In nucleophilic substitution, a nucleophile attacks a halogenoalkane and replaces the halogen atom. The halogen leaves as a halide ion (the leaving group).R-X + Nu− → R-Nu + X− Common nucleophiles are OH−, CN−, NH3 and H2O. Each has a lone pair of electrons.Mechanism (for a primary halogenoalkane, e.g. 1-bromopropane with OH−):
• A curly arrow goes from a lone pair on the O of OH− to the δ+ carbon.
• At the same time, a curly arrow goes from the C-Br bond to the Br atom. The bond breaks heterolytically: both electrons go to the bromine, which leaves as Br−.
• The product is propan-1-ol, CH3CH2CH2OH.The nucleophile attacks from the side opposite the halogen. For primary halogenoalkanes the bond forming and the bond breaking happen in a single step. (Tertiary halogenoalkanes react by a different, two-step route, met later in the course.)Exam points: show the δ+ and δ− on the C-X bond; start the first arrow at the lone pair, not at the negative charge; make the second arrow start on the C-X bond; and show the halide ion as a product.With the cyanide ion, CN−, the new C-C bond lengthens the carbon chain: bromoethane gives a product with three carbon atoms.
• A curly arrow goes from a lone pair on the O of OH− to the δ+ carbon.
• At the same time, a curly arrow goes from the C-Br bond to the Br atom. The bond breaks heterolytically: both electrons go to the bromine, which leaves as Br−.
• The product is propan-1-ol, CH3CH2CH2OH.The nucleophile attacks from the side opposite the halogen. For primary halogenoalkanes the bond forming and the bond breaking happen in a single step. (Tertiary halogenoalkanes react by a different, two-step route, met later in the course.)Exam points: show the δ+ and δ− on the C-X bond; start the first arrow at the lone pair, not at the negative charge; make the second arrow start on the C-X bond; and show the halide ion as a product.With the cyanide ion, CN−, the new C-C bond lengthens the carbon chain: bromoethane gives a product with three carbon atoms.