Oxidation of primary alcohols using potassium dichromate.

Primary and secondary alcohols are susceptible to oxidation reactions using K₂Cr₂O₇/H⁺. While tertiary alcohol are resistant to oxidation reactions. Secondary alcohols when oxidised, produces ketones. However primary alcohols can be oxidised to two different products; an aldehyde or a carboxylic acid. The type of product produced is dependent on the conditions of the reaction.

From the above illustration, it can be seen that if we heat the primary alcohol (example shows ethanol) with K₂Cr₂O₇/H⁺ under reflux conditions we will get our carboxylic acid. While if we heat the primary alcohol with K₂Cr₂O₇/H⁺ using a distillation setup we will be able to obtain the aldehyde.

The reflux apparatus is shown above. A condenser is attached about the round-bottom flask which contains the reaction mixture. Cold water is passed through the condenser from the bottom and the warmer water is removed from the top. In this set-up, the reactants/products will undergo a cycle of evaporation and condensation.

This setup is extremely purposeful for the formation of carboxylic acid when primary alcohols are oxidised. This is because the primary alcohols will first be oxidised to the aldehyde and because the reflux setup tracks the product, the aldehyde is able to be subsequently oxidised by K₂Cr₂O₇/H⁺ to produce the carboxylic acid.

Hence, in order to produce just the aldehyde, a distillation setup (shown below) is used. This setup allows the aldehyde to be distilled out of the reaction mixture. This is because the aldehyde has a lower boiling point than ethanol and water(water is the solvent) as in only has permanent dipole-permanent dipole interaction as its most predominant Inter-molecular forces. This interaction is weaker than the hydrogen bonding which both ethanol and water has.

Hence, the distillation process allows the aldehyde to escape, preventing it from being oxidised by the oxidising agent. This entry serves as an avenue to explain the difference in the oxidation reaction of the primary alcohol.

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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.}

Hydroxy Compounds - Reactivity of Phenols

Phenol like benzene undergoes electrophilic substitution reaction. The presence of the -OH substitutent enriches the electron density of the pi electron cloud. Hence, phenol undergoes electrophilic substitution readily.

The O atom of phenol lies on the same plane as the C atoms, this enables the p-orbital of O to overlap with the p-orbital of carbon. This results in a formation of tunnel where O lone pair of electron can flow despite O being electron-withdrawing.

If we were to draw the lewis structure of phenol, we can show the lone pair of electron on O can indeed be donated to the pi electron cloud, thus enriching the latter electron density. However, the converse is impossible (i.e. pi electron cloud donating an electron pair to O atom) as that would result in O to violate the octet rule.

In conclusion, I would like to highlight two reactions of phenol which helps to reiterate the concept that the -OH substitutent results in the pi electron cloud to be enriched hence enabling phenol to be more susceptible to attacks of electrophiles. Therefore, a milder reaction condition is required.

(1) Electrophilic substitution reaction - Nitration
Since the pi electron cloud is enriched, the electrophilic substitution of phenol do not require a halogen carrier. In fact, phenol's reaction are generally milder. Using of dilute HNO₃ results in the formation of the mono-substituted product (I showed the 4-substitution as it is favoured).

(2) Electrophilic substitution reaction - Bromination
While in the reaction between Br₂ and phenol, the use of Br₂ in CCl₄ results in the mono-substituted product. While the usage of aqueous Br₂ results in the formation of the tribromophenol product to be formed, which is is a useful reaction as it produces a white precipitate. This reaction helps distinguish benzene and phenol.

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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.}

Hydroxy Compounds - Acidity of Hydroxy compounds

Phenol are generally insoluble in water. The large phenyl substitutent is hydrophobic, hence despite the -OH functional group is able to form hydrogen bonding with water, the larger substituent restricts the solubility of phenol in water.

In order to dissolve phenol, NaOH is added. This is because there is the formation of the sodium phenoxide, a soluble ionic salt. However, when NaOH is added to an alcohol, a sodium salt is not formed. Hence, the difference in the reaction with NaOH, suggests that the acidity of the two hydroxy compound differs.

When we obtain separate aqueous solutons of phenol and an alcohol, the pH of the former is less than 7 while the pH of the latter is 7. Hence, this implies that alcohol does not like to dissociate in water, despite the polar OH present. Whereas, phenol likes to dissociate in water.

Acidity of the hydroxy compound can be explained from two fronts. (1) The polarity of the O-H bond which results in the ease of losing of the proton. (2) The stability of the conjugate base, thus affecting the equilibrium position.

(1) The polarity of the O-H bond.
The phenyl substitutent is electron-withdrawing. This results in the O to be even more electron-withdrawing and this in turn causes the O-H bond of phenol to be even more polarised. The H is now even more electron deficient and becomes more susceptible to be lost as a H₊. While the alkyl groups are electron-donating and the converse effect is observed.

(2) The stability of the conjugate base.
From the earlier illustration, the conjugate base of phenol is the phenoxide ion. This ion is stabilised by resonance (due to alignment of the p-orbtial, similar to aryl halides), hence the negative charge on O can be delocalised throughout the ring and not found on O only. (Note: pi electron pair does not go to O with the negative charge as it will cause the structure to violate octet rule.)

However, the conjugate base of alcohol, a alkoxide ion, is being destabilised. The alkyl group is electron-donating and hence, it intensifies the negative charge on O. This effect is a destabilising effect.

As a result, the stability of the phenoxide ion, results in the equilibrium position to favour the right hand side. Hence, phenol is a weak acid. (Note: Stability in chemistry is due to lower in the energy content of the substance (aka potential energy).)

In conclusion, I prefer to use reason (2) to account for the acidity of phenol. Although, reason (1) is find but I think it is actually more abstract than how I have explained. In addition, despite phenol is a weak acid, it is actually too weak to react with Na₂CO₃.
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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.}