Here’s the general mechanism for 1,2 and 1,4 addition to α,β-unsaturated compounds and some points on how to decide which one is going to happen.

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Direct and Conjugate Addition to α,β-Unsaturated Compounds by Ayraethazide is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Base Vs Nucleophile

While studying organic chemistry, we often use the terms base and nucleophile. For example, if we consider hydroxide ion (\(HO^-\)), it can act both as a base and as a nucleophile. Then, what is the difference between a base and a nucleophile? How are we going to decide if the outcome of a reaction between an alkyl halide and a negatively charged (at times neutral) species is elimination or substitution?

In simple words, the negatively charged species would have acted as a base if it had formed a bond with a proton at the end of the reaction and it would have acted as a nucleophile if it had formed a bond with a carbon atom at the end of the reaction.

It has to be noted here that along with the difference between a base and a nucleophile in the way they react with a substrate as stated above, the scale used to measure their strengths are also different.

To understand this, let us consider the following equilibrium,

\[AcO^- + H_2O \rightleftharpoons HOAc + HO^-\] 

In the above equilibrium, there are two bases, \(ACO^-\) and \(HO^-\). Which out of these two is a strong base? Is it \(ACO^-\) or \(HO^-\)? The answer for this question can be arrived at with a simple logic that if an acid is strong its conjugate base should be weak. Here in the above case, Acetic Acid is a stronger acid than Water and hence acetate ion \(ACO^-\) should be a weaker base than hydroxide ion \(HO^-\).

Now, consider the following equilibrium,

\[CN^- + H_2S \rightleftharpoons HCN + HS^-\]

Similar to the earlier case, this equilibrium is also having two bases \(CN^-\) and \(HS^-\). Which out of these two is a strong base? The situation now is not as simple as it was earlier. In the earlier case it was a comparison between Water and Acetic Acid and it was very obvious that acetic acid was the stronger acid among the two. Here the comparison is between \(HCN\) and \(H_2S\) and it not very obvious. We need the help of some physical constants.

Let us say we are allowed only one constant. What is that one constant that we will want to conclude which among the two is a strong base? Let us see. We want to know which out of the two is a strong base. A strong base should be capable of abstracting even the weakly acidic hydrogen from a molecule and hence would be present mostly in the form of its conjugate acid. So, if \(CN^-\) was a strong base among the two, the concentration of it will be very less compared to that of HCN. Similarly as \(CN^-\) abstracts the proton from \(H_2S\) the concentration of \(H_2S\) will also be less compared to that of \(HS^-\). As the concentrations of the reactants are less than that of the products, the equilibrium constant will be less than 1. This is the constant that we are looking for.

If we know the equilibrium constant we can tell the direction in which the equilibrium is highly shifted and hence will be able to conclude on the strength of the base. The equilibrium constant is a way of finding out about the stability of the species. The equation \(\Delta G^o = -RTlnK\) gives us the relation between the standard free energy change and the equilibrium constant. Higher the equilibrium constant stabler the products than the reactants.

The above discussion goes to show that Basic Strength is a Thermodynamic Property. it depends on the stability of the species. A weak base is a stable one and a strong base is an unstable one.

Let us now turn our focus on the Nucleophile. Let us say a nucleophilic substitution reaction is performed. In fact two of them are performed. They are as shown below,

\[1. \ RX + CN^- \rightarrow R-CN + X^-\]

\[2. \ RX + HO^- \rightarrow R-OH + X^-\]

If both the above reactions were performed in the same reaction conditions and if the reaction 1. gets over in 7 minutes and the reaction 2. gets over in 10 minutes, which out of \(CN^-\) and \(HO^-\) should be a strong nucleophile? Most of us will conclude that \(CN^-\) should be the strong nucleophile because it reacts faster while all other conditions are same. Most of us will be correct. \(CN^-\) indeed will be the strong nucleophile among the two.

A Strong Nucleophile is the one that is able to attack the electrophilic carbon faster and a Weak Nucleophile is the one that takes lot of time to attack the electrophilic carbon. This goes to show that Nucleophilic Strength is a Kinetic Property. It depends on how fast or slow the nucleophile is capable of attacking the electrophilic carbon.

There is another way to predict if a species will act as a base or a nucleophile in a reaction. Let us take the example of Fluoride, \(F^-\) and Iodide, \(I^-\). \(F^-\) is a smaller ion and \(I^-\) is a larger ion. \(F^-\) can not be easily polarized while \(I^-\) can easily be polarized. Since \(F^-\) can not be easily polarized it will be a hard species and since \(I^-\) can easily be polarized it will be a soft species. A hard species prefers to form a bond with another hard species and a soft species prefers to form a bond with another soft species.

An Alkyl Halide will have acidic hydrogen as well as electrophilic carbon. Hydrogen will be harder than carbon because of the size. Hydrogen being smaller can not be polarized while carbon can be polarized because of it being larger. So, when an alkyl halide and \(F^-\) ion are allowed to react, \(F^-\), a hard species, will prefer to form a bond with the hydrogen, a hard species. This will lead to elimination as \(F^-\) is preferring to act as a base and pull the acidic hydrogen out. On the other hand, when an alkyl halide and \(I^-\) are allowed to react, \(I^-\), a soft species, will prefer to form a bond with the carbon, a soft species. This will lead to substitution as \(I^-\) is preferring to act as a nucleophile and attack the electrophilic carbon.

There is another way of looking at this. Let us say we have \(C_2H_5O^-\) and \(C_2H_5OH\). Which out of these two is the stronger nucleophile? If you have answered \(C_2H_5O^-\) is the stronger nucleophile among the two, you are correct. If you have to explain why \(C_2H_5O^-\) is stronger than \(C_2H_5OH\), what will your answer be? Think for a couple of minutes and comment your answer either here or on our facebook page. We’ll continue this discussion in the next post.

Nucleophile vs Electrophile

Nucleophile: is negative so is attracted to positive parts of molecules
has an electron pair it can donate
example reaction: (OH)-ion added to halogenoalkane, displacing halide ion to for alcohol. 

Electrophile: is positive so is attracted to negative parts of molecules, like a double bond (electron lover)
wants to accept an electron pair 
example reaction: Hydrogen bromide (H is delta +ve due to dipolar bond) added to alkene (with double bond) to form halogenoalkane. 

DNA Alkylation

Cellular DNA is constantly subject to damage by various intracellular and extracellular chemicals. Among these chemicals are alkylating agents, which transfer an alkyl group to a position in DNA. These chemicals can be cytotoxic and mutagenic, as alkylation of DNA, from simple methylation to the transfer of bulky alkyl groups, can cause polymerase error as well as blocking or stalling of polymerases. This can result from distortions in the helical geometry brought about by the alkylated DNA lesion, or by an inability of the polymerase to replicate past the lesion, even in the absence of significant helical distortion. The nucleobase guanine is especially vulnerable to alkylation by electrophilic alkylating agents due to its electron-rich character.

The simplest alkylating agent is methyl methanesulfonate (MMS) which delivers an electrophilic methyl group to nucleophilic sites on DNA (A above). This occurs primarily at the N7 of deoxyguanine, but also occurs at other nucleophilic sites in DNA such as the N3 of deoxyadenosine and, to a lesser extent, at other nitrogens and oxygens on DNA bases. Alkylation by MMS is believed to exhibit its cytotoxicity by stalling the replication fork. Despite its widespread use in research settings, MMS is not used therapeutically due to its high mutagenicity and nonspecific cytotoxicity.

The first alkylating agents used in a therapeutic capacity were the nitrogen mustards. These agents were derived from the notorious chemical weapon, mustard gas, which contains a sulfur instead of a nitrogen. They were initially developed as chemical weapons, but their therapeutic potential was recognized by researchers at Yale University during World War Two. The accepted mechanism for cytotoxicity derives from DNA alkylation and crosslink formation by these agents. The electrophilicity of these agents is increased drastically by a phenomenon known as neighboring group participation. The intramolecular nucleophilic attack by the Nitrogen leads to the formation of the aziridinium cation pictured above in figure B. This increases the electrophilicity of the C1 carbon, making it more prone to nucleophilic attack and greatly enhancing the rate of nucleophilic attack by N7 of guanine. Remember that intramolecular reactions happen far more quickly than intermolecular reactions, meaning that formation of the aziridinium cation is much faster than nucleophilic attack of guanine or water on the primary alkyl halide. Nucleophilic attack by the N7 of guanine results in alkylation of the nucleobase. Subsequent formation of another aziridinium cation and attack by another guanine, results in the formation of a DNA crosslink. This can occur with a guanine on the same strand or on the opposite strand. These bulky crosslinks are a severe block to polymerases, resulting in a failure to replicate or transcribe DNA, leading to cell death via apoptosis if the lesion is not repaired. Many nitrogen mustards, such as cyclophosphamide and chlorambucil are still in use today.



Methylamine CH5N

  • Appearance: Colorless gas
  • Molar Mass: 31.06 g/mol
  • density: 694 kg/m3
  • Melting Point: -93.10 °C
  • Boiling Point: -6.6°C 
  • pKb: 3.36

Methylamine is a derivative of ammonia and is the simplest primary amine. It’s commonly sold as anhydrous gas in metal containers. It’s a relatively easy compound to synthesis in lab, being a simple reaction between methanol and ammonia. Methylamine is a great nucleophile that’s also highly basic with little to no hindrance. Liquid methylamine has solvent properties similar to ammonia. It is toxic and listed as a List 1 Precursor chemical by the DEA since it’s well known for it’s use in the production of methamphetamine. 

So here is the full solution to that problem that was sent in by cookiemonsteryuum (see my previous post)

Here’s the problem;

Here’s how to find reagent S;

So now the hydrogen cyanide will react with the butanone;

At about pH 12 the hydrogen cyanide will be 99% deprotonated and it will act as a good nucleophile to the carbonyl group. This nucleophilic addition then gives Compound T 
Compound T is a racemic mixture however, with the two enantiomers represented as follows;

Using the Newman projection allows us to view how the nucleophilic cyanide molecule attacks the carbonyl (if you are struggling to visualise the molecules in its different projections, use molecular models). I’ve colour coded the butanone molecule to help keep track of the different groups

Now we react the cyanide with the molecule, while seeing the two different approaches it may take;

So that’s why two enantiomers form from this nucleophilic addition reaction. In fact the butanone molecule can be called prochiral since it converts to a chiral product in a single step. 
So now we dehydrate the racemic mixture of Compound T, to give Compound U. The dehydration of alcohols gives alkenes;

In fact two alkenes may form and using Zaitsev’s Rule, the major and minor products can be predicted - giving us Compound U.