Alkyl Halides - Stability of Carbocation (SN1)

When an alkyl halide reacts with a nucleophile in a nucleophilic substitution, it has two possible pathway: S_N1 or S_N2. The former is a two steps mechanism, where an intermediate is formed. While the latter is a single step reaction, with no intermediate formed.

Hence, a simple tool to predict whether an alkyl halide reacts in a S_N1 or S_N2 manner, it is dependent on whether an carbocation intermediate can be formed. If the carbocation intermediate can be formed, it must be because it is stable. In an earlier article, we discussed about carbocation and you can read it here.

This entry intends to elaborate further about the possible types of carbocations that can be formed from different alkyl halides. The following illustration gives us some examples of such alkyl halides which undergoes S_N1 mechanism.

The tertiary alkyl halide tends to undergo S_N1 mechanism because it is able to form a tertiary carbocation (as shown below) which is made stable by the 3 alkyl groups it is attached to. The alkyl groups being electron donating, allow the positive charge to be delocalised by induction. The fewer the alkyl groups, the less stable the carbocation would be and hence the less likely for the alkyl halide to undergo S_N1.

However, there are other alkyl halides which are not tertiary and they are able to undergo S_N1. This is because they have other portion(s) of the ion which enables the positive charge to be delocalised and hence making the ion stable.

In the following two cases (carbon with positive charge is attached to a C=C and benzene ring), the carbocation are stable because the positive charge is being delocalised by resonance.  This happens when the p orbital of the carbon with the positive charge and the p orbital of the carbon that makes the alkene (or benzene) can overlap with each other.

The overlapping of all the relevant p orbital enables the pi electrons are able to flow towards the carbon with the positive charge, hence the flow of pi electrons cause the positive charge to be delocalised; This form of delocalisation is delocalisation by resonance. Therefore, it is because these two carbocations are made stable, their correspond alkyl halide would react with a nucleophile in a S_N1 manner.

As delocalisation of the positive charge into the benzene ring is quite extensive, if the carbon with the positive charge is attached to a cyclohexene (as shown below). The stabilising effect from benzene will make the carbocation more stable than the latter's case.

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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.}

Alkenes - Stability of intermediates.

In the electrophilic addition mechanism, you have observed that a carbo-cation intermediate is produced. In addition, in an earlier article, you would have read about another intermediate, the radical.

Reaction intermediates are formed when a chemical reaction occurs in at least two steps. They are usually not produced in the last step of the reaction, and have to generally reactive so as the reaction can go to completion.

The carbo-cation intermediate is similar to the radical intermediate. In the latter's case, the carbon is electron deficient because it is one less from fulfilling the octet. While in the former's case, the carbon is also electron deficient. But in this case, the carbon actually is two short of fulfilling the octet.

Let's examine the geometry for the carbo-cation. With 6 valence electrons; there are 3 bonded electron pairs. Hence, by VSEPR theory, the geometry of the carbo-cation will be planar. Therefore, there will be an empty p-orbital remaining.

Due the effect of hyperconjugation (Note: This term is not part of the GCE A syllabus), alkyl substituents are electron donating. Hence, increasing the number of alkyl substituent around the carbon which carries the positive charge will make the cation more stable.

Another effect which helps to stabilise the carbo-cation would be to delocalise the positive charge. This effect was seen in the stabilising of the carbo-cation intermediate formed in electrophilic substitution reaction(it is actually known as the Wheland intermediate).

In conclusion, a simple chemistry rule - localising of electrons, or charges will make the substance more reactive. Generally, a chemical reaction occurs when there is a flow of electrons, which results in bonds to be broken/formed and/or instances where the oxidation number to change.

Therefore, when you intensify charges or localise electrons, these will promote the flow of electrons, because it makes the electrons more attracted to a positive charge or more exposed to attacks, which promotes a chemical reaction - i.e. substance is more reactive.

Hence, despite a more reactive intermediate favours a faster reaction, it is not produced. This is because you actually need to first produce the reactive intermediate, which is dependent on the reaction energetics. Therefore, this actually results in a multi-step reaction to produce the more stable intermediate as oppose to a less stable intermediate.

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^{Article written by Kwok YL 2009.}

^{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.}

Alkanes - Stability of Radical Intermediate

Why is it that when excess propane reacts with chlorine gas under the presence of sunlight, the ratio of 1-chloropropane to 2-chloropropane is less than 3:1? This is despite there are 6 Hydrogen atoms can be substituted to give 1-chloropropane and 2 of those can be substituted to give the latter.

To examine the situation, let's focus on the propagation step of the Free Radical Substitution mechanism of the reaction between propane and chlorine.

In the propagation step, the rate determination step, is most likely the reaction between the alkane molecule and the halogen radical. This is because the bond broken is the C-H single bond; the strongest bond broken in all the steps in the mechanism.

If we see all the C-H bonds in the alkane molecule are equally strong, then there is no discrimination to which Hydrogen atom is favoured to be removed. This would have resulted in the above mentioned 3:1 ratio. But it is not true, there is a difference

From the illustration above and the definition of a radical, having substitutents (such as alkyl substituents) who are willing to donate some electron density to the carbon radical, help to make the carbon radical less electron deficient and hence more stable.

The resultant: A stable product would be more likely be produced than an unstable product. Hence, the CH₃CHCH₃ radical is more readily formed than then CH₃CH₂CH₂ radical. - (This is nature's phenomenon.)

However, as the halogen radicals are highly reactive (in fact almost all radical are reactive), probability is also an important factor, hence CH₃CH₂CH₂ radical is still formed in larger quantities since there is a higher chance of forming.

The different stability observe in the two radical intermediate implies that the E_a of the propagation step which forms the CH₃CHCH₃ radical is lower than that which is required to form the other radical.

Since in the formation of alkyl radical the enthalpy change for that reaction is different (due to different radical stability) while the same H-X bond is formed. Therefore, the C-H bond strength in the alkane molecule is definitely not identical.

In addition, there is another means to allieviate the electron deficiency of a carbon radical. We must first examine the geometry of the radical.

When the radical centre is adjacent to a carbon-carbon double bond, the following phenomenon (known as delocalisation by resonace) occurs, resulting the radical to become more stable.

Hence, the stabilisation of a radical intermediate can occurs through (i) electron donation or (ii) delocalisation - Both methods work because in their own way they help to allieviate the electron deficiency with a carbon radical has to endure.

In conclusion, this also further adds to the value that despite your Data Booklet giving you the bond energy of the C-H bond to be 410 kJ mol^-1, the C-H bonds in a molecule is definitely far from identical.

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^{Article written by Kwok YL 2009.}

^{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.}