Haloalkane reactions + CFCs and Organic Analysis Flashcards
Describe the products of nucleophilic substitution of haloalkanes with ammonia and cyanide ions
hydrolysis is an example of nucleophilic substitution
nucleophiles : ammonia + cyanide ion
Describe the products of nucleophilic substitution of haloalkanes with ammonia
ammonia and 1-bromopropane
Describe the mechanisms for these reactions
Reaction between ammonia and 1-bromopropane
ammonia + 1-bromopropane —> 1-aminopropane and ammonium bromide
This reaction takes place in two stages
In ammonia, the nitrogen atom, has a lone pair of electrons, so ammonia can act as a nucleophile
The lone pair of electrons on the nitrogen atom is attracted to the positively charged carbon atom in the haloalkane
The nitrogen atom donates the lone pair of electrons to form a covalent bond to the carbon atom
A carbon atom can have a maximum of four bonds
So the covalent bond between the carbon atom and the bromine atom breaks
This is heterolytic fission - with both electrons moving onto the halogen
At the end of stage 1, we have a molecule with a positively charged nitrogen atom
We also have a bromide ion
Now a second molecule of ammonia removes an H+ ion from the nitrogen atom
The covalent bond between the hydrogen and nitrogen breaks with both electrons moving onto the nitrogen
At the end of this stage, we have made the product 1-aminopropane (or propan-1-amine) which is a primary amine
We have also made the ammonium ion NH4+
The ammonium ion and bromide ion form ammonium bromide NH4Br
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Also acts as a base since the NH3 removes a hydrogen from the NH3 attached to the larger molecule, forming an amine and an ammonium ion
Conditions of nucleophilic substitution with ammonia
To carry out this reaction, we warm our haloalkane with a excess concentrated ammonia solution in ethanol
This reaction is carried out in a sealed tube
This increases the pressure of the reaction and the prevents the ammonia escaping as a gas
Important that we use an excess of ammonia
As you can see in the product amine, the nitrogen atom still has a lone pair of electrons
So this can also act as a nucleophile reacting with any unreacted haloalkane
By using an excess of ammonia, we make it more likely that our haloalkane reacts with ammonia rather than with our product amine
Describe the products of nucleophilic substitution of haloalkanes with the cyanide ion
cyanide ion and 1-bromopropane
1-bromopropane + cyanide ion —> butanenitrile + bromide ion
In this reaction, we mix the haloalkane with ethanol and an aqueous solution of potassium cyanide
We then heat this under reflux
The compound we form is called a nitrile, - butanenitrile
-C 3= N
The nitrile is named based on the new carbon chain, not the original haloalkane
This reaction is extremely useful because it allows us to increase the length of the carbon chain
Describe the mechanisms for these
reactions
nucleophilic substitution of haloalkanes with the cyanide ion
cyanide ion and 1-bromopropane
The cyanide ion has a lone pair of electrons on the carbon atom
This lone pair of electrons is attracted to the positively charged carbon atom on the haloalkane
The lone pair of electrons now forms a covalent bond
A carbo atom can have a maximum of four bonds
So the covalent bond between the carbon atom and the bromine atom breaks
Heterolytic fission
At the end we have produced our nitrile molecule (butanenitrile) and a bromide ion
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reaction conditions for nucleophilic substitution involving CYANIDE ION
Warm temperature
Aqueous ethanol solvent
KCN
Describe the elimination reaction between haloalkanes and hydroxide ions
2-bromopropane + KOH
KOH dissolved in ethanol
We can also carry out a different type of reaction using hydroxide ions
Reacting haloalkane with KOH dissolved in ethanol
NaOH dissolved in ethanol can also be used
NOT AQUEOUS + HOT
The reaction is heated under reflux
There is no water present in the reactants
In the absence of water, the hydroxide ion acts as a base
A hydrogen and halogen are eliminated from the haloalkane
This is an example of an elimination reaction
The hydrogen an bromine must be on adjacent carbon atoms
They cannot be bonded to the same carbon atom
The products are an alkene, propene, potassium bromide and water
Describe the mechanism for this reaction
2 bromopropane + hydroxide ion
elimination reaction
2 bromopropane + hydroxide ion
A lone pair of electrons on the oxygen atom forms a covalent bond to a hydrogen in the haloalkane
Hydrogen can have a maximum of one bond
So the electron pair between the hydrogen and carbon move down, between the two carbon atoms
(this releases H atom - to for water molecule)
This electron pair will form part of the double bond in the alkene
Carbon atoms can have a maximum of four bonds
So the electron pair between the carbon and bromine atom moves onto the bromine
This is heterolytic fission and this produces the bromine ion which is the leaving group (electron withdrawing group)
The bromide ion and potassium ion form potassium bromide
At the end, we have our product alkene, a water molecule and potassium bromide
In this reaction, the hydroxide ion is acting as a base
At the start of the mechanism, the hydroxide ion forms a covalent bond to a hydrogen in the haloalkane
At the same time, the electron pair between the carbon and hydrogen move down
What this means is that the hydroxide is actually removing a hydrogen ion from the haloalkane
A hydrogen ion is H+, in other words a proton
So because the hydroxide is accepting a proton, it is acing as a base
How to know if molecule will undergo nucleophilic substitution or elimination
Primary haloalkane - nucleophilic substitution (easier of the mechanisms to do)
Primary -> nucleophilic substitution
Tertiary haloalkane - elimination (C not positive enough so has to do elimination)
Secondary mostly affected by conditions
CH3 - electro donating groups
Pushes electrons closer to the C
This makes the central C atom less positie
Tertiary has more electron donating roups
Pushes electonrs closer to the C
Makes C less partially positive
C is not as attracted to nucleophile
So favours elimination
Describe the role of the ozone layer
The surface of the Earth is constantly bombarded with UV (ultraviolet) radiation in sunlight
UV radiation can damage DNA
Too much exposure to UV can be harmful to humans and is linked to skin cancer
In the atmosphere, there is a region called the ozone layer
This is found in the stratosphere around 20-40km above the Earth’s surface
The ozone layer contains relatively high concentration of the chemical ozone
Ozone - has symbol O3
The ozone layer absorbs a great deal of UV radiation from the sun before it can reach the Earth’s surface
The ozone layer protects living organisms from excessive UV exposure
How does the ozone is formed and absorbs UV
In the stratosphere, UV radiation causes the double bond in oxygen molecules to break
In this reaction, we form two oxygen radicals (species with an unpaired electron)
uv O=O --> O` + O` (2O') OXYGEN MOLECULE OXYGEN RADICAL
Now an oxygen radical reacts with an oxygen molecule to form a molecule of ozone
If this was the only reaction taking place, then ozone levels would continuously increase
However, an ozone molecule can absorb ultraviolet light turning back to a oxygen radical and an oxygen moleucle
This absorption of UV radiation is how the ozone layer protects us
Because ozone is formed and broken down at the same rate, the amount of ozone in the ozone layer should remain constant
However certain chemicals can cause ozone in the ozone layer to be broken down
What are CFCs
Because ozone is formed and broken down at the same rate, the amount of ozone in the ozone layer should remain constant
However certain chemicals can cause ozone in the ozone layer to be broken down
This chemicals are called chlorofluorocarbons or CFCs
e.g. trichlorofluromethane
CFCs are haloalkanes
CFCs are very stable molecules
That is because of the high bond enthalpy of the carbon to halogen bonds (strong bonds)
CFCs are also relatively non-toxic
In the past CFCs were produced in vast quantities and were used in a range of applications e.g. in fridges and freezers.
However, scientists later discovered that CFCs could cause the destruction of the ozone layer
Describw how chlorofluorocarbons are linked to ozone destruction
The reactions which can lead to the destruction of ozone
CFCs make their way up to the stratosphere
Once in the stratosphere, UV radiation causes a carbon-chlorine bond to break
UV
CFCL3 —-> CFCl2 + Cl (chlorine radical)
This process is called photodissociation
This is homolytic fission and produces a chlorine radical which is highly reactive
The next stage involves two steps
Called propagation step 1 and propagation step2
In propagation step 1, a chlorine radical reacts with a molecule of ozone
This produces an oxygen molecule and a highly reactive chlorine monoxide radical
In propagation step 2, the chlorine monoxide radical reacts with an oxygen radical (there are oxygen radicals or ozone constantly being formed in the sratosphere)
produces chlorine radical plus an oxygen molecule
now the chlorine radical from step 2 can go back and trigger step 1 again
in effect, the chlorine radicals are acting as a catalyst for this reaction
ClO` - intermediate
Cl + O3 --> ClO + O2
ClO + O3 ----> Cl + 2O2
Because of this cycle, one CFC molecule can lead to the destruction of many thousands of ozone molecules
what led to CFCs being phased out
many scientific organisations provided evidence that CFCs were leading to the destruction of the ozone layer
This led to CFCs being phased out and less harmful alternatives developed
interpret a mass spectrum showing fragmentation patterns for organic compounds
sample lost electron to form positive ion
positive molecular ion moves down mass spectrometer - produces a spectrum
m/z ratio - tells us the mass and charge of molecular ion
since every molecular ion has 1+ charge the m/z ratio tells us the mass of the molecular ion
M+ molecular ion peak - tells us Mr of molecule
Tiny peak - one unit to right of M+ peak
peak due to presence of c-13 isotope
represent 1% of c atoms
m+1 peak
when we carry out mass spectrometry on an organic molecule, get a range of peaks
that is because organic molecules break up/fragment in mass spectrometer
fragmentation - ion+ radical
Interpret an infrared spectrum for an organic compound
In an organic compound, the bonds are constantly vibrating
Types of vibration
stretching (symmetric and asymmetric) and bending
Stronger bonds vibrate faster than weaker bonds
If the bond is between two heavy atoms then the vibrations will be slower than if the bond is between two lighter atoms
These vibrating bonds can absorb radiation which has the same frequency as the bond vibration
For organic molecules, this radiation lies in the infrared region of the electromagnetic spectrum
When the bonds absorb radiation, the degree of stretching or bending increases
So if we pass infrared radiation through a sample of our organic molecule , then the bonds will absorb specific frequencies.
An absorption spectrum can be produced
Simplified IR absorption spectrum from ethanol
The percent transmittance is shown on the y-axis
This tells us how much infrared radiation passes through the sample
A peak shows us that IR has been absorbed (U shape quadratic - min point)
X-AXIS SHOWS THE WAVENUMBER
wave number is the number of wavelengths per cm
Details of IR spectrum
Below around 1500cm-1 we have the fingerprint region
The fingerprint region is a complex series of peaks which are specific to the molecule being studied
Because of its complexity, the fingerprint region is often analysed by computer
Organic compounds tend to have a peak around 300cm-1
This is caused by vibration of C-H bonds
However, sometimes this peak can be partially obscured by nearby peaks
3 characteristic peaks
In alcohols, vibration of the O-H bond in the OH group gives us a peak between 3200 to 3600 cm-1
Vibration of the C-O bond in the OH group can also cause a peak 1000-1300cm-1
However because this peak is in the fingerprint region, it can be difficult to see
Aldehydes and Ketones contain the carbonyl group
Vibration of the C=O double bond in the carbonyl group produces a peak between 1630 and 1820cm-1
Although typically, this peak centres near 1700cm-1
So this peak can indicate the presence of an aldehyde or ketones
Carboxylic acids also contain the carbonyl group
The peak around 1700cm-1 indicates the presence of the carbonyl group
However carboxylic acids also have an O-H bond in the OH group
This produces a very broad absorption peak
within the range 2500 - 3300cm-1
So the combination of these two peaks shows the presence of a carboxylic acid
C-O single bond can also produce a peak between 1000 and 1300cm-1
However because this peak is in the fingerprint region, it may not be easy to pinpoint
The absence of a peak can be a useful way to rule out certain compounds
E.g. the absence of the carbonyl peak around 1700cm-1 shows that the compound cannot be an aldehyde, ketone or carboxylic acid
Describe how the absorption of infrared by greenhouse gases leads to global warming
Vibrating covalent bonds can absorb infrared energy
This process takes place with the gases in the atmosphere
The sun emits infrared and UV radiation
However these can pass through the atmosphere and are absorbed by the earth’s surface
The surface of the Earth now re-emits the radiation as infrared with a longer wavelength
This infrared has the same frequency as the vibrational frequency of the bonds in greenhouse gases
Includes water vapour, methane and CO2 O=C=O
The vibrating bonds absorb the infrared energy and then reemit this into the atmosphere
This causes the temperature of the atmosphere to increase near the surface of the Earth
The combustion of fossil fuel is increasing the concentration of carbon dioxide in the atmosphere
This is leading to global warming
This is why there is a growing effort to decarbonise human activities
Describe the practical applications of infrared spectroscopy
Infrared spectroscopy has several practical applications
can be used to monitor air pollution such as CO and Nitrogen oxide from car exhausts
It can also be used in breathalysers to check the level of alcohol in the breath
