THE ADVANTAGE OF TERMOLOGY

Termology facilitates communication of a concept. In a discussion. When you use a well-used term your audience would immediately know what you are talking about. This tool is used in all disciplines - chemistry, mathematics, physics, biology, economics, religion, etc. Of course the disadvantage is that it is meaningless to those not familiar with the topic; sometimes even for chemists not in a specialised branch of chemistry. Here we will discuss two terms: Nucleophilic Substitution (S_N) Reaction and Elimination (E) Reaction.

NUCLEOPHILIC SUBSTITUTION REACTIONS, SN

Carbon bonded to a more electronegative atom will always be slightly positive owing to the unequal sharing of the valence electrons forming the bond. Because the carbon is slightly positive (δ‾) it is opened to attack by other anions present in the system.

Z ‾   +   RX   �   RZ   +   X ‾

Z ‾ is then known as a nucleophile; a Greek term equivalent to nucleus lover. The nucleus in this context refers to the carbon. This class of reaction is known as Substitution by a Nucleophile, or S_N in short.

ELIMINATION REACTIONS

In some reactions instead of replacing one nucleophile (X) by another (Z), the product obtained is the lost (or elimination) of HX. The name for this class of reaction is an Elimination Reaction, E. (What else do you want to call it).

Substitution and Elimination reactions dominant a major part of Organic Chemistry reactions.

UNIMOLECULAR REACTION

Energy profile for the formation of carbonium ion

In a reaction the elementary step in the mechanism with the highest activation is known as the rate determining step. This is because once the system has sufficient energy to climb this activation energy barrier (Eact), the other steps will not present a problem. So the rate of the reaction is very much dependent on how fast this step can proceed.

When the rate determining step is the dissociation of a substituent from the molecule (to form the carbonium ion), that is involving only one molecule, it is known as a unimolecular reaction.

• SN via UNIMOLECULAR REACTION, S_N1

  When the unimolecular reaction is followed by the bonding of the carbonium ion with a nucleophile, the reaction is known as a Substitution by a Nucleophile via a Unimolecular Reaction, SN1.

  Note: I will write S_N1 as SN1 and S_N2 as SN2

  The driving forces for the SN1 reaction has a push factor and a pull factor. What do I mean by a push-pull factor? Just imagine you go to a departmental store to shop. You saw an item you like. This can be considered a push factor. Then the item is on sale. This becomes a pull factor.

  The factor pushing for a SN1 reaction is a weak C−X bond. The pull factor is the stability of the carbonium ion. The production of a tertiary carbonium ion encourages a SN1 reaction, as compared to a primary carbonium ion.

• ELIMINATION via UNIMOLECULAR REACTION, E1

  However if the carbonium ion proceed to eject a proton to form an alkene, it is known as an Elimination via a Unimolecular reaction, E1.

Carbonium ion chemistry is encouraged by polar solvents since it would help to stablise the ion.

Another feature is the products formed via carbonium ion are not stereospecific. So all unimolecular reactions would give products of mixed configuration.

+ Z �

+

For SN1 reaction the product is formed when the carbonium ion reacts with the attacking nucleophile. The carbonium ion is planer and the empty orbital can be attacked from either side of the plane. So the products expected is a mixture of two enantiomers.

For E1 the alkene formed is determined more by its stereo stability. The trans-isomer is preferred over the cis-isomer.

• SN via BIMOLECULAR REACTION, S_N2

  Stereochemistry and energy profile for a SN2 reaction

  This class of reaction where a nucleophile forcefully substituted an attached substituent is known as a Bimolecular Nucleophilic Substitution, S_N2. Bimolecular because the rate determining step (the elementary step with the highest energy barrier) involves two molecules; RX and Z‾.

  The driving force for a SN2 reaction is the "brute force" of the attacking nucleophile. It has the philosophy of "I came, I saw, I take". Of course this also implies that X is not strongly bonded to the carbon.

  It should be clear that the attack from the rear would not be easy for tertiary alkyl halides. The nucleophile will have to squeeze through the small "tunnel" (the yellow portion in the diagram) surrounded by bulky R1, R2, and R3 bonded to the carbon. Always remember all CC σ−bonds are about 109� apart.

  CH3-X 1,500		CH3CH2-X 50		(CH3)2CH-X 1
  	CH3CH2CH2-X 20

  This problem increases with increase in size of any of the alkyl groups. The chain will fold over the tunnel and completely block it making it impossible for any nucleophile to squeeze through to the carbon. Of course this is a simply picture. We must also remember the σ−bond is constantly rotating about the carbon−carbon axis, and so chain will swing round and round providing windows of opportunity.

  The relative rates presented above (1,500 : 50 : 3 and 20) are experiment results.

  STEREOCHEMISTRY

  �

  For SN2 the attacking nucleophile has to force the X to leave by attacking the reverse side and so produces only one enantiomer.

  Note: We are assuming that the starting alkyl halide has a chiral carbon and only one of the enantiomeric alkyl halide is used. If a mixture of alkyl halide is used then the products will still be a mixture of enantiomers regardless of the type of SN reaction.

• BIMOLECULAR ELIMINATION REACTION, E2

  Stereochemistry and energy profile for an E2 reaction

  When there is a proton at the β-carbon to a nucleophile, a strong base would behave as a base, instead of a nucleophile. It will forcefully pull out the proton causing the substituent X to be eliminated.

  
  
  
  
  

  This reaction is classified as Bimolecular Elimination reaction, or E2 reaction.

  A common reagent used is a potassium hydroxide / alcohol mixture to give the potassium alkoxide, a strong base.

  STEREOCHEMISTRY

  It must be noted that the elimination would only take place when the nucleophile is anti to the proton. This is necessary as when both groups are leaving, their sp�-orbitals are simultaneously transformed into p-orbitals to form the π−bond. So E2 is a stereospecific reaction.

  Let us consider the elimination reaction for the secondary 1-methylbutyl bromide; CH₃CH₂CH₂CH(Br)CH₃. The various conformation possible for this compound would be.

  (A)

  (B)

  (C)

  (D)

  The most stable conformations are (A) and (B) as the bulky groups (Br and ethyl) are furthest away from each other. The least stable will than be conformation (C). (A) is not anti so E2 Reaction is not possible. So reaction can only occur with (B), (C) and (D).

  Experiments showed that the products for this reaction is in favour of the conformation (B) and (C) to give 2-pentene as the major product.

  CH3CH2CH2CH(Br)CH3

  EtOK / EtOH � 55�C

  CH3CH2CH2CH=CH2 1-pentene (31%)

  +

  CH3CH2−CH=CH−CH3 trans-2-pentene (52%) ; cis-2-pentene (17%)

  �

  trans-2-pentene

  �

  cis-2-pentene

  There are actually three (A) species since each proton is counted as one conformation. If the E2 Reaction is based strictly on anti conformation population than the major product should be 1-pentene. Thus the results showed that for this reaction the elimination is influenced more by the energetic of the products formed. An alkene with more alkyl groups at the carbons of the π−bond will have lower energy.

  A E2 Reaction on (B) will give a trans-2-pentene, while (C) will give cis-2-pentene. As expected there will be more trans-2-pentene then cis-2-pentene from steric consideration.

  The amount of 1-pentene can be increased if a t-BuOK / t-BuOH mixture is used. The bulky tertiary-butoxide would find it hardly to squeeze through to reach the proton at C3. The ratio is then increase to 1-pentene / 2-pentene = 8.1.

Quaternary ammonium hydroxide can also undergo a E2 Reaction

CH3CH2CHCH3              I            +N(CH3)3   OH‾

100�C �

CH3CH2CH=CH2 95%

+

CH3CH=CHCH3 5%

+

N(CH3)3   +   H2O

Unlike the alkyl halides the product is almost exclusively 1-butene.

(A)

(B)

(C)

(D)

An inspection of the conformation will show that (A) is the most stable but being non-anti it will not undergo E2 Reaction. (B) and (C) is now extremely difficult as the ammonium group is very very bulky compared with the bromide radical. So we can assume that most of the time the molecule is present as (A,D). This would explain why the product is almost exclusively 1-butene.

SN2 versus E2

Although under certain conditions the anionic (Z‾) will behave as a base, a chemist cannot ignore the fact that it can react as a nucleophile and take part in SN2 reaction.

For the above reaction the products obtained were.

1-methylbutyl bromide + EtOK / EtOH

� 55�C

trans-2-pentene (41%)  cis-2-pentene (14%)   +   CH3CH2CH2CH(OEt)CH3 (20%) 1-pentene (25%)

Strong base such as alkoxide favour elimination. The type of α-carbon and β-carbon is also important. For primary carbon even strong base like alkoxide would find it difficult to execute a E2 reaction. But if the α-carbon and β-carbon are secondary carbons E2 would be more prominent. However if the α-carbon is a tertiary carbon then SN1 might become the main reaction.

SUMMARY

 −CH−CX− + Z
    β     α

REACTION	α-carbon	β-carbon	Products
SN1	tertiary		racemic*
SN2	primary		stereospecific
E1	tertiary	tertiary	trans
E2		tertiary	stereospecific
*Racemic mixture is a mixture of enantiomers.

SOME NUCLEOPHILES / BASES

Hydroxide ion Alkoxide ion Terminal acetylene

Tutorial 6.6

When is an anion a base or a nucleophile?.      Answer