Factors affecting rate of SN2 reactions

Rate of reaction is the most important parameter in organic synthesis. It determines how fast the products are synthesized and what conditions make the reactions slower. The reaction conditions are set accordingly to have an adequate amount of product in minimum time.

The factors affecting the rate of S_N2 reactions are, therefore, important to understand in order to achieve the product in minimum possible time. The factors are:

• Strength of nucleophile
• Nature of solvent
• Nature of leaving group
• Nature of substrate

The general rate equation for S_N2 reactions is given below. It suggests that the rate of S_N2 reactions can be enhanced by increasing the concentration of substrate and nucleophiles. All the factors which enhance the availability of either substrate or nucleophile will increase the rate of S_N2 reactions.

Rate = k [Substrate] [Nu^–]

Nucleophilicity (strength of a nucleophile)

The nucleophilicity of a nucleophile is defined as its ability to attack an electrophilic center to displace the leaving group by substitution.

The strength of a nucleophile depends upon the following factors:

1. Charge on nucleophile
2. Electronegativity of nucleophile
3. Size of nucleophile
4. Steric hindrance
5. Resonance effect

1. Charge on nucleophile

Nucleophile plays an important role in maintaining the rate of reaction. If the nucleophile is not strong, the reaction will proceed slow or might not occur at all. The choice of a nucleophile is the principal factor for the reaction to take place at a fair speed. (Greater the negative charge on a nucleophile, greater will be its nucleophilicity).

For S_N2 reactions all processes take place in a single step so, the nucleophile must be strong enough to substitute the leaving group.

A general comparison of the nucleophilicities of some nucleophiles is given

OH^– > H₂O

RO^– > ROH

NH₂^– > NH₃

S^-2 > HS^– > H₂S

For example

The rate of hydrolysis of methyl bromide that acquires SN₂ mechanism is more than 5000 times greater when Hydroxyl (-OH) is used as nucleophile than H₂O because hydroxyl is a far stronger nucleophile than the latter one.

2. Electronegativity of nucleophile

Greater electronegativity of a nucleophilic atom, suggests weaker nucleophilic strength because the electronegativity of an atom decreases its ability to share electrons. When nuclear attraction force for the electrons rises, nucleophiles become weaker and unable to attack.

Along a period, nucleophilic strength decreases gradually as the electronegativity of atoms increases.

For example

Similarly, compounds containing these atoms show same trend of decreasing nucleophilicity.

CH₃^– > NH₂^– > OH^– > F^–

On the other hand, down the group, the size of atoms increases, decreasing the effective nuclear charge, which in turn decreases the electronegativity of atoms. Hence, nucleophilicity increases down the group.

For example

RSH > ROH

RS^– > RO^–

I^– > Br^– > Cl^– > F^–

3. Size of nucleophile

Nucleophilicity increases with the increasing sizes of atoms and decreases with decreasing sizes because the relative charge densities change directly with sizes.

For example:

Bromide ion (Br^– ) is a better nucleophile as compared to chloride ion (Cl^– ) because of more polarizability and availability of electrons to attack the electrophilic center.

4. Steric Hindrance

Sterically hindered nucleophile affects the rate of reaction. For the reaction to occur easily, the nucleophile must approach the electrophilic center. If the nucleophile is a bulky molecule e.g. ter. butoxide ion, it will face hindrance from the atoms of the substrate and the attack on the electrophilic center will be restricted. The nucleophile now acts as a base and attacks the hydrogen, leading the reaction to elimination instead of substitution.  Hence, bulky nucleophiles are not good nucleophiles.

Hindered atoms will be less nucleophilic, particularly attacking at crowded electrophiles.

(CH₃)₃CO^– < (CH₃)₂CHO^– < CH₃CH₂CO^– < CH₃O^–

The nucleophilic character of an atom decreases as it gets crowded by bulky groups.

For example

When a Lewis base attacks a proton, it does not feel any hindrance and easily forms a bond. In an SN2 reaction when a methoxide ion (CH₃O^–) approaches a tetravalent carbon atom, it readily reacts. However, if tertiary butoxide ion C(CH₃)₃O^– is allowed to react with the same tetravalent carbon atom, it will face severe steric hindrance. Thus, the ability of tertiary butoxide ion to function as a nucleophile is sufficiently low.

5. Effect of resonance on strength of nucleophile

Resonance is another factor that may increase or decrease the nucleophilicity of a nucleophile. For the reaction to occur the lone pair should be available on the electronegative atom of the nucleophile but if there is resonance, the lone pair gets involved in delocalization and it would not be available for the attack anymore and hence, the nucleophilicity of the nucleophile will decrease.

For example

Methoxide ion is more reactive as compared to carboxylate ion because the negative charge of oxygen in methoxide is localized and ready to attack over the substrate. On the other hand, the negative charge is delocalized and therefore, not available for attack in carboxylate ion.

Similarly, in case of amines (-CH₂NH₂), the lone pair is available on nitrogen, while in amides (-CONH₂) the lone pair of nitrogen is involved in delocalization with the lone pair of the oxygen atom and this makes amide (-CONH₂) a weaker base than amine (-CH₂NH₂).

Effect of solvent in SN2 reactions

Solvent plays an important role in determining the rate of S_N2 reaction. The nucleophile should be strong enough to attack at electrophilic carbon so that the leaving group would depart.

There are three types of solvents:

1. Polar protic solvents
2. Polar aprotic solvents
3. Nonpolar solvents

Polar aprotic solvents are best suited for S_N2 reactions as polar solvents create nucleophiles.

For example, the dissociation of NaCl in water will produce a nucleophile (Cl^–). Protic polar solvents will reduce the reactivity of this nucleophile. The partial positive charge on hydrogen atom cages the strong negative charge of nucleophile by forming hydrogen bonding. Hence, the rate of reaction decreases.

In case of polar aprotic solvents, these solvents have dipole moments, they solvate the nucleophile without affecting its nucleophilicity. These solvents do not alter the nucleophilicity of nucleophile and hence, the reaction goes without the interference of solvents.

Hence, Polar aprotic solvents are the best choice for S_N2 reactions.

Effect of leaving group

Leaving groups are the ions that have to leave the substrate upon attack of a relatively stronger nucleophile in nucleophilic substitution (S_N) reactions.

As discussed earlier, the leaving group is a weak base i.e conjugate base of a strong acid. The ability of a leaving group to leave is directly proportional to its stability. The stability of negatively charged species is explained by the SHAB concept.

For example

Iodide (I^–) is a good leaving group as compared to chloride (Cl^–) because the negative charge on iodide is more dispersed due to its larger size and is more stable than chloride.

Factors that affect the Leaving ability of leaving groups

1. Strength of carbon leaving group (C-LG) bond

If the leaving group is highly electronegative, the C-LG bond will not break easily.

For example, the order of bond strength of alkyl halides is

R-F > R-Cl > R-Br > R-I

2. Stability of leaving group (LG)

The stability of leaving groups is another factor effecting the leaving group departure. The greater the stability of leaving group, the easier would be the leaving.

When the leaving group leaves, it takes its electrons back from the carbon atom with which they were shared during bond formation. So, the leaving group must be able to accommodate these electrons, this assures the stability of halide atoms.

For example

Among the halides, iodide ion is the best leaving group because it has less electronegativity than other halides and its greater size makes it more able to retain its two electrons.

I^– > Br ^– > Cl^– > F^–

The stability of leaving groups also depends upon the type of structure.

Resonance stabilized structures are more stable and good leaving groups.

For example, the carboxylate ion (CH₃COO^–) is stable because its negative charge can be delocalized while this is not possible in alkoxide RO^–. So, carboxylate ion is a good leaving group but a bad nucleophile due to resonance.

Besides these, there are some poor leaving groups that are first protonated in order to convert them into good leaving groups. For example, H₂O is a weak base and a better leaving group than OH^–.

I^– + ROH  →  (NO REACTION)

I^– + R-O⁺H₂  →  RI + H₂O

Effect of substrate

For nucleophilic substitution (S_N) reactions, the substrate is a molecule in which an alkyl group containing an electrophilic carbon is linked with a leaving group.

In S_N2 reactions, the attack of nucleophile takes place to form a bond and leaving group leaves the substrate simultaneously. The attack of nucleophile must occur from the side opposite to the leaving group. This causes an ease in the formation of C-Nu bond.

There are five groups attached to the substrate molecule in transition state:

When a nucleophile approaches a carbon atom from backside, it will feel very less steric hindrance as hydrogens are small atoms. The activation energy will be low and the rate of reaction will be maximum.

Now, if we replace one hydrogen atom with an alkyl group, the incoming nucleophile will face comparatively greater steric hindrance. The steric hindrance will be maximum when all the hydrogens are replaced by alkyl groups as in tertiary carbon. This is why tertiary alkyl halides do not undergo the S_N2 mechanism.

The order of reactivity of alkyl halides to give S_N2 reaction is:

Methyl halide > Pri. alkyl halide > Sec. alkyl halide > Ter. alkyl halide

The question which arises here is how steric hindrance affects the rate of reaction?

The nucleophile when approaching a carbon atom, in a tertiary electrophilic center feels severe steric hindrance. The activation energy of the reaction increases, decreasing its rate.

Related topics

• SN1 vs. SN2

Key Takeaway(s)

The factors that favor S_N2 mechanism and increase the rate of S_N2 reactions are:

• A strong nucleophile (small in size, less electronegative, negatively charged species).
• Good leaving group ( large-sized atom, less electronegative, highly polarizable, weak conjugate base, mainly Br^– and I^– in halogens).
• An unhindered alkyl group mainly primary or secondary carbon.
• Polar aprotic solvent.

Concepts berg

What are the 3 key factors that affect the rate of the S_N2 reaction?

The 3 key factors that affect the rate of S_N2 reactions are:

1. Strength of nucleophile
2. Nature of solvent
3. Nature of leaving group

There is also a fourth factor that somehow affects the rate of reaction of S_N2 reactions, that is:

• Nature of the substrate

What increases the rate of S_N2 reaction?

A strong nucleophile, an uncrowded substrate, a good leaving group, and a polar aprotic solvent  are the factors that increases the rate of S_N2 reaction.

What decreases the rate of S_N2 reactions?

A weak nucleophile, crowded substrate, a bad leaving group, and polar protic or nonpolar solvent decreases rate of S_N2 reactions.

Is S_N1 faster or S_N2?

S_N2 reactions are faster as compared to S_N1.

S_N2 are single step reactions and S_N1 reactions occur in two steps. The first step is slow and is the rate determining.

What is the best solvent for S_N2 reaction?

Polar aprotic solvent is the best solvent for S_N2 reaction as it only solvates cations and it does not alter the reactivity of nucleophiles.

Is S_N2 rate governed by steric effects?

Yes, steric hindrance greatly affects the rate of reaction because steric hindrance increases the activation energy of reaction decreasing the rate of reaction.

What is the effect of leaving group on S_N2 reaction?

In SN2 reactions, the leaving group plays an important role in maintaining the rate of reaction as it is a single step mechanism. So, if the leaving group does not leave easily, it will effect the reaction rate. Two factors affect the activity of leaving group. The strength of C-LG bond and the stability of LG ion after leaving.

What are the effects of substrate on S_N1 and S_N2 reaction?

In S_N1 type reactions, the first step is carbocation formation. The carbocations are stable i.e. tertiary carbocations are most stable due to inductive effect of the three electron donating groups attached.

While in S_N2 type reactions, crowded substrate is avoided because back attack is severely hindered by bulky groups of substrate and the reaction goes slow.

The order of reactivity of alkyl halides for S_N1 reactions is as:

ter. alkyl halide > sec. alkyl halide > pri. alkyl halide.

And for S_N2 reactions:

ter. carbon < sec. carbon < pri. carbon.

What is the effect of solvent on S_N2?

S_N2 reactions are carried out in polar aprotic solvents because these solvents only solvate cations. They do not alter the reactivity of a nucleophile. While polar protic solvents have partial positive hydrogen. It cages (traps) the nucleophile and reduces its activity.

Why is a weak base not added in an S_N2 reaction?

Weak bases are not added in S_N2 reactions because these reactions require strong nucleophiles. Strong bases are used which may be strong nucleophiles and could attack carbon atoms. This makes leaving groups leave.

Why does S_N2 use a strong base while S_N1 uses a weak base?

S_N2 reactions require strong bases because attack of nucleophiles and departure of leaving groups takes place in a single step. While in S_N1 reaction, a carbocation is formed at first, followed by attack of nucleophile. So there is no need of a strong nucleophile as even weak nucleophiles can easily approach carbocations. Furthermore, strong nucleophile acting like a strong base will abstract hydrogen atom of substrate and the molecule can undergo elimination reaction.

What is the charge on the transition state during the S_N2 reaction?

The charge on the transition state during S_N2 reaction is zero. The extent of bond formation of nucleophile is equal to the extent of breakage of (C-LG) bond. So the negative charge coming inside is equal to the negative charge leaving the substrate. Hence the overall charge of the transition state becomes zero.

What are the factors which influence nucleophilicity?

Nucleophilicity is affected by the following factors in the following ways.

• Negative charge on the nucleophile increases its nucleophilicity.
• Electronegativity of the nucleophile decreases its strength.
• Size of the nucleophile decreases its nucleophilicity.
• Steric hindrance of substrate decreases its nucleophilic character.
• Resonance effect of nucleophile decreases its the strength of nucleophiles.

Is rearrangement possible in S_N2 reaction?

There is no chance of rearrangement in S_N2 reactions because it is a simultaneous process where no intermediate is formed. The rearrangement happens only in cases where reactive intermediates are formed.

Why the rate of reaction decreases in S_N2 mechanism with increase in polarity of solvent?

Increase in polarity of solvent decreases the reactivity of nucleophiles. As for S_N2 reactions, strong nucleophile is required, therefore, polarity of solvent decreases the rate of reaction.

Why is the first step of the S_N1 reaction considered to be the rate-determining step?

Because the first step of S_N1 reaction is slow i.e. the formation of carbocation. The slowest step of reaction mechanism is considered to be the rate-determining step because it controls the rate of reaction.

What effect would we expect if a concentration of substrate is tripled and the nucleophile is halved in an S_N2 reaction?

S_N2 reactions are bimolecular, meaning that they are concentration dependent reactions. The rate of reaction depends upon the concentrations of the nucleophile and the substrate both.

The nucleophile and substrate react in 1:1, as in the general equation, the reaction rate equation is

R = k [Nu] [Substrate]

So, when [substrate] is tripled and [Nu] halved, the rate will increase by 1.5x.

R = k [½][3]

R = 1.5k

Would an SN2 reaction still happen if the leaving group is a weak one like OH?

Yes, by protonation the bad leaving groups can easily be converted into good leaving groups because OH^– is a strong base and a bad leaving group while OH₂ is a weak base and a good leaving group.

References

• Organic chemistry-1 for dummies by Arther Winter