CAPITULO 2: MARCO TEÓRICO
2.6 Medio ambiente laboral
2.6.1 Iluminación
Organic analysis
You should know the tests for:
Alkenes
Test: Add bromine in hexane.
Observation: The brown bromine becomes colourless.
Halogenoalkanes
Test: Heat under reflux with sodium hydroxide solution, then acidify with dilute nitric acid. Add silver nitrate solution.
Observation: Chlorides give a white precipitate that is soluble in dilute ammonia.
Bromides give a cream precipitate that is insoluble in dilute ammonia but dissolves in concentrated ammonia.
Iodides give a yel low precipitate that is insoluble in concentrated ammonia.
OH group (in alcohols and acids)
Test: To the dry compound add phosphorus pentachloride.
Observation: Steamy fumes of hydrogen chloride produced.
Alcohols
Test: Warm with dilute sulphuric acid and potassium dichromate(VI) solution.
Observation: Primary (1 °) and secondary (2°)alcohols reduce the orange dichromate(VI) ions to green Cr3+ ions.
Tertiary (3°) alcohols do not affect the colour as they are not oxidised.
To distinguish between 1 o and 2° repeat the experiment, but distil the product into ammoniacal silver nitrate solution:
1 o alcohols are oxidised to aldehydes, which give a silver mirror.
2° alcohols are oxidised to ketones, which do not react.
Carbonyl, C=O group (aldehyde or ketone)
Test: Add a solution of 2,4-dinitrophenylhydrazine.
Observation: A red or orange precipitate is seen.
The solid must b e pu rified by recrystall isation in order to have a sharp and accu rate melti ng temperatu re.
When identifying species that cause lines in the spectru m, always give a structu ral form ula for the ion and its + charge.
Fig 5. 7 Mass spectrum of ethanol
The peaks given by 0-H and N-H are usually broad owing to hyd rogen bonding.
To distinguish between aldehydes and ketones:
Test: Warm with ammoniacal silver nitrate solution.
Observation: Aldehydes give a silver mirror, ketones have no reaction.
Iodoform reaction
Test: Gently warm with a solution of sodium hydroxide and iodine.
Observation: A pale yellow precipitate of CHI3
This test works with carbonyl compounds containing the CH3C=O and alcohols containing the CH3CH(OH) groups.
Carboxylic acids, COOH group
Test: Add to a solution of odimn hydrogen carbonate.
Observation: Bubbles of gas, which turn lime water milky.
Interpretation of data
You should be able to deduce structural formulae of organic molecules given data obtained from:
Chemical methods
Carry out tests as above to detern1ine th functional roup in the molecule. Measure the melting point of the ub tance. �fake a olid derivative (such as the 2,4-D �p deri ati e from carbonyl ompound ), purify it, and measure its melting temperature. Check the m ltin temperature values with a data bank to identify the substance.
Mass spectra (see Figure 5 . 7)
Observe the fragments obtained and look for the value of the molecular ion (M)+, if any.
See if there is a peak at (M-1 5)+. If so the substance probably had a CH3 group which is lost forming the (M- 1 5)+ peak.
If there is a peak at (M-28)+, the substance probably had a CO group.
Other fragments will help to identify the structure.
100
These are used to identify functional groups in the substance, and also to compare the spectrum of the unknown with a data bank of spectra ('finger printing').
110 T R A N S I T I O N M E TA L S , Q U A N T I TAT I V E K I N E T I C S A N D A P P L I E D O R G A N I C C H E M I S T R Y
Fig 5. 8 Infrared spectrum o f ethanol
The chem ical sh ift is the d ifference between the absorption frequencies of the hyd rogen nuclei in the compound and those in the refe rence compound.
Fig 5. 9 High resolution NMR spectrum of ethanol
4000 3000 2000 1 500 1 000 500 Wavenumber I cm-1
NMR spectra
1 1
The phenomenon of nuclear magnetic resonance occurs when nuclei such as 1H are placed in a strong magnetic field and then absorb applied radio frequency radiation.
The nuclei of hydrogen atoms in different chemical environments within a molecule wil l show up separately in a NMR spectrum . The values of their chemical shift, 8, are different.
The hydrogen nuclei in a CH3 group will have a different chemical shift from those in a CH2 or in an OH group. The value of o of the peak due to the hydrogen in OH (or in NH) depends upon the solvent.
In low resolution NMR, each group will show as a single peak, and the area under the peak is proportional to the number of hydrogen atoms in the same environment. Thus ethanol, CH3CH20H wil l have three peaks of relative intensities 3:2: 1 and methyl propane
CH3CH(CH3)CH3, wil l have two peaks with relative intensities of 9: 1 . I n high resolution NMR spin coupling i s observed. This i s caused by the interference of the magnetic fields of neighbouring hydrogen nuclei. If an adjacent carbon atom has hydrogen atoms bonded to it, they will cause the peaks to split as follows:
1 neighbouring H atom peak splits into 2 lines (a doublet) 2 neighbouring H atoms peak splits into 3 lines (a triplet) n neighbouring H atoms peak splits into (n + 1) lines.
Thus ethanol gives three peaks (see Figure 5 .9):
1 peak due to the OH hydrogen, which is a single line (as it is hydrogen bonded)
1 peak due to the CH2 hydrogens, which is split into four lines by the three H atoms on the neighbouring CH3 group.
1 peak due to the CH3 hydrogens, which is split into three lines by the two H atoms on the neighbouring CH2 group.
1 0 9 8 7 6 5
ppm
4 3 2 1 0
1 A common error is to think that KCN or HCN will react with alcohols to fo rm RCN. You m ust first convert ROH to R-halogen.
2 Many q uestions on organic synthesis req u i re, somewhere in the sequence, the conversion of the starting su bstance to a carbonyl compound or a halogenoalkane.
Recrystallisation is only suitable for pu rifying solids.
When a specific safety precaution is asked for, do not give the use of safety glasses, lab coats etc. as examples. The answer is probably 'no naked flames (ether solvent)' or 'carry out the experiment in a fume cupboard', in which case the specific hazard such as toxicity must be stated .
Visible and UV spectra
n bonds will absorb light energy as an electron is promoted from a bonding to an anti-bonding n orbital.
The energy gap is in the near UV or visible range, and the group causing this is called a chromophore.
If the n bonded electrons are linked to a conjugated system (alternate double and single carbon-carbon bonds), the absorption will shift to a lower frequency, and the sub tance may be coloured. The colour observed will be the complementar colour of that absorbed.
Organic synthesis
Reagents and conditions for the preparation of substance B from A
If the number of carbon atoms in the chain i : increased by one, consider:
a halogenoalkane with KCN b carbonyl with HCN
c the Grignard reagent CH3MgBr
d a Grignard reagent added to C02 or methanal increased by more than one, consider:
a a Grignard reagent with more than one carbon t m b Friedel-Crafts reaction for aromatic substan e decreased by one, consider:
a the Hofmann degradation b the iodoform reaction.
Practical techniques
Recrystallisation. Dissolve the olid in a mini mum f hot olvent.
Filter the hot solution through a preheated funnel u in fluted filter paper. Allow to cool. Filter under reduced pre ure (Buchner funnel), wash with a little cold olvent and allow to dry. The olid should have a sharp melting temperature.
Heating under reflux is nece ary when either the reactant has a low boiling temperature or the reaction is slow at room temperature.
Safety precaution. You must use a fume cupboard when a reactant or product is toxic, irritant or is carcinogenic.
Fractional distillation. This is used to separate a mixture of two liquids. These mixtures can usually be separated into pure samples, but if the boiling points are too close, a good separation will be difficult.
If a liquid of composition X is heated, it will boil at T1 oc to give a vapour of composition Y (see Figure 5. 10). If this is condensed and reboiled, it will boil at a temperature of T2 oc giving a vapour of composition Z. Eventually pure B will distill off the top and pure A will be left in the flask.
Applied organic chemistry
Targeting pharmaceutical compounds. Those which are ionic or have several groups that can form hydrogen bonds with water, will tend to be retained in aqueous (non-fatty) tissue. Compounds with long hydrocarbon chains and with few hydrogen bonding groups will be retained in fatty tissue. The latter will be stored in the body, whereas water-soluble compounds will be excreted.
Nitrogenous fertilisers are of three types:
1 quick release, such as those containing N03- and NH/ ions 2 slow release, such as urea, NH2CONH2
3 natural, such as slurry, compost and manure.
The first two are water-soluble and can be leached out. The last is bulky and contains little nitrogen.
T R A N S I T I O N M E TA L S , Q U A N T I TAT I V E K I N E T I C S A N D A P P L I E D O R G A N I C C H E M I S T R Y
Polymers usually soften over a range of temperatu res rather than have a sharp melting temperature.
The main reason for this is that the polymer is a mixture of molecules of different chain lengths.
Esters, oils and fats. Simple esters are used in food flavourings, perfumes and as solvents for glues, varnishes and spray paints. Animal fats are esters of propan- 1 ,2,3-triol and saturated acids such as stearic acid C17H35COOH. When hydrolysed by boiling with aqueous sodium hydroxide, they produce soap which is the sodium salt of the organic acid.
Vegetable oils are liquids which are esters of propan- 1 ,2,3-triol and unsaturated acids.
Margarine is made from polyunsaturated vegetable oils by partial hydrogenation (addition of hydrogen) so that only a few double bonds remain.
Polymers are of two distinct chemical types:
1 Addition. The monomers contain one or more C=C groups.
Polymerisation is the addition caused by breaking the n bonds, and the polymer and the monomer have the same empirical formula.
Examples are poly(ethene), poly(tetrafluoroethene) (PTFE) and poly(propene).
2 Condensation. Both the monomers have two functional groups, one at each end. Polymerisation involves the loss of a simple molecule (usually H20 or HCl) as each link forms. Examples are nylon, a polyamide (see Figure 5 . 1 1 ), and terylene, a polyester.
a diacid chloride a diamine a polyamide
Fig 5. 1 1 A polyamide
The answers to the numbered questions are on pages 1 38-1 39.
Synthetic polymers are not easily biodegraded and can cause an
environmental problem of disposal. One answer is for them to be recycled .
Checklis t
Before attempting the questions on this topic, check that you:
Can recall the tests for C=C, C-Hal, COH, C=O, CHO and COOH groups.
Can deduce the functional groups present from results of chemical tests.
Can interpret mass, I R, NMR and UV spectra.
Can deduce pathways for the synthesis of organic molecules.
Appreciate how the structure of a pharmaceutical affects its solubility.
Can distinguish between types of polymer and understand polymerisation.