Chemistry 1B - Lecture 8 Reaction of Organohalides PDF

Reactions of Haloalkanes It can be used to study the interaction between the electrophilic carbon and the halide leaving group and the bond forming between the carbon and the nucleophile They undergo only two main types of reactions. These are substitution and elimination reactions Und...

Reactions of Haloalkanes It can be used to study the interaction between the electrophilic carbon and the halide leaving group and the bond forming between the carbon and the nucleophile They undergo only two main types of reactions. These are substitution and elimination reactions Under substitution reactions there are the Unimolecular Nucleophilic Substitution (SN1) and the Bimolecular Nucleophilic Substitution (SN2) reactions 1  Reactions of Haloalkanes Under elimination reactions there are the Unimolecular Elimination (E-1) and the Bimolecular Elimination (E-2) reactions Unimolecular Nucleophilic Substitution (SN1) In this type of reaction, bond breaking between the carbon atom and the halide leaving group is entirely completed BEFORE bond formation of the carbon and the nucleophile happens 2  Reactions of Haloalkanes This means that the bond-breaking step is the rate-determining step because once the bond is broken, the nucleophile then attacks the generated electrophile in a very fast reaction The breaking of the carbon-halide bond is a very slow process Let’s take the conversion of 2-bromo-2- methylpropane to 2-methoxy-2-methylpropane as an example: 3  Reactions of Haloalkanes The mechanism… Step 1: This is a very slow rate determining step Step 2: This is a very fast step 4  Reactions of Haloalkanes Step 3: This is a very fast step Bimolecular Nucleophilic Substitution (SN2) In this type of reaction, the breaking of the carbon-halide bond and the formation of the nucleophile-carbon bond occur in a concerted or simultaneous way 5  Reactions of Haloalkanes In another sense, we could say that the two reactants are involved in the rate-determining step Take the reaction between bromomethane and hydroxide ion to form methanol and the bromide ion 6  Reactions of Haloalkanes Here the hydroxide ion (nucleophile) attacks the reactive (electrophilic) centre from the side opposite to the leaving group, i.e. the reaction involves a backside attack by the nucleophile 7  Reactions of Haloalkanes The important thing to note here is that the new O-C bond is forming as the old C-Br bond is breaking – this happens simultaneously! SO HOW CAN WE PREDICT WHETHER A REACTION IS GOING TO PROCEED VIA A SN1 OR SN2 MECHANISM? Several factors can be used to predict SN1 or SN2 is favoured – these can be grouped into four main categories: 8  Reactions of Haloalkanes 1. The Structure of the Substrate For SN2 reactions, the main factor here is that the nucleophile MUST ACCESS the electrophilic carbon (C-X) atom For this reason simple alkyl halides show the following general trend of reactivity: methyl>primary>secondary>tertiary In terms of how fast they react, methyl halides undergo SN2 reaction most rapidly followed by primary alkyl halides then come secondary alkyl halides 9  Reactions of Haloalkanes For tertiary alkyl halides, the reactions are so slow that they are considered unreactive in the SN2 sense The biggest factor, which makes tertiary alkyl halides unreactive is that the nucleophile does not have access to the electrophilic carbon atom This is because the carbon to which the halide is attached to has got bulky groups attached, stopping the nucleophile’s approach 10  Reactions of Haloalkanes This type of restricted access is known as steric hindrance In methyl, primary and secondary alkyl halides, the steric hindrance is not as great as it is in tertiary alkyl halides 11  Reactions of Haloalkanes In SN1, the primary factor that determines the reactivity of organic substrates is the relative stability of the carbocation that is formed This carbocation is an electron-deficient species so it needs electrons to stabilize it Alkyl groups are electron donors or pushers 12  Reactions of Haloalkanes This means they push electrons towards the electron-deficient site (carbocation) and in doing so, it stabilizes the positive charge The tertiary carbocation is a relatively very stable ion because three alkyl groups are pushing electrons to the electron-deficient site thus stabilizing it 13  Reactions of Haloalkanes 2. The Concentration and Reactivity of the Nucleophile In SN1, this is not an issue because the nucleophile is not involved in the rate- determining step In SN2 reactions, this is a major factor simply because the nucleophile is involved in the formation of the transition state Both the concentration and identity of the attacking nucleophile must be known 14  Reactions of Haloalkanes The general rule is that the higher the concentration of the nucleophile, the faster the SN2 reaction proceeds The identity of the nucleophile is also very important in the rate-determining step Nucleophiles can be classified as good or poor A good nucleophile reacts rapidly with the given substrate to give a product A poor nucleophile reacts very slowly with the same substrate 15  Reactions of Haloalkanes Take a look at the following examples: In general, the methoxide ion is a good nucleophile while methanol is a poor one 16  Reactions of Haloalkanes Two general rules are: 1. A negatively charged nucleophile is always a more reactive nucleophile than its conjugate acid 2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicity parallels basicity i.e. RO- > OH- > RCO2- > ROH > H2O 17  Reactions of Haloalkanes 3. The Effect of the Solvent The relative strength of the nucleophile can be diminished by being in a solvent, which will coordinate with it A molecule such as water or alcohol – called a protic solvent – has a hydrogen atom attached to an atom of a strongly electronegative element (oxygen) Such solvents can ‘bind’ to the nucleophile in the following manner 18  Reactions of Haloalkanes The protic solvent forms hydrogen bonds to the nucleophile thus diminishing the ability to acts as a nucleophile 19  Reactions of Haloalkanes 4. The Nature of the Leaving group The best-leaving groups are those that give the most stable ion after they depart This means when they cleave off from the carbon atom, the negative charge on them is stabilized and it exists in that form very favourably In the halides, the order of leaving is as follows: I- > Br- >Cl- > F- 20  Reactions of Haloalkanes Elimination Reaction: 1. b-Elimination In the presence of a strong base, such as a hydroxide or an ethoxide ion, haloalkanes can undergo an elimination reaction, which is known as dehydrohalogenation This means a halogen atom can be removed from one carbon (the a-carbon) and a hydrogen atom for an adjacent carbon atom (the b-carbon) 21  Reactions of Haloalkanes This is an elimination reaction because a small molecule such as HCl, HBr, or HI is removed from this larger haloalkane And because they are removed from adjacent carbon atoms, this type of elimination reaction is referred to as a b-elimination reaction 22  Reactions of Haloalkanes Take a look at the three examples below In this case, there is only one possible type of b-hydrogen atom so this reaction will lead to only one product as in 1-butene Now let us look at the next two examples: 23  Reactions of Haloalkanes In these two cases, there are three sets of b- hydrogen atoms which could lead to three possible products 24  Reactions of Haloalkanes However, in these cases, two of the b-hydrogen atoms are identical and as such would lead to two identical products Therefore, for these two cases, there will be only two possible (isomeric) products as shown above When such a possibility arises, the major product will come from the dehydrohalogenation which leads to the most substituted alkene 25  Reactions of Haloalkanes This generalization is Zaitsev’s rule In the reactions given, the major products will be 2-methylbut-2-ene and methylcyclopentene 26  Mechanism of the b-elimination b-elimination reactions proceed by two limiting mechanisms: E1 and E2 The fundamental difference between them is the timing of the bond-breaking and the bond-forming steps E1 mechanism At one extreme, breaking of the C-X bond is complete before any other reaction occurs 27  Mechanism of the b-elimination Once the C-X bond is broken, a carbocation is formed and this allows for the base now to abstract (remove) the hydrogen atom from the b-carbon which then leads to the formation of the double bond between the a- and the b- carbon atoms This is the E1 reaction where E stands for elimination and the 1 stands for the fact that only one reactant is involved in the rate- determining step – Unimolecular Elimination 28  Mechanism of the b-elimination The reactions of 2-bromo-2-methylbutane and 1-bromo-1-methylcyclopentane shown above proceed via this mechanism Step 1: This step involves the breaking of the C-Br bond in the rate-determining step to form the carbocation intermediate 29  Mechanism of the b-elimination Step 2: Here the proton on one of the b-carbon atoms from the carbocation intermediate is abstracted by the ethoxide ion (base) to give the alkene product 30  Mechanism of the b-elimination E2 mechanism At the other extreme is a concerted (one- step) process, designated as E2 E because it is an elimination reaction and 2 because two reactants, in this case, the haloalkane and the base are involved in the transition state of the rate-determining step In the E2 mechanism, there is only one step 31  Mechanism of the b-elimination The key step here is that, as the C-H bond is breaking, the new C-C p bond is forming and at the same time, the C-Br bond is also breaking 32