Problem 194 - A Complete Resource Book in Chemistry for JEE Main

Question

\(5.2 \mathrm{~g}\) of polyhydric alcohol was treated with an excess of methyl magnesium bromide to produce \(3.36\) litre of \(\mathrm{CH}_{4}\) at STP. Calculate number of OH groups present in polyhydric alcohol (molar mass of alcohol \(=104 \mathrm{~g}\) \(\left.\mathrm{mol}^{-1}\right)\)

Step-by-Step Solution

Answer

The polyhydric alcohol has 3 OH groups.

Step 1: Determine CH4 moles produced

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, to find the moles of \(\text{CH}_4\) produced, we use the formula: \[\text{Moles of } \text{CH}_4 = \frac{\text{Volume of } \text{CH}_4}{22.4} = \frac{3.36}{22.4} \approx 0.15 \text{ moles}.\]

Step 2: Relate OH groups to CH4 moles

Each \(\text{OH}\) group in the polyhydric alcohol reacts with \(\text{CH}_3\text{MgBr}\) to produce one mole of \(\text{CH}_4\). Therefore, the number of \(\text{OH}\) groups is equal to the number of moles of \(\text{CH}_4\) produced, which is 0.15.

Step 3: Calculate moles of polyhydric alcohol

Given that the mass of the polyhydric alcohol is 5.2 g and its molar mass is 104 g/mol, the moles of polyhydric alcohol is calculated by: \[\text{Moles of alcohol} = \frac{5.2}{104} = 0.05 \text{ moles}.\]

Step 4: Find the number of OH groups per molecule

Since 0.05 moles of polyhydric alcohol produce 0.15 moles of \(\text{CH}_4\), there must be 3 moles of \(\text{CH}_4\) produced per mole of alcohol (0.15 / 0.05). Hence, there are 3 \(\text{OH}\) groups in the polyhydric alcohol.

Key Concepts

Moles Calculation

Understanding how to calculate moles is essential in chemistry. It helps us measure chemicals accurately. The concept of moles allows chemists to count particles in a given mass.

To calculate moles, you can use the formula: \[ \text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass}} \]In this formula:

• **Mass of substance** is the amount of substance you have, measured in grams.
• **Molar mass** is the mass of one mole of the substance, also in grams per mole.

For example, with the original exercise, the calculation was about converting grams into moles using the molar mass of the polyhydric alcohol, which was given as 104 g/mol. This helps us understand how many molecules of alcohol we have to further analyze and react with other substances.

Grignard Reaction

Grignard reactions are a powerful tool in organic chemistry, named after Victor Grignard, who discovered them. These reactions involve organomagnesium compounds known as Grignard reagents. They are typically used in forming carbon-carbon bonds.

A Grignard reaction usually consists of three key components:

• **An organohalide**, like methyl magnesium bromide used in the exercise.
• **A carbonyl compound**, which reacts with the Grignard reagent.
• **A solvent/medium**, usually an ether, to carry the reaction.

In the given exercise, the Grignard reagent (methyl magnesium bromide) reacts with the polyhydric alcohol. Each \( \text{OH} \) group in the alcohol forms a new carbon-hydrogen bond, releasing methane \( \text{CH}_4 \). This transformation is useful as a method for creating different organic molecules.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a standard set of conditions for experimental measurements. STP is commonly used in chemistry to establish consistency in experiments.

• **Standard Temperature** is 0 °C or 273.15 K.
• **Standard Pressure** is 1 atm (atmospheric pressure).
• Under these conditions, 1 mole of an ideal gas occupies a volume of 22.4 liters.

In the context of the exercise, these conditions helped to determine the moles of methane gas \( \text{CH}_4 \). Since the problem states the volume of gas at STP, you can directly calculate the number of moles using the relation that 1 mole of any ideal gas is equivalent to 22.4 liters at STP. This provided a direct link between the volume of gas produced and the number of molecules—critical for solving the exercise.