/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q32P Use resonance forms of the conju... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use resonance forms of the conjugate bases to explain why methanesulfonic acid (CH3SO3H,pKa= -2.6) is a much stronger acid than acetic acid (CH3COOH,pKa =4.8)

Short Answer

Expert verified

Methane sulfonic acid is a stronger acid than acetic acid as it is highly stabilized by the resonance structures and the inductive effect.

Step by step solution

01

Resonance forms

Resonance structures constitute structures that explain the electron delocalization in a particular molecule. A molecule comprising several resonance structures possesses significant stability.

02

Acidic Strength

Numerous factors impact the acidic strength, and some of them include resonance and inductive effect. The conjugate base stability is more significant for a molecule comprising resonance effect, and such molecules are acidic.

03

Resonance forms to explain why methanesulfonic acid is a stronger acid than acetic acid

The acidic strength decides the stability of a conjugate base. The acid comprising a more stable conjugate base is highly stable.

The methanesulfonate ion is stabilized by resonance and induction. It has three resonance structures. The sulfur atom is more electronegative than carbon and comprises a small role in stabilizing the negative charge on carbon.

The acetate ion has two resonance structures. It does not have an inductive effect to stabilize the anion. The resonance structures of acetate ion and methylsulfonate ion can be given as:

Resonance structures of acetate and methylsulfonate ion

Acetic acid is a mild acid, but methanesulfonic acid is stronger than acetic acid.

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 91Ó°ÊÓ!

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

Under normal circumstances, tertiary alcohols are not oxidized. However, when the tertiary alcohol is allylic, it can undergo a migration of the double bond (called an allylic shift) and subsequent oxidation of the alcohol. A particularly effective reagent for this reaction is Bobbitt’s reagent, similar to TEMPO used in many oxidations. (M. Shibuya et al., J. Org. Chem., 2008, 73, 4750.)

Show the expected product when each of these 3° allylic alcohols is oxidized by Bobbitt’s reagent.

(a)

(b)

(c)

(d)

Predict the major products of dehydration catalyzed by sulfuric acid.

(a) heptan-1-ol

(b) heptan-2-ol

(c) pentan-2-ol

(d) 1-methylcyclohexanol

(e) Cyclohexylmethanol

(f) 2-methylcyclohexanol

Give the structure of the principal product(s) when each of the following alcohols reacts with (1) Na2Cr2O7/H2SO4 (2) PCC, (3) DMP, and (4) 1 equiv NaOCl-TEMPO.

(a)octan-1-ol (b)octan-3-ol

(c) 4-hydroxydecanal (d) 1-methylcyclohexan-1,4-diol

Both cis- and trans-2-methylcyclohexanol undergo dehydration in warm sulfuric acid to give 1-methylcyclohexene as the major alkene product. These alcohols can also be converted to alkenes by tosylation usingand pyridine, followed by elimination using KOC(CH3)3as a strong base. Under these basic conditions, the tosylate of cis-2-methylcyclohexanol eliminates to give mostly 1-methylcyclohexene, but the tosylate of trans-2-methylcyclohexanol eliminates to give only 3-methylcyclohexene. Explain how this stereochemical difference in reactants controls a regiochemical difference in the products of the basic elimination, but not in the acid-catalyzed elimination.

A student wanted to use the Williamson ether synthesis to make (R)-2-ethoxybutane. He remembered that the Williamson synthesis involves an SN2 displacement, which takes place with inversion of configuration. He ordered a bottle of (S)-butan-2-ol for his chiral starting material. He also remembered that the SN2goes best on primary halides and tosylates, so he made ethyl tosylate and sodium (S)-but-2-oxide. After warming these reagents together, he obtained an excellent yield of 2-ethoxybutane.

(a) What enantiomer of 2-ethoxybutane did he obtain? Explain how this enantiomer results from the SN2 reaction of ethyl tosylate with sodium (S)-but-2-oxide.

(b) What would have been the best synthesis of (R)-2-ethoxybutane?

(c) How can this student convert the rest of his bottle of (S)-butan-2-ol to (R)-2-ethoxybutane?

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.