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Sodium wire can be used to dry hexane solvent. (a) Write an equation for the reaction that occurs. (b) How might you destroy excess Na and what precautions should you take?

Short Answer

Expert verified
(a) 2 Na + 2 H₂O → 2 NaOH + H₂; (b) React excess Na with ethanol carefully, wearing safety gear and ensuring good ventilation.

Step by step solution

01

Identify the Reaction Components

Sodium (Na) is a strong reducing agent and reacts with moisture (water) in the hexane solvent. The reaction involves sodium and water to produce sodium hydroxide (NaOH) and hydrogen gas (H₂).
02

Write the Balanced Reaction Equation

The balanced chemical equation for the reaction between sodium and water is: \[ 2 \, \text{Na} + 2 \, \text{H}_2\text{O} \rightarrow 2 \, \text{NaOH} + \text{H}_2 \] This equation shows that sodium reacts with water to form sodium hydroxide and hydrogen gas.
03

Evaluate a Method to Destroy Excess Sodium

To destroy excess sodium, it should be reacted carefully with alcohol like ethanol, which reacts less violently than water. The reaction forms sodium ethoxide and hydrogen gas: \[ \text{Na} + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_2\text{H}_5\text{ONa} + \text{H}_2 \]
04

List Required Precautions

When reacting sodium with ethanol, it's important to perform the reaction under controlled conditions. Use small amounts of sodium and ethanol at a time, wear protective equipment such as gloves and goggles, and perform the reaction in a well-ventilated area or fume hood to safely manage the release of hydrogen gas.

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Key Concepts

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

Sodium as a Reducing Agent
Sodium is a fascinating and mighty element in the world of chemistry, renowned for its role as a strong reducing agent. A reducing agent is a substance that donates electrons in a chemical reaction, reducing other substances and being oxidized itself in the process. Sodium is highly reactive, making it efficient at transferring electrons. When sodium wire is used to dry hexane solvent, it reacts with any moisture present, tearing apart water molecules and resulting in sodium hydroxide and hydrogen gas through a vigorous process. This reaction satisfies the moisture deficiency in the solvent but must be handled with care due to the rapid and intense nature of sodium's reactivity.
  • Sodium donates its valence electron easily.
  • This ability makes sodium effective in reducing other substances.
  • Its reactivity can be harnessed but must be controlled carefully.
Remember, when wielding the power of sodium, it is crucial to anticipate its energetic eagerness to react!
Balanced Chemical Equations
Balanced chemical equations are essential in chemistry for representing the conservation of mass in a chemical reaction. They show the exact proportion of reactants and products involved. Let’s take a closer look at the balanced equation from our exercise: \[ 2 \, \text{Na} + 2 \, \text{H}_2\text{O} \rightarrow 2 \, \text{NaOH} + \text{H}_2 \]This equation denotes that two molecules of sodium (\( 2 \, \text{Na} \)) react with two molecules of water (\( 2 \, \text{H}_2\text{O} \)) to form two molecules of sodium hydroxide (\( 2 \, \text{NaOH} \)) and one molecule of hydrogen gas (\( \text{H}_2 \)). The principle of balancing chemical equations ensures that the number of atoms for each element is conserved from reactants to products. This is how chemists uphold the law of conservation of mass in every chemical transaction.
  • Each side of the equation has equal numbers of each type of atom.
  • Balancing equations maintain both the mass and charge balance.
  • It's vital for calculating quantities needed or produced in reactions.
This neat organization helps chemists predict results, ensuring accuracy and efficiency in scientific experiments and industrial applications.
Safety Precautions in Chemical Reactions
Safety is paramount whenever handling chemicals, especially in reactions involving volatile substances like sodium. Sodium reacts energetically with water and even more so with acids or alcohols, producing hydrogen gas which is highly flammable. Consequently, safety precautions are non-negotiable. When destroying excess sodium, it's advised to convert it into a less reactive form, such as sodium ethoxide using ethanol.
  • Conduct reactions with small quantities carefully to control the process.
  • Always wear protective gear, namely gloves and eye protection.
  • Ensure the workspace is well-ventilated or use a fume hood.
Staying safe means understanding the risks involved and taking steps to mitigate potential hazards. Never overlook these precautions, as they ensure your well-being and the smooth running of experiments!

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Most popular questions from this chapter

When we discussed oxidation of alcohols in Section \(30.5,\) we stated that the mechanism of oxidation of an aldehyde to carboxylic acid is analogous to that of the conversion of an alcohol to an aldehyde. Propose a mechanism for the oxidation of \(\mathrm{RCHO}\) to \(\mathrm{RCO}_{2} \mathrm{H}\) using \(\mathrm{KMnO}_{4}\) in acidic aqueous solution.

Suggest starting materials for the preparation of the following alcohols using a Grignard reagent as one precursor. In each case, state how you might make the Grignard reagent: (a) 2 -methylpropan- 2 \(\mathrm{ol}\) (b) heptan-I-ol; (c) pentan-2-ol.

What spectroscopic technique(s) would you choose to distinguish between (a) propan-1-ol and propan-2-ol; (b) diethyl ether and butan-1-ol; (c) cyclohexanol and hexan-1-ol?

An alcohol \(\mathbf{X}\) has a composition of \(64.8 \% \mathrm{C}\) and \(13.6 \%\) H. The mass spectrum shows a parent ion at \(m / z=74 .\) The \(^{1} \mathrm{H}\) NMR spectrum of \(\mathbf{X}\) dissolved in \(\mathrm{CDCl}_{3}\) has signals at \(\delta 3.71\) (sextet, \(1 \mathrm{H}\) ), 2.37 (singlet, \(1 \mathrm{H}\) ), 1.46 (multiplet, \(2 \mathrm{H}\) ), 1.17 (doublet, 3H), 0.93 (triplet, \(3 \mathrm{H}\) ) ppm; in the \(^{13} \mathrm{C}\) NMR spectrum, four resonances are observed. Use these data to suggest a structure of \(\mathbf{X}\) and comment on isomer possibilities that retain the \(\mathrm{OH}\) functionality.

(a) Explain why the displacement of \(\mathrm{OH}\) in an alcohol by Br is carried out under acidic conditions. (b) Suggest products at each stage in the following reaction scheme: Why does the OH group need to be protected before the following reaction is carried out? (c) Suggest how you would prepare the following \(^{13}\) C-labelled compound \(\left(\boldsymbol{\theta}=^{13} \mathbf{C}\right)\) starting from the precursor shown below:

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