Nucleophilic Substitution Reactions and the SN2 Mechanism

So far we have seen how haloalkanes are classified, named, prepared, and what their physical properties look like. Now comes the most important question: what happens when these compounds react? The halogen sitting on an $sp^3$ carbon is not just a label; it is a leaving group waiting to be replaced. The reactions of haloalkanes fall into three broad categories: nucleophilic substitution, elimination, and reaction with metals. This topic dives deep into the first of those: nucleophilic substitution, where an electron-rich species swoops in and takes the halogen’s place.

What Is Nucleophilic Substitution?

Think back to what makes the $C-X$ bond special. The halogen is more electronegative than carbon, so it pulls electron density toward itself. This leaves the carbon atom with a partial positive charge ($\delta+$) and the halogen with a partial negative charge ($\delta-$):

$Nu^- + \overset{\delta+}{C}-\overset{\delta-}{X} \longrightarrow C-Nu + X^-$

A nucleophile (from Greek: “nucleus-loving”) is any electron-rich species, an ion or molecule with a lone pair or negative charge ready to be shared. When a nucleophile encounters this electron-poor carbon, it donates its electron pair to form a new bond with the carbon. At the same time, the halogen departs, taking the bonding electrons with it and leaving as a halide ion ($X^-$). The halogen in this role is called the leaving group.

Because the reaction is kicked off by a nucleophile replacing an existing group, it is called a nucleophilic substitution reaction. This is one of the most broadly useful reaction types in all of organic chemistry. Alkyl halides serve as the substrate (the molecule being attacked), and the range of nucleophiles you can use is enormous, which is exactly what makes these reactions so versatile.

The Product Gallery: What Nucleophiles Can Make

One haloalkane, many possible products. The identity of the nucleophile determines what functional group you end up with. Here is a survey of the most common transformations:

Reagent	Nucleophile	Product formula	Product class
$NaOH$ or $KOH$	$OH^-$	$ROH$	Alcohol
$H_2O$	$H_2O$	$ROH$	Alcohol
$NaOR'$	$R'O^-$	$ROR'$	Ether
$NaI$	$I^-$	$RI$	Alkyl iodide
$NH_3$	$NH_3$	$RNH_2$	Primary amine
$R'NH_2$	$R'NH_2$	$RNHR'$	Secondary amine
$R'R''NH$	$R'R''NH$	$RNR'R''$	Tertiary amine
$KCN$	$^-C \equiv N$	$RCN$	Nitrile (cyanide)
$AgCN$	$Ag-CN$	$RNC$	Isocyanide
$KNO_2$	$O=N-O^-$	$R-O-N=O$	Alkyl nitrite
$AgNO_2$	$Ag-ONO$	$R-NO_2$	Nitroalkane
$R'COOAg$	$R'COO^-$	$R'COOR$	Ester
$LiAlH_4$	$H^-$	$RH$	Hydrocarbon
$R'^-M^+$	$R'^-$	$RR'$	Alkane (new C-C bond)

Notice the pattern: in every row, the nucleophile simply takes the halogen’s place. Change the nucleophile and you change the product. This single reaction type can convert a haloalkane into an alcohol, ether, amine, nitrile, ester, or even a plain hydrocarbon.

Ambident Nucleophiles: Two Faces, Two Products

Some nucleophiles are more interesting than others because they carry two different atoms that can each serve as the point of attack. These are called ambident nucleophiles (from Latin ambi, meaning “both sides”).

The cyanide case: $KCN$ vs $AgCN$

The cyanide ion ($CN^-$) is a resonance hybrid of two contributing structures:

$^{\ominus}C \equiv N \longleftrightarrow \; :C = N^{\ominus}$

Both carbon and nitrogen have lone pairs available for bonding. So which end attacks?

• $KCN$ is predominantly ionic. It dissociates in solution to give free $CN^-$ ions. The attack happens mainly through the carbon end, because the resulting $C-C$ bond is more stable than a $C-N$ bond. Product: alkyl cyanide (nitrile), $RCN$.
• $AgCN$ is largely covalent. The carbon end of the cyanide is tied up in a covalent bond with silver, so the nitrogen atom is the one free to donate its electron pair. Product: isocyanide, $RNC$.

The nitrite case: $KNO_2$ vs $AgNO_2$

The nitrite ion ($NO_2^-$) also has two nucleophilic centres: oxygen and nitrogen.

• $KNO_2$ (ionic) allows the oxygen to attack, giving an alkyl nitrite ($R-O-N=O$).
• $AgNO_2$ (covalent) directs attack through nitrogen, giving a nitroalkane ($R-NO_2$).

The key takeaway: with ambident nucleophiles, whether the reagent is ionic (K or Na salt) or covalent (Ag salt) determines which atom does the attacking, and that changes the product entirely.

Solved Example 6.5: Why $KCN$ and $AgCN$ Give Different Products

Problem: Haloalkanes react with $KCN$ to form alkyl cyanides as the main product, while $AgCN$ forms isocyanides as the chief product. Explain why.

Solution:

$KCN$ is an ionic compound. In solution, it fully dissociates to release free $CN^-$ ions. Both carbon and nitrogen atoms on this ion could potentially donate an electron pair, but the attack occurs preferentially through the carbon atom. The reason is thermodynamic: a $C-C$ bond is more stable than a $C-N$ bond, so the product formed through carbon attack (the nitrile $RCN$) is energetically favoured.

$AgCN$, on the other hand, is predominantly covalent. The carbon end of the cyanide group is locked in a covalent bond with silver and is not available for nucleophilic attack. The nitrogen atom, however, retains a free lone pair. It is this nitrogen lone pair that attacks the alkyl halide, forming a bond through nitrogen and producing the isocyanide $RNC$.

In short: ionic $KCN$ frees the carbon to attack (nitrile), while covalent $AgCN$ frees only the nitrogen to attack (isocyanide).

The $S_N2$ Mechanism: Two Molecules, One Concerted Step

Now that we know what nucleophilic substitution achieves, the natural next question is: how does it actually happen at the molecular level? Two distinct mechanisms have been discovered. The first is the $S_N2$ mechanism, which stands for Substitution, Nucleophilic, Bimolecular. (The “2” refers to the fact that two species, the nucleophile and the substrate, are both involved in the rate-determining step.)

This mechanism was proposed by Edward Davies Hughes and Sir Christopher Ingold in 1937. Hughes worked under Ingold at University College London and earned a D.Sc. degree from the University of London for his research on reaction mechanisms.

How the reaction proceeds

Consider the reaction between chloromethane ($CH_3Cl$) and hydroxide ion ($OH^-$) to produce methanol:

$OH^- + CH_3Cl \longrightarrow CH_3OH + Cl^-$

Experiments show that the rate of this reaction depends on the concentration of both the hydroxide ion and the chloromethane. Double the concentration of either reactant, and the rate doubles. This is second-order kinetics, confirming that both molecules participate in the single rate-determining step.

Here is what happens at the molecular level:

Step 1: Backside approach. The hydroxide ion approaches the carbon atom from the side directly opposite the chlorine (the leaving group). It does not come in from the same side as the chlorine, and it does not come from the side. It must attack from directly behind.

Step 2: Simultaneous bond-making and bond-breaking. As the oxygen of $OH^-$ begins to share its electron pair with the carbon, the $C-Cl$ bond starts to weaken at exactly the same time. This is not a two-step process; the bond to the nucleophile forms as the bond to the leaving group breaks. The two events are perfectly synchronised.

Step 3: The transition state. At the midpoint of this process, the carbon is partially bonded to five groups: the three hydrogens, the incoming $OH$, and the outgoing $Cl$. The three hydrogen atoms lie flat in a single plane, with the $OH$ partially attached on one side and the $Cl$ partially attached on the other. Both the oxygen and the chlorine carry partial negative charges ($\delta-$) at this moment. This arrangement is called the transition state. It is extremely unstable, sits at the very top of the energy barrier, and cannot be isolated or captured.

Step 4: Completion. The $OH$ group closes in fully, the $Cl$ departs completely as $Cl^-$, and the three hydrogens have swung all the way to the other side.

Fig 6.2: Space-filling model of the $S_N2$ reaction. The red ball represents the incoming hydroxide ion and the green ball represents the outgoing halide ion

The entire process happens in one single step with no intermediate. There is only a transition state, not a distinct chemical species that exists between reactant and product.

Inversion of Configuration: The Umbrella Analogy

What is configuration?

Configuration refers to the three-dimensional spatial arrangement of groups around an atom. Consider a carbon atom bonded to four different groups. You can arrange those groups in two distinct ways that are mirror images of each other:

Structure (A) and structure (B) have the same four groups, but their spatial arrangement is different. One is the mirror image of the other. We say the configuration of carbon in (A) is the mirror image of the configuration in (B).

How $S_N2$ flips the configuration

Now imagine you are watching the $S_N2$ reaction in slow motion. The nucleophile pushes in from behind. The three substituents that are not the leaving group start to move, tilting away from the incoming nucleophile. At the transition state, they pass through a flat plane. Then they continue moving to the other side as the leaving group exits.

The result? Every group that was on the left is now on the right, and vice versa. The configuration has been completely inverted. This is exactly like what happens when a strong gust of wind catches an umbrella and flips it inside out: the ribs that pointed downward are now pointing upward, and the entire shape is reversed.

This process is called inversion of configuration (sometimes also called Walden inversion). It is a signature feature of the $S_N2$ mechanism: every $S_N2$ reaction produces a product whose configuration is the mirror image of the starting material’s configuration at the reaction centre.

Steric Effects: Why Size Matters in $S_N2$

The $S_N2$ mechanism demands that the nucleophile reach the back side of the carbon bearing the leaving group. Anything that blocks that approach will slow the reaction down. This is where steric hindrance (physical crowding by bulky groups) becomes the deciding factor.

Fig 6.3: Steric effects in the $S_N2$ reaction. Relative rates are given in parentheses

Consider the four classes of alkyl halides:

Substrate type	Groups on the reacting carbon	Relative $S_N2$ rate	Why?
Methyl ($CH_3X$)	Three small H atoms	30	Back side is wide open
Primary ($RCH_2X$)	Two H atoms, one alkyl group	1	Slight crowding from one alkyl group
Secondary ($R_2CHX$)	One H atom, two alkyl groups	0.02	Two alkyl groups significantly block approach
Tertiary ($R_3CX$)	No H atoms, three alkyl groups	~0	Nucleophile simply cannot get through

The trend is dramatic. A methyl halide reacts 30 times faster than a primary halide in $S_N2$, and a primary halide reacts 50 times faster than a secondary halide. Tertiary halides are essentially unreactive by this pathway.

The reactivity order for $S_N2$ is therefore:

$\text{Methyl halide} > \text{Primary halide} > \text{Secondary halide} > \text{Tertiary halide}$

The physical picture is straightforward: each alkyl group attached to the reacting carbon is like a shield blocking the nucleophile’s path. With three such shields (tertiary), the nucleophile has no way to reach the carbon from behind, and the reaction does not proceed via $S_N2$ at all.