Topic 20: Organic Chemistry HL Flashcards
What are the reactions of the halogenoalkanes?
Nucleophilic substitution reactions (with sodium hydroxide to produce an alcohol). SN1 or SN2.
Describe SN1.
Occurs with tertiary halogenoalkanes > secondary > primary (very rarely). Favoured in protic solvents (H bonded to FON).
Tertiary halogenoalkanes have three alkyl groups around the carbon with the carbon-halogen bond which causes steric hindrance. It is difficult for an incoming group to attack this carbon atom.
1) The halogenoalkane breaks its carbon-halogen bond heterolytically (all electrons to one species). The carbon atom is left with a temporary positive charge and is called a carbocation intermediate.
2) The carbocation intermediate is strongly attracted to the positive charge so attacks. It can do so from either direction due to the carbocation’s open structure (it is not stereospecific).
SN1 is unimolecular in that the rate-determining step is determined by one factor: the concentration of the halogenoalkane.
Describe SN2.
Occurs with primary halogenoalkanes > secondary > tertiary (very rarely). Favoured in aprotic solvents.
The hydrogen atoms around the carbon with the carbon-halogen bond are small so the carbon atom is relatively open to attack (there is less steric hindrance than SN1).
1) The nucleophile attacks the carbon atom. An unstable transition state is formed in which the carbon is weakly bonded to both the halogen and the nucleophile. The carbon-halogen bond breaks heterolytically releasing the halogen. The nucleophile must attack the carbon atom on the opposite site from the leaving group (the reaction is stereospecific).
SN2 is bimolecular in that the mechanism is determined by the concentration of both the halogenalkane and the hydroxide ion (nucleophile).
Compare SN1 with SN2 for its favoured halogenoalkane, nature of mechanism, relative rate and favoured solvent.
SN1: Favoured by tertiary halogenoalkanes. Two step reaction with a carbocation intermediate. Higher rate of reaction. Favoured solvents are polar and protic.
SN2: Favoured by primary halogenoalkanes. One step concerted mechanism with an unstable transition state. Lower rate of reaction. Favoured solvents are polar and aprotic.
What are the reactions of the alkenes? (HL)
Electrophilic addition reactions. This is because the pi bond represents an area of electron density so is attracted to electrophiles.
Define Markovnikov’s rule
The hydrogen will attach to the carbon that is already bonded to the greater number of hydrogens. This occurs 90% of the time and forms a major product. The other 10% a minor product is formed. Occurs with asymmetric alkenes.
How do we determine the major product for an electrophilic addition reaction with an interhalogen compound eg. BrCl?
Chlorine is more electronegative than bromine. Clδ- and Brδ+. Therefore the carbocation formed would involve Brδ+.
How are the nitronium ions formed for the nitration of benzene?
They are generated by using a nitrating mixture composed of concentrated nitric and sulfuric acid at 50°C. The sulfuric acid is stronger so protonates the nitric acid. The nitric acid will then lose a water molecule and form nitronium ions.
Give the equation for the nitration of benzene.
Benzene + nitric acid -> Nitrobenzene + water
C₆H₆ + HNO₃ -> C₆H₅NO₂ + H₂O
Catalysed by sulfuric acid
What reducing agents are used to reverse the oxidation of alcohols?
Sodium borohydride (NaBH₄) which can reduce aldehydes and ketones. Conditions: in aqueous or alcoholic solutions.
Lithium aluminium hydride (LiAlH₄) which is the stronger reducing agents and is able to reduce carboxylic acids. Conditions: dry ether. Sodium borohydride is not strong enough to do so.
How does the reduction of nitrobenzene work?
Nitrobenzene (C₆H₅NO₂) can be converted into phenylamine (C₆H₅NH₂) in a two stage reduction process. It is reacted with tin (Sn) and concentrated hydrochloric acid while being heated under reflux in a boiling water bath.
1) Substitution reaction. Phenulammonium ions are formed.
2) Removal of H+. Phenylamine is formed.
Rules of E/Z isomers.
Rule 1) The atom bonded to the carbon of the double bond with the higher atomic number takes priority.
Rule 2) If the atoms are the same, the longer hydrocarbon chain takes priority.
Rule 3) If two higher priority groups are on the same side of the double bond then the isomer is Z. If they are on opposite sides then the isomer is E.
Define optical isomers.
Molecules that are non-superimposable mirror images of each other.
Define enatiomers.
Non-superimposable mirror images. They have identical physical/chemical properties apart from their optical activity and their reactivity with chiral compounds.
Define diastereomers.
Non-superimposable but NOT mirror images.
