/*! 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} Problem 6.3 By taking into account electrone... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

By taking into account electronegativity differences, draw the products formed by heterolysis of the carbon-heteroatom bond in each molecule. Classify the organic reactive intermediate as a carbocation or a carbanion.

Short Answer

Expert verified

Answer

a. Intermediate formed is the carbocation.

b. The reactive intermediate formed is the carbocation.

c. The reactive intermediate formed is carbanion.

Step by step solution

01

Step-by-Step SolutionStep 1: Electronegativity

The ability/potential of an atom to pull electron density from a bond shared with another atom towards itself is referred to as its electronegativity.

The electronegative atom bears a partial negative charge and the less electronegative one bears a partial positive charge.

02

Carbocation and Carbanion

A carbon atom bearing a positive charge on itself is termed as carbocation and the one bearing a negative charge is termed as carbanion.

A carbon atom bonded to atoms having electronegativity higher than carbon leads to the generation of a carbocation and the ones bonded to atoms having less electronegativity than carbon result in the formation of a carbanion.

03

Heterolysis

Heterolysis means the breakage of a bond in which one atom acquires both the shared pair of electrons.

The atom attaining excess electrons will bear a negative charge and the other will acquire a positive charge.

04

Heterolysis in the given compounds

a.

Heterolysis of a carbon-oxygen bond

The heterolysis of the carbon-oxygen (C-O) bond in the given compound generates a carbocation as the electrons shared between carbon and oxygen get transferred to the more electronegative oxygen atom.

b.

Heterolysis of carbon-bromine bond

The bromine atom is more electronegative than carbon and hence the electrons are transferred onto the bromine atom.

Thus, the heterolysis of the carbon-bromine bond generates a carbocation.

c.

Heterolysis of carbon-lithium bond

The carbon atom is more electronegative compared to lithium and hence it will pull the electrons towards itself thereby generating a carbanion.

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

Consider the following energy diagram

  1. How many steps are involved in this reaction?
  2. Label ∆H°and Ea for each step, and label∆H°overall .
  3. Label each transition state.
  4. Which point on the graph corresponds to a reactive intermediate?
  5. Which step is rate-determining?
  6. Is the overall reaction endothermic or exothermic?

As we learned in Chapter 4, propane (CH3CH2CH3)has both 1°and 2°hydrogens.

  1. Draw the carbon radical formed by homolysis of each type of C-H bond.
  2. Use the values in Table 6.2 to determine which C-H bond is stronger.
  3. Explain how this information can be used to determine the relative stability of the two radicals formed. Which radical formed from propane is more stable?

As we learned in Chapter 4, monosubstituted cyclohexanes exist as an equilibrium mixture of two conformations having either an axial or equatorial substituent. When R=CH2CH3, Keqfor this process is 23. When R=C(CH3)3,Keqfor this process is 4000.

a. When , which conformation is present in higher concentration?

b. Which R shows the higher percentage of equatorial conformation at equilibrium?

c. Which R shows the higher percentage of axial conformation at equilibrium?

d. For which R is more negative?

e. How is the size of R related to the amount of axial and equatorial conformations at equilibrium?

(a) Add curved arrows for each step to show how A is converted to the epoxy ketone C. (b) Classify the conversion of A to C as a substitution, elimination, or addition. (c) Draw one additional resonance structure for B.

Follow the curved arrows and draw the products of the following reaction.

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.