21 - Organic Chemistry
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21.1 Determination of structure

21.1.1 : The structure of a chemical compound can usually not be determine accurately with information form only one source. This stems firstly form the great number of possible organic compounds possible, 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 possible techniques which go beyond both chemical and physical properties, and by combining the information form 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.

21.1.2 : Covalent bonds are not generally of a static length, but rather they resonate in and out rapidly. This effect allows such bonds to absorb electro-magnetic energy if it is of a certain wavelength (equivalent to a certain energy). The wavelength corresponding to many organic molecules just 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.

21.1.3 : We already did how it works...so skipping 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.

21.1.4 : H NMR spectrum are based on some complex nuclear magnetic resonance stuff, but basically it gives information about what atoms H atoms are bonded to (though other nearby functional groups 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.

21.2 Hydrocarbons

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.

21.2.2 : Homolytic (the same) fission ... A° °B ... each atoms remains with one lone electron -> highly reactive. Hetrolytic (different) fission ... A :B ... One atom gets a lone pair of electron, the other gets none.

Reaction of alkanes with halogens (ie Cl-Cl)...(Initiation) In the presence of UV light, the Cl-Cl bond breaks homolytically resulting in two Cl° free radicals ie Cl₂ -- UV light --> 2Cl°. (Propagation) These radicals then react with alkane (ie CH₄) to form HCl and CH₃° ie Cl° + CH₄ -> HCl + CH₃°. This CH₃° free radical can then react with a chlorine molecule to form another chlorine free radical ie CH₃° + Cl-Cl -> CH₃Cl + Cl°. (Termination) This reaction is continued until two free radicals react to form a single molecule ie Cl° + Cl° -> Cl2 or Cl° + CH₃° -> CH₃Cl or CH₃° + CH₃° -> C₂H₆.

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. 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 has substitution where the hydrogens are replaced by other electrophiles.

21.2.4 : Octane rating is a scaled devised to measure how smoothly a fuel burns in a combustion engine. The scale is bases 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.

21.3 Halogenoalkanes

21.3.1 : Nucleophillic substitution by S_N1 and S_N2 mechanisms

S_N1 mechanism ... First, due to the electron donating effect of the alkyl groups, the carbon-halogen bond breaks hetrolytically, resulting in (using CH₃Cl as an example) CH₃⁺ and Cl^- produced. This is the rater determining step (hence the 1st order reaction). The nucleophile then attacks the positive carbon atom and forms CH₃N.

S_N2 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 is enough to attract the nuleophile to form an intermediate with carbon forming an intermediate with effectively 5 bonds...one to the nucleophile, on 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 by a different mechanism ... Some good nucleophiles are ROH, CN^-, OH^-, and RNH₂.

I think Markolnikov's rule 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 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.)

21.3.2 : S_N1 reactions tend to occur with tertiary carbon (ie 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. S_N2 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.

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 does so most easily therefore has the highest rate, while H-F is the slowest.

21.4 Alkanols

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. ie - 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 par 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 S_N2 reaction though S_N1 is possible under other conditions...ie tertiary alcohols). Water and a proton are then split off producing an ether (alkoxyalkane) and water (and regenerating the acid as a catalyst). ie CH₂OH + H⁺ -> CH₂OHH⁺. then CH₂OHH⁺ + CH₂OH -> CH₂OCH₂ + H₂O + H⁺.

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 replace the C-OH, but this bond will either be on the end (an aldehyde) or in the middle (a keytone). Aldehydes can be reoxadized to form carboxillic acid. Tertiary alcohols will not oxidize.

Primary ... CH₃CH₂OH --Cr₂O₇^2---> CH₃CHO (aldehyde/alkanal) + H₂O --Cr₂O₇^2---> CH₃COOH (alkanoic acid)

Secondary ... CH₃-CH(OH)CH₃ --Cr₂O₇^2---> CH₃-CO-CH₃ (keytone/alkanone)

21.5 Alkanals and alkanones

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.

21.5.2 : This is the same as with the alcohols alkanals will react to for carboxillic acid as follows.

CH₃CHO (aldehyde/alkanal) --Cr₂O₇^2---> CH₃COOH (alkanoic acid)

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)

21.6 Alkanoic acids

21.6.1 : Alkanoic acids can be formed by oxidizing primary alkanols with acidified dichromate (IV) as follows.

CH₃CH₂OH --Cr₂O₇^2---> CH₃CHO (aldehyde/alkanal) + H₂O --Cr₂O₇^2---> CH₃COOH (alkanoic acid)

21.6.2 : The OH group in alkanols doesn't act as an acid, but that is alkanoic acids do...why ? ... 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 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).

21.6.3 : Soaps are formed basically of a long hydrocarbon chain ending in a COO^-Na⁺, or similar, head -- ie CH_3-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 in fats, non-polar dirt etc). this means the molecules position themselves around small 'blobs' of non-polar dirt (called micielles). these are pulled out of the fabric etc by these molecules and are then held in suspension in the water and eventually washed away.

Other Notes in this Category

1. 01 - Stoichiometry
2. 02 - Atomic Theory
3. 03 - Periodicity
4. 04 - Bonding
5. 05 - States of Matter
6. 06 - Energetics
7. 07 - Kinetics
8. 08 - Equilibrium
9. 09 - Acids and Bases
10. 10 - Oxidation and Reduction
11. 11 - Organic Chemistry
12. 12 - Atomic Theory
13. 13 - Periodicity
14. 14 - Bonding
15. 15 - States of Matter
16. 16 - Energetics
17. 17 - Kinetics
18. 18 - Equilibrium
19. 19 - Acids and Bases
20. 20 - Oxidation and Reduction
21. 21 - Organic Chemistry

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