Topic 21 - Organic Chemistry
From KstructIB
[edit] 21.1 Determination of structure
[edit] 21.1.1
The structure of a chemical compound can usually not be determine accurately with information from only one source. This stems firstly form the great number of organic compounds possible, many of which can have very similar chemical and physical properties. There are a variety possible techniques which go beyond both chemical and physical properties, and by combining the information from all these sources it is generally possible to deduce the structure of a compound through the individual pieces of evidence given by each of the sources.
[edit] 21.1.2
Covalent bonds are not usually of a static length, but rather resonate in and out rapidly. This effect allows such bonds to absorb electromagnetic energy if it is of a certain wavelength (equivalent to a certain energy). The wavelength corresponding to many organic molecules happens to be within the infrared spectrum. As a result, bonds can be identified in a compound by the wavelengths they absorb. This, however, is usually insufficient as it does not offer enough information about the relative placement of the bonds, or their quantity. The spectrum information is given in the data book and can be matched to any given data.
[edit] 21.1.3
We already covered how it works, so skipping straight on.
There will usually be a peak in the spectrum at the mass of the entire molecule. There may also be, if present, a peak at Mr-15 representing the loss of a CH3 fragment (so there must be one in the molecule). Mr-18 represents the loss of water, Mr-31 represents CH3O and Mr-45 is COOH. Various other fragments could be lost, but the IB only seems to be worried about those listed here.
[edit] 21.1.4
H NMR spectrums are based on some complex nuclear magnetic resonance stuff, but basically give information about what atoms hydrogen atoms are bonded to (though other nearby functional groups can have an effect, so really it identifies 'bits' of the molecule), and how many bonds of each type there are. The information is in the data book.
[edit] 21.2 Hydrocarbons
[edit] 21.2.1
The C-C and C-H bonds are very unreactive due to their lack of significant polarity and the high bond energy.
[edit] 21.2.2
Homolytic (the same) fission (A° °B) : Each atom remains with one lone electron and is highly reactive.
Hetrolytic (different) fission (A :B) : One atom gets a lone pair of electron, the other gets none.
[edit] Reaction of alkanes with halogens (i.e. Cl-Cl)
[edit] Initiation
In the presence of UV light, the Cl-Cl bond breaks homolytically resulting in two Cl° free radicals. Cl2 -- UV light --> 2Cl°.
[edit] Propagation
These radicals then react with alkane (i.e. CH4) to form HCl and CH3°.
Cl° + CH4 -> HCl + CH3°
This CH3° free radical can then react with a chlorine molecule to form another chlorine free radical.
CH3° + Cl-Cl -> CH3Cl + Cl°
[edit] Termination
This reaction is continued until two free radicals react to form a single molecule.
Cl° + Cl° -> Cl2 or Cl° + CH3° -> CH3Cl or
CH3° + CH3° -> C2H6.
[edit] 21.2.3
The structure of Benzene was originally thought to be a ring of alternating double and single bonds, however this does not fit for several reasons.
For one thing, we can consider the enthalpy of combustion of Benzene as compared with this model (an unstable compound called cyclohexatriene). The enthalpy of combustion can be projected from the enthalpies of cyclohexadiene and cyclohexene, but benzene is significantly lower (and therefore more stable) than this projected value.
Also significant are the reactions which benzene undergoes. Double bonds, as seen with alkenes, tend to undergo addition reactions where a double bond breaks forming a single bond between the carbons and two new bonds. Benzene, however, does not undergo such reactions, but rather undergoes substitution, where the hydrogen atoms are replaced by other electrophiles.
[edit] 21.2.4
Octane rating is a scaled devised to measure how smoothly a fuel burns in a combustion engine. The scale is based around two measuring points, heptane which has an octane rating of 0, and 2,2,4-trimethylpentane which has an octane rating of 100. A fuel with an octane rating of 60, for example, would be the same as a mixture of 60% 2,2,4-trimethylpentane and 40% heptane.
In the past, tetraethyl lead (IV) was added to fuels to retard it's ignition and make the fuel burn more smoothly, however this has caused significant problems with lead concentrations in the atmosphere.
The other way to increase octane rating is by using highly branched chains, or aromatic compounds (benzene rings) which also burn more smoothly, thus producing high octane, lead free fuels.
[edit] 21.3 Halogenoalkanes
[edit] 21.3.1
Nucleophillic substitution can occur via two different mechanisms (at least as far as the IB is concerned)
[edit] SN1 mechanism
First, due to the electron donating effect of the alkyl groups, the carbon-halogen bond breaks hetrolytically, resulting in the production of (using CH3Cl as an example) CH3+ and Cl-. This is the rate determining step (hence the 1st order reaction).
The nucleophile then attaches to the positive carbon atom and forms CH3N.
[edit] SN2 mechanism
Rather than completely breaking the bond, a polar bond is formed between the halogen and carbon, producing a delta+ve charge on the carbon.
This charge is enough to attract the nuleophile to form an intermediate in which carbon effectively forms 5 bonds (one to the nucleophile, one with the halogen and 3 others). This is the rate determining step, hence the second order reaction.
The halide ion then breaks off hetrolytically forming again CH3N + Cl-, but via a different mechanism.
Some good nucleophiles are ROH, CN-, OH-, and RNH2.
Markolnikov's rule probably fits in here. In this example, when HX adds across an asymmetrical double bond, the major product formed is the molecule where the less electronegative atom adds to the carbon with the most hydrogens already on it. This is because hydrogen adds on, and produces a carbocation intermediate. The intermediate where the C+ has the most electron donating groups around it will be most common (i.e. the most alkyl groups).
[edit] 21.3.2
SN1 reactions tend to occur with tertiary carbon (i.e. with 3 alkyl groups off the carbon connected to the halogen) for two reasons. Firstly, there is not room for a nucleophile to attack due to the steric hindrance caused by the bulky alkyl groups, and secondly the inductive electron donating effect of these groups mean it is much more likely that the C-X bond will break rather than just becoming highly polar.
SN2 reactions, however, occur on primary carbons since there is plenty of space between small hydrogen atoms for the nucleophile to attack, and it is unlikely that the C-X bond will break on it's own.
[edit] 21.3.3
Different halogens will obviously have different bond strengths. the F-H bond is strongest (and shortest) while I-H is the longest (and weakest). As a result, the H-I bond is easiest to break, and therefore has the highest reaction rate, while H-F has the slowest.
[edit] 21.4 Alkanols
[edit] 21.4.1
Dehydration to form alkenes and alkoxyalkanes:
Alkenes : In the presence of H2SO4 and the proper temperature (hot for primary, warm for secondary and cool for tertiary) alcohols can lose a water molecule and form an alkene.
CH2H-CH2OH -- H2SO4 and heat --> CH2CH2 + H2O.
Alkoxyalkanes : In acidic conditions alcohols act as a base, and accepts a proton. This produces a positive charge on the oxygen (and inductively a delta+ve on the carbon). The spare pair of electrons on the oxygen atom of another alcohol are attracted to the delta+ve carbon making it act like a nucleophile, and so it attacks the carbon atom (this assumes an SN2 reaction though SN1 is possible under other conditions i.e. tertiary alcohols). Water and a proton are then split off producing an ether (alkoxyalkane) and water (and regenerating the acid as a catalyst).
CH2OH + H+ -> CH2OHH+. then CH2OHH+ + CH2OH -> CH2OCH2 + H2O + H+.
[edit] 21.4.2
Oxidation of alkanols is different depending on whether they are primary, secondary or tertiary.
For primary and secondary, a C=O bond replaces the C-OH, but this bond will either be on the end (an aldehyde) or in the middle (a keytone). Aldehydes can be reoxidised to form carboxillic acid. Tertiary alcohols will not oxidize.
Primary :
CH3CH2OH
-- Cr2O72- --> CH3CHO (aldehyde/alkanal) + H2O
--Cr2O72---> CH3COOH (alkanoic acid)
Secondary :
CH3-CH(OH)CH3
--Cr2O72---> CH3-CO-CH3 (keytone/alkanone)
[edit] 21.5 Alkanals and alkanones
[edit] 21.5.1
Carbonyl compounds are reactive because they contain a delta+ve carbon atom, and are unsaturated (both due to the C=O bond). Thus, the Pi electrons can be relatively easily shifted to form a new bond on both the carbon and oxygen atoms, and since nucleophiles are attracted to the carbon atom, the happens relatively quickly.
[edit] 21.5.2
This is the same as with the alcohols. Alkanals will react to for carboxillic acid as follows.
CH3CHO (aldehyde/alkanal) --Cr2O72---> CH3COOH (alkanoic acid)
[edit] 21.5.3
This is the reverse of the process shown in the alcohols above. Alkanals will be reduced to primary alkanols, alkanones will be reduced to secondary alkanols by LiAlH4.
Alkanals : CH3CHO --LiAlH4--> CH3CH2OH (primary alkanol)
Alkanones : CH3-CO-CH3 --LiAlH4--> CH3-CH(OH)CH3 (secondary alkanol)
[edit] 21.6 Alkanoic acids
[edit] 21.6.1
Alkanoic acids can be formed by oxidizing primary alkanols with acidified dichromate (IV) as follows.
CH3CH2OH
--Cr2O72---> CH3CHO (aldehyde/alkanal) + H2O
--Cr2O72---> CH3COOH (alkanoic acid)
[edit] 21.6.2
The OH group in alkanols doesn't act as an acid, but that in alkanoic acids do. This is a result of the inductive effect of the C=O group.
In alkanols, the R groups are electron donating, resulting in a negative charge being inductively pushed along the chain, creating a large -ve charge on the oxygen atom. The C=O bond, however, is electron withdrawing which results in a delta+ve carbon atom. This inductively increases the polarity of the O-H bond, and also produces a more stable anion when the proton is lost (because electron density is being pulled away, creating a smaller negative charge on the oxygen).
[edit] 21.6.3
Soaps are formed of a long hydrocarbon chain ending in a COO-Na+, or similar, head (i.e. CH3-CH2-[-CH2-]n-COO-Na+). They work because the head is hydrophilic (dissolves in water) while the tail is hydrophobic (doesn't dissolve in water, but does dissolve in fats and non-polar dirt). This means the molecules position themselves around small 'blobs' of non-polar dirt (called micielles). These are pulled out of the fabric (or whatever they're in) by these molecules, and are then held in suspension in the water eventually being washed away.
