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IB Chemistry - Organic: Reaction Pathways

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hugh hathaway

on 13 January 2016

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Transcript of IB Chemistry - Organic: Reaction Pathways

= Alkene Dihalogenoalkane Trihalogenoalkane

Tetrahalogenoalkane Halogenoalkane Alkane Poly(alkene) Alcohol Aldehyde Carboxylic Acid Ketone Ester Amine Amide Reaction Pathways: IB Chemistry Blue = Standard Level Red = Higher Level Nitrile e.g. Propene e.g. Propane Alkene to Alkane. C = C C H H H H H H H H - + C C C H H H H H H H + H - C C C H H H H H H H H Yellow = Option G Electrophilic Addition A temporary dipole is induced in the H molecule. 2 Propene Carbonium ion Propane Hydrogen e.g. 1-bromopropane Alkene to Halogenoalkane. Electrophilic addition to unsymmetric alkenes C = C C H H H H H H Br H - + C C C H H H H H H H + Br - C C C H H H H H H H Br C C C H H H H H H H + Br - C C C H H H H H H H Br Major product:
2-bromopropane Minor product:
1-bromopropane Hydrogenbromide Markovnikov's Rule When adding to an unsymmetric alkene, due to the positive inductive effect of alkyl groups, the transition state where the C is surrounded by the most alkyl groups will be the most stable. Thus, the product which this intermediate proceeds will be the major product. + Alkane to Halogenoalkane Hetrolytic and Homolytic fission (under UV light) CH + Cl CH Cl + HCl 4 (g) 2 (g) 3 (g) (g) Heterolytic fission: The bond on the halogen / hydrogen halide molecule breaks unevenly. This results in one atom having a negative charge and one having a positive charge. From this point onwards negatively charged ions will pull H ions from the alkene and the positivley charged ions will be grabbed by the negatively charged carbons in the alkanes. + Homolytic fission: The bond on the halogen molecule breaks evenly. This results in both atoms regaining the electron they invested in the bond. Thus both atoms have a neutral charge, and yet they have not obtained a full valence shell of electrons. This makes these 'free fadicals' very reactive. Free radicals will 'grab' H atoms from the alkane, thus becoming a stable biatomic molecule; however the alkane will now become a free radical. CH + Cl HCl + CH 4 * 3 * (g) (g) (g) (g) This alkane free radical can react with the diatomic halogen, thus creating a halogenoalkane and another free radical, which may proceed to react as before. As can be seen, these reactions act as a sort of chain reaction, only terminating when two free radicals react with each other. CH + Cl * 3 (g) 2 (g) CH Cl + Cl 3 (g) (g) * Initiation: (under UV light) Cl 2Cl 2 (g) (g) * Continuation: Termination: Cl + Cl * * (g) (g) Cl 2 (g) CH + Cl (g) 3 * * (g) CH Cl 3 (g) CH (g) 3 * CH (g) 3 * + C H 2 6 (g) Halogenoalkane to Alkene. Elimination Reaction Primary Halogenoalkanes Tertiary Halogenoalkanes OH - C C C H H H H H H H Br OH C = C C H H H H H H + H O + Br 2 Conditions: Refluxing a primary halogenoalkane with NaOH dissolved in ethanol results in the elimination of the halogen and hydrogen to form an alkene, water along with a halogen anion. Mechanism: E2 (Elimination dimolecular), because two species are involved in the slow step that forms the transition state. Note: The hydroxide ion (OH ) is acting as a base (Brønsted-Lowry & Lewis) as it accepts a hydrogen ion by donating a lone pair of electrons to it. - C C C H H H CH H CH H Br 3 3 C C C H H H CH H CH H Br 3 3 + + - C C C H H H CH H CH H 3 3 + C C C H H CH H CH H 3 3 = + H + Conditions: Same as for primary halogenoalkanes Mechanism: E1 (elimination unimolecular), because only one species is involved in the slow step (the heterolytic fission of the C-X bond). Secondary Halogenoalkanes undergo both reaction mechanisms Electrophilic addition C = C C H H H H H H Br Br - + C C C Br H H H H H H + Br - C C C Br H H H H H H Br Hydrogenbromide Alkene to dihalogenoalkane. 1,2-dibromopropane A temporary dipole is induced in the Br molecule. 2 Note: Although the halogen species used here is Br, this mechanism is applicable to all other diatomic halogens. e.g. Propanol Alcohol to Carboxylic acid. Oxidation of Primary Alcohols C C C H H H H H H = O Propan-1-ol Propanal C C C H H H O H H = O H Propanoic acid Conditions: Common oxidising agents are potassium manganate (VII) or potassium dichromate (VI) in dilute sulfuric acid. To obtain the maximum yeild of carboxylic acid the alcohol can be heated under reflux with an excess of the oxidising agent. To obtain the maximum yeild of the aldehyde, the reaction can be . Alcohol to Ketone Oxidation of secondary alcohols. C C C H H H H H H = O Common oxidising agents are potassium manganate (VII) or potassium dichromate (VI) in dilute sulfuric acid. To obtain the maximum yeild of ketone the alcohol can be heated under reflux with an excess of the oxidising agent. e.g. propanone C C C H H H H H H H = O Halogenoalkane to Alcohol Nucleophilic Substitution of Primary Halogenoalkanes - S 2 N OH : - C C C H H H H H Br H H --- --- OH -
+ - Br Nucleophilic Substitution of Tertiary Halogenoalkanes - S 1 N Slow step C C C H H H CH H CH H Br 3 3 C C C H H H CH H CH H Br 3 3 + + - C C C H H H CH H CH H 3 3 + OH : - C C C H H H CH H CH H 3 3 O H Note: Secondary halogenoalkanes react by both mechanisms Alkene to Alcohol Electrophilic Addition C = C C H H H H H H + + H O H : : - + OH - : C C C H H H H H H H O H Major product: Propan-2-ol C C C H H H H H H H C C C H H H H H H H + OH - : C C C H H H H H H H O H Minor product: Propan-2-ol For the S 1 mechanism, it is important to show the OH ion approaching from the opposite side of the Br. This is done by having the curly arrow reach the C from behing the Br. This is important because it demonstrates the produced by the Br atom. N steric hindrance Halogenalkane to Amine Nucleophilic Substitution of primary halogenoalkanes - S 2 NH 3 : C C C H H H H H Br H H --- --- C C C H H H H H H H NH + HBr 2 H H H N + - + + N Nucleophilic Substitution of primary halogenoalkanes - S 1 N C C C H H H CH H CH H Br 3 3 C C C H H H CH H CH H Br 3 3 + + - C C C H H H CH H CH H 3 3 + NH : 3 C C C H H H CH CH H H NH 2 + H + Nitrile to Amine Reduction of Nitriles C C C H H H N H H + 2H 2 C C C H H H N H H H H H H Propanenitrile Propanamine e.g. Propanenitrile Halogenoalkane to Nitrile N C C C H H H CH H CH H Br 3 3 C C C H H H CH H CH H Br 3 3 + + - C C C H H H CH H CH H 3 3 + CN C C C H H H CH CH H H CN CN C C C H H H H H Br H H --- --- C C C H H H H H H H CN + Br C Nucleophilic Substitution of primary halogenoalkanes - S 2 Nucleophilic Substitution of secondary / tertiary halogenoalkanes - S 1 N - N - 1-bromopropane Butanenitrile - 2-dibromobutane-1-nitrile 3 3 3 3 Substitution and Elimination As the same reactants are involved in both reaction methods the conditions must be manipulated in order to get the desired reactions to take place. In order to raise the ratio of elimination to substitution one can: - Use ethanol as a solvent rather than water.
- Increase the temperature
- Increase the number of 'R' groups on the halogenoalkane. That is to say, that tertiary halogenoalkanes are more likely to react by elimination than primary halogenoalkanes. Esterification - Condensation reaction Carboxylic acid and Alcohol to Ester C C C H H H O H H = O H e.g. Propanoic acid C C C H H H O H H = O H C C C H O H H H H H H H + C C C H O H H H H H H C C C H H H H H = O + H O 2 This is really slow unless a is used, such as condensation reaction catalyst concentrated sulfuric acid e.g. Propanamine C C C H H H H H H H N H H Amines and Carboxyllic acids to Amides Condensation Reaction = C C C H H H N H H C C C H H H H H H H N H H C C C H H H O H H = O H + C C C H H H H H = O C C C H H H H H H H N H This thing here? we call this a peptide link + H O 2 propanamine propanoic acid N-propylpropanamide Propylpropanoate Propanoic acid Propanol Primary alcohol to Alkene Elimination Reactions : : H + H - + Protonated propanol Propanol H - + C C C H H H H H H H + C = C C H H H H H H + H + H O H : : Conditions: Water can be eliminated from primary alcohols when heated with a such as at . dehydrating agent phosphoric acid 180 degrees celcius Note: Phosphoric acid is a as the H ions are regenerated. catalyst + Cyanohydrin C CN R R' OH Aldehyde / Ketone to Cyanohydrin Nucleophilic Addition of HCN to Aldehydes or Ketones R R' - + O C CN - R R' O C CN - + H R R' O C CN H As HCN is very dangerous and deadly, the CN is created from
in dilute in situ KCN sulfuric acid. One key differentiator between aldehydes and ketones is that:

Aldehydes are very easy to oxidise, due to the H attatched to the same C as O. This makes aldehydes excellent reducing agents.
Ketones resist further oxidation as they are more stable and carbon to carbon bonds would have to be broken. One key differentiator between aldehydes and ketones is that:

Aldehydes are very easy to oxidise, due to the H attatched to the same C as O. This makes aldehydes excellent reducing agents.
Ketones resist further oxidation as they are more stable and carbon to carbon bonds would have to be broken. Oxidation of primary alcohols. C C C H H H H H H = O Common oxidising agents are potassium manganate (VII) or potassium dichromate (VI) in dilute sulfuric acid. To obtain the maximum yeild of aldehyde the alcohol can be heated under reflux with an excess of the oxidising agent. Aldehyde to Carboxcylic acid Oxidation of Aldehydes C C C H H H H H H = O + H O 2 C C C H H H O H H = O H +2H + +2e - Common oxidising agents are potassium manganate (VII) or potassium dichromate (VI) in dilute sulfuric acid. To obtain the maximum yeild of aldehyde the alcohol can be heated under reflux with an excess of the oxidising agent. distilled Ni Catalyst
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