Organic Chemistry
Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
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Chapter 15, Problem 15.81P
Interpretation Introduction

(a)

Interpretation: An explanation for higher stability of triphenylmethyl radical than other radicals is to be stated.

Concept introduction: The delocalization of lone pair or free electrons from one place to another is known as resonance. The stability of compound depends upon the number of resonating structures. More the resonating structures of compound more will be its stability.

Interpretation Introduction

(b)

Interpretation: The dimerization of two triphenylmethyl radicals to form A is to be represented by the use of curved arrow notation.

Concept introduction: The delocalization of lone pair or free electrons from one place to another is known as resonance. The stability of compound depends upon the number of resonating structures. More the resonating structures of compound more will its stability.

Interpretation Introduction

(c)

Interpretation: A reason for the formation of A rather than hexaphenylethane is to be proposed.

Concept introduction: The chemical reaction which proceeds through the formation of radicals is known as chain reaction. It involves three steps, initiation, propagation and termination.

Interpretation Introduction

(d)

Interpretation: An explanation of the use of 1H and 13C NMR spectroscopy to distinguish between hexaphenylethane and A is to be stated.

Concept introduction: Spectroscopy method is used to identify the structure of the molecule. It is based on the interactions between matter and electromagnetic radiations. Proton NMR spectroscopy identifies the number of hydrogen atoms present in a molecule and the nature of the functional group. The value of chemical peaks depends upon the chemical environment around the hydrogen atom.

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The triphenylmethyl radical is an unusual persistent radical present insolution in equilibrium with its dimer. For 70 years the dimer was thoughtto be hexaphenylethane, but in 1970, NMR data showed it to be A. a.) Why is the triphenylmethyl radical more stable than most otherradicals?b.) Use curved arrow notation to show how two triphenylmethyl radicalsdimerize to form A.c.) Propose a reason for the formation of A rather than hexaphenylethane.d.) How could 1H and 13C NMR spectroscopy be used to distinguishbetween hexaphenylethane and A ?
IV B. Answer only a and b
2. For each of the following molecules, draw all possible mono-brominated products of an allylic bromination. (hint: Don't forget to look for resonance forms of the allylic radical initially formed.) a. b. f a

Chapter 15 Solutions

Organic Chemistry

Ch. 15 - Prob. 15.11PCh. 15 - Synthesize each compound from (CH3)3CH. a....Ch. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Draw all constitutional isomers formed when each...Ch. 15 - Draw the structure of the four allylic halides...Ch. 15 - Which compounds can be prepared in good yield by...Ch. 15 - Which CH bond is most readily cleaved in linolenic...Ch. 15 - Prob. 15.23PCh. 15 - Draw the products formed when each alkene is...Ch. 15 - Problem 15.24 When adds to under radical...Ch. 15 - Prob. 15.26PCh. 15 - Draw an energy diagram for the two propagation...Ch. 15 - Prob. 15.28PCh. 15 - Problem 15.27 Draw the steps of the mechanism that...Ch. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Why is a benzylic CH bond labeled in red unusually...Ch. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - What alkane is needed to make each alkyl halide by...Ch. 15 - Which alkyl halides can be prepared in good yield...Ch. 15 - Prob. 15.41PCh. 15 - 15.40 Explain why radical bromination of p-xylene...Ch. 15 - a. What product(s) (excluding stereoisomers) are...Ch. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - 15.44 Draw all constitutional isomers formed when...Ch. 15 - Draw the organic products formed in each reaction....Ch. 15 - Prob. 15.49PCh. 15 - 15.47 Treatment of a hydrocarbon A (molecular...Ch. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - 15.53 Consider the following bromination: . a....Ch. 15 - 15.54 Draw a stepwise mechanism for the following...Ch. 15 - Prob. 15.57PCh. 15 - An alternative mechanism for the propagation steps...Ch. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Devise a synthesis of each compound from...Ch. 15 - Devise a synthesis of each target compound from...Ch. 15 - Devisea synthesis of each target compound from the...Ch. 15 - Devise a synthesis of each compound using CH3CH3...Ch. 15 - Prob. 15.65PCh. 15 - 15.63 As described in Section 9.16, the...Ch. 15 - 15.64 Ethers are oxidized with to form...Ch. 15 - Prob. 15.68PCh. 15 - Prob. 15.69PCh. 15 - 15.67 In cells, vitamin C exists largely as its...Ch. 15 - What monomer is needed to form each...Ch. 15 - Prob. 15.72PCh. 15 - Prob. 15.73PCh. 15 - 15.71 Draw a stepwise mechanism for the following...Ch. 15 - 15.72 As we will learn in Chapter 30, styrene...Ch. 15 - Prob. 15.76PCh. 15 - 15.74 A and B, isomers of molecular formula , are...Ch. 15 - Prob. 15.78PCh. 15 - Radical chlorination of CH3CH3 forms two minor...Ch. 15 - 15.76 Draw a stepwise mechanism for the...Ch. 15 - Prob. 15.81PCh. 15 - Prob. 15.82PCh. 15 - Prob. 15.83P
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