Halogenoalkanes Flashcards
What is the general formula of a halogenoalkane. And state how halogens are represented In organic chemistry.
CnH2n+1 X Where the halogen atom is an X.
How can halogenoalkanes be classified.
Halogenoalkanes can be classified as primary, secondary or tertiary depending on the number of carbon atoms attached to the C-X functional group.
Draw 3-chloro-2-iodo-4-methylpentane and state what this shows about the naming principles for haloalkanes.
SEE PIC
The substituents should me named alphabetically. So if there is both alkyl and halogen substituents then you need to name before the name and alphabetically.
How does polarity of the C-X bond change down the group and why.
The C-X bond becomes less polar down the group as the atoms become less electronegative. Hence the C-Fl bond is the most polar and the C-I bound is the least polar.
Comment on the solubility of Halogenalkanes. In terms of the intermolecular forces present.
The C-X bond is not polar enough for them to be soluble in water. The main intermolecular forces are dipole dipole and van der Waals forces of attraction. ( they are soluble in Hydrocarbons so they can be used in dry cleaning fluids to remove oily stains).
What are the three possible reactions of halogenoalkanes.
Haloalkanes undergo either substitution or elimination reactions. You have the two organic mechanisms(displayed formula) for the same reagent and you also have free radical structural formula reactions.
Give the definition of a nucleophile and give three nucleophiles for nucleophilic substitution reactions with haloalkanes. Also define substitution in this context.
Nucleophile: electron pair donator e.g. :OH-, :NH3, CN-Substitution: swapping a halogen atom for another atom or groups of atoms.
What property do nucleophiles always have and how can they be represented in general terms.
:Nu represents any nucleophile – they always have a lone pair and act as electron pair donators.
Give the two possible factors that could affect the rate of substitution , and an argument for both and then experimentally the factor that has been shown to be the more important factor for reactivity (rate of substitution) of haloalkanes. State which bond has the greatest rate of substitution.
Both the bond polarity and the bond enthalpies are possible factors. The bond polarity decreases down the group as the atoms get less electronegative and the bond enthalpy follows the same trend since the shared electrons in the C-X bond are more strongly attracted to smaller atoms and going down the group the shared C-X electrons get further from the halogen nucleus . A lesser bond enthalpy predicts a faster rate of substitution,
whereas a greater bond polarity would predict a faster rate of substitution since the carbon atom would be more easily attacked by nucleophiles as would be more positive.
Experimentally the bond enthalpy is a greater factor so the rate of these substitution reactions depends on the strength of the C-X bond
The weaker the bond, the easier it is to break and the faster the reaction.
C-Fl is slowest.
C-I is fastest.
What is the factor you need to be considering for rate of substitution. Hence comment on Iodo and fluoro alkanes.
The bond enthalpy is the important factor that decreases down the group as atomic radius of halogens increases and going down the group the shared C-X electrons get further from the halogen nucleus. The iodoalkanes are the fastest to substitute and the fluoroalkanes are the slowest. The strength of the C-F bond is such that fluoroalkanes are very unreactive.
Draw a nucleophilic substitution mechanism for a haloalkane in general terms using :Nu for a compound with 2 carbon atoms.
See picture.
Define a hydrolysis reaction in the context of a halogenoalkane.
What acts as a poor nuceophile in this reaction and hence in what terms should the hydrolysis reaction be considered.
A hydrolysis reaction is defined as the spllitting
up of a molecule, in the case a halogenoalkane molecule, by a reaction with water. Water acts as a poor nucleophile reacting slowly in a substitution reaction with the halogenoalkane. If aqueous is mentioned.
Draw a mechanism for the reaction of bromoethane with water.
See picture.
Write , in structural formula, the reaction that occurs between bromoethane and water.
CH3CH2Br- + H2O INTO CH3CH2OH + Br- + H+.
For 6 marks outline a method in which you could compare the reactivity of different halogenoalkane compounds. Use equations to illustrate your answer.
Define what the first type of reaction that occurs is and what it is similar to.
Aqueous silver nitrate is added to a halogenoalkane. The halide leaving group combines with a silver ion to form a silver halide precipitate.
Aqueous hydrolysis : CH3CH2X + H2O INTO CH3CH2OH + X- + H+
The hydrolysis is just the splitting of a molecule so this is just the same as nucleophilic substitution with OH- ions except in this case you get the extra H+ as water adds on not OH- to make the alcohol , a neutral nucleophile.
Formation of precipitate : Ag+ (aq) + I-(aq) INTO AgI (s)
The precipitate only forms when the halide ion has left the halogenoalkane and so the rate of formation of the precipitate can be used to compare the reactivity of the different halogenoalkanes.The quicker the precipitate is formed, the faster the substitution reaction and the more reactive the halogenoalkane.
The rate of these substitution reactions depends on the strength of the C-X bond . The weaker the bond, the easier it is to break and the faster the reaction.
The iodoalkane forms a precipitate with the silver nitrate first as the C-I bond is weakest and so it hydrolyses the quickest.
AgI (s) - yellow precipitate (Fastest , lowest bond enthalpy )
AgBr(s) – cream precipitate
AgCl(s) – white precipitate ( Slowest , greatest bond enthalpy)
What two things is the boiling point of a halogenoalkane dependent on. And hence what does the boiling point come down to.
The B.P depends on the chain length and the halogen atom that is present. Both will increase the B.P if they increase the Mr and hence B.P will increase down the halogen group and with increased chain length coming down to the strength of the van Der Waals.
Give the conditions , change in functional group and
possible reagents for a nucleophilic substitution mechanism for a haloalkane.
Change in functional group: halogenoalkane INTO alcohol.
Reagent: potassium (or sodium) hydroxide.
Conditions: In aqueous solution; Heat under reflux.
It will only be substitution if the reaction is with AQEOUS HYDROXIDE IONS !
Both aqueous OH- and the haloalkane are mixed in ethanol.
Name and outline a mechanism for the reaction of 1-bromopropane with aqueous potassium hydroxide heated under reflux. State the leaving group.
See picture
Leaving group is :Br-
Although you don’t need to learn or to use this mechanism what in reality is different about the mechanism for substitution with tertiary halogenoalkanes and explain why by drawing this mechanism for 2-bromo-methylpropane.
See picture for mechanism.
Forms via carbocation intermediate in 2 steps rather than the standard easy 1 step substitution reaction.
1)The Br first breaks away from the halogenoalkane to form a carbocation intermediate.
2)The hydroxide nucleophile then attacks the positive carbon.
Explanation:Tertiary halogenoalkanes undergo this mechanism as the tertiary carbocation is stabilised by the electron releasing methyl groups around it. (See alkenes topic for another example of this). Also the bulky methyl groups prevent the hydroxide ion from attacking the halogenoalkane in the same way as the mechanism above.
Draw out the reaction ( not mechanism) in displayed formula for 1-bromopropane with aqueous potassium hydroxide heated under reflux.
See picture.
For the reaction of halogenoalkanes with cyanide give the reagent and conditions but also state what is slightly different about the reagent that is used. Give also the change in functional group and what is synthetically significant about this reaction.
Change in functional group: halogenoalkane Into nitrile
Reagent: KCN dissolved in ethanol/water mixture
This reagent is significant since as opposed to just aqueous , the potassium cyanide is an aqueous ALCOHOLIC solution not just aqueous like the hydroxide ions should be for substitution reactions.
Conditions: Heating under reflux
Mechanism: Nucleophilic Substitution
Type of reagent: Nucleophile, :CN-
This reaction increases the length of the carbon chain (which is reflected in the name) so is important when the length of the chain needs to be increased by 1.
How would you name a nitrite giving this in context to nucleophilic substitution using KCN. Also name CH3CH2CN and (CH3)2CHCH2CN
Nitrile groups have to be at the end of a chain. Start numbering the chain from the C in the CN for substituents.
The root name of the nitrite can be simply identified as being 1 more than the halogenoalkane it was formed from e.g for example 1-bromopropane would become butanenitrile with a root name that is one greater. Should include the C from the nitrile group in your root name.
To prevent a clash of N’s you need to keep the e in the name for example butanenitrile not butannitrile.
CH3CH2CN is propanenitrile
(CH3)2CHCH2CN is 3-methylbutanenitrile
What is the change in functional group for a nucleophilic substitution reaction where the halogenoalkane is reacted with Ammonia and state the reagent and the conditions for this reaction.
Change in functional group: halogenoalkane Into amine. Reagent: Exess concentrated solution of NH3 dissolved in ethanol. (not just aqueous like hydroxide ions for substitution).
Conditions: Heating under pressure (in a sealed
tube).
:NH3 nucleophile.
What are the two ways that you can name amines. Draw a molecule of propyl amine and state what else this could be called.
Most commonly done by the traditional names where the alkyl group attached to the nitrogen will form the prefix followed by amine as the suffix where the nitrogen is always on the end of the chain so numbering for the amine isn’t needed.
Can also apply IUPAC nomenclature for naming so propylamine could also be called propane-1-amine in IUPAC terms. SEE picture for displayed formula.
Draw , in displayed formula the reaction, between 1-bromopropane and NH3 dissolved in ethanol and heated under pressure in a sealed tube.
Comment on the balanced equation you have given and why In relation to the mechanism it is like this. Also name the main organic product from this reaction.
See Picture for the mechanism. Propylamine is the main organic product with the bi-product being ammonium bromide.
There is 2 moles of NH3 in the reactants since in the second step of the mechanism another NH3 nucleophile will attack a hydrogen forming NH4+ which will go on to combine with the halogen leaving group as shown by the ammonium halide in the products.
Name and outline a mechanism for the reaction between ethanoic ammonia and 1-bromopropane.
State why this mechanism is different from the other two substitution mechanisms.
See picture.
Nucleophilic substitution.
Ammonia is a neutral nucleophile but when it transfers its lone pair to form a bond with the partially positive carbon it forms a positive intermediate with the positive charge on the nitrogen since it is now sharing the electrons as opposed to them both being situated on the Nitrogen nucleophile. Since the nucleophile is neutral this means that the product formed must also be neutral and so a proton must be lost to form a neutral amine product. The lost proton will react with a second molecule of ammonia to form the ammonium ion which in a balanced reaction will go on top react with the halogen leaving group.
Why is an excess of ammonia used in the nucleophilic substitution reaction of ammonia with a halogenoalkane.
Further substitution reactions can occur between the halogenoalkane and the amines formed leading to a lower yield of the amine. Using excess ammonia helps minimise this.
Define elimination.
The removal of a small molecule (often water) from the parent organic molecule.
What is the only nucleophile that can undergo an elimination reaction with a halogenoalkane and given that often a mixture of products from both substitution and elimination will occur state the two factors that influence the predominating reaction.
1)The conditions of the reaction: the solvent
Aqueous: substitution Cold OH- in water Room temp
Alcoholic: elimination Hot OH- in ethanol High Temp
2) The structure of the halogenoalkane:
Primary haloalkanes will react via substitution.
Secondary will do both.
Tertiary haloalkanes have bulky methyl groups and so the hydroxide nucleophile is prevented from attacking the partially positive carbon.
So tertiary will favour Elimination.
Give the change in functional group and the conditions for a reaction where an elimination takes place between hydroxide ions and a haloalkane.
Change in functional group: halogenoalkane Into alkene
Reagents:Potassium (or sodium) hydroxide.
Conditions Hot OH- In ethanol ; Heat no reflux.
Mechanism: Elimination
What is the extra bi-product that you get from an elimination reaction between haloalkane and OH- ion.
You will also need to include the water molecule that is eliminated as a product in both the mechanism and the displayed formula reaction. And the leaving group in the products, note dont include the metal atom in a mechanism since it doesn’t feature in the reactants so can’t be in the products.
What does the OH- act as in elimination compared to substitution.
Substitution: Nucleophile
Elimination: Base (forms water with H).
Write, in displayed formula, the reaction between 1-bromopropane and pottasium hydroxide where the hydroxide ions are dissolved in ethanol at a high temperature.
See picture.
Give the mechanism for the reaction between 1-bromopropane and pottasium hydroxide where the hydroxide ions are dissolved in ethanol at a high temperature.
See picture.
State the type of reaction involving halogenoalkanes that would lead to the formation of structural and stereo isomerism(specifically geometrical).
Elimination (not substitution) reactions will produce a mixture of elimination products if the halogenoalakne is secondary or tertiary as this means the OH- base can attack the hydrogen on either neighbouring carbon swinging in a double bond from either side.
Give all of the structural isomers formed from an elimination reaction between 2-methyl-2-chlorobutane and basic ethanoic OH- ions. Draw this out using arrows , without the mechanism, and state what is Eliminated to give both products. Name the products you have drawn.
Position isomer 1 ) 2-methylbut-1-ene
Position isomer 2) 2-methylbut-2-ene
neither have different groups on both carbons and so in this example there is no third isomer that is an E/Z stereoisomer.
State the uses of long and short chain CFC’s. And also comment on the reactivity of CFC’s under normal conditions.
Short chain CFC’s are used for aerosol propellants and refrigerants.
Long chain CFC’s are used as dry cleaning and degreasing solvents.( Will dissolve in organic compounds)
They are very unreactive under normal conditions.
Give two contrasting effects of Ozone depending on weather it is in the upper or lower atmosphere.
The naturally occurring ozone (O3) layer in the upper atmosphere is beneficial as it filters out much of the sun’s harmful UV radiation. However:
Ozone in the lower atmosphere is a pollutant and contributes towards the formation of smog.
Outline the two main reasons as to why many CFC’s have been stopped.
Many of these uses have now been stopped due to the toxicity of halogenoalkanes and also their detrimental effect on the atmosphere.
Explain , with the aid of equations, how one molecule of CFCL3 could potentially destroy many thousands of Ozone molocules , and also state why it is that the radical you have mentioned is formed as opposed to the other possible radical. (6 marks)
`Note this is the reaction that releases chlorine radicals , not the general reaction of chlorine with an alkane to form a haloalkane.
Name this molocule.
Chlorine radicals are formed in the upper atmosphere when energy from ultra-violet radiation causes C–Cl bonds in chlorofluorocarbons (CFCs) to break.
Formation of radical:CF2Cl2 —> CF2Cl* + Cl* (is a radical)
The C-F bond is stronger than the C-Cl bond and is not affected by UV this is why the chlorine and not fluorine radical is formed. The chlorine free radical atoms catalyse the decomposition of ozone, due to these reactions, because they are regenerated. (They provide an alternative route with a lower activation energy)
These reactions contributed to the formation of a hole in the ozone layer.
Destruction of ozone: (is a radical)
Cl+ O3 —> ClO+ O2.
ClO+O3—–> 2O2 +Cl Overall 2O3 —–> 3O2
The regenerated Cl radical means that one Cl radical could destroy many thousands of ozone molecules.
Molecule is called trichlorflouromethane.
Explain what actions have been taken to combat the destruction of ozone and why the alternatives to CFC’s are chemically less harmful.
Give examples of some less harmful alternatives.
Legislation to ban the use of CFCs was supported by chemists and they have now developed alternative CHLORINE FREE compounds.
Initially HCFC’s where used since the addition of a C-H bond meant that they decomposed lower in the atmosphere meaning the chlorine radicals where released lower in the atmosphere where they couldn’t destroy the ozone layer. e.g CHFCl2
Then second generation replacements that where introduced where HYDROFLUROCARBONS (HFC’s) this did away with the chlorine atom all together since they did not produce any chlorine radicals or anything to damage ozone since the C-Fl bond is too strong so not affected by UV. e.g CHF2CF3
Scientists are still seeking for a third generation replacement since hydrofluorocarbons are not entirely free of environmental problems. they’re considering going back to use NH3 for refrigerants that was used even before CFC’s.
