SUBSTITUTION REACTION
In a substitution reaction one atom or group of atoms in a molecule is replaced by another. They are known to proceed by a free radical or ionic mechanism.
Free Radical Substitution :
CH4 + Cl2 →hν CH3Cl + HCl + Br2 →hν Br+ HBr CH2 = CHCH3 + NBS → CH2 = CHCH2Br
C—H bond is replaced by a C—X bond in free radical halogenation. Reaction takes place through free radical intermediates and is thus a homolytic substitution reaction.
CH3CH3 + Cl2 → CH3CH2Cl + HCl Step I (initiation) Cl—Cl → Cl + Cl • • Step II (propagation)
CH3CH2—H + Cl → CH• 3CH• 2 + HCl CH3CH• 2+Cl—Cl → CH• 3CH• 2Cl + Cl• Step III (termination)
Cl + Cl → Cl2
• • CH
3CH• 2 + Cl → CH• 3CH2Cl CH3CH• 2 + CH• 2CH3
→
coupling CH2CH2CH2CH3
CH3CH• 2 + CH3CH• 2
→
disproportionation
CH2 = CH2 + CH3 —CH3
Reactivity of the halogens for free radical substitution is in order
F2 > Cl2 > Br2 > I2
For a given halogen, abstraction of H is in the following order :
allylic 3º > allylic 2º > 2º > 1º (CH3) > vinylic
For
H H H H
H CH2—H 1º H
H H H Allylic
2º
vinylic
Allylic 2º 3º
Abstraction of H from allylic carbon or benzylic carbon takes place using NBS (N-bromosuccinimide) in which weak nitrogen-bromine bond can be cleaved homolytically into radical upon warming or exposure
∆
→
NBS Br
→hν
NBS
Br
IONIC SUBSTITUTION
Substitution reaction may be brought about by (a) replacement by electrophiles-called Electrophilic Substitution reaction (SE)
(b) replacement by nucleophiles-called Nucleophilic Substitution reaction (SN)
Electrophilic Substitution Reaction (SE)
This SE reaction takes place in benzene nucleus (aromatic compounds) in which π elelctrons are highly delocalised and an electrophile can attack this region of high elelctron density.
+ E+ →
E
↔
E
≡
E
+ CH3Cl →AlCl3 CH3
+ HCl
Step I. Formation of an electrophile CH3Cl + AlCl3 → AlCl–4+C⊕H3
Step II. Attack of electerophile on benzene when resonance-stabilised σ complex is formed.
+ CH3
⊕
↔
⊕ H CH3
↔
H
⊕ CH3
↔ (carbocation-an
arenium ion) H
CH3
⊕ ≡
H CH3 +
σ-complex
Step III. Loss of H+ from σ complex to form end product
⊕ H
CH3 →
CH3 + H+ Step IV. AlCl4– + H+ → AlCl3 + HCl
Organic Chemistry Fundamentals
REACTION MECHANISM
KEY CONCEPT
An additional reagent (called Lewis acid) is always required that can help in the formation of an electrophilic
+ NHO3 →H2SO4
Nitration would not take place in the absence of H2SO4 . It helps in the formations NO2+ (nitronium ion). Similarly Br+ electrophile is formed when Fe or FeCl3 is present.
⊕ H
CH3 + NΘu → H CH3 H Nu
If there is attack of nucelophile on positive site, aromaticity is lost, hence this is not desirable.
Nucleophilic Substitution Reactions
When a substitution reaction is brought about by a nucleophile, the reaction is called SN (nucleophilic (N) substitution (S) reaction.
R—X + OHΘ → R—OH + XΘ Substrate nucleophile leaving group
N1
S Reaction
If rate of subsititution depends on the concentration of the substrate, then it is said to be the Unimolecular (1) Nucleophilic (N) Substitution (S) reaction, written as SN1 determining step.
R⊕ + OHΘ →fast R — OH
Carbocation formed can undergo rearrangement to give more stable carbocation before attack of the nucleophile
CH3CCH2Cl
The fact that the rate law depends only on the concentration of tert-butyl chloride means that only tert-butyl chloride is present in the transition state that determines tha rate of the reaction. There must be more than one step in the mechanism because the acetate ion must bnot be involoved until after the step with this transition state. Because only one molecule (tert-butyl chloride) is present in the step involving the transition state that determines the state of reaction, this step is said to be Unimolecular. The reaction is therefore, described as a 1
SN reaction. recemisation and inversion.. When (–) 2-bromo-butane having chiral centre is treated with low [OH–] such that SN1 reaction is followed, (+) 2-butanol is obtained. There is also loss in activity. This loss in optical activity is due to formation of d-and l-isomers by SN1 reaction.
Nucleophilic reagent attacks both (I) backside and (II) front side of the carbocation. In back side attack (I), configuration is retained but in front attack (II), inversion takes place. There can be recemisation if d-and l-are formed in equal amounts. Attack is preferred on the side opposite to where leaving group Br– exists since it shields the passage of nucleophile for attack. (In this case front attack is preferred, hence mixture is not purely racemic and some optical actvity exists)
In SN1 reaction the order of reactivity of RX is Allyl or benzyl > 3º > 2º > 1º > CH3X
SN2 Reaction
If substrate nucleophile bothe are involved in the rate-determining step then this is called as
Bimolecular (2) Nucleophilic (N) Substitution (S) reaction indicated as SN2.
R—X + OH– → ROH + X–
dt
dx = k[R—X] [OH–]
Step I. OH– attacks the side opposite to that where halide exists to give an intermediate (transition state)
H—C—Br + OH| –
| H
H
→
slow
HO C Brδ–
δ– H H
H Intermediate (A) Step II. HBr is stronger acid than H2O therefore, Br– is a better leaving group than OH–, hence Br–, is lost from the intermediate
Intermediate (A) fast→ HO—C H H H
How do we confirm this type of attack? Again we take (+)-2-bromo-butane (with one chiral carbon). It is treated with conc. aq. KOH (OH– being strong nucleophile). We get (–)-2-butanaol and no (+) isomer. There is thus complete stereochemical inversion.
H—C—Br + OH–
|
| CH2CH3
CH3
→
slow HO C Brδ–
δ– H CH3
CH2CH3
Θ
→
Br –
fast HO—C—H |
| CH2CH3
CH3
(–)
For SN2 reaction the order of reactivity is CH3X > 1º > 2º > 3º (alkyl halide)
CH2Br > CHBr CH3
> CBr CH3
CH3
Thus, in SN1 reaction, recemisation as well as inversion is observed, while in case of SN2 reaction, completee inversion takes place (where chiral carbon exists).
Rearrangement of the carbocation (formed in SN1 reaction) leading to more stable carbocation is also observed in SN1 reaction (solved example)
SN2 reaction
RX = CH3X 1º 2º 3º SN1 reaction CH3X > 1º > 2º > 3º SN2 SN2 mixed SN1
High concentration of the nucleophile favours SN2 reaction while low concentration favours SN1 reaction.
CH3CCH| 2Br
| CH3
CH3
shows SN2 reaction with C2H5O–, but
SN1 reaction with C2H5OH.
The higher the polarity of the solvent, the greater is the tendency for SN1 reaction
Elimination VS Substitution
CH3CCH2CH3
|
⊕ CH3
OH–
SN CH3CCH2CH3
CH| 3
OH |
CH3C=CHCH3
CH| 3
E2 OH–
In the above example we find that a carbocation can show SN2 as well a E2 reaction. Where substitution and elimination are competing reaction, the proportion of elimination increases as the structure of alkyl halide is changed from primary to secondary to tertiary.
Many tertiary alkyl halides yield exclusively alkenes under these conditions
Elimination increases Substitution increases RX = 1º 2º 3º
CH3CBr + C| 2H5ONa
| CH3
CH3
→
E2 CH3C ||
| CH3
CH2
CH3C—ONa + C2H5Br
|
| CH3
CH3
1º
→
SN2 CH3C—OC2H5
|
| CH3
CH3
The general pattern of reactivity expected from various structural classes of alky halides in reactions with a representative range of nucleophiles (which may behave as bases).
Effect of Solvent
SN1 reaction are faster in the more polar solvents.
SN2 reaction involving a negative nucleophile is slower in more polar solvent and that involving a neutral nucleophile is faster in more polar solvent.
In additions to these polarity effects, the ability of certain solvents to form hydrogen bonds to the nucleophile also affects the rate of the SN2 reaction.
Such solvents are termed protic solvents and have a hydrogen bonded to nitrogen or oxygen (H2O, ROH, RCOOH are some examples of protic solvents).
Density of Cubic Crystals :
The density based on the structure can be calculated from the mass contained in a unit cell and its volume.
If N is the number of molecules per unit cubic cell of edge length a, then the mass and volume per unit cell are Mass =
NA
M N Volume = a3
Therefore, Density = volume
mass =
3NA
a NM
The value of N for the three cubic cells can be calculated as follows :
Primitive cubic cell : In a primitive cubic cell, atoms are present at the corners of the cube. There are eight corners of a cube and thus eight atoms are present at these corners. Now, any particular corner of the cube is actually shared amongst eight such cubic unit cells placed adjacent to one another. Thus, the contribution of the atom placed at one of the corners to the single cubic unit cell is 1/8. Since there are eight corners of a cube, the number of atoms associated with a single primitive unit cell is 8/8 = 1.
Body-centred cubic cell : In a body-centred cubic unit cell, besides atoms being present at the corners, there is one atom in the centre of the cube which belongs exclusively to this cubic unit cell. Therefore, number of atoms per unit cell are two.
Face-centred cubic cell : Here, atoms, besides being at the corners, are also present at the centre of the six faces. Each of these atoms is shared between two such unit cells. Thus, their contribution to the unit cell is 6/2 = 3 atoms, making a total of 4 atoms per cubic unit cell
Packing in a simple Cubic Lattice :
In a lattice of this type, the spheres are packed in the form of a square array by laying down a base of spheres and then piling upon the base other layers in such a way that each sphere is immediately above the other sphere, as shown in fig.
Packing in a simple cubic lattice
In this structure, each sphere is in contact with six nearest neighbours (four in the same base, one above and one below). The percentage of occupied volume in this structure can be calculate as follows:
The edge length a of the cube will be twice the radius of the sphere, i.e. a = 2r. Since in the primitive cubic lattice, there is only one sphere present in the unit lattice, the volume occupied by the sphere is
V =
3
4πr3 or V =
3 4π
3
2 a
The fraction of the total volume occupied by the sphere is
φ = 3
3
a 2 a 3
4
π
= 6
π = 0.5236
or 52.36 percent
Thus, the structure is relatively open since only 52.36% (π/6) of the total volume is occupied by the spheres. The remainder, i.e. 0.4764 of the total volume is empty space or void volume.
No crystalline element has been found to have this structure.
Closest Packing :
In closest packing arrangements, each sphere is in contact with the maximum possible number of nearest neighbours. Fig. shows a closest packed layer of spheres. Each sphere is surrounded by six nearest neighbours lying in the plane, three spheres Just above it and three below it, thus making the total number of nearest neighbours equal to twelve.
If the spheres are packed in the same plane, then just above these spheres
A B
A A A
B B B
C C C C
A A A
A A A
A
A
A
A
B B
Fig. (a) Closest packed layers of spheres B
C C C C