KWOK The Chem Teacher

Bimolecular Nucleophilic Substitution

noreply@blogger.com (Mr Kwok) — Tue, 08 Nov 2011 23:23:00 +0000

The following video clip is an updated version of the bimolecular nucleophilic substitution reaction. It makes use of a generic nucleophile which is an anion reacting with a generic alkyl halide.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Ionic Equilibrium - use of solids as indicators

noreply@blogger.com (Mr Kwok) — Fri, 19 Aug 2011 00:14:00 +0000

Indicators are used for you to help us know that a particular reactant is completely used up for a reaction. Through acid-base titration, we have seen that indicators inform us that a reactant is completed used through a change in colour of the solution. These indicators first create a colour in the original solution and when the end-point is reached, they will show a change in colour.

However, a colour change in the solution is not the only way to indicate that the reaction is completed. Another way is to form a coloured solid while doing a titration. An example would be to determine the concentration of Cl^- ions present in a solution by titrating the solution against a standard solution of AgNO₃.

(A) Scenario: How it is used.

When you add Ag⁺ ions to a solution containing Cl^-ions, a white precipitate is seen. This happens because the K_sp value of AgCl is very small and hence the product of the concentrations of Ag⁺ ions and Cl^- ions (also known as ionic product) can easily be larger than the K_sp. This results in precipitation of the AgCl solid. Chromate ions (CrO₄^2-) are added so that when a red precipitate of Ag₂CrO₄ is formed, you would that all the AgCl has been precipitated out. This enables us to determine how much Ag⁺ ions was added into the reaction mixture and as a result allow us to know the amount of Cl^- ions present initially.

(B) Explanation: How this works.

The numerical values of the K_sp of AgCl and Ag₂CrO₄ are 1.77 x 10^-10 and 1.12 x 10^-12. Although the K_sp value of Ag₂CrO₄ is smaller, only a small concentration of CrO₄^2- ions is added while a larger concentration of Cl^- ions is present. Hence, it is easier for the ionic product of AgCl to exceeds its K_sp than that of Ag₂CrO₄.

Furthermore, when the red precipitate of Ag₂CrO₄ is formed, we can be quite certain that all of the Cl^- ions are precipitated out. This is because of the following:

Although K_sp of the Ag₂CrO₄ is of a smaller value, the K_sp expression of silver chromate is essentially a concentration³ term. This is in comparison to silver chloride whose K_sp expression is a concentration². Hence, the solubility of silver chromate is greater than that of AgCl. This implies that the former is more soluble than the latter and it is quite fair to say it is more difficult to precipitate out too.

In addition, for the same small concentration of Ag⁺ ions present (since both solids are insoluble you do not expect a large concentration of Ag⁺ ions), a larger concentration of CrO₄^2- ions is present in the solution than Cl^- ions. This would mean that a greater percentage of the Cl^- ions is precipitated out.

(C) The steps taken: The mathematics behind solving for the percentage of Cl^- ion precipitated.

When Ag₂CrO₄ is first formed, we have obtained a saturated solution of Ag₂CrO₄. Hence, by solving the K_sp of Ag₂CrO₄, we can obtain the concentration of Ag⁺ ions left in the solution. As both the equilibria which show the dissociation of AgCl and Ag₂CrO₄ exist in the same system, the concentration of Ag⁺ ions is the same for both equilibrium. It is not possible to obtain two distinct concentration of Ag⁺ ions (one for each insoluble solid doing slight dissociation) when both solids exist in the same solution. Therefore, the concentration of Ag⁺ ions is common for both ionic solids.

Hence, with this piece of information, when we can solve for the concentration of Cl^- ions in the solution using the K_sp expression of AgCl. This turns out to be a very small number which suggest almost all of the Cl^- ions has been precipitated out of the solution, therefore giving us an endpoint.

(D) Conclusion: Final thoughts.

Curiously, is it possible to have absolutely no Cl^- ions present in this context? The answer is no . As the solid AgCl remains in the reaction mixture, some would definitely dissolve giving rise to a concentration of Cl^- ions in the solution.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Patterns in Chemistry

noreply@blogger.com (Mr Kwok) — Sun, 14 Aug 2011 11:42:00 +0000

I must confess that I was never fascinated by Chemistry; at least not when I was a 15 years old. I was more intrigued by numbers and I was fascinated by how the lung and the heart works. However, it was during my years in junior college that made me fond of the subject as I saw that beneath the massive number of equations and facts, lies interesting patterns for me to identify and be acquainted with. This was a challenge that I enjoyed: It grew to become a fervent interest and ultimately a career.

As a 15 years old student, I remembered that the colours and the smells were something that made this subject memorable. But as much as there are many things that I could rely on my senses to help me appreciate the subject, there were even more that were beyond the capacity of my senses to reach out to. This made it hard for me to relate with the abstract facts. However, I discovered that learning Chemistry often require one to be able to spot that pattern; for most of the time, we are not faced with questions on the factual information but rather questions that required us to apply the knowledge. Hence, recognising patterns is important.

I could remember the difficulties that my peers faced in studying Organic Chemistry. It was a similar fear that some of my students experienced. I empathise their difficulty as this topic has numerous equations, conditions and reagents to put into memory. Simply, this topic was a string of roman letters which form words that seemingly look English but is not English; yet it has its own meaning. Essentially, Organic Chemistry could be seen as a language on its own.

However, the success in learning this “language” is simply being able to identify the patterns that are involved and relate them to other patterns or principles that one has prior knowledge of. There are only polar and non-polar covalent bonds. The polar bonds are broken during “nucleophilic substitution” and “nucleophilic addition” reactions. While, the non-polar ones are broken in either “free radical substitution”, or “electrophilic substitution” or “electrophilic addition” reactions. With the aid of the “dot and cross” diagram and the definitions of “nucleophile”, “electrophile” and “free radical”, I was able to identify that the first has a lone pair of electrons available for donation, the second has an empty and energetically accessible orbital to accept a pair of electrons, and the last has an unpaired electron. Lastly, if a stronger covalent bond is broken, we would expect more severe conditions needed for the reactions to occur.

When I am given an unknown equation, instead of trying to dig out when and where I memorized this equation, I chose to ask
What is the polarity of the bond that is most likely to be broken?
Could the other species be either a nucleophile, or an electrophile or a free radical (It can’t be all three and if it were, you will be stuck.)?
Is the covalent bond broken relatively strong or weak to those that I know?

Hence, the subject becomes less frustrating and more student-friendly.

However the beauty of this subject lies in the twists that it provides. I guess this was ultimately the reason that drew me toward its. I was fascinated by (instead of frustrated by) the weakness of the patterns which allows me to predict all the cases but yet seemingly insufficient to include absolutely everything within it. That motivated me to improve the definition of my pattern to make it more concise (maybe even precise) and hopefully more accurate until the next factual information that weakens this pattern, and the cycle repeats.

noreply@blogger.com (Mr Kwok) — Sat, 06 Aug 2011 02:53:00 +0000

The post with regards to drawing a titration curve was updated. Click here to view it.

Ideal Gas - Assumptions

noreply@blogger.com (Mr Kwok) — Sat, 16 Jul 2011 16:13:00 +0000

The ideal gas laws resulted in the following ideal gas equation:

PV = nRT

The basis of the equation is on the following three points. For the same number of moles of gas particles:

The pressure of the gas is directly proportional to temperature, provided volume of gas is constant.
The volume of the gas is directly proportional to the temperature, provided temperature of the gas is constant.
The pressure of the gas is inversely proportional to volume of the gas, provided temperature of the gas remains constant. This makes sense as when I increase pressure, the volume of space that the gas particles can move around in becomes smaller.

These three laws result in the two assumption of ideal gases:

The intermolecular forces of the gas is insignificant.
The size of the gas particles is negligible (or the volume of the gas molecule is negligible)

However, real gas deviates from these assumption because they have intermolecular forces, either Van der Waals' interactions or Hydrogen-bonding. In addition, the size of the gas particles is not negligible - the gas molecule has a volume and it is not a point particle. Therefore, when ideal gas is subjected to increasing pressure, the PV term remain constant. However, when real gas is subjected to increasing pressure, we first observe a dip and then subsequently the PV value soars.

(1) The dip in PV value for real gas.

This can be reasoned. When the pressure starts to increase and it is not sufficiently high, the gas would come close together. This provides the opportunity for the individual gas particles to interact with each other as they are able to cause their intermolecular forces to act upon each other. Once the intermolecular forces act that serves as an additional pull that brings the gas particles closer together and hence the volume of the gas is smaller than predicted hence the PV term is smaller, thus resulting in the dip.

The sheer fact that real gases have intermolecular forces, this is the reason why gases can be condensed to liquid state. The absence of intermolecular force will mean that it is impossible for gases to liquidfy.

(2) The PV value soars.

At a given high pressure imposed on an ideal gas, the PV term remains the same as when the gas experiences a lower pressure. As gases can be compressed, the volume of the gas (which can be seen as the volume of the space the gas particle is traveling in) cannot be directly obtained. It is actually approximated from the volume of the container that contains the gas. Although, the volume of the container consists of the volume of the space the gas can move in and the volume of the molecules, this "guess" is perfectly fine for ideal gas as the particles are point particles (hence the particles have no volume). Thus, the volume of the container is a good approximate to the volume of space that gas particles can travel in.

However, for real gases when it is subjected to high pressure (and if it still remains as a gas), the size of the particles become significant as compared to volume of the space the particles move in. Hence, for a fixed container, the significant size of the particles reduces the volume of the space that the particles can travel in and hence the volume of the gas is actually smaller than the actual volume of the container. Since the volume of the container is used in the calculation, this results in the PV to be larger than expected.

The molecular interpretation of this increase can be seen when we have the gas particles so close together that their respective electron cloud comes into contact with each other. This results in a repulsion and "pushes" the particles away from each other, increasing their volume.

As we deal with the environment which makes up of a mixture of different gases, it useful to remind ourselves the purpose of Dalton's Law of partial pressure. The law states that when we have a mixture of gases which do not interact chemically with each other, the total pressure of the gas mixture is the sum of the individual partial pressure of the different gases that make up the mixture.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Physical Periodicity - Application of 1st Ionisation Energy trend

noreply@blogger.com (Mr Kwok) — Sun, 01 May 2011 12:06:00 +0000

The periodic table often provides us with much information, such as reactivity and physical properties of the different elements. The grouping of these elements is really an enlightened and detailed classification of the numerous elements found on earth.

In physical periodicity, much is mentioned about the trends in physical properties along a period. The following video aims to show how the first ionisation energy trend can be used to predict an unknown species which is found on a nth ionisation energy trend. You will be pleasantly surprised by the pattern(s) that we can observe from trends in the ionisation energies.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Expansion of Octet. Why do elements do it?

noreply@blogger.com (Mr Kwok) — Sun, 17 Apr 2011 12:54:00 +0000

Most of the elements do not have the noble gases' electronic configuration. Since the latter is generally unreactive, we have come to the conclusion that the octet configuration is stable and that the other elements would aim to achieve it. These elements can only achieve the octet configuration via chemical bonding; either forming of a covalent bond or an ionic bond with another atom.

However, we also noticed that there are compounds which clearly show that the element can form more bonds such that they have more than the 8 electrons around it. Compounds such as PCl₅ and H₂SO₄; where phosphorous and sulfur have 10 and 12 electrons around them respectively. Glaringly, all the elements which are found in the first and second period of the periodic table does not form compounds in which they have more than 8 electrons around them.

Therefore, these observations together with the knowledge of quantum shells, subshells and orbitals, we are able to come to the conclusion that elements found from Period 3 and beyond, have available empty orbital which must be energetically accessible to accept electrons for bonding. Thus, these elements are able to accept more electrons to bond with other elements. This results in the expansion of octet. In the case of PCl₅, phosphorous must have caused one of its valence electron to be excited to the 3d subshell (its energetically accessible orbital), so that it has 5 orbitals available for bonding and hence able to form a single bond each with 5 chlorine atoms.

However, is it that random, is it simply that elements which have energetically accessible orbital will exercise that privilege to use them? How do we account for compounds such as XeF₂? Xe a clear noble gas element and yet it is willing to expand its octet, why would it do so? The expansion of the octet results in the elements to be able to make use of their energetically accessible empty orbitals. These orbitals are energetically slightly higher than the orbitals which the valence electrons are found. Hence, if the element wants to expand its octet, it will excite some of its valence electrons to the empty and energetically accessible orbital, so that they can form a bond with another atom via equal sharing of electrons or accept a pair of electrons from the donor to form a dative bond. If this is a favourable process, the energy from the formation of the bond must have compensated the energy needed to move a valence electron out of it the orbital that it resides in.

This is the complete reason to why the Period 1 and Period 2 elements do not expand their octet. Their next empty orbital belongs to a different principal quantum shell. Hence, to make use of it for bonding would require a large amount of energy, and it is not compensated by the energy formed from the bond formation. Therefore, the combination of these two reasons allow us to observe expansion of octet in some elements and not all of them.

Interestingly, the cation of an ionic compound generally do not favour expansion of octet which results in formation of another neutral compound. (For example adding more F₂ to MgF₂ to form MgF₄.) This is because it results in Mg to use more than 2 valence electrons for bonding. This is energetically meaningless for the atom as it will have to invest in large amount of energy to make use of an inner shell electrons for bonding. Hence, if cation wants to expand its octet, then it needs to have energetically accessible empty orbital to accept electron pairs from a donor.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Theory: Bond energy of A - B is the average of Bond energies of A-A and B-B

noreply@blogger.com (Mr Kwok) — Sun, 10 Apr 2011 13:25:00 +0000

Let's try to make sense to one of the many theories in chemical bonding. We understand that the bond energy represents the amount of energy that is needed to break one mole of a covalent bond. Hence, the larger the bond energy, the stronger the bond. Interestingly, one of the theories in chemical bonding suggests that the bond energy of A-B is the average of the sum bond energy of A-A and the bond energy of B-B.

In this write-up, I will propose my interpretation to how this theory is obtained and hence using this as an angle to account for the anomalous situations. The Molecular Orbital (MO) theory best explains how the covalent bond is formed and this theory succinctly focuses on the overlapping of atomic orbital (Click here for some explanation to MO theory). While, at the GCE "A" level, we learn about overlapping of orbital for the formation of covalent bond; this is basically the fundamental and not the complete picture of MO theory. Nonetheless, this simple idea used in the "A" level is still competent to account for the theory of this topic.

Two assumptions are made when we talk about overlapping of atomic orbital to form covalent bond using the "A" level idea. These will help to explain the theory in discussion:

We assume that only a certain proportion of the atomic orbital will be used for bonding. Hence, when small atoms overlap with each other, the proportion of the overlap over the total area of the orbital is larger than for big atoms - that is where we obtain the term "effective overlap of atomic orbitals" to determine our strength of our covalent bonds. Strength of covalent bond is then told to us via bond energy.
We assume that the atoms will always only make use of that same particular portion of their atomic orbital for bonding, regardless of the identity of the other atom.

These two propositions will suggest that if we half the bond energy of A-A, we can infer the area of the atomic orbital that the atom is using for bonding. This would be a fair suggestion, as a bond is made up of overlapping between two atomic orbital and the effectiveness the overlap (and hence the proportion used for overlapping) is told to us by the bond energy.

This explanation will nicely allow us to agree that the average of the sum of the bond energy of A-A and B-B will give us the bond energy of A-B.

However, when A and B is of different electronegativity, the electron cloud (which contains the bonded electrons) between A and B is pulled towards the atom which is more electronegative. You maybe incline to conclude that the area of the atomic orbital used for overlapping between A and B, may not be the same as A bonds with A and when B bonds with B (we can only ascertain this through theoretical calculations). More importantly, the difference in electronegativity results in an added electrostatic attraction between partial charges that are formed in A and B; not accounted by the merely overlapping of atomic orbital. The added electrostatic attraction creates our polar covalent bond. A covalent bond with partial charges at each end.

Deviation to this theory is due to the weakness of the theory. Essentially, it only accounts for a covalent bond formed between two atoms due to only the overlapping of atomic orbital. While, in the case of polar covalent bonds, we now know that there are two factors that result in this covalent bond to be formed:

Overlapping of atomic orbital and the constituent atoms; and
Electrostatic attraction of partial charges due to the electronegativity difference of the constituent atoms.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Atomic Structure - Aufbau Principle

noreply@blogger.com (Mr Kwok) — Fri, 25 Mar 2011 02:19:00 +0000

Writing electronic configuration of the various elements require us to make use of the Aufbau Principle. This is because it is not all the electrons found in the principal quantum shell have the same amount of energy.

In each principal quantum shell (except when n=1), there are subshells. These subshells do not have the same energy level. Hence, they are not filled at the same time.

Below is the video which summaries the different energy level as well as a short description to how Aufbau Principle works. This video will make use of nitrogen and manganese to demonstrate how the principle works in the writing of electronic configuration.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Double Indicator Titration

noreply@blogger.com (Mr Kwok) — Fri, 11 Feb 2011 04:11:00 +0000

When we have a mixture containing Na₂CO₃ and NaOH and you titrate it with a standard solution of HCl, you will need to use two indicators; one after another. The video below will explain how double indicators work in titration, using the above mentioned as the example.

An interesting question to take note. In what circumstances would the use of double indicator titration be limited?

Note: This video will assume that you do have an awareness the meaning of the following terms such as "working pH" and "equivalence point". They are not explicitly stated in the video.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Volumetric analysis - Titration

noreply@blogger.com (Mr Kwok) — Sat, 22 Jan 2011 08:06:00 +0000

Here is a video that shares about volumetric analysis, with a focus on titration calculations and choice of indicators.

Note: You may wish to take note that the equivalence point is the mid-point of the vertical portion of the titration curve. It was not explicitly stated in the video.

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^{Article written by Kwok YL 2011.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Commenting on enthalpy change of neutralisation.

noreply@blogger.com (Mr Kwok) — Sat, 23 Oct 2010 16:48:00 +0000

The definition of the enthalpy change of neutralisation is the heat evolved when one more of water is formed between an acid and an alkali. Essentially, it is the reaction between 1 mole of H⁺ and 1 mole of OH^- to give 1 mole of H₂O.

Typically, enthalpy change of neutralisation between a strong acid and a strong base is -57 kJ mol^-1. While, if it is between a weak acid and a strong base, or a strong acid and a weak base, the enthalpy change of neutralisation will be slightly less exothermic than -57 kJ mol^-1. This is largely because some amount of heat is taken in my the molecule to allow for complete acid (or base) dissociation.

1. Interesting question:

However, when we react 1 mole of H₂SO₄ with 2 moles of NaOH, do we get a more exothermic enthalpy change of neutralisation? The answer is NO.

This is because the reaction between sulfuric acid and sodium hydroxide produces 2 moles of water. This implies that the enthalpy change for this reaction is twice of that of enthalpy change of neutralisation; since the latter is defined to be per mole of water formed.

Therefore, when we obtain -114 kJ mol^-1 (which is more exothermic than -57 kJ mol^-1) for the reaction between 1 mole of sulfuric acid and 2 moles of sodium hydroxide, this value refers is the enthalpy change of reaction. Hence, when this number is divided by two because two moles of water is obtained, we will get enthalpy change of neutralisation.

2. Application - in terms of planning an experiment:

In conclusion, we often use calorimetry method to typical to determine the enthalpy change of neutralisation of an acid and a base. However, in using energetics in determining strength of acid, we are essentially trying to determine enthalpy change of reaction.

For example, when the reaction between sulfuric acid and sodium hydroxide is compare it with ethanoic acid and sodium hydroxide (and both acids are equimolar), it is important to ensure that the amount of hydroxide used is in excess. It is also good to ensure the volumes of hydroxide used are the same - this facilitates the comparison.

The former will produce an enthalpy change of reaction that is slightly more than twice of that of the latter; resulting in the former to have a temperature change that is twice as much as the latter.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Ionic equilibrium - deciding on end point

noreply@blogger.com (Mr Kwok) — Sat, 23 Oct 2010 04:33:00 +0000

Carbonic acid, H₂CO₃, is an example of a weak dibasic acid. As such, carbonic acid has two end points when titrated with NaOH using the a pH meter. Hence, it can be slightly confusing when deciding on which equilibrium to use to calculate the pH of the first end point.

1. Question:

Do we take equilibrium B, which shows us HCO₃^- dissociating into H⁺ and CO₃^2-? Or do we take equilibrium C, which shows us HCO₃^- functioning as a conjugate base?

The choice is simple, in equilibrium C, the equilibrium constant, K_b is larger than the equilibrium constant for equilibrium B. Therefore, it is likely that HCO₃^- will function as a base rather than an acid. Hence, the pH of the end point A, will give us an alkaline pH.

However, when we add more NaOH, such that we obtain point B. At that point, we have only HCO₃^- and CO₃^2-, thus to calculate the pH, we are essentially making use of equilibrium B. At that point, pH = -lg(4.8 x 10^-11) > 9. In addition, the addition of NaOH will inhibit the forward reaction of equilibrium C. Hence, only equilibrium B is occuring.

2. Application to similar patterns:

Therefore, when we are dealing with a species that has two possible equilibria, (i.e it can function as an acid or a conjugate base), we will need to see which equilibrium (the acid dissociation or the base dissociation) has the higher equilibrium constant. The equilibrium which has the higher equilibrium constant will be the one that we choose to calculate the pH at that equivalent point.

In conclusion, do you notice that I mentioned we will be able to see two equivalent points when we use a pH meter. Why do you think it is not suitable to use phenolphathelein to detect the two end points?

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

November Papers

noreply@blogger.com (Mr Kwok) — Mon, 27 Sep 2010 03:52:00 +0000

Here are some of my thoughts for Nov 2005 Paper 3.

Here are my thoughts for Nov 2004 Paper 3.

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Kwok YL 2010.

ionic equilibrium titration curves

noreply@blogger.com (Mr Kwok) — Fri, 16 Jul 2010 16:50:00 +0000

Titration curves are informative. They are able to give us information about the bascity of the acid, the volume of base required for neutralisation. the pH of the solution and the type of indicator which should be used There are a few situations to take note when we sketch a titration curve.

1. General points to take note.

In drawing the titration curve, we will need to take note what solution would be in the conical flask and what will be in the burette. If an acid is in the conical and the base is in burette, the pH curve will show an increasing pH as the solution in the burette is added into the conical flask.

In sketching the titration curves, the following are points to take note.

The shape of the curves (e.g is it moving from low pH to high? the number of equivalence points?)
The volumes where neutralisation has occurred (i.e. the verticals of the graph).
The volumes and the pH of the effective buffer region. This requires you to check which point the number of moles of weak acid is the same as its conjugate base (or vice versa).
For the final point, do not assume that the effective buffer capacity is before neutralisation.

2. Examples of weak dibasic acid and weak diacidic base.

Organic bases and acids are generally examples of weak base and weak acid respective. For example CH₃NH₂ and CH₃COOH respectively. The former is a base because N has a lone pair of electrons to form a dative bond with H⁺, hence making it a proton acceptor (thus a base). While the latter has a -COOH which enables a H⁺ (thus an acid) to be donated. The above are examples of organic moles which are dibasic acid and diacidic base.

Using dibasic acids as an example, these substances are able to donate two H⁺. However, the protons are not donated at the same time. The acid (malic acid = HOOCCH₂CH(OH)COOH) will dissociation once first to form HOOCCH₂CH(OH)COO^-, then the acid will dissociate a second time (from HOOCCH₂CH(OH)COO^-) to give ^-OOCCH₂CH(OH)COO^- and H⁺.

Each dissociation has its own acid dissociation constant with the first acid dissociation constant generally larger than the second. The easiest reason is that the second acid dissociation is the removal of a H⁺ from an anion; which will be more difficult than removing a proton from a neutral acid molecule.

3. Titrating a weak monobasic acid against a strong base

When CH₃COOH placed in a conical flask and NaOH is added to the acid. The pH increases as more NaOH is added and that is due to decreasing amount of the weak acid CH₃COOH and increasing amount of conjugate base CH₃COO^-. At the point where n(CH₃COOH) = n(CH₃COO^-), that is the maximum buffer capacity. At 2A cm³ of NaOH added, all the acid is used up and only the conjugate base is left, hence the pH is greater than 7.

4. Titrating a weak diacidic base against a strong acid.

When titrating a weak diacidic base against a strong monobasic acid, with both having the same concentration. The volume required for the monobasic acid to completely react with the two basic groups in the diacidic base is twice the volume of the base used. In addition, the effective buffer capacity will be 14 - pK_b values.

However, when we titration against a strong dibasic acid, the volume required by the dibasic acid is the same as the volume of diacidic base used. This is when both acid and base have the same concentration. The pH of the effective buffers remain the same.

5. Titrating a weak dibasic acid against a strong base.

When titrating a weak dibasic acid against a strong monoacidic base, with both having the same concentration. The volume required for the monoacidic base to completely react with the two acidic groups in the dibasic acid is twice the volume of acid used. In addition, the effective buffer capacity will be the pK_a values.

However, when we titration against a strong diacidic base, the volume required by the base is the same as the volume of acid used. This is when both acid and base have the same concentration. Note that the pH of the effective buffers remain the same.

6. Conclusion

Hence, when sketching the titration curves, it is good advice to think about how the titration is done. Do not rush through with fixed model, e.g. the point where maximum buffer capacity exists is always between the starting point and the equivalence points. Always fist ask yourself what is in the solution at any particular significant point on the curve.

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^{Article written by Kwok YL 2010. (edited in 2011)}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Transition metals - Colour of complex

noreply@blogger.com (Mr Kwok) — Sun, 11 Jul 2010 01:58:00 +0000

One of the properties of transition metals is its ability to form complexes that are coloured. The rich colours of blue in copper (II) sulfate and deep purple in potassium permanganate are distinctive characteristics of a transition metal.

Hence, the property is a resultant of the definition that transition metals are elements that is able to form at least one ion or compound that has an incomplete d subshell.

Essentially, the colours which we see from transition metal complexes are due to an electronic transition. An electromagnetic wave in the UV-visible frequency was absorbed such that an electron can be transferred from a lower energy level to a higher energy level. The complementary colour would thus be reflected and it is this colour that we observed.

In addition, there are two routes for electron transfer. (1) d-d transition and (2) charge transfer.

1. d-d transition.

As most of the transition metal complexes (for example [Cu(H₂O)₆]²⁺) which you see is in an octahedral geometry, we will be using that as our starting point for the discussion of d-d transition.

The 6 ligands are placed on the x, y and z axis. Hence, this results in the d_{x² - y²} and the d_z² to receive the most electronic repulsion from the ligands, hence they are moved to a higher energy level.

This splitting of d-subshell is essential. There is now an energy gap between the lower set of d orbital and the higher set of d orbital. In addition, there must be an incompletely filled d orbital at the higher energy level so that an electron from the d orbital that has the lower energy level can be transferred.

An electromagnetic wave within the visible frequency is absorbed and that must correspond the the energy gap between the set of d orbital. This will result in the complementary colour to be reflected.

Do note that splitting of the d subshells to two different energy levels is still relevant in non-octahedral geometries (e.g tetrahedral). It is just that it may not be in this fixed arrangement of three d orbital are in the lower energy level and two d orbital are in the higher energy level.

2. Charge transfer.

Essentially, to enable d-d transition, there must be electrons in the 3d subshell. However, for potassium permanganate, the Mn is in +7 state. The electronic configuration of Mn (VII) is 1s²2s²2p⁶3s²3p⁶. There is no electron in the 3d subshell. Hence, the colour of potassium permanganate is due to charge transfer.

An electromagnetic wave within the UV-visible frequency is absorbed and that must correspond to the energy gap between a filled orbital of lower energy level and an empty orbital of higher energy level. This results in an electron to be excited from the lower energy level to the higher energy level, causing a complementary colour to be reflected.

The details to which electron is being excited and where it is excited to is beyond to scope of this discussion.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Group II - Thermal decomposition

noreply@blogger.com (Mr Kwok) — Fri, 09 Jul 2010 16:05:00 +0000

Group II metal carbonates, nitrates and hydroxides are capable of decomposition to give the corresponding metal oxide and release CO₂, NO₂ and O₂, and H₂O respectively. The following are the equations.

MCO₃ (s) -> MO (s) + CO₂ (g)

M(NO₃)₂ (s) -> MO (s) + 2NO₂ (g) + ½O₂ (g)

M(OH)₂ (s) -> MO (s) + H₂O (g)

The ease of decomposition decreases as we go down the group. I will use the metal carbonates as an example to illustrate this trend. The reason used will be applicable to describe the ease of decomposition of Group II nitrates and hydroxides.

1. Charge density and Polarising power of cation.

The Group II metals share a common charge of +2, but down the group the ionic radius increases. Hence, the charge density of the cation decreases and its polarising power becomes weaker. Hence, the cation is less able to polarise the anion and more importantly is less able to distort the covalent bond found in the carbonate ion.

This distortion weakens the covalent bond and hence making the decomposition easier.

Interestingly, this reason is why Group I nitrates decompose to give MNO₂ and O₂ instead of forming MO, NO₂ and O₂ as products. Only LiNO₃ decomposes like a Group II nitrate. This is not surprising as Li and Mg have a diagonal relationship. Their small size and charge will lead to a high charge density which make Li⁺ more polarising than the other Group I cations.

2. Percentage change in lattice energy.

This reason is not often used as it requires some mathematical conceptualisation.

The decomposition of metal carbonates form metal oxides. The different between the lattice energy (LE) of the two ionic compounds is the sum of ionic radius. Since both compound have the same charges for the cation and that for the anions, we can conclude that the difference in lattice energy will determine whether the decomposition is more favourable or less.

A smaller cation will have a larger percentage difference in the two LE while the bigger cation will have a smaller difference. You can make use of the above picture to conceptualise this statement.

Therefore, down the group the difference in lattice energy becomes smaller and hence decomposition becomes more difficult.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Transition Metal - Trend in atomic radius.

noreply@blogger.com (Mr Kwok) — Fri, 09 Jul 2010 03:46:00 +0000

As we move across Period 4, moving from K to Cu, we observe the graph below which shows the trend of how atomic radius changes with an equal increase in proton and a corresponding increase in electron.

1. Decreasing size of atom across the period (general trend).

Like in every period we have observed. The general trend we observe is that there is a general decrease in the size of the atomic radius. This trend is similar to what we see in the red box.

To account for this trend, the basic idea is that the effect of the increased nuclear attraction due to the increase in number of protons is more significant than the effect of shielding due to the adding of electrons. Normally, across the period, we add electrons in the same shell and the shielding due to this addition is relatively insignificant.

Hence, the shielding of the outermost electrons (4s electrons) is due to the inner shell. However, in the case of the transition metals, it is the addition of an electron in the 3d subshell. Although, the 3d subshell is in the inner shell, it is relatively weak in shielding as compared to a quantum shell (which is what K and Ca experience). Therefore, it is not surprising that the transition metals are smaller than K or Ca.

2. Relatively constant atomic radius observed amongst transition metals.

From the diagram below, you can see that the 3d subshell is an inner shell which can shield the 4s electrons.

Hence, despite there is an increase number of protons which results in an increased nuclear attraction, the addition of an electron in a 3d subshell shields the 4s electrons better than when an electron is added to a subshell which is found in shell with the quantum number 4. This happens because the 3d electrons are in an inner shell.

Therefore, the effect of increasing nuclear charge will somewhat be neutralised by the increase in shielding effect due to the addition of an electron to the 3d subshell. This results in the effective nuclear charge to remain relatively constant as we move across the period. Therefore, the size of atoms to be approximately the same.

In addition, the gradual filling of the 3d subshell, improves its ability to shield the 4s electrons from the nucleus.

Moreover, this diagram which shows how the 3d subshell is formed. It actually forms an asymmetrical sphere that shields the 4s electrons. A quantum shell forms a more symmetrical sphere that shields the outer shell. Hence, this is why the 3d subshell remains relatively poor in shielding.

3. Application.

The application of the above explanation will also help to account the 1st ionisation energy trend across the entire period 4 as well as the trend seen for just the transition metals.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Thoughts about Group VII

noreply@blogger.com (Mr Kwok) — Thu, 08 Jul 2010 12:50:00 +0000

So is it intermolecular forces or is it inter-atomic forces?

One of the more puzzling thing in the learning of Group VII elements is trying to understand when do we use intermolecular forces and when do we use inter-atomic forces to explain phenomenon observed in this group.

1. Intermolecular forces at work.

When dealing with hydrogen halides (HX), when we are trying to account the different boiling points, we are explaining in terms of differing strength of intermolecular forces. Changing the state of HX from liquid to gas, ensures that the substance retain its molecular form - it is still HX when it changes from liquid to gas. As energy is needed to change the state, some form of attraction must be broken. Hence, it is the Van der Waals interaction between the HX molecules.

Down the group, HX gets larger, the electron cloud becomes more polarisable, hence induced dipole interaction becomes stronger.

2. Inter-atomic forces at work.

However, when we discuss about the thermal stability and the acidic strength of HX, we are basically talking about the following pairs of equation:

2HX (g) -> H₂ (g) + X₂ (g) &
HX -> H⁺ + X^-

The equation shows clearly that the H-X bond has to be broken. Hence, the ease of thermal decomposition and the acidic strength HX is determined by the strength of the H-X single bond. The stronger the bond, the more thermal stable the HX will be and the weaker acid it will be.

Although, in conclusion, although there are differing strength in the acidic strength of HX, they are generally strong acids as they do dissociate completely.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Organic Chemistry - Functional Group Conversions.

noreply@blogger.com (Mr Kwok) — Sat, 08 May 2010 15:47:00 +0000

Worried that you are given a large molecule and you do not know how to handle the functional group conversions? Have no fear! The solution requires two main points:

Identifying the functional groups on the molecule. Be certain of the reactions that the functional group has.
Identify the types of reactions that the reagent can undergo.

Below is a video which highlights a phenyl molecule which contains phenol, primary alcohol, alkyl benzene and alkene as functional groups. This video will highlight how the above two points can be applied.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Answers to poll

noreply@blogger.com (Mr Kwok) — Sun, 02 May 2010 13:47:00 +0000

Q1: Which reagent can be used to distinguish between methanol and ethanol.

Answer: (1) KMnO₄/H⁺ can be used. Both decolourises purple KMnO₄, but only methanol will produce a colourless and odourless gas which forms a white precipitate with Ca(OH)₂. This is not observed for ethanol.

(2) NaOH, I₂, warm. Ethanol produces a pale yellow crystal with antiseptic smell. While methanol doesn't. (Note: As I₂ is a mild oxidising agent, both primary alcohols will be oxidised, hence we will see the purple decolourising for both).

Q2: CH₃MgBr is a source of a the nucleophile CH₃^-. This nucleophile is similar to the cynaide ion. To which of the following will CH₃MgBr react and produce a chiral compound? (i) CH₃CHO (ii) CH₃CH₂CHO (iii) CH₃COCH₃ (iv) CH₃COCH₂CH₃

Answer: (1) Look for the asymmetrical carbonyl. Thus, (iii) is out. As the nucleophile is CH₃^-, when added to the carbonyl it is common to an R group found in (i) and (iv), hence despite both (i) and (iv) are asymmetrical, addition of the nucleophile does not produce a chiral compound. Hence answer is (ii).

Q3: Given that the numerical value of K_a of H₂CO₃ and phenol is 4.5 x 10^-7 and 1.3 x 10^-10. Which of the following reactions will take place? (i) phenol and carbonate (ii) carbon dioxide and phenoxide

Answer: (ii). This is because carbon dioxide is a stronger acid than phenol. Hence, if phenol and carbonate reacts, we will get phenoxide (anion of phenol) and CO₂ as the products. BUT, CO₂ is a stronger acid (than phenol) hence, it will react with the phenoxide ion (since phenol is a weaker acid than CO₂, this also implies that phenoxide is a stronger conjugate base than carbonate), therefore returning to the original reactants. Hence, this implies that phenol will not react with carbon dioxide.

Q4. What would you expect when CH₃CH(Cl)CN is boiled with water in the presence of an acid?

Answer: The first reaction will be the acid hydrolysis of the -CN functional group to give -COOH. However, since water is a nucleophile and heat is added, there is nucleophilic substitution for the alkyl halide. This alkyl halide is a secondary alkyl halide, which has potential to undergo S_N1 mechanism.

This mechanism suggest that even if a weak nucleophile is used, the nucleophilic substitution can occur since the rate of reaction via S_N1 is independent on the nucleophile.

Hence, it is also possible for the Cl to be substituted by the OH. Hence, we would expect that CH₃CH(OH)COOH is produced. This is because we assume that S_N1 has taken place.

Fortunately, for most of the time we expect student to consider about acid hydrolysis. This is because water is a weak nucleophile and the reaction between a primary alkyl halide is not very favourable.

Q5. The more acidic the substituted benzoic acid, the less susceptible it is toward electrophilic attacks. True or False? Explain

Answer: True. This is because when a substituted benzoic acid is more acidic, it implies that its conjugate base is more stable. The conjugate base is made more stable because the substitutent must have been an electron withdrawing substituent that cause benzene ring to be even more electron withdrawing. This results in a greater extent of the delocalisation of the negative charge.

Since, the substituent is electron withdrawing, it pulls the pi electron cloud electron density making it less available for benzene to make use of it for electrophilic substitution.

Hope you got them right. :)

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Strength of Organic Bases

noreply@blogger.com (Mr Kwok) — Sun, 02 May 2010 01:46:00 +0000

In the earlier articles on ionic equilibrium, I wrote about the definition of bases and how to make use of the base dissociation constants to determine the strength of the base. In Organic Chemistry, the Nitrogen Compounds are substances that act as bases. N atom when bonded contains a lone pair of electrons and this lone pair of electrons is available to accept protons.

Many would wonder if alcohol whose Oxygen atom also contains a lone pair of electrons, will be base. Generally, because N is less electronegative than O, this result in the N containing compounds to be more likely the base than O containing compounds. Hence, amines and ammonia are usually bases, while alcohols and water are generally not.

Like Organic acids, Organic bases are generally weak bases. They partially dissociates in water, form a cation and the OH^-. This cation is formed because the base donates its lone pair of electron to the proton and hence accepts a protons; resulting in an overall positive charge.

Hence, this article is about how we can use the properties of the substituents and predict the strength of the Organic base. When the substituent is electron donating, e.g. alkyl substituents, the substituents donate electron density via the induction and hence enriches the electron density of N, making it more willing to donate its electrons.

While, electron withdrawing substituents have two means of electron-pulling. If the N is attached to an electronegative atoms, e.g. F, the electron density from N can be pulled away via induction.

Alternatively, the electron withdrawing substituents can pull electron pairs from N via delocalisation due to resonance. This is apparent in phenylamine, where the p orbital of N and the p orbital of the C in benzene ring are overlapping. Hence, the lone pair on N is delocalised into the benzene ring via resonance. Do note that this process will in-turn make the benzene ring more susceptible to electrophilic substitution. (You can make use of this picture from phenol to conceptualise the delocalisation. It is via the same process, making use of the same principle.)

Finally, the amides are N-compounds that are neutral. This is because their lone pair of electrons have been delocalised to the O of C=O and hence making it so unavailable that it does not have basic properties.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Alkyl halides, aryl halides, acyl halides

noreply@blogger.com (Mr Kwok) — Sun, 25 Apr 2010 13:00:00 +0000

Acid chlorides (aka Acyl Chlorides), alkyl chlorides and aryl chlorides are halogen containing compounds. Generally, when we are asked to perform a chemical test to these three compounds, the aim of the chemical test is to see which of these compounds will be able to produce a AgCl precipitate. Hence, this test presupposes that a nucleophilic substitution must happen to release the Cl^- so that it can be precipitated with a Ag⁺

When comparing alkyl chloride and aryl chloride. The chemical test steps consist of the following:

NaOH, heat
Cool
Add HNO₃
Add AgNO₃

The first step is the nucleophilic substitution where chloride ion leaves. Aryl chloride does not undergo nucleophilic substitution reaction.

While when comparing alkyl chloride and acid chloride. The chemical test is.

Add AgNO₃.

Acid chloride has a very electron deficient carbon, because it is attached to both the O and the Cl. That makes this carbon very willing for nucleophiles to attack it. Hence, in this chemical test, H₂O is the nucleophile that causes a nucleophilic substitution to occur. Acid chloride is so susceptible to nucleophilic substitution that a weak nucleophile can be used (H₂ vs OH^-) and a milder condition is required (room temperature vs heat).

In addition, this post would also like to serve us a good entry which teaches us how to describe the nucleophilic substitution using acid chloride and a nucleophile. The mechanism is different from S_N1 and S_N2. The picture below describes the mechanism while the video below describes how to draw the mechanism.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Strength of Organic Acids

noreply@blogger.com (Mr Kwok) — Mon, 19 Apr 2010 15:28:00 +0000

An acid is a substance which donates a proton. By the sheer fact that a proton is a charged particle, you would expect that a polar bond is more susceptible to be broken to release a proton than a non-polar bond. There are two main factors that will affect how readily a substance donates a proton.

Strength of the bond broken to release the proton.
Stability of the anion that is formed when the proton is released.

In organic chemistry, the organic acids break the O-H bond. Hence, with a common bond broken, to discuss about the differing acid strength in terms of bond strength will be meaningless. Hence, we will be discussing about the strength of acid through the stability of the anion formed from the dissociation. Since, the more stable the anion, the less incline for it to accept the proton, the more willing for the substance to remain in its dissociated form.

Since organic acids are weak acids, we would expect partial dissociation of the acid. Hence, the magnitude of the acid dissociation constant will provide good information about the strength of the acid. Therefore, a acid that produces a stable anion (conjugate base) would imply that the acid is stronger than another acid which does not form a stable anion.

There are two factors that affects the stability of the anion.

Delocalisation of charge by resonance
Induction effect.

In the delocalisation of charge by resonance, we requires the atom that has the negative charge to be attached to another atom which is part of a double bond. The ethanoate ion is a suitable example to illustrate this pattern.

In this case, the p orbital of C and O can overlap with the p orbital of the O that carries the negative charge. Hence, the electron can flow and therefore causing the anion to be stabilised.

While, in the induction effect, the substitutents can cause an overall electron donating or electron withdrawing effect. A substituent which is electron withdrawing, will pulls the electron density away from the atom that carries the negative charge. This effect spreads out the negative charge and hence creating a delocalisation.

When we compare the effect of the phenyl substitutents. There are two effects:

If the phenyl ring is attached directly to an atom that carries a negative charge, it will be able to delocalise the charge via resonance. If the phenyl ring is attached indirectly to the atom with the negative charge, it will act as an electron withdrawing substituent.

Although the phenyl ring is capable of delocalising the charge via resonance, into its ring of C atoms. It is important to realise that the extent of this delocalisation is smaller than the delocalising of the charges which we see in the case of the carboxylate ion (i.e. ethanoate's case). This is becaue the latter delocalises the charge to O, a more electronegative atom than C.

In addition, electron donating substituents on benzene ring will diminish the electron withdrawing effect of the ring. Converse is true for electron withdrawing substituents.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}

Formation of racemic mixture in Nucleophilic Addition.

noreply@blogger.com (Mr Kwok) — Sun, 11 Apr 2010 13:40:00 +0000

A nucleophilic addition reaction involves a carbonyl and HCN in the presence of NaCN as the catalyst. Details of the mechanism can be found here.

However, this reaction is capable of forming a racemic mixture. This is possible when the following rules are fulfilled:

The carbonyl is asymmetrical.
The product formed causes the C on C=O to change from sp² to sp³.
The product is an optically active product.

Isn't it slightly ironic that an optically active product is formed and yet we actually obtain a racemic mixture? Before my explanation, it is important to note that a racemic mixture does not mean that an optically active compound is absent. In fact, a racemic mixture implies that the pair of enantiomers exist in the same mixture.

In the reaction between an asymmetrical carbonyl and HCN results in the cyanide ion able to attack the carbonyl above or below the plane of the carbonyl.

If the cyanide ion attacks above the plane, an optically active addition product is formed. If the cyanide attacks below the plane of the carbonyl, the mirror image is formed.

As it is equally probable for the nucleophile to attack above or below the plane of the carbonyl, both enantiomers are formed in equal proportions and hence a racemic mixture is obtained.

This manner of forming racemic mixture, is also applicable the reaction between an alkyl halide and nucleophile, in which, this reaction must occur in the S_N1 mechanism. The carbocation being planar is equally probable for it to be attacked above and below its plane.

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^{Article written by Kwok YL 2010.}

^{Disclaimer and remarks:}

^{If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.}
^{This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.}