Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 27: Problem 73

The iodide ion cannot displace the \(-\mathrm{OH}\) group in ethanol, but excess HI will react to produce ethyl iodide. Explain.

Short Answer

Expert verified
The weak nucleophilic nature of iodide ions prevents them from displacing the OH group in ethanol. However, excess hydroiodic acid (HI) provides a more concentrated source of iodide ions, hence overcoming the steric hindrance to the reaction, and leads to the formation of ethyl iodide.

Step by step solution

01

Understand the Reactivity of Iodide and Hydroxyl Ions

The iodide ion is a weak nucleophile because it carries a negative charge. Negative charges repel the electron cloud of the organic substrate, causing reduced reactivity. On the other hand, the hydroxyl (OH) group in ethanol is a strong base due to its ability to donate an electron pair. Therefore, it is less likely for the iodide ion to displace the hydroxyl group.
02

Consider the Effect of Excess HI

When there is excess HI present, the reaction can go forward because the increased concentration of iodide ions helps to overcome the steric hindrance to its reaction with the ethanol. Hence, the reaction with excess HI ends up forming ethyl iodide.
03

Formulate a Comprehensive Explanation

The iodide ion cannot displace the OH group in ethanol due to its weak nucleophilic nature as opposed to the strong basic nature of OH. However, in the presence of excess HI, the concentration of iodide ions is increased, which enables the reaction to go forward, leading to the formation of ethyl iodide.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Answer the following questions for this \(S_{\mathrm{N}} 2\) reaction: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{NaOH} & \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+ \mathrm{NaBr} \end{aligned}$$ (a) What is the rate expression for the reaction? (b) Draw the reaction profile for the reaction. Label all parts. Assume that the products are lower in energy than the reactants. (c) What is the effect on the rate of the reaction of doubling the concentration of \(n\) -butyl bromide? (d) What is the effect on the rate of the reaction of halving the concentration of sodium hydroxide?

What is the major product expected from the monobromination of methylcyclohexane?

Write the formulas of the products expected to form in the following situations. If no reaction occurs, write N.R. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(\mathrm{aq})+\mathrm{HBr}(\mathrm{aq}) \longrightarrow\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow\)

Write the formulas of the products formed from the reaction of 2 -butene with each of the following substances: (a) \(\mathrm{H}_{2} ;\) (b) \(\mathrm{Cl}_{2} ;\) (c) \(\mathrm{HCl} ;\) (d) \(\mathrm{H}_{2} \mathrm{O}\) (in acid).

Indicate the principal product(s) you would expect in (a) treating \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) (b) exposing a mixture of chlorine and propane gases to ultraviolet light; (c) heating a mixture of isopropyl alcohol and benzoic acid; (d) oxidizing sec-butyl alcohol with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) in acidic solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free