1. 20.2 Nucleophilic Substitution Reactions

2. Starter Outline the differences between the Sn1 and Sn2 Mechanism.

3. Assessment Statemen ts 20.2.1 Explain why the hydroxide ion is a better nucleophile than water. 20.2.2 Describe and explain how the rate of nucleophilic substitution in halogenoalkanes by the hydroxide ion depends on the identity of the halogen. 20.2.3 Describe and explain how the rate of nucleophilic substitution depends on whether the halogenoalkane is primary, secondary or tertiary. 20.2.4 Describe, using equations, the substitution reactions of halogenoalkanes with ammonia and potassium cyanide. 20.2.5 Explain the reactions of primary halogenoalkanes with ammonia and potassium cyanide in terms of the S N 2 mechanism. 20.2.6 Describe, using equations, the reduction of nitriles using hydrogen and a nickel catalyst.

4. Substitution Mechanism Review In halogenoalkanes the C-Hal bond is polar and the C is slightly positive. This makes it susceptible to attack by NUCLEOPHILES. A nucleophile is a reagent that attacks a center of positive charge by donating an electron pair. This makes halogenoalkanes more reactive than alkanes as they will undergo NUCLEOPHILIC SUBSTITUTION reactions where the halogen is replaced by another atom or group of atoms. A general formula for halogenoalkanes is R-X where R is the alkane chain and X is F, Cl, Br or I.

5. HALOGENOALKANES The bond between the C and the X is slightly negative towards the halogen and the C is slightly positive or electron deficient. Common nucleophiles that would be attracted to this slightly positive C are: H 2 O, OH -, NH 3 and CN -. All these species have at least one lone pair of electrons. The shorthand for these reactions is S N which stands for substitution nucleophile.

6. HOMOLYTIC VS. HETEROLYTIC FISSION In the free radical reactions with alkanes we saw bonds breaking and leaving one electron from the bond with each species – two free radicals formed. This is an example of HOMOlytic fission. As the halogenoalkanes have polar bonds they split by HETEROlytic fission where both the shared electrons in the bond go with one of the products. In this case the electrons go with the halogen and the halogen is referred to as the LEAVING GROUP. R-X + Y -  R-Y + X - where Y - is the nucleophile.

7. NUCLEOPHILIC SUBSTITUTIONS Halogenoalkanes warmed with aqueous alkali undergo hydrolysis to form an alcohol: R-X(l) + OH - (aq)  R-OH(aq) + X - (aq) Ex. C 4 H 9 Br(l) + OH - (aq)  C 4 H 9 OH(aq) + Br - (aq) In this reaction the nucleophile is OH -. There are two mechanisms by which this can occur, S N 1 and S N 2.

8. SN1SN1 The S N 1 mechanism occurs with tertiary halogenoalkanes ex. CH 3 C(CH 3 )ClCH 3 The 3 alkyl groups get in the way of the nucleophile attack – this is called STERIC HINDRANCE. So the first step involves the halogenoalkane breaking the C-X bond heterolytically to form a positively charge CARBOCATION INTERMEDIATE, R-C +. This is a slow (rate determining) step. R-C+ reacts quickly with OH - to make the R-OH.

9. SN1SN1 The carbocation is stabilized by the other alkyl groups. They have an electron donating or positive inductive effect. S N 1 is unimolecular which means the slow step, or rate determining step, is determined by the concentration of only one molecule, in this case the halogenoalkane. The rate expression is rate = k[R-X]. So the rate is first order with respect to the halogenoalkane and zero order in the OH -. This is why it is called S N 1: substitution nucleophilic unimolecular.

10. S N 1 MECHANISM

11. SN2SN2 In S N 2 the mechanism is bimolecular and the attack of the OH - ion on the halogenoalkane is the rate determining step. The reaction has a transition state (activated complex) in which the bond to the OH - ion starts to form as the bond to the halogen is breaking. The overall reaction is R-X + OH -  R-OH + X -. The rate expression is: rate = k[R-X] 1 [OH] 1. The rate is dependent on the concentration of both reactants as it is done in one steps so it is S N 2, substitution nucleophilic bimolecular.

12. S N 2 MECHANISM

13. HALOGENOALKANES Like alcohols, halogenoalkanes can be primary, secondary or tertiary. With primary halogenoalkanes the S N 2 mechanism is more common. S N 1 is more common with tertiary halogenoalkanes. Either mechanism can occur with secondary halogenoalkanes.

14. NUCLEOPHILIC SUBSTITUTIONS A nucleophile can be denoted using Nu -. During S N 1 mechanism the intermediate has a finite existence and occurs at a potential energy minimum.

15. NUCLEOPHILIC SUBSTITUTIONS In the S N 2 mechanism the transition state occurs at a potential energy maximum.

16. Assessment Statement 20.2.1 Explain why the hydroxide ion is a better nucleophile than water.

17. NUCLEOPHILE STRENGTH The rate at which these reactions occur depend on the strength of the nucleophile. Some substances will more readily attack the slightly positive carbon than others. Ex. OH - is a stronger nucleophile than H 2 O as the negative charge on the OH - will be attracted to the slightly positive charge on the C. This means S N 2 hydrolysis will occur faster in aqueous alkali than in a neutral solution. Water will react quite effectively with tertiary halogenoalkanes as the S N 1 mechanism is faster than the S N 2.

18. OH - Vs H 2 O as a Nucleophile The nucleophile is attracted to the partially positive carbon atom that is bound to the electronegative halogen. Think… Draw the Lewis structures of CN -, OH -, NH 3 and H 2 O What would make an effective nucleophile? Rank them in order in decreasing electrophile strength.

19. Neutral or negatively charge containing a lone pair of electrons. The more dense the charge, the better. The less electronegative nucleophiles are better, thus NH 3 is better than H 2 O. OH- has three lone pairs while H2O has only 2. (the less electronegative the less likely it wants to hold on to its electrons)

20. Assessment Statement 20.2.2 Describe and explain how the rate of nucleophilic substitution in halogenoalkanes by the hydroxide ion depends on the identity of the halogen.

21. Two things is going to effect the rate of reaction with respect to the halogen attached Bond Strength Polarity of the C-halogen bond. Think…… Looking at the periodic table of group 7 halogens, which one would have the greatest bond strength with carbon? Which one would produce the greatest positive charge on the carbon.

22. NUCLEOPHILIC SUBSTITUTIONS Which halogen is being replaced will also affect the rate of reaction. The polarity of the C-Hal bond will decrease with halogens further down the group. This means the C will be less positive when bonded to I than to F so we would expect the rate to be faster with F. On the other hand, the bond with I will be weaker than with F so we would expect the rate to increase with I. Overall what is seen is a greater rate with iodoalkanes than with halogenoalkanes further up the group so bond strength is thought to be the overriding factor.

24. Assessment Statement 20.2.3 Describe and explain how the rate of nucleophilic substitution depends on whether the halogenoalkane is primary, secondary or tertiary. Which one would produce the most stable carbocation?

25. NUCLEOPHILIC SUBSTITUTIONS Tertiary halogenoalkanes usually react with an S N 1 mechanism because tertiary carbocations are relatively stable. This stability comes from the INDUCTIVE EFFECT of the alkyl groups which reduces the charge on the central carbon. This stabilizes the carbocation intermediate needed for S N 1.

27. NUCLEOPHILIC SUBSTITUTIONS The change from tetrahedral to trigonal planar geometry when the carbocation is formed increases the bond angle from 109° to 120°. In tertiary halogenoalkanes this means the alkyl groups can move further apart which stabilizes the carbocation by reducing steric stress. In the S N 2 mechanism nucleophiles usually attack the central carbon from the direction opposite to the halogen. In tertiary compounds the alkyl groups are too big to allow this so only S N 1 is seen.

28. NUCLEOPHILIC SUBSTITUTIONS Secondary halogenoalkanes can react by either mechanism. As S N 1 mechanisms are usually faster the rate of hydrolysis of halogenoalkanes is usually in the order: Tertiary > secondary > primary (All things being equal) These rates can be compared experimentally by adding AgNO 3 (aq) to the reaction mixture. As the halide ion forms it will react with the Ag + ion to form a silver halide precipitate. The silver halides have different colors that change in the presence of light (refer back to periodicity).

29. NUCLEOPHILIC SUBSTITUTIONS Halogenated aromatic compounds (contain a benzene ring) are much less reactive than other halogenoalkanes. This is because the C-Hal bond is stronger and more difficult to break as one of the halogen’s lone pairs interacts with the delocalized electrons on the benzene ring. The Nu - cannot attack from the opposite side as benzene is too big. The partial positive charge on the C is reduced due to the mobile electrons in the benzene ring.

30. Assessment Statement 20.2.4 Describe, using equations, the substitution reactions of halogenoalkanes with ammonia and potassium cyanide. 20.2.5 Explain the reactions of primary halogenoalkanes with ammonia and potassium cyanide in terms of the S N 2 mechanism.

33. AMINES Many substances can act as nucleophiles besides the hydroxide ion. Others that we will look at are ammonia, NH 3 and the cyanide ion, CN -. When ammonia reacts with a primary halogenoalkane a PRIMARY AMINE forms: R-X(l) + NH 3 (l)  R-NH 2 (l) + HX(l) This occurs by an S N 2 mechanism. After the transition state breaks down into the product a further H + ion must be lost to form the amine. The H + bonds with the halogen to form HX.

34. AMINES A secondary halogenoalkane will make a secondary amine, and a tertiary halogenoalkane will make a tertiary amine. The mechanisms followed will be the same as with OH -. Usually these reactions are carried out by adding a concentrated ammonia solution in a sealed tube that raises pressure. Heat is also added. Amines themselves can act as nucleophiles and form further substituted products. Increasing the ammonia concentration decreases these other products though.

35. Assessment Statement 20.2.6 Describe, using equations, the reduction of nitriles using hydrogen and a nickel catalyst.

37. NITRILES If CN - reacts with a primary halogenoalkane a NITRILE is formed. This occurs by an S N 2 mechanism: R-X(l) + C=N - (aq)  R-C=N(l) + X - (aq). This reaction adds a new C-C bond and is a useful way to lengthen a hydrocarbon chain. The triple bond in the nitrile may be readily reduced using H 2 and a Ni catalyst to form a primary amine, this amine would have one more carbon than the one made by reacting with NH 3. R-C=N(l) + 2H 2 (g) -> R-CH 2 -NH 2 (l)