SN2 Ionic Substitution Reactions - University of...

SN2 Ionic
Substitution
Reactions

Chem 14D
Winter 2005

SN2 Ionic Substitution Reactions

Substitution can occur in organic compounds that have an electronegative atom or group bonded to an
sp3 hybridized carbon.

S stands for Substitution General Mechanism: Nuc = Nucleophile
N stands for Nucleophilic R-LG + Nuc  R-Nuc + LG LG = Leaving Group

Nucleophile enters as leaving group leaves.

2 stands for Bimolecular Rate = k[alkyl halide][nucleophile] (2nd order reaction)
Rate = k[CH3Cl][HO-] for the reaction above

- Concerted reaction because nucleophile attacks and leaving group leaves simultaneously. No intermediates

are formed.

“Gedunker” experiment: More often molecules collide, the faster the reaction.

- Kinetics – factors that effect reaction rate Example: 1. Cars on a freeway. The more cars, the

HO- + CH3Cl  reaction more accidents. 2. Recall throwing molecule balls in
class.
collision

Chance of collision increases as concentration increases.
2 x [HO-], 2 x rate linear relationship

[CH3O] / 10, rate / 10 Measured rate is related to the mechanism.
Rate _ [HO-][CH3O] Kinetics can disprove a mechanism.
Rate = k [HO-][CH3O]
Caution: No Trimolecular collisions!
A+B+CD

  or Audi-Benz-Cadillac collide altogether?
Audi  Benz  Cadillac  HOCH3 + CL
More likely!
Moment of Collision collision

HO- + CH3-Cl 

Lifetime is 10-15s (femto second); doesn’t last very long

Transition state studied by Zewail
H ≠ Bond getting weaker, further.
Getting closer. Partial _- _-
bond lone pair HO-----------------C-----------------Cl Cl starting to leave.

beginning to become -
O-C bond.
- Halogens are more electronegative than
HH carbon, so they have a larger share of the

electrons. This polar C-halogen bond causes

alkyl halides to undergo substitution and

elimination reactions.

Trigonal bipyramidal is the best way to arrange 5 atoms around a central atom.
Uh oh…pentavalent carbon? No, central C still has 8 e-, so the rule is not violated.

- In Backside Attack, the nucleophile attacks from the backside of the carbon-leaving group
bond) due to:
o Electrostatics (Negative charges on nucleophile and leaving group repel)
o Sterics, steric hindrance (Crowding; leaving group blocks approach of nucleophile to
the front)
o Hughes, Ingold noticed that the stereocenter will change (won’t happen in front side
attack)
front

Retention of
Stereochemistry

back Inversion of
Stereochemistry
99% SR or RS,
but not always true

- See virtually 100% Inversion of stereochemistry (also called a Walden inversion, this is good
and bad
o When a chiral alkyl halide undergoes an SN2 reaction, only one substitution product is
formed

Bruice, p. 366
o Good: says backside attack is an accurate model
o Bad: questions the 2 reasons above. There must be another factor.
- Real reason for backside attack?

Greatest stabilization occurs when orbitals overlap end to end. The overlap between the
orbital containing the pair of donated e- by the nucleophile and the _* carbon-leaving
group antibonding orbital is maximized.

Bruice, p. 364

Practice Problem from Thinkbook:
4) Provide the organic product(s) of this reaction. If more than one product is formed, indicate
which product (if any) is the major one. If no reaction occurs, write “NR.”
Answer:

- SN2 Energetics
o _G = _H-T_S
 Gibbs free energy (_G) – energy of whole system
 Enthalpy (_H) – differences energy due to bond changes (usually tens of

kcal/mol)
 Entropy (_S)- freedom of the system, molecules like to be floppy or more
molecules; entropy is a small factor (perhaps only one cal/mol), and even

when multiplied by T (temperature in Kelvin), it is still smaller than _H
o At reasonable temperatures, _G ≈ _H; Products and reactants aren’t the only thing
that matter, the transition state has its own energy (really important).

o Energy Profile: TS≠, can go forward or backward, matters

because need to get over this hill for reaction to

CHOH3- Cl _G≠ occur.

Example: breakfast @ Bombshelter

If you have enough energy, you will leave the
classroom and eat breakfast.

Energy _G

CCHl- 3OH _G is negative, so it wants to
spontaneously go to products

Reaction Coordinate

- Transition state (TS≠): the highest energy point in the energy reaction profile (due to partial bonds)
- Energy of Activation (_G≠): Energy needed to reach transition state; controls the rate of a reaction

- 2 partial bonds don’t make up for 1 full bond

- 2 partial bonds in ≠ is energetically expensive
- Breakfast metaphor: Rate of room emptying is a function of:

1) Energy of the students
2) Height of the hill of stairs leading up to the doors

- _G: influences the position of the equilibrium, but not the rate
- _G≠: influences rate of reaction, but not equilibrium

- Example: Cells  amino acids

Can’t wait for years for a reaction to complete, so enzymes make the [TS]≠ more stable.
This _G≠, so reactions go faster.

- Arrhenius Equation: (figured out before ≠) A0 - constant, as fast as reaction can
possibly go
k = A0e-_G≠/RT _G≠ - energy of activation (height of
hill)
rate R – gas constant
T – temperature in Kelvin

- Rate and _G≠ are inverses, so _G≠, rate

- Exponential factor: small changes in _G≠ can lead to large rate changes
- Temperature: temp, rate

- “Spontaneity”: If _G < 0, spontaneous reaction
S + O2  SO2

_H 0 0 -72 kcal/mol
- Very spontaneous thermodynamically but in terms of rate (kinetics), it’s extremely slow.
(Block of sulfur isn’t going to SO2. If you heat block of sulfur, it will convert to SO2. Like a match, heating
initiates a rxn. There are lots of hills in the rxn that convert sulfur, _G≠ of one of the hills is probably very large).
- So if _G < 0, then thermodynamically spontaneous.
- If _G < 25kcal/mol, then kinetically spontaneous.

Practice Problem from Thinkbook:
5) Consider this reaction:

a) Write the rate expression for this reaction.
b) Write a curved arrow mechanism for this reaction.

c) Draw the transition state.

Answer:

SN2 Variables:

Nucleophile: how does the nucleophile influence the rate [TS]≠?

R≠

Nuc------------------C-----------------LG
-

-
RR

Role of nucleophile in partial bonds:
- Share electrons (it’s making the partial bond)
- The more complete the bond is, the more stable the [TS]≠

The single most important factor which controls the nucleophilicity or basicity of any molecule or ion is the ability,
desire, or driving force to share an e- pair.

- Nucleophilicity: ability to share e- pairs with electrophile
- Basicity: ability to share e- pair with H
- Stronger bases are better nucleophiles

* When comparing molecules with attacking atoms that vary greatly in size, the polarizability (ability to skew the
electron cloud) of the atom and the reaction conditions determine whether the greater polarizability of the larger
atoms makes up for their decreased basicity.
* The relationship between basicity and nucleophilicity becomes inverted when the reaction is carried out in a
protic solvent (the solvent molecules have a hydrogen bonded to a nitrogen or oxygen). This will be explained in
the solvents section.

4 Factors influencing Nucleophilicity (similar to basicity factors because both nucleophiles and bases share e-):
1) Resonance (can increase or decrease e- density at the atom that shares e- with the electrophile, but
generally decreases nucleophilicity)

CH3O- vs.

Methoxide

Stronger nuc. b/c neg charge Weaker nuc. b/c neg charge spread out
more concentrated over 2 oxygens

No res.  no res.
Loses res. when reacts so more hesitant to do so.
Doesn’t want to lose resonance, it’s stabilizing like $

2) Atomic Size (refers to size of atomic radius of atom doing e- sharing, not molecule as a whole, just the

“business” end - the end that forms a new bond with electrophiles):

CH3O- vs. CH3S-

O is smaller so charge is more concentrated, greater drive to share e-, better nuc.
O and S both have the same # of valence e- and formal charge of -1, but O has a smaller atomic radius.

Smaller atoms are better nucleophiles because they have a more concentrated e- density, and thus a stronger
driving force to share e-.

3) Electronegativity – measure of e- greediness, think of it as “anti-nucleophilicity”

Higher electronegativity means lower nucleophilicity, because the role of a nucleophile is to share e-. If the atom

is more electronegative it is less willing to share its e- and wants to hold onto them.

F- vs. HO-

EN: 4.0 3.5

Poorer nucleophile better nucleophile

4) Inductive Effect – the electronic effect of atoms other than the atom that is sharing e- density; can increase

or decrease e- density on the atom sharing e- with the carbon

Electron withdrawing groups decrease nucleophilicity.

Electron donating groups increase nucleophilicity.

CH3CH2O- vs. CF3CH2O-

Since F is very EN, it draws e- density away from O-, making for a poorer nucleophile

X-Factor (doesn’t consistently fit into this sequence of decreasing importance)

Formal Charge: If you have more e- density, better nucleophile

HO- vs. H2O

More e- density (formal neg. charge) neutral

Better nucleophile

F- vs. NH3 (A Toss Up) Here’s a case where EN
More e- density stronger nucleophile overrides formal charge

But high EN

Practice Problem from Thinkbook:
8) Rank the nucleophilicity in aprotic solvent and briefly outline your reasoning.

Answer: (consider 4 factors) No res. No res. Has res.
(Least nucleophilic, least willing to share)
1) Resonance:

2) Atomic Size: O is smaller S is larger

(better nuc, concentrated charge)

So overall answer: Best nuc. Medium nuc. Poorest nuc.

11) Select the poorest nucleophile: HO, CH3CO2-, CH3O-
hydroxide acetate methoxide

Answer: No res. Has res. No res.

Poorest nucleophile because res. reduces e- density

- Leaving Group – portion of molecule that leaves with the pair of e- that was the bond between the leaving
group and some other atom; role of leaving group is to accept e- and leave, the more easily the leaving group
can accept and spread out e- density, the better it is

R ≠ Metaphor: Your significant other packs your bags
(weakens your bond), encouraging you to leave.
Nuc--------------------C-------------------LG
- Want to encourage LG to accept e- and leave.
-

RR

*Relatively good leaving groups make alkyl halides convenient to study for substitution reactions.

*Cells of plants and animals exist in mostly aqueous environments. Since alkyl halides are insoluble in water,

biological systems use compounds in which the group that is replaced is more polar than a halogen and thus
more soluble in water.

Consider the same 4 nucleophile factors, just reverse your logic.
The weaker the basicity of a group, the better is its leaving ability because weak bases readily bear the electrons

they formerly shared with a proton. Better bases are better nucleophiles, which are poorer leaving groups.

Bruice, p. 367

Since basicity and leaving group ability are inversely related (weaker conjugate bases = better leaving groups),
alkyl fluorides are least reactive while alkyl iodides are the most reactive of the alkyl halides.

Bruice, p. 367
1) Resonance: acetate is a better LG b/c it accommodates e- density better due to res.

vs. acetate

2) Atomic Size: the larger the size of the “business” atom, the better the LG because its e- shell is less

concentrated and can accept more e- more readily and leave.

*Alkyl iodides are least basic, while halide fluorides are most basic because larger atoms are better able to

stabilize their negative charge.

3) Electronegativity: the higher the EN, the better the LG, because it has a higher affinity for e-

4) Inductive Effect: can be e- withdrawing (usually) or e- donating, so can work either way

Withdraws e-: better LG

Donates e-: worse LG

X Factor: Formal Charge

R-LG+  LG Best leaving group, charge being quenched

R-LG  LG- Bad leaving group

Neutral  charged (unfavorable)

R-LG-  LG2- Horrible, very unlikely (Almost never see 2- charge except in metals)

Charged  more charged

- Good Leaving Groups: Iodides and sulfonates are the best leaving groups.

Iodides (I-) good because of large size

Sulfonates good LGs due to resonance.

Diazonium good because it’s a gas, so once you make it, it
leaves. Also good because +  neutral.
*Metaphor: It’s hard to get a dozen cats out the door because
they come back in, but a gas only leaves.

Moderate Leaving Groups: Like I-, but smaller
Br- Cl- Like sulfonates, but less resonance

H20, ROH O can accept e-, but it’s small

+
R---CH2

Bad Leaving Groups: (not leaving groups unless special circumstances)

F- Too Small

HO-, CH3O- O- not stabilized by resonance

Never Leaving Groups: Poor EN, very small
Negative charge on C, not stabilized by resonance
H:-
H3C:- (carbon anion)

Practice Problems from Thinkbook:

17) Label these leaving groups as best, middle, or poorest:

CF3SO3-, CH3SO3-, CH3CO2-

Answer: 3 res. contributors 3 res. contributors 2 res. contributors

Inductive effect e- donation by CH3 e- donation by CH3
Medium e- dispersion Worst e- dispersion
Effective e- dispersion Middle Poorest
Best

Steric Effects: results from repulsion by two or more atoms or atom groups; decreases reactivity when groups

are in the way at the reaction site

*Note: Steric effects affect nucleophilicity, but not basicity. (Strength of a base only depends on its willingness to
share its electrons). A bulky nucleophile cannot approach the back side of a carbon as easily as a less sterically

hindered nucleophile.

R3C:

SN2 Rate: SN2 reactions are sensitive to increasing steric hindrance at the electrophilic carbon

H3C-LG > RCH2-LG > R2CH-LG >>> R3C-LG
Methyl 1° 2° 3° (no reaction)

Fastest slowest

- Tertiary alkyl halides cannot undergo SN2 reactions because the three alkyl groups make it impossible for the

nucleophile to come within bonding distance of the tertiary carbon.

- The larger the group attached to the C attached to the LG, the more hindrance (harder for nucleophile to reach

electrophilic carbon), the slower the reaction.

- SN2 reaction rates decrease as the number or size of substituents rises.

Figure 10.3 Reaction coordinate diagrams for a) the SN2 reaction of methyl bromide with hydroxide ion; b) an

SN2 reaction of a sterically hindered secondary alkyl bromide with hydroxide ion. Bruice, p.365

Lecture Supplement, Hardinger’s Thinkbook, p. 5:

Practice Problems from Thinkbook
19) Select the slowest reaction. Briefly explain your choice.

a) OR b)

Answer:
Reaction a) is more sterically hindered at the electrophilic carbon, so it is slower than reaction b).

Solvent Effects:

What is the role of a solvent in a reaction?

- Serve as buffer, absorb heat in reaction so temp doesn’t rise too quickly

- Dissolve stuff in solution so particles can react with each other. Can very significantly influence the rate of an

SN2 reaction H2O
+-

KF K+ (aq) + F- (aq)

Solid

K+ has + charge, wants to be neutral
F- has – charge, wants to be neutral

Quantity vs. Quality
Lots of F-H bonds, but quality not that good

However, quantity makes it strong.

≈40-50 H2O molecules around K+ F- is a nucleophile.

H bonding decreases ability of F- to share e-,
solvent occupies some of its e- density.

How does the solvent effect the reaction rate?
- The energy of activation determines the rate of an SN2 reaction.
- If solvent stabilizes reactants more than the transition state, reaction is slower because Eact (energy of
activation) increases.

- If solvent stabilizes transition state more than reactants, reaction is faster because Eact decreases.

SN2 reactions need polar solvents to:
1) generate nucleophile (dissociate salts to give cations and anions)
2) help leaving group leave and stabilize the LG-
3) stabilize transition state (partial charges)

H Bonding: In general H bonding decreases nucleophilicity because it ties up e- density.
Protic: capable of donating H for H bonding (attach to O, N, F)
X-H OH bond most common!

_+
- Separate cation from anion: enhances nucleophilicity
*Example: a referee in a boxing match separates boxers, then they’re more exposed
- If you increase _, increase separation, increase nucleophilicity

Dielectric constant (_): measure of ability of substance to separate ions

Nucleophilicity is highest when the solvent is polar but cannot H bond with the nucleophile. (polar, aprotic
solvents are the best for SN2 reactions).

*The relationship between basicity and nucleophilicity becomes inverted when the reaction is carried out in a
protic solvent (the solvent molecules have a hydrogen bonded to a nitrogen or oxygen.

Definitions from Hardinger’s Thinkbook and added details from Bruice
Dielectric constant (_): measure of ability of substance to separate ions

Polar solvent: has high dielectric constant, _≈20; stabilizes reactants or transition states with large charges
better than those with small charges do

Nonpolar solvent: has low dielectric constant, _≤20; stabilizes species with smaller or no charge; In most
nonpolar solvents, ionic compounds are insoluble, but they can dissolve in aprotic polar solvents such as DMF
(dimethylformamide) or DMSO (dimethylsulfoxide).

Protic solvent: capable of H bonding (usually needs an N-H or O-H bond); solvent molecules arrange themselves
so their partially positively charged hydrogens point toward the negatively charged species, causing an ion-
dipole interaction. The solvent shields the nucleophile, so at least one of these interactions must be broken
before the nucleophile can undergo an SN2 reaction. It is easier to break the ion-dipole interaction between an
iodide ion (a weak base) and the solvent than between a fluoride ion (a stronger base) and the solvent because
weak bases interact weakly with protic solvents while strong bases interact more strongly.

Aprotic solvent: does not have an H atom that can participate in H bonding (No N-H or O-H bonds)

Aprotic polar solvents: Do not have an H attached to an O or N, so there are no positively charged H’s to form
ion-dipole interactions. Have partial negative charge on their surface that can solvate cations, but partial positive
charge is on inside of molecule, making it less accessible. Thus, a “naked” anion like fluoride is a better
nucleophile in DMSO than it is in water.

Lecture Supplement, Hardinger’s Thinkbook, p. 6

Lecture Supplement, Hardinger’s Thinkbook, p. 7

Practice Problem from Thinkbook:
13) Select the slowest reaction. Briefly explain your choice.

OR
Answer:
CH3OH is a protic solvent. Because of hydrogen bonds, protic solvents decrease nucleophilicity. Therefore,
CH3S- is a better nucleophile than CH3O- because smaller atoms have more concentrated electron density, so
they hydrogen bond better. As a result, the reaction involving CH3O- is slower.
30) Consider this reaction:

a) Draw the product of the reaction.
b) Write the curved arrow mechanism for this reaction, including the transition state for each step.
c) Changing only the electrophile, write a complete reaction that is clearly slower than the reaction given above.
Briefly explain why your new reaction is slower.
d) Changing only the nucleophile, write a complete reaction that is clearly faster than the reaction given above.
Briefly explain why your new reaction is faster.
Answer:

a) Note inversion of stereochemistry.

b)

c)

F is a poorer leaving group than Br, so this reaction is slower. You can also increase steric hindrance at the
carbon undergoing substitution to make the reaction slower.
d)

Iodide does less hydrogen bonding than methanethiolate, so iodide a better nucleophile, making for a faster
reaction. *Changing the solvent doesn’t change the nucleophile itself, although it may alter nucleophilicity.

How do you decide if a reaction is a “reasonable SN2” reaction? (Really asking about rate, which is asking about
TS≠)

R ≠ Has to be reasonably stable.

Nuc---------------------------C---------------------------LG TS is energetically expensive
because of partial bonds.
-
-

RR

Three Main Requirements for an SN2 reaction: work together, if one is really good,
1) moderate or better leaving group team;

2) good nucleophile other can be weaker

3) Carbon undergoing nucleophilic attack cannot be tertiary

Is the following reaction reasonable?

I OCH3 OCH3


LG: I– good, large size  … so overall OK
Nuc: - OCH3 no res. , small , neg. charge , highly EN 
Carbon is 1°
CH3OH solvent (polar, but protic)

Met all requirements, so yes, it’s reasonable.

The Reversibility of an SN2 Reaction (Information from Bruice, p.372)

- SN2 reactions usually take place in one direction, but not the other (even though they seem like they could by

another nucleophilic substitution) due to leaving tendency of leaving groups.
CH3CH2Cl + HO-  CH3CH2OH + Cl-

Stronger base, worse LG Weaker base, better LG

Therefore, HO- can displace Cl- in the forward direction, but Cl- cannot displace HO in the reverse direction.

- If the difference between the basicities of the nucleophile and the leaving group is not very large, the reaction
will be reversible, because the pKa values of the conjugate acids of the two leaving groups are similar.

(pKa of HBr=-9; pKa of HI=-10)

CH3CH2Br + I- CH3CH2I + Br-

- Le Chatelier’s principle: If an equilibrium is disturbed, the system will adjust to offset the disturbance, so you
can drive a reversible reaction toward desired products by removing one of the products as it is formed.

A+B C+D

Keq = [C][D]
[A][B]

To maintain the value of the equilibrium constant, if [C] is decreased, A and B will react to form more C and D.