NMR Spectrum Analysis

The question asks for the molar ratio of chloroform to acetone in a mixture, given the information from its ¹H NMR spectrum. Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful technique used to determine the structure and composition of molecules. In ¹H NMR, the signals observed correspond to the different types of protons present in the sample.

The area under each signal (peak) in an NMR spectrum, known as the integral height, is directly proportional to the number of protons responsible for that signal. This allows us to determine the relative number of each type of proton in the sample.

Identifying Signals in the Spectrum

The spectrum shows two singlet signals:

• A singlet at $\delta$ 7.25 ppm with an integral height of 12 mm.
• A singlet at $\delta$ 2.1 ppm with an integral height of 18 mm.

We need to assign these signals to chloroform and acetone.

• Chloroform ($\text{CHCl}_3$) has one proton ($\text{-CH}$). This proton is deshielded due to the electronegative chlorine atoms and is expected to appear at a higher chemical shift value (downfield). The singlet at $\delta$ 7.25 ppm corresponds to the proton in chloroform. There is 1 proton per molecule of chloroform.
• Acetone ($\text{CH}_3\text{COCH}_3$) has six equivalent protons ($\text{-CH}_3$ groups). These protons are less deshielded than the chloroform proton and are expected to appear at a lower chemical shift value (upfield). The singlet at $\delta$ 2.1 ppm corresponds to the six protons in acetone. There are 6 protons per molecule of acetone.

Calculating Proton Ratio from Integral Heights

The ratio of the total number of protons giving rise to each signal is equal to the ratio of their integral heights.

Ratio of total protons (Chloroform : Acetone) = Ratio of integral heights (Chloroform : Acetone)

Ratio of total protons = 12 mm : 18 mm

Simplifying the ratio: $\frac{12}{18} = \frac{2}{3}$

So, the ratio of the total number of protons from chloroform to the total number of protons from acetone in the mixture is 2:3.

Determining Molar Ratio

We know the number of protons per molecule for each compound:

• Chloroform: 1 proton per molecule
• Acetone: 6 protons per molecule

Let $n_C$ be the number of moles of chloroform and $n_A$ be the number of moles of acetone in the mixture. The total number of protons from chloroform is $n_C \times 1$. The total number of protons from acetone is $n_A \times 6$.

The ratio of the total protons is given by:

$\frac{\text{Total protons from Chloroform}}{\text{Total protons from Acetone}} = \frac{n_C \times 1}{n_A \times 6}$

We found this ratio from the integral heights to be 2:3.

So, $\frac{n_C \times 1}{n_A \times 6} = \frac{2}{3}$

Now, we can solve for the molar ratio $\frac{n_C}{n_A}$:

$\frac{n_C}{n_A} \times \frac{1}{6} = \frac{2}{3}$

Multiply both sides by 6:

$\frac{n_C}{n_A} = \frac{2}{3} \times 6$

$\frac{n_C}{n_A} = \frac{12}{3}$

$\frac{n_C}{n_A} = 4$

This means the ratio of moles of chloroform to moles of acetone is 4:1.

Let the number of moles of chloroform be $N_C$ and the number of moles of acetone be $N_A$.

Total protons from chloroform = $N_C \times 1$

Total protons from acetone = $N_A \times 6$

Ratio of total protons = $(N_C \times 1) : (N_A \times 6) = 12 : 18$

$\frac{N_C}{6 N_A} = \frac{12}{18} = \frac{2}{3}$

$\frac{N_C}{N_A} = \frac{2}{3} \times 6 = 4$

Thus, the molar ratio of chloroform to acetone is 4:1.