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Organic chemistry |
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Organic chemistry (hl)
20.1 - Determination of structure 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 possible organic compounds, many of which have very similar chemical properties but different physical properties, similar physical properties but different chemical properties or very similar properties in both categories. There are a variety of 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. Most of the techniques involve either the absorption of electromagnetic energy of specific frequency or the emission of electromagnetic radiation of specific frequencies. Summary
Covalent bonds are not of a static length, but rather they vibrate rapidly. This vibration may take the form of stretching or bending and the amount by which they bend or stretch is determined by the quantity of energy they contain. his energy can be absorbed in discrete packets (quanta) in the infrared region of the spectrum. Consequently if a range of wavelengths (in the IR region of the spectrum) is passed through a sample there will be absorption corresponding to the specific energies that can be absorbed by the covalent bonds. As the number of bonds is usually large and the number of different ways that these bonds can stretch and bend is also large then the spectrum obtained by absorption of energy is complex. The spectrum can be divided into two regions.
Absorptions occuring in the first region can sometimes be identified by consulting the data book. Typical strong absorptions in this region include C=O (the carbonyl group) stretches at around 1600 - 1800cm-1 (see example below) The fingerprint region can be matched to other previously taken spectra to see if an unknown sample corresponds to one previously scanned IR Spectra, 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. Example: IR Spectrum of propanone CH3COCH3 The embedded FTIR of acetone was recorded as a thin film using a PE1605 FTIR Mass spectra See Atomic Theory section 12.1 There will usually be a peak in the spectrum at the mass of the unbroken molecule - the molecular ion - corresponding to the whole molecule which has lost just one electron (and consequently still has the same mass). Other peaks appear as the fragments of the main molecule are broken off in the bombardment process. These fragment masses will correspond to the most obvious parts of the molecule. Example: Propan-1-ol If Propan-1-ol is used for the sample, fragments will appear at
Example: Pentan-2-ol
The molecular ion is designated "m+" , the next fragment m with one mass unit removed etc.
More examples here Nuclear Magnetic Resonance NMR The nuclei of atoms with an odd number of nucleons (protons and neutrons) can absorb energy when in the presence of an external magnetic field and change their spin states. Hydrogen has only one nucleon and its nucleus can have spin states of +1/2 and -1/2. When the hydrogen nuclei are in a magnetic field this can be detected as absorption of radiation at specific frequencies depending on the actual environment (position in the molecule) in which the hydrogen atoms find themselves. If a hydrogen atom is attached to an electronegative atom (such as oxygen) the electrons are drawn away from the hydrogen leaving it more exposed to the externally applied magnetic field. It is in a specific environment and will absorb radiation of a specific frequency. The amount of absorption (area under the curve) is related to the number of hydrogens within the molecule having the same environment and so comparison of the peaks (called integration) gives the relative numbers of hydrogens in each part of the molecule. The spectrum is taken in conjuction with an internally added standard compound called TetraMethylSilane (TMS) which sets the zero on the spectrum. Example: Ethanol CH3CH2OH has three different environments for hydrogen CH3 -Three hydrogens in a methyl group not attached to an electronegative atom CH2 -Two hydrogens in a group not attached to an electronegative atom but much closer to an oxygen OH - A hydrogen attached to an electronegative Consequently there will be peaks corresponding to three hydrogens at 1,2 (compared to TMS=0), two hydrogens at 3,4 and one hydrogen at 3,7. See examples of the spectrum here 
20.2 - Hydrocarbons Hydrocarbons are defined as compounds containing only carbon and hydrogen. This definition covers the homologous series:
The chemical properties of the alkanes are dominated by their low reactivity. This is down to the fact that the C-C and C-H bonds are very strong due to their lack of significant polarity and the high bond energy. For an alkane to react it must undergo homolytic fission of the C-C or C-H bonds Homolytic fission The Bond breaks evenly with one electron from the bonding pair going to each atom A° °B This produces species with single electrons called free radicals. Heterolytic fission When the bond breaks, one atom gets a lone pair of electron, the other gets none. Reaction of alkanes with halogens This is a free radical reaction which must be initiated by chlorine breaking apart homolytically in the presence of UV light to give chlorine free radicals (initiation step) Cl2 -- UV light --> 2Cl° These radicals then react with the alkane (eg CH4) to form HCl and CH3° (propagation step) 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° This reaction is continued until two free radicals react to form a single molecule (termination step) Cl° + Cl° -> Cl2 or Cl° + CH3° -> CH3Cl or CH3° + CH3° -> C2H6 The reactions of alkenes Electrophilic addition across the double bond 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 since 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...ie the most alkyl groups.)
The structure of benzene This was originally thought to be a ring of alternating double and single bonds, however this does not fit for several reasons.
Firstly, there 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 reacts by electrophilic substitution where the hydrogens are replaced by other electrophiles. Octane rating of fuels This is a scale devised to measure how smoothly a fuel burns in a combustion engine. The scale is based around two measuring points, n-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. Another way to increase the octane rating is by using highly branched chains, or aromatic compounds (benzene rings) which also burn more smoothly, thus giving rise to high octane, lead free fuels.
20.3 - Halogenoalkanes The most common reaction of the halogenoalkanes is nucleophilic substitution (SN1 and SN2 mechanisms) The type of mechanism depends on the nature of the halogenoalkane - primary halogenoalkanes react via SN2 and tertiary halogenoalkanes via SN1. The reason for the different mechanisms lies with the stability of the intermediate tertiary carbocation which forms in the case of the SN1 mechanism, whereas a primary carbocation would not be stable encouraging SN2. Another factor encouraging the SN1 mechanism in tertiary halogenoalkanes is the steric hindrance an aproaching nucleophile experiences preventing easy access to the partially positive carbon atom SN1 mechanism First, due to the electron withdrawing effect of the halogen, the carbon-halogen bond breaks heterolytically, resulting in Example: (CH3)3C-Cl
This is the rate determining step (hence the 1st order reaction). The nucleophile then attacks the positive carbon atom and forms (CH3)3C-Nu. SN2 mechanism Rather than completely breaking the bond, the polar bond between the halogen and carbon produces a partial +ve charge on the carbon. This is enough to attract a nuleophile to form an intermediate with effectively 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 heterolytically forming CH3Nu + Cl-. Some good nucleophiles are ROH, CN-, OH-, and RNH2. Rates of nucleophilic substitution
The type of halogen determines the bond strength between the carbon and the halogen. F-C is the strongest and consequently fluoro alkanes are the least reactive (slowest rate) Primary (1º) halogenoalkanes tend to react via the SN2 mechanism which is slower. Consequently the order of reactivity is 3º > 2º > 1º.
20.4 - Alkanols Undergo several types of reactions
Dehydration to form alkenes or alkoxyalkanes The products formed depend on the conditions used: Alkenes are formed in the presence of H2SO4 (or H3PO4 better as it doesn't produce as many by-products) and the correct temperature (hot for primary, warm for secondary and cool for tertiary) alcohols lose a water molecule. Example: CH2H-CH2OH
-- H2SO4 (and heat, 170ºC) Alkoxyalkanes are produced under moderate temperature conditions (140ºC for primary alcohols) Water and a proton are then split off producing an ether (alkoxyalkane) and water (and regenerating the acid as a catalyst). Example: CH2H-CH2OH
-- H2SO4 (and heat, 170ºC)
Oxidation The products depend on the type of alcohol used primary, secondary or tertiary. For primary and secondary, a C=O bond replaces the C-OH, but this bond will either be at the end of the carbon chain (an aldehyde) or in the middle (a ketone). Aldehydes can be further oxidized to form carboxylic acids. Tertiary alcohols will not oxidise. Examples Primary alcohol ... CH3CH2OH
+ Cr2O72- Secondary alcohol ... CH3-CH(OH)CH3
--Cr2O72 Tertiary alcohol ... No reaction
Substitution Not very important for the alcohols except for halogenation using PCl5 which may be used as an identifying test for the alcohol group. Example: CH3CH2OH +
PCl5 The hydrogen chloride gas produced is seen as misty white fumes
Reaction with active metals Alcohols react with active metals releasing hydrogen - once again this is used as an identifying test for alcohols Example: CH3CH2OH +
Na The sodium compound formed is a very strong base (more so than NaOH) Esterification This is the reaction between an alcohol and a carboxylic acid which forms a -COO- linkage (ester link) holding two carbon chains together. The conditions required are concentrated sulphuric acid and elevated temperatures. Some polymers such as terylene contain ester linkages, these are called polyesters and may be formed by reaction of a dicarboxylic acid and a diol (compound with two -OH alcohol groups) The ester linkage ( -COO- ) can be broken by heating with dilute acid or dilute base. This is an example of hydrolysis (where a molecule is broken up under aqueous conditions) 20.5 - Alkanals and alkanones Carbonyl compounds are reactive because they contain a delta+ve carbon atom (caused by polarisation of the C=O bond), and are unsaturated. 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, this happens relatively quickly. Alkanals are oxidised in the same way as the alcohols and form carboxylic acid as follows. CH3CHO (aldehyde/alkanal)
--Cr2O72- Reduction 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. Examples: Alkanals ... CH3CHO --LiAlH4
Alkanones ... CH3-CO-CH3
--LiAlH4 20.6 - Alkanoic acids Alkanoic acids can be formed by oxidising primary alkanols with acidified dichromate (IV) as follows. CH3CH2OH --Cr2O72-
Comparitive acidity of carboxylic acids and alcohols The OH group in alkanols doesn't act as an acid, but the OH hydrogen in alkanoic acids does. This is a result of a combination of two factors:
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 increase 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). The extra stablity of the anion formed in the case of carboxylic acids pulls the equilibrium more to the right hand side: CH3COOH Soaping action Soaps are made up of a long hydrocarbon chain ending in a COO-Na+, or similar, head Example: CH3[-CH2-]n-COO-Na+ dissolves in water to give CH3[-CH2-]n-COO- and Na+ They work because the ion has a "head" that is hydrophilic (dissolves in water) while the tail is hydrophobic (doesn't dissolve in water, but does in fats, non-polar materials, dirt etc). This means the ions position themselves around small 'blobs' of non-polar dirt (called micelles). Eventually these structures have so many negative charges sticking out of them that they become soluble. The grease or water insoluble part is then held in solution in the water and eventually washed away. Resources The following excellent animations are by Johnathon Hopton (Knockhardy Science) Electrophilic addition to ethene Electophilic adition to ethene Nucleophilic substitution with the cyanide ion E2 elimination of a halogenoalkane by base E2 elimination of 2 bromobutane Nucleophilic addition to an alkanone (or alkanal)
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Copyright: 2003 Isis Publication
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(CH3)3C+ + Cl- 
CH3COO-
+ H+
