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Chapter 1: Q22P (page 19)

Change the following condensed structure to Kekul岷� structures:

a. CH3NH(CH2)2CH3

b.(CH3)2CHCl

c.(CH3)3CBr

d.(CH3)3C(CH2)3CHO

Short Answer

Expert verified

Organic structures are represented as structural formulas with bonds indicated by short lines. Such structural representations are known as the Kekule structures. Unlike the Lewis structure, lone pairs are omitted in the Kekule structure.

Step by step solution

01

Definition of Kekule structure

Organic structures are represented as structural formulas with bonds indicated by short lines. Such structural representations are known as the Kekule structures. Unlike the Lewis structure, lone pairs are omitted in the Kekule structure.

02

Subpart (a) Drawing the Kekule structure of CH3NH(CH2)2CH3 .

From the condensed formulas, omit the lone pair of electrons that represent each bond with short straight lines, and form the respective Kekule structure.

03

Subpart (b) Drawing the Kekule structure of .(CH3)2CHCl

The Kekule structure of the given compound is shown below.

04

Subpart (c) Drawing the Kekule structure of .CH3)3CBr

The Kekule structure of the given compound is shown below.

05

Subpart (d) Drawing the Kekule structure of .(CH3)3C(CH2)3CHO

The Kekule structure of the given compound is shown below.

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Most popular questions from this chapter

Predict whether He2+ exist?

a. Draw two Lewis structures for C2H6O.

b. Draw three Lewis structures for C3H8O.

(Hint:The two Lewis structures in part a are constitutional isomers鈥攎olecules that have the same atoms but differ in the way the atoms are connected. The three Lewis structures in part b are also

constitutional isomers.)

a. Which bond will be longer?

b. Which bond will be stronger?

  1. C-Cl or C-l
  2. C-C or C-l
  3. H-Cl or H-F

a. What are the relative bond lengths and strengths of the bonds in Cl2 and Br2 ?

b. What are the relative lengths and strengths of carbon-halogen bonds in CH3F,CH3Cl and CH3Br ?

A tertiary amine reacts with hydrogen peroxide to form a tertiary amine oxide.

Tertiary amine oxides undergo a reaction similar to the Hofmann elimination reaction (Section 10.10), called a Cope elimination. In this reaction, a tertiary amine oxide, rather than a quaternary ammonium ion, undergoes elimination. A strong base is not needed for a Cope elimination because the amine oxide acts as its own base.

Does the Cope elimination have an alkene-like transition state or a carbanion-like transition state?
See all solutions

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