Question: What do the following pKa values tell you about the electron-withdrawing abilities of nitro, cyano, chloro, and hydroxy groups?

Short Answer

Expert verified

Answer

The magnitude of the electron-withdrawing effect in the order of increasing manner is, OH < Cl < CN < . Thus, the nitro group is the strongest electron-withdrawing group of all.

Step by step solution

01

Electron-withdrawing groups

An electron-withdrawing group withdraws electrons present in the reaction centre, and the electron density of a conjugated pi system decreases.Examples of electron-withdrawing groups are nitro (-NO2), cyano (-CN), carboxy (-COOH) groups etc.

02

The electron-withdrawing abilities of cyano, chloro, nitro and hydroxyl groups 

The acetic acid derivatives are often used to test the electronic effects of substituents series. The synthesis of these compounds is easy, and pKavalues can easily be measured by the titration method.

The substituents present on carbon 2 of acetic acid can have only an inductive effect, and no resonance effect is possible. This is because theCH2 group is sp3 hybridized, and there is no possibility of pi overlapping.

So, the conclusions can be drawn by the givenpKa values. Firstly, all the four different substituents present on CH2 group are electron-withdrawing groups because all are stronger than acetic acid. Secondly, the magnitude of the electron-withdrawing effect in the order of increasing manner is, OH < Cl < CN < NO2. Thus, the nitro group is the strongest electron-withdrawing CH2 group of all.

03

Magnitude of electron-withdrawing effect

The magnitude of the electron-withdrawing effect in the order of increasing manner is, OH < Cl < CN <NO2 . Thus, the nitro group is the strongest electron-withdrawing group of all.

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.

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

Question:

  1. Why do most long-chain fatty acids show a large peak in the mass spectrum at m/z 60?
  2. Use equations to explain the prominent peaks at m/z 74 and m/z 87 in the mass spectrum of 2-methylpentanoic acid.
  3. Why doesn’t the mass spectrum of 2-methylpentanoic acid show a large peak at m/z 60?

Question:An unknown compound gives a molecular ion of m/z 70 in the mass spectrum. It reacts with semicarbazide hydrochloride to give a crystalline derivative, but it gives a negative Tollens test. The NMR and IR spectra follow. Propose a structure for this compound, and give peak assignments to account for the absorptions in the spectra. Explain why the signal at 1790 cm-1 in the IR spectrum appears at an unusual frequency.

Question: Propose a mechanism for conversion of the dianion to the ketone under mildly acidic conditions.

Question: A carboxylic acid has two oxygen atoms, each with two nonbonding pairs of electrons.

  1. Draw the resonance forms of a carboxylic acid that is protonated on the hydroxy oxygen atom.
  2. Compare the resonance forms with those given previously for an acid protonated on the carbonyl oxygen atom.
  3. Explain why the carbonyl oxygen atom of a carboxylic acid is more basic than the hydroxy oxygen.

Question: A carboxylic acid has two oxygen atoms, each with two nonbonding pairs of electrons.

  1. Draw the resonance forms of a carboxylic acid that is protonated on the hydroxy oxygen atom.
  2. Compare the resonance forms with those given previously for an acid protonated on the carbonyl oxygen atom.
  3. Explain why the carbonyl oxygen atom of a carboxylic acid is more basic than the hydroxy oxygen.
See all solutions

Recommended explanations on Chemistry Textbooks

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