaving the other Br attached to the carbon
- The other carbon that was once part of the double bond and that does not have the bromine bonded to it is a highly reactive carbocation intermediate, which the Br⁻ ion readily bonds to to form dibromoethane
For all mechanisms, it is crucial that you can draw them
What is the mechanism of the reaction between C₃H₆ and HBr?
- The double bond cleaves heterolytically
- As HBr has a strong dipole, the partially positively charged H atom simultaneously accepts a pair of electrons from the breaking double bond, forming a C-H bond
- As this happens, the HBr molecule breaks heterolytically, forming a Br⁻ ion and leaving the H bonded to the carbon
- The other carbon that was once part of the double bond and that does not have the H bonded to it is a highly reactive carbocation intermediate, which the Br⁻ ion readily bonds to to form bromopropane
- This mechanism is similar to the addition of Br₂ to an alkene
- The mechanism shown in the image involves ethene as oppose to propene
What determines the stability of the carbocation intermediate in the reaction between C₃H₆ and HBr?
- The carbocation with the most alkyl groups bonded to it will be the most stable
- The terms tertiary, secondary and primary carbocation are used to indicate the number of attached alkyl groups
- Tertiary carbocations are the most stable and primary carbocations are the least stable because of the inductive effect
- The inductive effect explains how alkyl groups push electron density toward carbocations (illustrated by arrowheads on bonds), stabilising them
In the reaction between C₃H₆ and HBr, what determines which product forms the most?
- When HBr is added to an asymmetric alkene (like propene, but not ethene), multiple products are possible depending on which carbon in the double bond the H bonds to and which one the Br bonds to
- The major product, which is the product that forms the most, will be the one where the carbocation intermediate needed to form it is the most stable
- The H will therefore bond preferentially to the carbon in the double bond with the most hydrogens already attached (i.e. fewest alkyl groups) so the carbocation intermediary that the bromine bonds to is the most stable
- Hydrogen can bond to the other carbon, but will do so less readily as the carbocation will be less stable so this product will be the minor product
This uses Markovnikov’s rule, so the electrophilic addition of HBr to an asymmetric alkene to form the major product is called Markovnikov addition
