Coursework Plan For Halogenalkanes Essay Sample
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Coursework Plan For Halogenalkanes Essay Sample
To investigate the rate of hydrolysis of different haloalkanes and to see which C-X bond has the fastest rate
(X = any haloalkane bonded to a carbon atom) under a precipitation reaction acting as a monitor.
Haloalkanes are compounds which contain at least one halogen atom bonded to a carbon atom and have a general formula of CnH2n+1X.
The hydrogen atom in the alkane molecule gets replaces by a halogen atom in a nucleophillic substitution reaction. Nucleophiles are electron rich molecules which have lone pairs of electrons and therefore attack an electron loving carbon atom.
The C-X bond is polarised (C ?+-X ?-) due to the halogen atom being more electronegative then the carbon atom. The electrons in the covalent bond are nearer to the halogen atom therefore the halogen atom carries a partial negative charge. This leaves the carbon atom with a partial positive charge causing the whole bond to become polar but not the whole molecule. In a hydrolysis reaction the negative side of the bond causes the attraction of a hydroxyl ion (OH-) which acts as the nucleophile. It attacks the carbon atom of the haloalkane forming a halogen halide and an alcohol.
I will monitor the rate of hydrolysis by using silver nitrate (AgN03). The time it takes for the silver halide precipitate to form will indicate the rate of hydrolysis of each haloalkane.
C2H5X + OH – C2H5OH + X –
Below is a diagram showing a nucleophilic substitution reaction:-
Below is a SN1 mechanism:-
Below is a SN2 mechanism:-
In this reaction the haloalkane 1-Bromobutane undergoes a nucleophilic substitution reaction. The same will happen for 1-Chlorobutane and 1-Iodobutane.
A SN1 mechanism is when the bond between the carbon atom and halogen atom breaks due to heterolytic fission forming a carbocation. The carbocation is highly positively charged and so is attacked by the negatively charged nucleophile.
A SN2 mechanism occurs in primary haloalkanes where the 2 stands for there being 2 species (molecules or ions) involved in the initial stage of the reaction.
Below shows three symbol equations showing nucleophilic substitution taking place for each hydrolysis reaction that I will carry out:-
AgNO3 & CH3CH2OH
1. CH3CH2CH2CH2Cl (l) + :OH – (aq) CH3CH2CH2CH2OH (aq) + :Cl – (aq)
AgNO3 & CH3CH2OH
2. CH3CH2CH2CH2Br (l) + :OH – (aq) CH3CH2CH2CH2OH (aq) + :Br – (aq)
AgNO3 & CH3CH2OH
3. CH3CH2CH2CH2I (l) + :OH – (aq) CH3CH2CH2CH2OH (aq) + :I – (aq)
Below shows three ionic equations for each hydrolysis reaction that I will carry out:-
1. Ag + + Cl – AgCl (white precipitate)
2. Ag + + Br – AgBr (pale yellow precipitate)
3. Ag + + I – AgI (darker yellow precipitate)
Factors affecting the rate of hydrolysis
1] Polarisation of the C-X bond
Bond polarity occurs due to the difference in electronegativity in the C-X bond. The carbon atom is less electronegative then the halogen atoms so the electrons in the bond are more nearer to the halogen. The carbon atom will now have a partial positive charge known as ?+ (delta plus) and the halogen atom will have a partial negative charge known as ?- (delta minus). As you go down group 7 on the periodic table, polarisation of the C-X bond decreases. This is because the attraction between the two gets weaker due to a greater shielding effect because the number of electrons increase and a weaker nuclear attraction between the central atom of the carbon atom and the outer shell electrons of the halogen atom.
C-I > C-Br > C-Cl
Electronegativity decreases as you
go down group 7 in the periodic table
When a hydrolysis reaction takes place, it’s always the carbon atom which gets attacked by a nucleophile. The halogens are more electronegative than the carbon therefore pull electrons in the bond leaving the carbon atom electron deficient, in other words seeking for electrons. Now a negatively charged nucleophile with lone pairs attacks the carbon atom (or carbcation). I think the C-I bond will have the fastest rate of hydrolysis compared to the C-Br and C-Cl bonds. Not only that I can say the C-I bond has more shielding and a weak nuclear attraction, but because the electrons in the covalent bond are quite far away from the Iodine atom, less energy will be required to break the bond and therefore it will take less time for hydrolysis to be completed.
2] Bond Enthalpy
Bond enthalpy is basically the energy needed to break covalent bonds. As you go down group 7 in the periodic table, bond enthalpy decreases due to polarisation also decreasing. The nuclear attraction gets weaker between the carbon atom and halogen atom due to greater shielding and a larger atomic radius. The halogens which are at the top of the group will need much more energy to break their bond with carbon; however the ones which are at the bottom will need less energy. In conclude the rate of hydrolysis will increase down the group because the bond will break quicker and therefore form new bonds faster.
Shielding will increase because of more electron shells being present and so weaken the attraction between the outer shell bonding electrons and the central nucleus in the halogen atom. The bigger the bond enthalpy, the greater the amount of energy needed to break the bond and the lower the bond enthalpy, the lower the amount of energy needed to break the bond. This means that if more energy is needed to break the bond then the rate of hydrolysis will decrease and if less energy is needed to break the bond the rate of hydrolysis will increase.
Bond enthalpy (kJmol-1)
A table showing the average
bond enthalpies for the halogens.
Relating all this to the halogens states that the bond enthalpy decreases from chlorine to iodine so this means that the bonds will get weaker. Chlorine will have stronger bonds than bromine and iodine, bromine will have stronger bonds than iodine but not chlorine and iodine will have weaker bonds than chlorine and bromine.
When monitoring the experiment, the silver precipitate of an iodine haloalkane will form quicker compared to the silver precipitate of a chlorine and bromine haloalkane.
Prediction of order of reactivity: C-I > C-Br > C-Cl
The order of the induction effect is as follows: – Primary haloalkane > Secondary haloalkane > Tertiary haloalkane.
Induction occurs when haloalkanes shift their electrons to the carbon atom in the C-X bond. Because the halogen atom is ?- (delta minus), it has electrons to give and when it transfers electrons to carbon atom which is ?+ (delta plus) it makes it less positive. The rate of hydrolysis increases if the electron giving group is attached to the carbon which is being attacked, but the rate will get slower if a donating group such as CH3 has been attached. In this case the positive charge on the carbon will not increase but will decrease.
Solvation has a key role in affecting the rate of hydrolysis because it increases the solubility of the haloalkane.
Solvents have a variety of electronegativity strengths (the ability to pull electrons in a covalent bond) therefore what ever solvent is being used, the strength of the nucleophile will vary when it attacks the carbon atom in ethanol. When polar solutions are present in a reaction involving a haloalkane, it will make it easier for a nucleophillic substitution reaction to take place. This is because polar solutions contain charges which stabalise the carbocation (C+). The polar molecules surround the carbocation and help dissolve the haloalkane.
I will be using ethanol to provide nucleophiles and as a solvent, but there are also others possible solvents that can also be used such as ammonia, methanol, propylene carbonate, acetonitrile, acetone and water which is commonly used polar solvent. This factor will be controlled.
5] Stability of Carbocation
CH3Cl CH3+ + Cl-
CH3CH2Cl CH3CH2+ + Cl-
(CH3)2CHCl (CH3)2CH+ + Cl-
(CH3)2CCl (CH3)2C+ + Cl-
The order of “stability of carbocation” is as follows:-
Haloalkanes do not contain any double bonds and so can rearrange themselves into 3 ways primary, secondary and tertiary. A primary haloalkane has two hydrogen atoms and just one alkyl group attached to the carbon atom, a secondary haloalkane has only got one hydrogen atom and two alkyl groups attached to the carbon atom and finally a tertiary haloalkane has no hydrogen atoms attached to the carbon atom but three alkyl groups. Stability of the carbocation increases with the number of alkyl groups attached to the positively charged carbon atom. Tertiary carbocations are more stable and readily formed than secondary carbocations and so if the carbocation is more readily formed, a halide ion is more likely to leave. SN1 reactions and elimination reactions normally do not occur if a primary carbocation is formed.
To control this factor, I will have to keep the carbocation in my experiment constant. I will do this by using three primary haloalkanes known to be 1- Chlorobutane, 1-Bromobutane and 1-Bodobutane. This will make my experiment more reliable by testing each haloalkane under the same hydrolysis reaction and to be more specific, a SN2 mechanism will take place for each of the three reactions.
The number of alkyl groups on the positive carbon ion can have an affect on the rate of hydrolysis. This is because when alkyl groups are bonded to the carbon ion, they stabilise the positive charge into their sigma bonds. They delocalise a positive charge over a wider area. When the numbers of alkyl groups on the carbon ion are increased, this will lead to an increase in stability.
Controlling this factor won’t be a problem because I am already using the same primary haloalkane
(1-halobutane) and therefore hyperconjugation will have not effect on my experiment.
7] Leaving group
Order of leaving halide: – Iodine > Bromine > Chlorine
The SN1 reaction will go much faster if the carbocation is further stable and the better the leaving group then the stable the carbocation. Iodine comes out to be the best leaving group out of the three halogens. This is because it is at the bottom of the group and so has more electron shells causing the outer shell electrons to be further away from the halogen. The rate of hydrolysis does increase as you go down group 7 in the periodic table because of the fact that bond enthalpy decreases due to the atomic radius increasing, producing a weaker bond. Using this information I can say that iodine will be able to stabilise the carbon ion better than bromine and chlorine. In a general fact, the weaker the C-X bond then the better the leaving group because the halogens are more electronegative than carbon making the halide ion leave the alkyl group. For instance the iodine ion will be more stable than the fluorine ion.
I think bond enthalpy has the greatest influence on the rate of hydrolysis because it relates to bond polarisation and electronegativity. The C-X bond is polarised due to a difference in electronegativity where the halogen (X) atom pulls the electrons towards it self. This causes the carbon atom to gain a partial positive charge making in electron deficient. Because of the bond being polar, the haloalkane now has a permanent dipole therefore causing the carbon atom to be attacked by a nucleophile. As you go down the halogen group, electronegativity decreases. The halogens at the top of the group have a higher electronegativity and so have a higher bond polarity and the halogens at the bottom of the group have lower electronegativity and so have a lower bond polarity. C-F will more polar than the rest and therefore have the slowest rate of hydrolysis.
C-F > C-Cl > C-Br > C-I
Going back to my prediction on bond enthalpy, I have explained above how polarisation and electronegativity affect the C-X bond and now will talk about why it relates to bond enthalpy. The more polar the bond, the harder it is to break it because the bond is becoming ionic. This is because a higher dissociation enthalpy is required to break the bond. If more energy is needed to break the C-X bond, than the hydrolysis reaction will occur slower meaning that it will take longer for reactants to turn into products. I predict that the C-Cl bond is stronger than the C-Br and C-I bond because of the fact that C-Cl has a higher bond enthalpy and electronegativity compared to the C-Br and C-F bond. This therefore means that it will take longer periods of time to break.
My prediction can be backed up by what I mentioned before about the shielding effect and nucleur charge. As you go down group 7, the electronegativity decreases and shielding increases due to the number of energy levels also increasing. The C-Cl bond will need a higher amount of energy because the chlorine atom has less energy levels causing a greater nuclear attraction. However the C-I bond will need a lower amount of energy because the iodine atom has more energy levels and so has a greater shielding effect causing a weaker nucleur attraction. The ability for halogens to depart electrons in a covalent bond decreases down the group. The C-Cl bond holds on to the electrons in the covalent bond so strongly so that it takes a lot of energy to break the bond, and the opposite occurs in a C-I bond.
In my experiment, I will predict that 1-iodobutane will take the shortest time to form a silver precipitate and 1-chlorobutane will take the longest time to produce a silver precipitate. Therefore 1-iododbutane will undergo hydrolysis much faster than 1-chlorobutane because it acts as a better leaving group. It is a better leaving group because an iodine atom has more shielding, weaker electronegativity and a lower bond enthalpy. Iodine is less electronegative and therefore has a smaller bond polarity with carbon compared to chlorine.
* LR Ethanol (250cm3 reagent bottle) – The ethanol will act as the nucleophile in the reaction and so will be kept constant
* 0.1M Silver Nitrate (250cm3 reagent bottle) – This will help monitor the rate of hydrolysis by forming a silver halide precipitate
* Water bath – This will be needed so that all haloalkanes that are being used in the experiment are at a constant high temperature so a fair test could be achieved. The temperature needs to be high because it will increase the rate of reaction due to more particles moving around with greater kinetic energy.
* 305mm Thermometer x4 (-10ï¿½C/-110ï¿½C) – This helps measure the temperature of water and solutions when all 3 haloalkanes are in the water bath ensuring that the temperature stays the same.
* 250cm3 Beaker x2 – To take out an excess of ethanol and silver nitrate.
* 125 x 16mm Test tube x3 (with rim) – To help mix the haloalkane, ethanol and silver nitrate together.
* Test tube rack – To help keep hold of all the test tubes in the water bath so that they don’t tip over.
* 5cm3 Graduated pipette (bulb form) – Because small quantities of solutions are being, the readings of all measurements will be more accurate and therefore reduce the chances of errors being made.
* Tile and Permanent marker – So an (X) can be marked on the tile using a permanent marker.
* Digital Stopwatch – So the time taken (seconds) can be measured as the cross on the side of the test tube has vanished.
* Goggles – Should be kept on at all times for safety.
* Gloves – To reduce the chances of hands getting stained by chemicals.
* Pipette filler – Used to help run solution from test tube or beaker up a graduated pipette. Pulling up or pushing down helps getting accurate as possible while measuring.
* Distilled water (250cm3 bottle) – To clean equipment and to make sure no contamination takes place before experiment starts.
* Propanane – So contamination doesn’t take place if distilled water is not dried.
* 1-chlorobutane (comes with a micropipette)
* 1-bromobutane (comes with a micropipette)
* 1-iodobutane (comes with a micropipette)
1] To avoid contamination, clean all glassware with distilled water before carrying out the experiment. Use propanane to clean off all distilled water remaining on equipment so no other side reactions take place.
2] Set out all equipment, put goggles on and make sure you are working in a safe environment.
3] Add water into an electric water bath and make sure the water has a temperature of 55ï¿½C. (Make sure a 305mm thermometer is in the water bath to check the temperature)
4] Collect two 250cm3 beakers and add LR ethanol and 0.1M silver nitrate to each one making sure that there are in excess.
Collect 3 test tubes and put them in a test tube rack.
5] Label them 1, 2 and 3.
6] Using a micropipette add 2 drops of 1-chlorobutane in the first test tube, 2 drops of 1-bromobutane in the second test tube and 2 drops of 1-iodobutane in the third test tube.
7] Put the test tube rack with the 3 test tubes in the water bath and add a 305mm thermometer in each one. Wait until the temperature for all 3 test tubes is 55ï¿½C.
8] Using a pipette filler and 5cm3 graduated pipette pour 5cm3 of 0.1M silver nitrate into each of the 3 test tube.
9] Using a permanent marker mark a black cross on to a tile and place it on the side of the test tubes. The point for doing this is that so the time take can be measured when the black cross in longer visible to see when the silver precipitate halide forms.
10] Using pipette filler and 5cm3 graduated pipette pour 1cm3 of LR ethanol into each of the 3 test tube. As soon as the ethanol is in contact with the haloalkane and silver precipitate solution, start the stopwatch. When the black cross has disappeared, stop the watch and this will be the end point determination.
11] Gather new set of equipment and prepare for second experiment in order to get a second set of results.
12] Label test tubes 1, 2 and 3.
13] Using a micropipette add 2 drops of 1-chlorobutane in the first test tube, 2 drops of 1-bromobutane in the second test tube and 2 drops of 1-iodobutane in the third test tube.
14] Put the test tube rack with the 3 test tubes in the water bath and add a 305mm thermometer in each one. Wait until the temperature for all 3 test tubes is 55ï¿½C.
15] Using a pipette filler and 5cm3 graduated pipette pour 5cm3 of 0.1M silver nitrate into each of the 3 test tube.
16] Using a permanent marker mark a black cross on to a tile and place it on the side of the test tubes. The point for doing this is that so the time take can be measured when the black cross in longer visible to see when the silver precipitate halide forms.
17] Using pipette filler and 5cm3 graduated pipette pour 1cm3 of LR ethanol into each of the 3 test tube. As soon as the ethanol is in contact with the haloalkane and silver precipitate solution, start the stopwatch. When the black cross has disappeared, stop the watch and this will be the end point determination.
18] Finally gather another set of new equipment and prepare for a third experiment to a get a third set of results and total of 3.
19] Label test tubes 1, 2 and 3.
20] Using a micropipette add 2 drops of 1-chlorobutane in the first test tube, 2 drops of 1-bromobutane in the second test tube and 2 drops of 1-iodobutane in the third test tube.
21] Put the test tube rack with the 3 test tubes in the water bath and add a 305mm thermometer in each one. Wait until the temperature for all 3 test tubes is 55ï¿½C.
22] Using a pipette filler and 5cm3 graduated pipette pour 5cm3 of 0.1M silver nitrate into each of the 3 test tube.
23] Using a permanent marker mark a black cross on to a tile and place it on the side of the test tubes. The point for doing this is that so the time take can be measured when the black cross in longer visible to see when the silver precipitate halide forms.
24] Using pipette filler and 5cm3 graduated pipette pour 1cm3 of LR ethanol into each of the 3 test tube. As soon as the ethanol is in contact with the haloalkane and silver precipitate solution, start the stopwatch. When the black cross has disappeared, stop the watch and this will be the end point determination.
Why I chose this method
I have decided to carry out my experiment following this method because it tells me what to do in steps starting from the important instructions such as preparing the experiment, and the very important instructions such as health and safety. It’s good for me to prepare myself before I touch the chemicals because it will reduce error and make sure I don’t forget anything. I chose to add ethanol followed by silver nitrate to the 1-haloalkane because a reaction would not occur between the silver nitrate and 1-haloalkane. If I chose to carry out my method differently such as adding ethanol before silver nitrate, a reaction would have already started between the ethanol and 1-haloalkane. My method is accurate because as soon 0.1M of ethanol is added to 1-haloalkane, a nucleophillic reaction would start to occur.
The dependant variable in my experiment will be the time it takes for the silver precipitate to form (AgX).
I will get my results in seconds by using a stopwatch which will show my results correct to 2 decimal places after a colourless solution has appeared. I will then repeat the experiment 2 more times to obtain accurate reliable data.
This is the variable which I will change during my experiment. It will be the haloalkane that I use in my solutions that will be altered three times. As I am only doing three experiments to measure the rate of hydrolysis, the three haloalkanes that I will use are 1-Chlorobutane, 1-Bromobutane and 1-Iodobutane. This is because the whole point of the experiment is to find out which haloalkane will hydrolyse the reaction the fastest.
My experiment will be repeated and I should obtain a total of 3 results for each haloalkane test. This will provide me reliable results after an average has been worked out. Apart from the independent variable, all the key factors that I will control in order to obtain reliable results are listed below:-
1. When measuring my different solutions, I will either use a different clean graduated pipette for each one or clean the same graduated pipette with distilled water before I pour the solutions into the test tubes. This will avoid contamination.
2. The amount of ethanol I use will always be 1cm3 and therefore will be kept constant.
3. The number of 1-chlorbutane, 1-bromobutane and 1-iodobutane drops will be kept constant at only 2 drops in each test tube.
4. The amount silver nitrate will always be 1cm3 and therefore be kept constant.
5. When taking out any chemicals for extra use, I will always take out the chemicals from the same bottle as I did previously because different bottles vary in concentration.
6. When measuring the temperature I will make sure all the test tubes have the same temperature of 50ï¿½C. This will be achieved by making sure that a thermometer in kept inside the water bath and all test tubes so I can see when the experiment is ready to start as soon as all thermometers reach 50ï¿½C.
* Tie back long hair
* Make sure all bags and coats are kept far away from working area. Can cause accidents such as other pupils tripping over
* Stools should be kept tucked underneath the table and out of the way
* Goggles should be worn at all times during the experiment
* Haloalkanes are very flammable substances and should be handles with care
* If any chemicals have been split make sure they have been wiped up very quickly
* If any have been split on clothing, take off the solid substances quickly and pour plenty of water on the area where a liquid chemical may have dropped.
* When making the water bath, keep chemicals away from the electrical appliances that are used to warm the water up. This will reduce the chances of getting electrical shocks and fires.
If this chemical has been inhaled by a pupil then can cause irritation to the whole respiratory tract and the pupil will start to cough and get shortness of breath. Eating or drinking this chemical will cause irritation to the gastrointestinal tract because it’s very toxic. It may cause diarrhoea and make the pupil vomit. If in contact with the skin or eye, then may cause redness, itching and soreness. It can be very dangerous with sodium.
– If in contact with skin, pour plenty of cold soapy water on the part where chemical has landed for about 15 minutes
– If in contact with eyes, pour plenty of cold water on eyes for about 15 minutes making sure you lift up the top and bottom eye lids at times.
If this chemical has been inhaled by a pupil then can cause irritation to the upper respiratory tract and mucous membranes. The pupil will start to get a headache, cough, vomit, internal burning and get shortness of breath. Taking in this chemical either by drinking or eating will cause irritation to the gastrointestinal tract. If in contact with the skin can cause irritation such and itching and redness. If in contact with eyes or area around the eyes can also cause irritation and redness but also a little bit of pain.
– If the chemical has been inhaled and finding it difficult to breath, go out side and get some fresh air.
– If the chemical has been taken in by eating or drinking it the drink plenty of water.
– If the chemical has been in contact with the skin the remove any spillage clothing and run that area of skin under cold water for about 15 minutes.
– If the chemical has been in contact with eyes then add a lot of water to them making sure you lift the top and bottom eye lids occasionally.
This chemical may be toxic but the properties have not yet been fully discovered yet. If this chemical has been inhaled then can be harmful and can be irritating inside the body if taken in either by eating or drinking it. If in contact with skin or eye, it can also be irritating causing redness and itching.
– If the chemical has been inhaled then get some fresh air or if finding it difficult to breath provide oxygen.
– If the chemical has been taken in by eating or drinking then drink plenty of water.
– If chemical has been in contact with eyes then add a lot of water to them for at least 15 minutes.
– If chemical has been in contact with skin the pour a lot of soapy cold water on that area.
– Highly flammable
– Very toxic and if inhaled then can cause internal damage to either organs or tissues
– Can react very quickly and vigorously with other chemicals such as oxidising agents, potassium and bromine
– Danger of serious irreversible affects when in contact with skin
– Very corrosive
– Silver nitrate solutions can be harmful to eyes and cause the skin to blacken
– Can cause burns when in contact with skin
– Very reactive with other chemicals such as ammonia, magnesium and ethanol
– If ingested then can cause internal damage due to the absorption in the blood leading to decomposition of silver in tissues
6] www.philipharris.co.uk (equipment measurements)
1] Cambridge Chemistry 1 – Brian Ratcliff, Helen Eccles, David Johnson, John Raffan
– Page numbers: 137, 138 & 139
– Headings: – The classification of haloalkanes
– Nucleophillic substitution
– Published 2000, University of Cambridge
2] Advanced Chemistry for You – Headings: – Bond strengths
– Page numbers: 187
– Lawrie Ryan
3] Revise AS Chemistry – Helen Eccles, Mike Wooster
– Headings: – The halogenalkanes
– Reactions of the halogenalkanes
– Page numbers: 60, 61, 62 & 63
– Published by Heinemann educational publishers, 2000