1. Reactions of haloalkanes.

2. Nucleophilic substitution Halogens are relatively electronegative. So the carbon/halogen bond will be polarised. δ-δ- δ+δ+ The δ+ on carbon will be attractive to nucleophiles, resulting in nucleophilic substitution. (CH 3 ) 2 CHCl + Nuc: → (CH 3 ) 2 CHNuc + Cl _

3. Nucleophilic substitution

4. Reactions with sodium hydroxide. If haloalkanes are heated with aqueous sodium hydroxide the corresponding alcohol is formed. CH 3 CH 2 Cl + :OH - → CH 3 CH 2 OH + :Cl _ Chloroethane ethanol NB As haloalkanes are not very soluble in water they are dissolved in alcohol. Water by itself can also act as a nucleophile, but at a slower rate; CH 3 CH 2 Cl + :OH 2 → CH 3 CH 2 OH + HCl

5. Reaction with cyanide Cyanide ions can act as a nucleophile to form the corresponding nitrile. CH 3 CH 2 Cl +:CN - → CH 3 CH 2 CN + : Cl _ Chloroethane propanitrile An alcoholic solution of the reactant is refluxed with potassium or sodium cyanide.

6. NB The product has one more carbon than the haloalkane. So this reaction is important to organic synthesis by increasing chain length. + →

7. Reaction with ammonia. Ammonia has a lone pair of electrons and can act as a nucleophile. CH 3 CH 2 Br + :NH 3 → CH 3 CH 2 NH 2 + HBr _ Bromoethane ethylamine An alcoholic solution of the reactant is heated with excess ammonia under pressure.

8. If ammonia is not in excess further reactions are possible. The nitrogen of a primary amine itself has a lone pair and can act as a nucleophile, producing a secondary amine. (CH 3 CH 2 ) 2 N(H): This secondary amine also has a lone pair and can act as a nucleophile producing a tertiary amine. (CH 3 CH 2 ) 3 N : The tertiary amine still has a lone pair, and soacts as a nucleophile to give a quarternary ammonium salt. (CH 3 CH 2 ) 4 N+ The salt has no lone pair, so the reaction stops.

9. Write equations for the following reactions; 1) 1 bromobutane with sodium hydroxide. 2) 2 chloro 2 methyl propane with sodium cyanide. 3) 2 iodo propane with ammonia. 1) C 4 H 9 Br + NaOH → C 4 H 9 OH + NaBr 2) CH 3 CCl(CH 3 ) 2 + NaCN→ CH 3 CCN(CH 3 ) 2 + NaCl 3) CH 3 CHICH 3 + NH 3 →CH 3 CH(NH 2 )CH 3 +HI

10. Mechanisms of nucleophilic substitution There are two types of mechanism; 1) SN1 (Substitution nucleophilic 1) 2) SN2 (Substitution nucleophilic 2) SN1 occurs in tertiary haloalkanes. SN2 occurs in primary haloalkanes Secondary haloalkanes react with either, but not both, mechanism.

11. SN1 Reactions have two steps; Step 1 The C/halogen bond breaks heterolytically forming a carbocation; + XX - → +

12. Step 2 The carbocation reacts with the nucleophile. + XX + →

13. δ-δ- δ+δ+ + X-X- xx → xxNuc + → Step 1 Step 2

14. Substitution Nucleophilic 2 reactions SN2 reactions have only one step. The C/halogen bond breaks As the C/nucleophile bond forms.

15. 1) A nucleophile attacks the haloalkane… 2) … forming an intermediary complex… 3) … which breaks down to complete the reaction. δ+δ+ δ-δ-

16. Why the difference? Carbocations are unstable, but can be stabilisded by alkyl groups via their electron releasing inductive effect. + Tertiary haloalkanes have three alkyl groups. Giving three +I inductive effects. The combined +I effect is sufficient to stablise the ion. So tertiary haloalkanes react SN1.

17. But primary haloalkanes have only one alkyl group on the carbon with the halogen; Primary haloalkanes have only one alkyl group. Giving one +I inductive effect. The +I effect is insufficient to stablise the ion. So primary alkanes react SN2. +

18. Elimination reactions If haloalkanes are refluxed with solid sodium hydroxide elimination occurs instead of substitution. CH 2 HCH 2 Br + OH _ ↓ 2 HC=CH 2 + HOH + Br- Effectively eliminating a small molecule, HBr.

19. Elimination

20. 1) The hydroxide ion acts as a base and accepts a proton. 2) The electrons are passed on… 3) A halide ion leaves and a double bond is formed. Forming water Overall a hydrogen halide molecule is elminated.