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Write equations for the reaction of 1 -butanol, a primary alcohol, with these reagents. (a) \(\mathrm{H}_{2} \mathrm{SO}_{4},\) heat (b) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{H}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
(a) 1-butanol forms butene. (b) 1-butanol oxidizes to butanoic acid.

Step by step solution

01

Understanding the Reaction Conditions (a)

In the first case, you are given 1-butanol and sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)) with heat. Sulfuric acid is a dehydrating agent, and when heated with a primary alcohol, it typically leads to dehydration to form an alkene.
02

Write the Equation for Reaction (a)

1-butanol \((\mathrm{C}_4\mathrm{H}_9\mathrm{OH})\) undergoes dehydration in the presence of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) and heat to form butene. The chemical equation can be written as: \( \mathrm{C}_4\mathrm{H}_9\mathrm{OH} \xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}, \text{heat}} \mathrm{C}_4\mathrm{H}_8} + \mathrm{H}_2\mathrm{O} \).
03

Interpreting the Reaction Conditions (b)

In the second case, you are using a strong oxidizing agent, potassium dichromate (\( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)), in the presence of sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)). This combination is used to oxidize primary alcohols to carboxylic acids.
04

Write the Equation for Reaction (b)

For 1-butanol \((\mathrm{C}_4\mathrm{H}_9\mathrm{OH})\), the oxidation in the presence of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) and \( \mathrm{H}_{2} \mathrm{SO}_{4} \) produces butanoic acid \((\mathrm{C}_3\mathrm{H}_7\mathrm{COOH})\). The reaction can be written as: \( 3\mathrm{C}_4\mathrm{H}_9\mathrm{OH} + \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 4\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 3\mathrm{C}_3\mathrm{H}_7\mathrm{COOH} + \mathrm{Cr}_{2} \mathrm{(SO}_{4})_3 + 2\mathrm{K}_2\mathrm{SO}_{4} + 7\mathrm{H}_{2}\mathrm{O} \).

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Key Concepts

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

Dehydration of Alcohols
In the world of organic chemistry, dehydration of alcohols is a crucial reaction. When an alcohol is subjected to certain conditions, it can lose water (hence "dehydration") and transform into an alkene. This happens commonly with the help of a dehydrating agent like sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \)).

For example, when 1-butanol, a primary alcohol, is heated in the presence of \( \mathrm{H}_{2} \mathrm{SO}_{4} \), it undergoes dehydration to form butene, an alkene. Here’s how this works:

1. **Protonation of the Alcohol:** The alcohol (1-butanol) reacts initially with \( \mathrm{H}_{2} \mathrm{SO}_{4} \) by getting protonated, which converts the hydroxyl group into a better leaving group.

2. **Loss of Water:** Following protonation, water is removed, forming a carbocation intermediate.

3. **Formation of Alkene:** The final step involves the removal of a proton yielding the alkene, butene.

This transformation is illustrated by the reaction:
\[ \mathrm{C}_4\mathrm{H}_9\mathrm{OH} \xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}, \text{heat}} \mathrm{C}_4\mathrm{H}_8 + \mathrm{H}_2\mathrm{O} \]
This represents a key synthetic route to create alkenes in organic synthesis.
Oxidation of Alcohols
Oxidation reactions are processes where electrons are transferred, often resulting in the addition of oxygen or the removal of hydrogen from a molecule. Alcohols frequently undergo oxidation, and the products depend on the type of alcohol and the reagent used.

When we look at primary alcohols like 1-butanol, they oxidize to form carboxylic acids. A commonly used reagent for this is potassium dichromate (\( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)), especially when used in combination with sulfuric acid.

Here’s the oxidation process of 1-butanol:
- **Initial Formation of Aldehyde:** First, the primary alcohol is oxidized to an aldehyde. This stage involves the removal of hydrogen atoms and the formation of a carbonyl group.
- **Further Oxidation to Carboxylic Acid:** Once the aldehyde is formed, it is further oxidized to a carboxylic acid (in this case, butanoic acid), through the addition of an oxygen atom.

This reaction can be represented as follows:
\[ 3\mathrm{C}_4\mathrm{H}_9\mathrm{OH} + \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 4\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 3\mathrm{C}_3\mathrm{H}_7\mathrm{COOH} + \mathrm{Cr}_{2}\mathrm{(SO}_{4})_3 + 2\mathrm{K}_2\mathrm{SO}_{4} + 7\mathrm{H}_{2}\mathrm{O} \]
This comprehensive change from an alcohol to an acid is indicative of the powerful oxidizing capacity of dichromate.
Primary Alcohol Reactions
Primary alcohols like 1-butanol exhibit unique behaviors in chemical reactions, largely distinct from secondary and tertiary alcohols. The presence of a single alkyl group bonded to the hydroxyl-bearing carbon atom is pivotal in determining their reactivity pattern.

Here are some key reactions involving primary alcohols:

- **Dehydration:** As seen with 1-butanol, primary alcohols can be dehydrated to form alkenes using acids like \( \mathrm{H}_{2} \mathrm{SO}_{4} \). This results in the generation of a double bond between carbon atoms.

- **Oxidation:** Primary alcohols are prone to oxidation to form aldehydes and then further to carboxylic acids. Using agents like \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \), the alcohol group is transformed into the corresponding acidic group.

- **Esterification:** Though not directly covered, it's worth noting that primary alcohols can also undergo esterification when reacted with acids, forming esters. This is an essential method in organic synthesis for creating a variety of ester compounds.

These fundamental reactions underscore the versatility of primary alcohols in synthesis and industrial applications, highlighting their importance in organic chemistry.

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Most popular questions from this chapter

Explain why the boiling point of ethylene glycol \(\left(198^{\circ} \mathrm{C}\right)\) is so much higher than that of 1 -propanol \(\left(97^{\circ} \mathrm{C}\right)\), even though their molecular weights are about the same.

Explain why methanethiol, \(\mathrm{CH}_{3} \mathrm{SH}\), has a lower boiling point \(\left(6^{\circ} \mathrm{C}\right)\) than methanol, \(\mathrm{CH}_{3} \mathrm{OH}\left(65^{\circ} \mathrm{C}\right)\), even though methanethiol has a higher molecular weight.

Draw structural formulas and write common names for the six isomeric ethers with the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\).

Arrange these compounds in order of increasing boiling point. Use the boiling point values of \(-42^{\circ} \mathrm{C}, 78^{\circ} \mathrm{C}\) \(138^{\circ} \mathrm{C},\) and \(198^{\circ} \mathrm{C}\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) \((\mathrm{c}) \quad \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) \((\mathrm{d}) \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Answer true or false. (a) The two most important reactions of alcohols are their acid-catalyzed dehydration to give alkenes and their oxidation to aldehydes, ketones, and carboxylic acids. (b) The acidity of alcohols is comparable to that of water (c) Water-insoluble alcohols and water-insoluble phenols react with strong bases to give water-soluble salts. (d) Acid-catalyzed dehydration of cyclohexanol gives cyclohexane. (e) When the acid-catalyzed dehydration of an alcohol can yield isomeric alkenes, the alkene with the greater number of hydrogens on the carbons of the double bond generally predominates. (f) The acid-catalyzed dehydration of 2-butanol gives predominantly 1-butene. (g) The oxidation of a primary alcohol gives either an aldehyde or a carboxylic acid depending on experimental conditions. (h) The oxidation of a secondary alcohol gives a carboxylic acid. (i) Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), can be prepared from ethylene, \(\mathrm{CH}_{2}=\mathrm{CH}_{2}\), by treatment of ethylene with \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}_{2} \mathrm{SO}_{4},\) followed by treatment with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (j) Treatment of propene, \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2},\) with \(\mathrm{H}_{2} \mathrm{O} /\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) followed by treatment with \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4}\) gives propanoic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\)

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