 One of the important reactions undergone by haloalkanes is nucleophilic substitution reactions. Haloalkanes form excellent substrate for these reactions and it is primarily because of the polar nature of the Cx-pond. You see, halogens are highly electronegative and they draw electron density towards themselves. And because of this, the carbon gets a partial positive charge and halogens get a partial negative charge. And that is why the carbon here becomes a very good electrophilic center and a suitable substrate for a nucleophilic attack. Any strong nucleophile or electron species can attack this carbon and eliminate halogen as a leaving group and form a substituted product. The major driving force for this reaction is that halide ions are good leaving groups. Okay, but what makes halide ions good leaving groups? Well, you see, halogens are highly electronegative in nature and when you break the Cx-pond, halogen atom leaves with an electron pair as halide ions. And because they are highly electronegative, halogens are comfortable holding on to these electrons. Now, this leaving group ability varies in the order F- less than Cl- less than Br- less than I-. As we go down the periodic table, the atomic size increases. And that means this negative charge can now disperse over a much larger volume. And this is what makes iodide ions much more stable than fluoride ions. Now, nucleophilic substitution reactions are an extremely powerful tool to convert one substrate to the other. For example, let's take a primary alkyl halide, let's say CS3, CH2, Br. Now, depending on what our nucleophile is, the nucleophile can attack this carbon and eliminate the bromide ion and form the final substrate product. For instance, if our nucleophile is a hydroxide ion, then the hydroxide ion can substitute the bromide ion and result in the formation of a primary alcohol. Now, if our nucleophile is a methoxide ion, CS3O-, then our final product would be an ether CS3, CH2, OCS3, correct? And if our nucleophile is, let's say, ammonia, in that case a lone pair of electrons on the nitrogen atom can attack the carbon atom, eliminate the bromide ion and result in the formation of a primary amine. And if we have CN-ion as a nucleophile, which is cyanide, then our substituted product would be a primary nitrile, CH3, CH2, CN, or a cyanide. Okay, let's take one last example. Let's take hydride ion, which is usually the reactive species in reducing agents like lithium aluminium hydride or sodium borohydride and so on. So these reducing agents have hydride ion as a reactive species and this hydride ion can act as a nucleophile and substitute the bromide ion to give us a hydrocarbon. In this case, our hydrocarbon formed would be ethane. As you can see, depending on what a nucleophile is, we can obtain a wide range of products simply by substitution. And this makes nucleophilic substitution reactions an extremely powerful tool in synthetic chemistry. Now, this nucleophilic substitution reaction can proceed via two different mechanisms. One is SN1 and the other one is SN2. Let's begin with SN1 reaction mechanism. SN1 stands for substitution nucleophilic unimolecular. Unimolecular means there is only one molecule in the transition state of the rate limiting step or rate determining step. For example, terbutyl bromide reacts with water to give a tertiary alcohol and HBr. So this is a typical SN1 reaction and the rate of this reaction was found to depend entirely on the concentration of the substrate, which is CS3 thrice CBr. That is, the rate equation is first order in the concentration of our alkyl halide and a zero order in the concentration of the nucleophile, which is water. So because the rate does not depend on the concentration of the nucleophile, we can safely conclude that the nucleophile does not play any role in the transition state of the rate determining step. So to understand this better, let's look at the mechanism of the reaction. Now, the first step is the formation of the carbocation, which is also the rate limiting step. So tertiary butyl bromide ionizes to form tertiary carbocation and bromide ion. The bromine atom leaves with an electron pair and gives the bromide ion. As you can see, because this step generates an intermediate, a reactive species, this is the slow step and also the rate determining step. The next step is the nucleophilic attack on the carbocation. So our nucleophile here is water. The lone pair of electrons on the oxygen atom attacks the electrophilic carbon. Now do you think this step would be a slow step or a fast step? You see, carbocation is a highly reactive intermediate, right? It is unstable because it is electron deficient. So it will not waste any time in reacting with whatever nucleophilic or electron species is available in its vicinity. So this nucleophilic attack of oxygen on the carbocation will be very fast. And the last step is the loss of proton to the solvent. Now water here also acts as a solvent. So this is an example of a solvo less reaction where the nucleophile and solvent are both the same. Here again in the last step, oxygen abstracts a proton and gives us a final neutral molecule which is a tertiary alcohol. Now this step is also a fast one. So as you can see here, SN1 mechanism is a multi-step process. The first step is a slow ionization to form carbocation. The second step is the fast nucleophilic attack on the carbocation. And the last step is a loss of proton and formation of a neutral species. Now the last step is mostly relevant when our nucleophile is water or alcohol where we need to deprotonate the product. Now to better appreciate the mechanism of SN1 reaction, let's look at the reaction energy diagram. So here you can see that the first step where the alkyl halide ionizes to give a carbocation is a highly endothermic step and has a large activation energy. Obviously this step which has a very large activation energy determines the overall reaction rate. Now the second step is the attack of the nucleophile on the carbocation which is strongly exothermic because here an unstable reactive intermediate attains stability by combining with a nucleophile. So basically the nucleophile reacts with the carbocation as soon as it forms. And that's why this step is a very fast step. Now you can see that there is a relative minimum here right? This represents the formation of the intermediate carbocation. So from this reaction energy diagram we can once again see why the rate does not depend on the concentration of the nucleophile. It simply depends on the substrate because the rate determining step or the step which has the largest activation energy is the formation of the carbocation which has nothing to do with the nucleophile. So from here we can easily conclude that those reagents and reaction conditions that favor the formation of a carbocation or a stable carbocation will accelerate the rate of the reaction and those conditions or reagents that hinder the formation of a carbocation will slow down or retard the reaction. Let's look at other factors that affect the rate of an SN1 reaction like substituent effects, solvent effects and leaving group effects in the upcoming videos.