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Organic I (Reactions) Alkanes Free-radical substitution Halogenation of alkanes is a free radical reaction. It results in –H bonds being replaced for –Hal bonds. The reagents and conditions are to mix the alkane with bromine or chlorine and expose to ultra-violet light. The ultra-violet light provides enough energy to break the halogen bond, which forms species called free radicals. When the Hal-Hal bond breaks, it does so with one electron from the bond going with each atom. This is what leads to production of the free-radicals mentioned. Definition: Homolytic fission is when a bond breaks and each atom in the bond takes one electron from the bond. Definition: A free-radical is a species with an unpaired electron. Definition: A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms. Free radicals are very reactive species, as energy is released when they pair up their electrons to make new covalent bonds. CH4 + Cl2 è CH3Cl + HCl CH3CH3 + Cl2 è CH3CH2Cl + HCl Free radical reactions such as this one are often very random and unpredictable. For example, there is no way on controlling which –H bonds are replaced, and further substitution may also occur… CH3Cl + Cl2 è CH2Cl2 + HCl So, a mixture of products is obtained, which makes this an inefficient synthesis of halogenoalkanes. In addition, mixtures of alkanes and chlorine are often explosive when radiated with ultra-violet light. Combustion Complete combustion of alkanes gives CO2 and H2O. CH4 + 2O2 è CO2 + 2H2O These reactions are very exothermic and burning alkanes and other hydrocarbons are commonly used as fuels. Page 1 of 10 thisisrevision.com Organic I (Reactions) Naming of alkanes Alkanes, and all other organic compounds, are named according to the longest unbroken chain of carbon atoms they possess. The suffix ‘ane’ indicates that the compound is an alkane. Compound CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2 CH2CH2CH2CH3 Name methane ethane propane butane pentane hexane heptane octane nonane decane NB Octane, as here named systematically, is not the same as octane in the motor industry. That ‘octane’ is actually ‘iso-octane’, a chain isomer of the octane shown above. If alkanes have side chains, then these are denoted with prefixes and, if necessary, numbers to indicate where they are on the chain. Sidechain CH3CH3CH2CH3CH2CH2etc Prefix methyl ethyl propyl etc The prefixes only have to be numbered if leaving them out would make the name ambiguous. It would be very helpful to draw these out as displayed formulae (where you can see every bond, or at least the C-C bonds). Compound CH3CH(CH3)CH3 (CH3)3CH (CH3)4C CH3CH2C(CH3)2CH3 CH3CH(CH3)CH(CH3)CH3 Name methyl propane also methyl propene dimethyl propane 2,2-dimethyl butane 1,2-dimethyl butane The carbon chain should be numbered so as to give smaller numbers to the substituents. Compound CH3CH2CH2CH2C(CH3)3 Name 2,2-dimethylhexane Alkanes can also exist as cyclic compounds, in which the atoms join up in a ring. If so, the prefix ‘cyclo’ is used to indicate them. Page 2 of 10 thisisrevision.com Organic I (Reactions) Alkenes Addition reactions Alkenes undergo addition reactions with hydrogen, halogens (such as bromine), hydrogen halides and potassium manganate(VII). Definition: An addition reaction is one in which two species combine to make single product. The reactions with bromine and potassium manganate are used as tests for alkenes. Tests for alkenes • Test 1 – Mix the alkene with a solution of bromine. ü Decolourisation, from brown to colourless. Bromine can be used either in organic solvents such as tetrachloromethane or hexane, alternatively it may be used in aqueous solution. In any case, the positive result is decolorisation of the solution. It would be good to indicate some practical awareness of the mixing would occur: § § § gaseous alkenes – bubble alkene through bromine solution. liquid alkene – shake alkene with bromine solution. solid alkene, dissolve alkene in inert solvent, then shake with bromine solution. If bromine water is used, then the product obtained is a bromo-alcohol, as water participates in the reaction as a nucleophile in the second step of the mechanism. CH3CH=CH2 + Br2 + H2O è CH3CH(OH)CH2Br + HBr If bromine in an inert solvent is used, then the product obtained is a dibromoalkane. CH3CH=CH2 + Br2 è CH3CHBrCH2Br These addition reactions could also be classified as an oxidation of the alkene. • Test 2 – Mix the alkene a cold dilute aqueous solution of alkaline potassium manganate (VII). ü Decolourisation, from purple to green to colourless. A brown precipitate of MnO2 forms. In this test, the alkene is turned into a diol. CH3CH=CH2 + H2O + [O] è CH3CH(OH)CH2OH KMnO4 reacts with many substances, primary and secondary alcohols, aldehydes and even some ketones. What makes this test specific to alkenes is the conditions. When the KMnO4 is alkaline, and cold, and dilute, it is such a poor oxidising agent that only alkenes are sufficiently reducing to react with it. This addition reaction can also be classified as an oxidation of the alkene. Page 3 of 10 thisisrevision.com Organic I (Reactions) Other reactions of alkenes: • Hydrogen adds to alkenes to form alkanes. Reagents and conditions are pass the gases over a heated nickel catalyst. CH3CH=CH2 + H2 è CH3CH2CH3 This can also be classified as a reduction of the alkene. • Hydrogen halides add to alkenes to form haloalkanes. Reagents and conditions are mix the gases at room temperature. CH3CH=CH2 + HBr è CH3CHBrCH3 When the substance attacking the double bond is unsymmetrical (like HBr), the hydrogen atom goes on the atom with more H atoms to begin with. This will be studied more in a later unit, just remember it for now. It is Markovnikov’s rule. Substances that add to C=C bonds, such as H2, Br2, and HBr are termed electrophiles. Definition: An electrophile is an electron deficient species that can accept a pair of electrons from an electron rich region of another molecule. Naming of alkenes The suffix ‘ene’ is used to indicate that a compound contains a C=C bond, sometimes with a number to indicate where it is in the carbon chain. The direction of numbering along the chain is chosen so as to give smaller numbers to the carbons containing the double bonds. Compound CH2=CH2 CH3CH=CH2 CH3CH2CH=CH2 CH3CH=CHCH3 Name ethene propene but-1-ene but-2-ene Draw ethene, showing all the bonds. If there are alkyl groups or halogen atoms to denote in the name, then the position of the double bond takes precedence. Compound CH2=CHCH2CH2CH2C(CH3)3 Name 6,6-dimethyl hept-1-ene Multiple double bonds are named as follows: Compound CH2=CHCH=CH2 CH2=C=CHCH2 Name buta-1,3-diene buta-1,2-diene This a in buta is just to make the name easier to say, and you can probably forget about it. Page 4 of 10 thisisrevision.com Organic I (Reactions) Halogeno alkanes Substitution reactions Definition: A substitution reaction is one in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms. Definition: A nucleophile is an electron rich species that is attracted to an electron deficient region of a molecule and can donate a pair of electrons to form a new covalent bond. The important point is the ability to donate a pair of electrons. • Substitution with hydroxide (hydrolysis) The reagents and conditions for hydrolysis of halogenoalkanes are boil under reflux with dilute aqueous sodium hydroxide. CH3CH2Br + NaOH • è CH3CH2OH + NaBr Substitution with cyanide (cyanolysis) This just like hydrolysis but the nucleophile is CN-. The reagents and conditions are to boil the halogenoalkane under reflux with sodium or potassium cyanide in a solvent such as ethanol and water, or propanone. CH3CH2Br + NaCN • è CH3CH2CN + NaBr Substitution with ammonia (aminolysis) This is just like hydrolysis but with ammonia as the nucleophile. The reagents and conditions are heat the halogenoalkane with excess concentrated ammonia in ethanol in a sealed tube. CH3CH2Br + NH3 è CH3CH2NH2 + HBr, or CH3CH2Br + 2NH3 è CH3CH2NH2 + NH4Br In the above reactions, the OH-, CN- and NH3 molecules are behaving as nucleophiles. Page 5 of 10 thisisrevision.com Organic I (Reactions) Halogeno alkanes Elimination reactions Definition: An elimination reaction is one where the elements of a small molecule are removed from a larger one. § § § § § Halogenoalkanes undergo elimination by removal of the halogen atom and a hydrogen atom. The hydrogen atom must come from a carbon atom that is adjacent to the carbon atom that loses the halogen atom. The C=C is formed between the two atoms that lose the hydrogen and halogen atoms. Often there is more than one hydrogen atom that can be removed, in which case the product will be a mixture of molecules. Sometimes the products can exist as cis and trans isomers. The reagents and conditions for elimination are heat strongly under reflux with concentrated potassium or sodium hydroxide in ethanol. Under these conditions, the OHion behaves as a base by removing a hydrogen atom from the molecule. CH3CH2Br + NaOH è CH2=CH2 + H2O + NaBr In the above equation, the “elements of a small molecule” that have been removed are those of “HBr”, although an actual molecule of HBr is not created. When more than one product is possible the one with the greater number of alkyl groups present on the C=C is formed in greater yield. When geometrical isomers are possible, cis and trans are formed in equal yield. CH3CH2CHBrCH3 è è CH3CH2CH=CH2 (smaller yield) CH3CH = CHCH3 (greater yield, cis and trans 1:1) Halogenoalkanes are more prone to elimination if they are more branched, i.e. tertiary eliminates more easily than secondary than primary. Branching provides more hydrogen atoms for the reaction to start with, and also sterically hinders the carbon atom thus preventing the OH- from acting as a nucleophile and causing a substitution reaction instead. Very important point: In substitution the OH- behaves as a nucleophile (electron pair donor), whereas in elimination it behaves as a base (proton acceptor). Draw 2-bromobutane showing all bonds. Label the H atoms that can be lost to form but1-ene and those that can be lost to form but-2-ene. Page 6 of 10 thisisrevision.com Organic I (Reactions) Test for halogen atoms in halogeno alkanes, Cl, Br and I Step one - Free the halide ion from the organic molecule. How this is done depends on the type of molecule, but for simple halogeno alkanes it is done by heating the halogenoalkane under reflux with dilute, aqueous sodium hydroxide solution. Step two – add excess dilute nitric acid, to neutralise excess NaOH or other alkaline impurities. Any alkaline impurities will give a grey precipitate when silver nitrate is added, which could be interpreted as a false-positive result. Step three – add silver(I)nitrate solution. ü Chloride ions will give a white precipitate ü Bromide ions will give a cream precipitate ü Iodide ions will give a yellow precipitate Step four – filter off the precipitate and test its solubility in ammonia solution: 1. AgCl will dissolve in dilute ammonia solution 2. AgBr will dissolve in concentrated ammonia solution 3. AgI will not dissolve in any concentration of ammonia solution. See group VII notes for more details of this test for halide ions. Naming of halogenoalkanes Halogen atoms are specified with a prefix and a sometimes a number to specify where the halogen is on the carbon chain. The chain is numbered from the direction that gives the smaller number of set of numbers for atoms which have halogen atoms attached. Halogen F Cl Br I Prefix Fluoro Chloro Bromo Iodo Compound CH3Cl CH3CHBrCH3 CH3CHBrCH2CH2CH3 Name chloromethane 2-bromopropane 2-bromopentane The prefixes are modified if more than one halogen is present. They are given alphabetically according the name of the halogen. Alkyl groups can also be included. Compound CH3CBr2CBrCl2 CH3CH(CH3)CHF2 Page 7 of 10 Name 1,2,2-tribromo-1,1-dichloro propane 1,1-difluoro-2-methyl propane thisisrevision.com Organic I (Reactions) Alcohols Oxidation of alcohols Reagents and conditions are to heat with potassium dichromate solution acidified with dilute aqueous sulphuric acid. You must specify the acid, writing H+ will not get you full marks. • Primary alcohols are oxidised to aldehydes then carboxylic acids. CH3CH2OH è CH3CHO è CH3CO2H • In order to obtain the aldehyde, the reaction is carried out with excess alcohol and the aldehyde is distilled off as it is formed. This is possible because the aldehyde has the lowest boiling point of the three compounds, because it does not have hydrogen bonds between its molecules. Reagents and conditions are add potassium dichromate(VI) in dilute sulphuric acid to the hot alcohol and allow the aldehyde to distil off. • In order to obtain the carboxylic acid, the reaction is carried out with excess oxidising agent and the heating is carried out under reflux. This is in order to maximise yield. Reagents and conditions are boil the alcohol with excess potassium dichromate(VI) in dilute sulphuric acid under reflux. In organic oxidation reactions, equations can be balanced using the symbol [O]. CH3CH2OH + [O] è CH3CHO + H2O CH3CHO + [O] è CH3CO2H • Secondary alcohols are oxidised to ketones. CH3CH(OH)CH3 + [O] • è CH3COCH3 + H2O Tertiary alcohols are not oxidised under these conditions. Reaction of alcohols with dehydrating agents Alcohols can be dehydrated to form alkenes. The reactions can be classified as dehydration or elimination. Reagents and conditions are heat under reflux at 170oC with excess conc. H2SO4, or heat at 300oC over Al2O3. You should be able to predict the alkenes that would form from a typical alcohol, such as butan-2-ol. The OH group is removed, along with any hydrogen atom attached to a carbon next to the carbon with the OH group. Note that geometrical isomers may form. CH3CH2CH(OH)CH3 è Page 8 of 10 CH3CH=CHCH3, major product, cis and trans. CH3CH2CH=CH2, minor product. thisisrevision.com Organic I (Reactions) Reaction of alcohols with halogenating agents Halogenating agents replace the –OH group with a halogen. You need to be familiar with three different ones for this unit. § Method one – PCl5(s) Reagents and conditions are carefully add solid PCl5 at room temperature, under dry conditions. CH3CH2OH + PCl5 § è CH3CH2Cl + POCl3 + HCl Method two – NaBr(s) and conc. H2SO4 NaBr(s) and conc. H2SO4 react to form HBr (but see group VII notes). Reagents and conditions add a mixture of the alcohol and conc. H2SO4 to NaBr at room temperature. CH3CH2OH + HBr § è CH3CH2Br + H2O Method three – phosphorous and iodine Phosphorous and iodine react together to form PI3. Reagents and conditions are add the alcohol to moist red phosphorous and iodine at room temperature. 3CH3CH2OH + PI3 è 3CH3CH2I + H3PO3 Test for –OH groups in alcohols (and carboxylic acids). The PCl5 reaction above is used to test for an OH group. Note that it works for any compound that contains an OH group, so both alcohols and carboxylic acids, so it is not a test for an alcohol specifically. Water also contains an –OH group, so the conditions must be anhydrous to avoid a false-positive result. • Carefully add a small amount of solid PCl5 to a small amount of the sample in a clean dry test tube, in a fume cupboard at room temperature. The sample must be dry. ü Effervescence of steamy fumes (HCl) which turn moist blue litmus paper red and give a dense white smoke (NH4Cl) with an ammonia stopper. PCl5 + ROH è POCl3 + RCl + HCl(g) HCl(g) + NH3(g) è NH4Cl(s) Using dichromate as a test for alcohols. If heating a test substance with excess potassium dichromate(VI) in dilute sulphuric acid turns it green, then the test substance would be a primary or secondary alcohol, but remember, it could also be an aldehyde. Page 9 of 10 thisisrevision.com Organic I (Reactions) Naming of alcohols and related compounds Alcohols are denoted with the suffix ‘ol’, with a number, if necessary, to indicate where the –OH group is in the molecule. Compound CH3OH CH3CH2OH CH3CH2CH2OH CH3CH(OH)CH3 Name methanol ethanol propan-1-ol propan-2-ol Draw methanol, showing all the bonds. Substituents can be included by using prefixes with numbers, if necessary, to indicate their positions. Compound CH3CH(CH3)CH2OH (CH3)3COH C6H5CH2CH2OH CCl3CF2OH Name 2-methyl propan-1-ol 2-methyl propan-2-ol 2-phenyl ethanol 2,2,2-trichloro-1,1-difluoro ethanol When primary alcohols are oxidised they initially form aldehydes, which contain the –CHO group. Aldehydes are denoted with the ‘al’ suffix. Compound HCHO CH3CHO CH3CH2CHO CH3CH2CH2CHO Name methanal ethanal propanal butanal Draw ethanal, showing all the bonds. Secondary alcohols are oxidised to ketones, which contain the >C=O group. Compound CH3COCH3 CH3COCH2CH3 CH3CH2COCH2CH3 CH3COCH2CH2CH3 Name propanone butanone pentan-3-one pentan-2-one Draw propanone, showing all the bonds. Aldehydes are further oxidised to carboxylic acids, which contain the –CO2H group. The –CO2H group is denoted with the suffix ‘oic acid’. Compound CH3CO2H CH3CH2CO2H Name ethanoic acid propanoic acid Draw ethanoic acid, showing all the bonds. Substituents can be indicated with prefixes and numbers, numbering from and including the C in the –CO2H group. Compound CH3CH(CH3)CH2CH2CO2H Page 10 of 10 Name 4-methyl pentanoic acid thisisrevision.com