Jun 11, 2020
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Nucleophilic Substitution - SN1
When tert-Butyl iodide was treated with sodium acetate a substitution reaction occured in which the acetate ion (\(CH3CO_2^-\)) substituted the iodide ion (\(I^−\)). Kinetic study of this reaction showed that the rate of this reaction depends only on the concentration of alkyl halide.
\[Rate \ \alpha \ [(CH_3)_3CI]\]
As the kinetic study suggests, the concentration of the nucleophile is not affecting the rate of the reaction. This means that there could be more than one step in the reaction mechanism and the step that involves the nucleophile does not affect the rate of the reaction. The other step involving the alkyl iodide could be the slower rate determining one.
So, a straight forward two step mechanism could be thought of, first step of which involves a slow ionization of the alkyl iodide into a carbocation and iodide ion followed by the fast attack on the cation by the nucleophile. As the first step involves only bond cleavage, it must be slower than the second step which involves only bond formation. This is called SN1 (Unimolecular Nucleophilc Substitution).
The reaction proceeds through the formation of a carbocation. The carbon atom carrying the positive charge is \(sp^2\) hybridized. Its geometry is trigonal planar in which the empty p-orbital is perpendicular to the plane of the molecular ion.
The nucleophile that attcks this can do so from either side leading to the formation of two products with different stereochemistry. For example, if the alkyl halide with which the reaction is started with is optically active, then the product of its SN1 substitution will be racemized i.e. will have retention as well as inversion about the stereochemical centre. If the incoming nucleophile occupies the same position as the leaving group the configuration will be retained and if it occupies a position opposite to that of the leaving group the configuration will be inverted.
It has been found that inversion is slightly more than retention of configuration in reactions proceeding through SN1 mechanism. Further investigation of the reaction mechanism reveals that before the formation of the free carbocation, there are two short lived stages which mainly undergoes inversion.
Question on Alkyl Halides
2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields
A) (a) and (c)
B) (c) only
C) (a) and (b)
D) All of these
Nucleophilic Substitution - The Mechanisms
The functional group in Alkyl Halides is the polar C - X bond. The electronegativity difference between the halogen and the carbon atom is the key factor that determines the chemical properties of alkyl halides.
Alkyl halides are generally found to undergo elimination or substitution reactions. When they undergo elimination reaction they form an alkene and when they undergo substitution they form a variety of compounds based on the nucleophile used.
A negatively charged species has two roles to play; 1. the role of a base and 2. the role of a nucleophile. So, when an alkyl halide is treated with a negatively charged species either elimination or substitution will be the major outcome. If the negatively charged species acts as a base, the major outcome will be an elimination reaction and if the negatively charged species acts as a nucleophile the major outcome will be a substitution reaction.
During the course of our discussion we will talk about the differences between as base and a nucleophile. For now, we will talk about the substitution reactions. In a substitution reaction, our main focus is on the nucleophile and the leaving group.
In a typical substitution reaction of an alkyl halide the nucleophile forms a bond with the electrophilic carbon and the leaving group, the halide ion leaves the molecule.
When methyl iodide was treated with sodium acetate a substitution reaction occurred in which the acetate ion (\(CH_3CO_2^-\)) substituted the iodide ion (\(I^-\)). Kinetic study of this reaction showed that the rate of this reaction depends both on the concentration of the alkyl halide and the nucleophile.
\[Rate \propto[CH_3I][CH_3CO_2^-]\]
When tert-Butyl iodide was treated with sodium acetate as substitution reaction occured in which the acetate ion (\(CH_3CO_2^-\)) substituted the iodide ion (\(I^-\)). Kinetic study of this reaction showed that the rate of this reaction depends only on the concentration of alkyl halide.
\[Rate \propto [(CH_3)_3CI]\]
Two types of kinetic observation gives us the idea that there should at least be two reaction mechanisms for the nucleophilic substitution reaction.
What are they?
Nucleophilic Substitution - Introduction
Nucleophile can be broadly described as a species which is nucleus loving and is in search of electron deficient sites to bond with. There are many organic molecules that have an electron deficient site, usually a carbon atom. Following are some examples of organic compounds having an electrophilic carbon atom.
Let us focus on Alkyl Halides and have a closer look at it.
There is an electronegativity difference between the halogen atom and the carbon atom. Here chlorine is more electronegative than carbon and hence it shows the negative inductive effect. It pulls the electron density in the bond, it forms with carbon, towards it. This results in a small positive charge on the carbon atom and a small negative charge on the chlorine atom.
The carbon atom directly connected to the chlorine atom thus becomes electron deficient and call be called as an electrophilic centre. Similarly, if we look at the other examples that we have taken above, the carbon atom connected to the electronegative atom in them (Oxygen and Nitrogen) is electron deficient and can be called as electrophilic centre. This electrophilic centre can attract a suitable nucleophile and if the conditions permit may substitute the halogen atom.
One suitable condition for such nucleophilic substitution is that the group that is being substituted, referred to as “leaving group“ be ready to leave when the nucleophile approaches. Halogen atoms are good leaving groups and hence alkyl halides readily undergo nucleophilic substitution reactions.
Ionic Reactions (Part I)
Ionic Reactions
Nucleophilic Substitution and Elimination of Alkyl Halides
Alkyl halides – has a halogen atom (Br, Cl, F, I) bonded to an sp3-hybridized (tetrahedral) carbon atom.
- The C-H bond is a polar bond. This is because the halogen atom hogs the electrons towards its side due to its electronegativity trait (it is more electronegative than carbon). This results to the carbon having a δ+ (partially positive charge) and the halogen a δ- (partially negative charge).
- Can be classified as primary, secondary, or tertiary:
o Primary (10) – has one R group (carbon group) directly bonded to the carbon bearing the halogen atom.
o Secondary (20) – has two R groups directly bonded to the carbon bearing the halogen atom.
o Tertiary (30) – has three R groups directly bonded to the carbon bearing the halogen atom.
- As the halogen atom decreases in size, the C-X (X=halogen) bond decreases and the bond strength increases, and the polarity increases. (Therefore C-F is most polar and has the strongest C-X bond and C-I is the least polar and has the weakest C-X bond)
Alkenyl halides – a compound in which the halogen atom is bonded to an alkene
Aryl halides – a compound in which a halogen atom is bonded to an aromatic ring
- Phenyl halides – if the aromatic ring is a benzene ring
Physical Properties of Alkyl Halides
1. Most alkyl halides have very low solubilities in water and other polar solvents.
2. Miscible with each other and other relatively nonpolar solvents.
3. Many chloroalkanes have cumulative toxicity and are carcinogenic.
Nucleophilic Substitution Reactions
- Among the most fundamental types of organic reactions.
- A nucleophile (Nu:) displaces a leaving group (LG) in the molecule that undergoes the substitution (substrate).
o Nucleophile – always a Lewis base, may be negatively charged or neutral.
o Leaving group – always a species that takes a pair of electrons with it when it departs.
- Often, substrate – alkyl halide (R-X) and the leaving group – (X-).
- Nu- + R – LG → R – Nu + LG-
Nucleophiles – reagent that seeks a positive center
- Any negative ion or uncharged molecule with an unshared electron pair.
1) Negatively-charged nucleophile – results in a neutral product. Formation of the covalent bond between the negative nucleophile and the substrate neutralizes the formal charge of the nucleophile
o H-O- + R-X → H-O-R + X-
2) Neutral nucleophile – results initially in a positively-charged product.
o H2-O- + R-X → H2O-R + X-
§ H2O-R ↔ HO-R + H3O
Base Vs Nucleophile
While studying organic chemistry, we often use the terms base and nucleophile. For example, if we consider hydroxide ion (\(HO^-\)), it can act both as a base and as a nucleophile. Then, what is the difference between a base and a nucleophile? How are we going to decide if the outcome of a reaction between an alkyl halide and a negatively charged (at times neutral) species is elimination or substitution?
In simple words, the negatively charged species would have acted as a base if it had formed a bond with a proton at the end of the reaction and it would have acted as a nucleophile if it had formed a bond with a carbon atom at the end of the reaction.
It has to be noted here that along with the difference between a base and a nucleophile in the way they react with a substrate as stated above, the scale used to measure their strengths are also different.
To understand this, let us consider the following equilibrium,
\[AcO^- + H_2O \rightleftharpoons HOAc + HO^-\]
In the above equilibrium, there are two bases, \(ACO^-\) and \(HO^-\). Which out of these two is a strong base? Is it \(ACO^-\) or \(HO^-\)? The answer for this question can be arrived at with a simple logic that if an acid is strong its conjugate base should be weak. Here in the above case, Acetic Acid is a stronger acid than Water and hence acetate ion \(ACO^-\) should be a weaker base than hydroxide ion \(HO^-\).
Now, consider the following equilibrium,
\[CN^- + H_2S \rightleftharpoons HCN + HS^-\]
Similar to the earlier case, this equilibrium is also having two bases \(CN^-\) and \(HS^-\). Which out of these two is a strong base? The situation now is not as simple as it was earlier. In the earlier case it was a comparison between Water and Acetic Acid and it was very obvious that acetic acid was the stronger acid among the two. Here the comparison is between \(HCN\) and \(H_2S\) and it not very obvious. We need the help of some physical constants.
Let us say we are allowed only one constant. What is that one constant that we will want to conclude which among the two is a strong base? Let us see. We want to know which out of the two is a strong base. A strong base should be capable of abstracting even the weakly acidic hydrogen from a molecule and hence would be present mostly in the form of its conjugate acid. So, if \(CN^-\) was a strong base among the two, the concentration of it will be very less compared to that of HCN. Similarly as \(CN^-\) abstracts the proton from \(H_2S\) the concentration of \(H_2S\) will also be less compared to that of \(HS^-\). As the concentrations of the reactants are less than that of the products, the equilibrium constant will be less than 1. This is the constant that we are looking for.
If we know the equilibrium constant we can tell the direction in which the equilibrium is highly shifted and hence will be able to conclude on the strength of the base. The equilibrium constant is a way of finding out about the stability of the species. The equation \(\Delta G^o = -RTlnK\) gives us the relation between the standard free energy change and the equilibrium constant. Higher the equilibrium constant stabler the products than the reactants.
The above discussion goes to show that Basic Strength is a Thermodynamic Property. it depends on the stability of the species. A weak base is a stable one and a strong base is an unstable one.
Let us now turn our focus on the Nucleophile. Let us say a nucleophilic substitution reaction is performed. In fact two of them are performed. They are as shown below,
\[1. \ RX + CN^- \rightarrow R-CN + X^-\]
\[2. \ RX + HO^- \rightarrow R-OH + X^-\]
If both the above reactions were performed in the same reaction conditions and if the reaction 1. gets over in 7 minutes and the reaction 2. gets over in 10 minutes, which out of \(CN^-\) and \(HO^-\) should be a strong nucleophile? Most of us will conclude that \(CN^-\) should be the strong nucleophile because it reacts faster while all other conditions are same. Most of us will be correct. \(CN^-\) indeed will be the strong nucleophile among the two.
A Strong Nucleophile is the one that is able to attack the electrophilic carbon faster and a Weak Nucleophile is the one that takes lot of time to attack the electrophilic carbon. This goes to show that Nucleophilic Strength is a Kinetic Property. It depends on how fast or slow the nucleophile is capable of attacking the electrophilic carbon.
There is another way to predict if a species will act as a base or a nucleophile in a reaction. Let us take the example of Fluoride, \(F^-\) and Iodide, \(I^-\). \(F^-\) is a smaller ion and \(I^-\) is a larger ion. \(F^-\) can not be easily polarized while \(I^-\) can easily be polarized. Since \(F^-\) can not be easily polarized it will be a hard species and since \(I^-\) can easily be polarized it will be a soft species. A hard species prefers to form a bond with another hard species and a soft species prefers to form a bond with another soft species.
An Alkyl Halide will have acidic hydrogen as well as electrophilic carbon. Hydrogen will be harder than carbon because of the size. Hydrogen being smaller can not be polarized while carbon can be polarized because of it being larger. So, when an alkyl halide and \(F^-\) ion are allowed to react, \(F^-\), a hard species, will prefer to form a bond with the hydrogen, a hard species. This will lead to elimination as \(F^-\) is preferring to act as a base and pull the acidic hydrogen out. On the other hand, when an alkyl halide and \(I^-\) are allowed to react, \(I^-\), a soft species, will prefer to form a bond with the carbon, a soft species. This will lead to substitution as \(I^-\) is preferring to act as a nucleophile and attack the electrophilic carbon.
There is another way of looking at this. Let us say we have \(C_2H_5O^-\) and \(C_2H_5OH\). Which out of these two is the stronger nucleophile? If you have answered \(C_2H_5O^-\) is the stronger nucleophile among the two, you are correct. If you have to explain why \(C_2H_5O^-\) is stronger than \(C_2H_5OH\), what will your answer be? Think for a couple of minutes and comment your answer either here or on our facebook page. We’ll continue this discussion in the next post.
[ALKYL HALIDE INTENSIFIES]