Effect of pH on rate of ketal formation between propanone & ethanol

Understanding how scientific principles governs us and regulates us has always been an interesting task for me. Being a skilled IB learner, I have always been intrigued to understand the way various chemical reaction occurs and the mechanism they follow. I have developed a special interest for this especially during my classes for Topic-10 (Organic Chemistry) in my second year of Diploma program. I always wanted to know more examples of organic reactions especially when I understood the concept of synthetic routes. This interest became stronger when I came across a research paper (Wright et al.) that spoke about the use of ‘ketals’ and ‘acetals’ as a precursor to synthesize resins. Exploring some organic chemistry websites and text books, I got to know more about them and could identify a similarity between this reaction and esterification which I studied in my class. The fact that intrigued me the most is that what factors could influence the way this reaction occurs. I realized the importance of this fact more when I understood the importance of knowing the mechanism of an organic reaction. I decided to study how certain conditions at which the reaction takes place would impact the reaction. To narrow down my thoughts into a precise platform, I faced two major challenges – Which parameter should I study to delineate the effect of some conditions on an organic reaction? The simplest choice I could get was rate of reaction. The next challenge was to design a process that allows me to collect the adequate data for this investigation. Going back to the concepts, I studied in Topic-6 (Kinetics), I could recollect about using colorimetry or spectroscopy as an analytical method to study rates of reaction. This led me to the research question given below:

How does the average rate of the reaction (measured in abs s^-1 ) of the intermolecular condensation between propanone and ethanol to produce 2,2-diethoxy propane in presence of HCl as catalyst depends on the pH of the medium, determined using UV-Vis spectrophotometer?

Propanone is an organic liquid having the functional group ketone. It has the displayed formula – CH₃COCH₃. Ethanol is an organic liquid belonging to the functional group of alcohol. This has a displayed formula of CH₃CH₂OH. In presence of dilute HCl as a catalyst, they undergo inter molecular condensation to produce a ketal – 2,2-diethoxy propane. Condensation products of ketones and alcohols are termed as ketals while condensation products of aldehyde and alcohols are known as acetals.

This reaction exemplifies condensation as it joins the two molecules – propanone and ethanol by elimination of a simple molecule-water.

The mechanism of the reaction (MacKENZIE and Stocker) is demonstrated below using skeletal structures of the reactants, intermediates and the products and curly arrows to indicate the movement of the pairs of electron.

Step - 1: The O atom of the propanone donates a lone pair to the Hydrogen (H⁺) ion furnished from dilute HCl and gets protonated. As a result, it gets a positive charge. Here, O atom behaves as a nucleophile and the H⁺ ion behaves as an electrophile.

Step - 2: The O atom of the ethanol molecule attacks the carbonyl C atom of the propanone group by using it’s lone pair. Here, the O atom of the ethanol is a nucleophile and the C atom of the carbonyl group is an electrophile. As a result, the pi bond of C=0 is broken and the electrons are given to the O atom which eventually neutralizes the positive charge acquired by the O atom is Step-1. This is also termed as ‘nucleophilic addition to C = O’ (Dong et al.).

Step - 3: The intermediate thus formed undergoes an intra molecular proton transfer. A H atom from the O atom of the OH group in ethanol moiety is transferred to the O atom of the OH group of the propanone moiety. This is an example of 1-3 proton shift causing the molecule to rearrange itself. The shift is 1-3 as if the O atom the H is shifting from is considered as 1 while the O atom, the H is shifting to is 3.

Step - 4: Another nucleophilic addition is done by the second ethanol molecule in a way similar to that in Step - 2. The product formed finally deprotonates itself to yield the main organic product – 2,2-diethoxy propane and water as a by-product.

A = ∈ cl

If molar absorptivity (∈) and path length are constant, then absorbance and concentration is directly related. Thus, these two terms become interchangeable

As the sample remains same and the path length remains unaltered (because the same spectrophotometer has been used), ∈ and path length (l) are assumed to be constant.

 \(Therefore, rate of reaction =\frac{change\ of\ concentration}{time\ taken}= \frac{change\ of\ absorbance}{time\ taken}\) \(\)

Here, propanone is a reactant and thus,

\(Rate \,of \,disappearance \,of propanone = \frac{change\ is\ concentration\ of\ propanone}{time\ taken\ for\ the\ change} =\frac{change\ is\ concentration\ of\ propanone}{time\ taken\ for\ the\ change}\) \(\)

 \(=\frac{Absorabnce\ of\ propanone\ at\ start-Absorabnce\ of\ propanone\ at\ end}{Time\ for\ which\ the\ reaction\ was\ carried\ out}=\frac{ΔA}{time\ taken}abs s-1\)  \(\) 

There is no correlation between the rate of the reaction and the pH of the medium.

There is no correlation between the rate of the reaction and the pH of the medium. This can be justified based on the fact that H⁺ is acting as a catalyst in this reaction and change in concentration of catalyst has an effect on rate of reaction.

The independent variable is the pH of the medium. The pH will be varied from 1.00 to 7.00. For the values of pH from 1.00 to 6.00, dilute HCl of appropriate concentrations, 1.0 × 10^-1 mol dm^-3, 1.0 × 10^-2 mol dm^-3, 1.0 × 10^-3 mol dm^-3, 1.0 × 10^-4 mol dm^-3, 1.0 × 10^-5 mol dm^-3, 1.0 × 10^-6: mol dm^-3 will be used. The pH will be calculated using the formula:

pH of the medium = - log (molar concentration of HCl used)

It has been assumed that HCl dissociates completely in aqueous solution.

The dependent variable is the rate of the reaction in abs s^-1.

The rate of the reaction will be calculated as the rate of disappearance of propanone. To do this, the absorbance of the propanone will be recorded at the start of the reaction and at the end of the reaction, change of absorbance will be calculated and the rate will be calculated using the formula:

 \(Rate \,of \,the \,reaction =\frac{difference\ in\ absorbance}{time\ taken}=\frac{ΔA}{600.00}abs s-1\) 

A UV-Visible spectrophotometer was used to record the absorbance of the sample at a wavelength at which propanone shows maximum absorbance.

• Propanone – 10 cc
• Ethanol – 10 cc
• Concentrated HCl (11.00 moldm^-3) – 10 cc