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Chapter 24: Problem 28

Explain why hydrolysis of 4-chloro-1butanol occurs much faster than the hydrolysis of 3 -chloro-1- propanol. Provide mechanisms that account for the products formed in each case.

Short Answer

Expert verified
4-chloro-1-butanol hydrolyzes faster due to intramolecular hydrogen bonding stabilizing the transition state.

Step by step solution

01

- Identify the Structures

First, identify the structures of 4-chloro-1-butanol and 3-chloro-1-propanol. For 4-chloro-1-butanol, the chlorine atom is on the 4th carbon of a 4-carbon chain with a hydroxyl group on the 1st carbon. For 3-chloro-1-propanol, the chlorine atom is on the 3rd carbon of a 3-carbon chain with a hydroxyl group on the 1st carbon.
02

- Understand Hydrolysis

Hydrolysis involves breaking a chemical bond by the addition of water. In these cases, it will involve the chlorine atom being replaced by a hydroxyl group. This will occur via an Sn2 mechanism for both compounds, where the nucleophile (water or hydroxide ion) attacks the carbon atom bonded to the chlorine atom.
03

- Mechanism for 4-chloro-1-butanol

In 4-chloro-1-butanol, the hydroxyl group on the 1st carbon can stabilize the transition state by forming an intramolecular hydrogen bond with the departing chlorine. This makes the reaction proceed more quickly.
04

- Mechanism for 3-chloro-1-propanol

In 3-chloro-1-propanol, there is no option for intramolecular hydrogen bonding as the hydroxyl group is not in a suitable position relative to the chlorine atom. Thus, the transition state is less stabilized compared to 4-chloro-1-butanol, making the hydrolysis slower.
05

- Compare the Reaction Rates

The hydrolysis of 4-chloro-1-butanol occurs faster due to the intramolecular hydrogen bonding which stabilizes the transition state. This factor is absent in 3-chloro-1-propanol, resulting in a slower reaction rate.
06

- Products Formed

Both 4-chloro-1-butanol and 3-chloro-1-propanol undergo hydrolysis to form corresponding alcohols. 4-chloro-1-butanol forms 1,4-butanediol and 3-chloro-1-propanol forms 1,3-propanediol.

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Key Concepts

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

4-chloro-1-butanol
4-chloro-1-butanol is a compound with a four-carbon chain. Here, the chlorine atom is attached to the 4th carbon, while a hydroxyl (-OH) group is attached to the 1st carbon. When this compound undergoes hydrolysis, it forms 1,4-butanediol.

Understanding the structure helps in predicting its reactivity. The location of the chlorine and hydroxyl groups allows for intramolecular interactions, which can significantly influence chemical reactions.
3-chloro-1-propanol
3-chloro-1-propanol is a three-carbon molecule with a chlorine atom on the 3rd carbon and a hydroxyl group on the 1st carbon. When hydrolysis occurs, 1,3-propanediol is formed.

Unlike 4-chloro-1-butanol, the spatial arrangement in 3-chloro-1-propanol does not allow for intramolecular hydrogen bonding, which affects its hydrolysis rate.
Intramolecular Hydrogen Bonding
Intramolecular hydrogen bonding occurs within a single molecule, where a hydrogen atom forms a bond with an electronegative atom (like oxygen or nitrogen) within the same molecule.

In the case of 4-chloro-1-butanol, the hydroxyl group on the 1st carbon can form a hydrogen bond with the chlorine atom on the 4th carbon during the hydrolysis reaction. This helps stabilize the transition state, making the reaction occur faster.

However, in 3-chloro-1-propanol, the positioning of the hydroxyl and chlorine atoms doesn’t allow for this type of bonding, thereby not offering the same stabilization during the reaction.
Sn2 Mechanism
Both 4-chloro-1-butanol and 3-chloro-1-propanol undergo hydrolysis via the Sn2 (bimolecular nucleophilic substitution) mechanism. This involves a nucleophile attacking the electrophilic carbon atom bonded to the leaving group (chlorine in this case) and displacing it.

The rate of an Sn2 reaction depends on several factors:
  • The strength of the nucleophile
  • The ability of the leaving group to leave
  • The structure of the substrate (steric hindrance)

The presence of intramolecular hydrogen bonding in 4-chloro-1-butanol makes this reaction proceed faster compared to 3-chloro-1-propanol, where such bonding is absent.
Reaction Rate
The hydrolysis of 4-chloro-1-butanol is quicker than that of 3-chloro-1-propanol.

This difference in reaction rates is mainly due to the stabilization provided by intramolecular hydrogen bonding in 4-chloro-1-butanol. Such stabilization is not present in 3-chloro-1-propanol, making it slower.

To sum up, the reaction rate is influenced by:
  • Transition state stabilization
  • The presence or absence of intramolecular hydrogen bonding
  • The overall structure and spatial arrangement of the molecules

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Most popular questions from this chapter

Treatment of the following two compounds with \(\mathrm{HCl}\) leads to the same chloride. Explain, mechanistically Hint: Watch the sulfur atom. What can sulfur do after an initial protonation of the oxygen atom?

The \(50-\mathrm{MHz}{ }^{13} \mathrm{C}\) NMR spectrum of the 2 -norbornyl cation in \(\mathrm{SbF}_{5} / \mathrm{SO}_{2} \mathrm{ClF} / \mathrm{SO}_{2} \mathrm{~F}_{2}\) solution exhibits an interesting temperature dependence. At \(-159{ }^{\circ} \mathrm{C}\), the spectrum consists of five lines at \(\delta 20.4[\mathrm{C}(5)], 21.2[\mathrm{C}(6)]\), \(36.3[\mathrm{C}(3)\) and \(\mathrm{C}(7)], 37.7[\mathrm{C}(4)]\), and \(124.5[\mathrm{C}(1)\) and \(\mathrm{C}(2)]\). However, at \(-80^{\circ} \mathrm{C}\), the spectrum consists of three lines at \(\delta 30.8[\mathrm{C}(3), \mathrm{C}(5)\), and \(\mathrm{C}(7)], 37.7[\mathrm{C}(4)]\), and \(91.7\) \([\mathrm{C}(1), \mathrm{C}(2)\), and \(\mathrm{C}(6)]\) ppm. Explain. Hint: Start with the low- temperature spectrum.

Heating \(\beta\)-hydroxyethylamide (1) with thionyl chloride affords \(\beta\)-chloroethylamide (2). In turn, amide 2 yields oxazoline 3 upon treatment with \(\mathrm{NaOH}\). (a) Write arrow formalism mechanisms for the formations of \(\mathbf{2}\) and \(\mathbf{3}\). Things are not quite as simple as they first appear. For example, if \(\mathbf{1}\) is treated with thionyl chloride below \(25^{\circ} \mathrm{C}\), compound 4 is obtained instead of 2 . Compound \(\mathbf{4}\) has the same elemental composition as \(\mathbf{2}\) but, unlike \(\mathbf{2}\), is soluble in water. Upon heating, \(\mathbf{4}\) gives \(\mathbf{2}\), and compound 4 yields 3 upon treatment with aqueous sodium carbonate. (b) Propose a structure for 4 and show how 4 is formed and converted into 2 and 3 .

When \(\mathrm{Z}=\mathrm{OCH}_{3}\), about 93\% of the products of solvolysis of \(\mathbf{1}\) can be attributed to neighboring phenyl group participation. When \(Z=\mathrm{NO}_{2}\), essentially none of the products come from neighboring participation by phenyl. Explain.
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