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\(5.2 \mathrm{~g}\) 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 \(\mathrm{OH}\) groups present in polyhydric alcohol (molar mass of alcohol \(=104 \mathrm{~g}\) \(\left.\mathrm{mol}^{-1}\right)\)

Short Answer

Expert verified
The polyhydric alcohol has 3 OH groups.

Step by step solution

01

Understanding the Reaction

Polyhydric alcohol reacts with methyl magnesium bromide to produce methane gas. For each hydroxyl (OH) group present in the alcohol, one molecule of CHâ‚„ is produced.
02

Use of Given Conditions

We know that 3.36 liters of CHâ‚„ at STP is produced. At STP, 1 mole of any gas occupies 22.4 liters. Therefore, moles of CHâ‚„ = \( \frac{3.36}{22.4} \).
03

Calculate Moles of CHâ‚„

Moles of CHâ‚„ are calculated as follows: \( \frac{3.36}{22.4} = 0.15 \text{ moles of } CH_4 \). This means 0.15 moles of hydroxyl groups reacted.
04

Calculate Moles of Alcohol

Given 5.2g of alcohol and its molar mass is 104 g/mol. The moles of alcohol are given by \( \frac{5.2}{104} = 0.05 \text{ moles} \).
05

Determine OH Groups per Molecule

Since each mole of alcohol produced 0.15 moles of CHâ‚„, we know each molecule of alcohol has \( \frac{0.15}{0.05} = 3 \) hydroxyl groups.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Methyl Magnesium Bromide Reaction
Methyl magnesium bromide, commonly known as a Grignard reagent, reacts with alcohols in specific ways to produce alkanes. In this case, each hydroxyl group (OH) in the polyhydric alcohol interacts with methyl magnesium bromide. This interaction displaces one molecule of methane (CHâ‚„) for each hydroxyl group present.

This chemical process is quite significant in organic chemistry as it provides insights into alcohol analysis and is often used to determine the number of OH groups in a compound like polyhydric alcohol. Knowing this, we can observe that the total volume of methane production is directly correlated with the number of hydroxyl groups involved in the reaction.
  • Methyl magnesium bromide is both a strong nucleophile and a strong base, making it reactive toward alcohols.
  • The displacement of methane offers a straightforward method to quantify OH groups.
  • Understanding this reaction is essential for analysis and synthesis of alcohols in organic chemistry.
Understanding how methyl magnesium bromide reacts with alcohols is crucial for chemists, especially those interested in synthetic approaches and compound analysis.
Calculating OH Groups
The number of hydroxyl groups in a polyhydric alcohol can be determined by measuring the amount of methane produced when reacted with excess methyl magnesium bromide. Typically, each OH group in the alcohol will produce one molecule of methane. In the exercise, we see that 3.36 liters of CHâ‚„ were generated. Since one mole of any gas at STP occupies 22.4 liters, the moles of methane produced can be calculated using this ratio.

By this calculation, we find that \( \frac{3.36}{22.4} \) yields 0.15 moles of CHâ‚„, indicating that 0.15 moles of hydroxyl groups have reacted. Knowing the relation:
  • Each OH group directly correlates with one molecule of CHâ‚„ produced.
  • The moles of CHâ‚„ can be used to determine the number of hydroxyl groups.
  • The division of total moles of gas produced by moles of alcohol gives the OH groups per molecule.
This approach simplifies the process of determining structural information about alcohol compounds.
Molar Mass of Alcohol
Molar mass is a fundamental concept when dealing with reaction stoichiometry. In the given exercise, the molar mass of polyhydric alcohol is provided as 104 g/mol. This is key for converting the mass of alcohol to moles, which is a pivotal step in the analysis and understanding of the reaction with methyl magnesium bromide.

By dividing the mass of the alcohol sample, 5.2 grams, by the molar mass, 104 g/mol, we find the number of moles of alcohol to be 0.05. Moles make calculations regarding reaction ratios more precise and are essential for accurate stoichiometric assessments.
  • Molar mass is used to convert the given mass into moles, providing a starting point for calculating reactant amounts.
  • Knowing moles is crucial to comparing against reaction products, like CHâ‚„ in this situation.
  • This allows us to determine the number of OH groups within each molecule of the alcohol.
Understanding the molar mass and its use in stoichiometry is critical for analyzing chemical reactions and deducing molecular composition.

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Most popular questions from this chapter

In the reaction, \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\), the reagent used would include (a) \(\mathrm{Hg}^{2+} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) Na in liquid \(\mathrm{NH}_{3}\), followed by \(\mathrm{CH}_{3} \mathrm{Cl}\) (c) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} / \mathrm{Pt}\), followed by \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{H}_{2} \mathrm{O}\) (d) \(2 \mathrm{~mol}\) of \(\mathrm{H}_{2} / \mathrm{Pt}\), followed by aq. \(\mathrm{NaOH}\)

Ethyl ester \(\stackrel{\mathrm{CH}_{3} \mathrm{MgBr} \text { (excess) }}{\longrightarrow} \mathrm{P}\) The product ' \(\mathrm{P}\) ' will be

Match the following: List I 1\. absolute alcohol 2\. denatured alcohol 3\. \(80 \%\) proof alcoholic liquor 4\. rubbing alcohol List II (i) alcohol made undrinkable by the addition of toxic materials or poisons. (ii) alcoholic liquor containing \(40 \%\) ethanol by volume (iii) anhydrous alcohol (iv) isopropyl alcohol The correct matching is:

Product C in the following reaction, will be \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br} \stackrel{\mathrm{aq} \cdot \mathrm{NaOH}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{Na}}{\longrightarrow \mathrm{B}} \stackrel{\mathrm{CH}_{3} \mathrm{I}}{\longrightarrow} \mathrm{C}\) (a) ethane (b) ethyl methyl ether (c) ethyl iodide (d) propane

\(\mathrm{A} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{B} \underset{\text { vigrous oxidation }}{[\mathrm{O}]} \mathrm{CH}_{3} \mathrm{COOH}\) Here, B is propanone so A will be (a) Ethyl alcohol (b) Isopropyl alcohol (c) n-propyl alcohol (d) tertiary butyl alcohol

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