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Chapter 2: Problem 25

Draw the products formed when 2-propanol [(CH\(_3)_2\)CHOH], the main ingredient in rubbing alcohol, is treated with each acid or base: (a) NaH; (b) H\(_2\)SO\(_4\); (c) Li\(^{+-}\)N[CH(CH\(_3)_2]_2\); (d) CH\(_3\)CO\(_2\)H.

Short Answer

Expert verified
2-propanol forms an alkoxide with NaH and LDA, propene with H2SO4, and isopropyl acetate with CH3CO2H.

Step by step solution

01

Identifying Reaction with NaH

2-propanol reacts with sodium hydride (NaH). NaH is a strong base and will deprotonate the hydroxyl group (-OH) of 2-propanol to form an alkoxide ion. The alkoxide ion is (CH\(_3\))\(_2\)CHO\(^{-}\) and hydrogen gas (H\(_2\)) is released as a byproduct.
02

Identifying Reaction with H2SO4

2-propanol reacts with sulfuric acid (H\(_2\)SO\(_4\)). Concentrated H\(_2\)SO\(_4\) can dehydrate alcohols to form alkenes. When 2-propanol is treated with H\(_2\)SO\(_4\), it undergoes dehydration to form propene, CH\(_3\)CH=CH\(_2\), with the elimination of water (H\(_2\)O).
03

Identifying Reaction with LiN[CH(CH3)2]2

LiN[CH(CH\(_3\))\(_2\)]\(_2\) is known as lithium diisopropylamide (LDA), a strong, non-nucleophilic base. LDA will deprotonate 2-propanol's -OH, forming the same alkoxide ion, (CH\(_3\))\(_2\)CHO\(^{-}\), as in the reaction with NaH. No alkene is formed because LDA does not promote elimination.
04

Identifying Reaction with CH3CO2H

2-propanol reacts with acetic acid (CH\(_3\)CO\(_2\)H). In the presence of acetic acid, 2-propanol participates in an esterification reaction to form an ester, isopropyl acetate (CH\(_3\)COO(CH\(_3\))\(_2\)CH), and water. This reaction is typically slow and may require a catalyst such as sulfuric acid for completion.

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Key Concepts

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

Acid-Base Reactions
Acid-base reactions are fundamental in organic chemistry. They involve the transfer of protons (H\(^+\)) between molecules.
In the case of 2-propanol reacting with a strong base like sodium hydride (NaH), an acid-base reaction occurs as follows: the -OH group in 2-propanol donates a proton to the NaH.
The result is the formation of an alkoxide ion,
  • Alkoxide Ion: The ion is (CH\(_3\))\(_2\)CHO\(^{-}\). It represents deprotonated 2-propanol, making it more reactive for other reactions.
  • Hydrogen Gas: Produced as a byproduct, highlighting the efficiency of NaH as a strong base.
This process shows the ability of alcohols to behave as acids under the right conditions, despite typically being quite weak acids. Acid-base reactions are essential for creating reactive intermediates, like the alkoxide ions, which are vital in various synthesis pathways.
Dehydration Reactions
Dehydration reactions are a type of elimination reaction where water is removed from a molecule. These reactions can change alcohols into alkenes under the right conditions.
When 2-propanol is treated with sulfuric acid (H\(_2\)SO\(_4\)), a typical dehydration occurs. Here's how:
  • Formation of Carbocation: 2-propanol converts into a carbocation (a positively charged ion) by losing water (H\(_2\)O), facilitated by the acidic environment.
  • Creation of Propene: The carbocation loses a further proton to form propene (CH\(_3\)CH=CH\(_2\)).
Dehydration reactions are crucial in the synthesis of alkenes, widely used in polymer production and as starting materials for further chemical transformations.
Esterification
Esterification is an important chemical process where alcohol reacts with an acid to form an ester. This reaction is often facilitated by a catalyst to speed up the process.
For 2-propanol, reacting with acetic acid (CH\(_3\)CO\(_2\)H) results in an esterification reaction, leading to:
  • Formation of Isopropyl Acetate: 2-propanol and acetic acid combine to form isopropyl acetate (CH\(_3\)COO(CH\(_3\))\(_2\)CH), a sweet-smelling ester.
  • Water as a Byproduct: Water is released during the process.
Catalysts like sulfuric acid can enhance this reaction by providing the necessary acidic environment. Esterification reactions hold significance in the production of flavors, fragrances, and polymers, as esters are common components in many everyday products.
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds and their properties. It explores how these compounds interact and transform under various conditions.
2-propanol provides a great example for exploring various reactions in organic chemistry:
  • Acid-Base Reactions: As seen with NaH and LDA, where proton exchange plays a central role in creating more reactive species.
  • Dehydration: Transforming alcohols into alkenes, which are foundational compounds in organic synthesis.
  • Esterification: Showcasing how alcohols can be transformed into valuable ester compounds.
By understanding these reactions, you gain insight into how complex organic molecules are built and altered, laying the groundwork for advancements in pharmaceuticals, materials science, and biotechnology.

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Most popular questions from this chapter

What is the conjugate acid of each base? a. H\(_2\)O b. \(^-\)NH\(_2\) c. HCO\(_3{^-}\) d. CH\(_3\)CH\(_2\)NHCH\(_3\) e. CH\(_3\)OCH\(_3\) f. CH\(_3\)COO\(^-\)

Rank the following compounds in order of increasing acidity. a. NH\(_3\), H\(_2\)O, HF e. CH\(_3\)OH, CH\(_3\)NH\(_2\), CH\(_3\)CH\(_3\) b. HBr, HCI, HF f. HCI, H\(_2\)O, H\(_2\)S c. H\(_2\)O, H\(_3\)O\(^+\). Ho- g. CH\(_3\)CH\(_2\)CH\(_3\), CICH\(_2\)CH\(_2\)OH, CH\(_3\)CH\(_2\)OH d. NH\(_3\), H\(_2\)O, H\(_2\)S h. HC = CCH\(_2\)CH\(_3\), CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\), CH\(_3\)CH = CH CH\(_3\)

Consider two acids: HCO\(_2\)H (formic acid, p\(K_a\) = 3.8) and pivalic acid [(CH\(_3)_3\)CCO\(_2\)H, p\(K_a\) = 5.0). (a) Which acid has the larger \(K_a\)? (b) Which acid is the stronger acid? (c) Which acid forms the stronger conjugate base? (d) When each acid is dissolved in water, for which acid does the equilibrium lie further to the right?

Which compound in each pair of isomers is the stronger acid? a. CH\(_3\)CH\(_2\)CH\(_2\)NH\(_2\) or (CH\(_3)_3\)N b. CH\(_3\)CH\(_2\)OCH\(_3\) or CH\(_3\)CH\(_2\)CH\(_2\)OH

Draw the product formed when the Lewis acid (CH\(_3\)CH\(_2)_3C^+\) reacts with each Lewis base: (a) H\(_2\)O; (b) CH\(_3\)OH; (c) (CH\(_3)_2\)O; (d) NH\(_3\); (e) (CH\(_3)_2\)NH.
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