EP ORGANIC CHEMISTRY -MOD.MASTERING 18W
EP ORGANIC CHEMISTRY -MOD.MASTERING 18W
9th Edition
ISBN: 9780136781776
Author: Wade
Publisher: PEARSON CO
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Chapter 16, Problem 16.44SP

(a)

Interpretation Introduction

Interpretation:

The information regarding the conjugation in the molecule is given. The nature of the given molecule as aromatic, antiaromatic or non-aromatic is to be predicted.

Concept introduction:

The conditions that a molecule must follow to be an aromatic are shown below.

  • The given compound must be cyclic in nature.
  • It must be planar.
  • There must be no sp3 hybridized carbon in the given compound.
  • The compound must follow Huckel’s rule of aromaticity that is (4n+2)π electron system must be present.

If the given compound follows first three conditions but it follows 4nπ electron system, then it is antiaromatic in nature. If the given compound fails to follow any of the above conditions then it is non-aromatic in nature.

(b)

Interpretation Introduction

Interpretation:

An explanation corresponding to the synthesis of molecule with three tert-butyl substituents and reason for not synthesizing unsubstituted compound is to be stated.

Concept introduction:

The tendency of a group to increase the electron density of the chain or ring by the transfer of electron density is known as +I effect.

(c)

Interpretation Introduction

Interpretation:

An explanation corresponding to the possibility of basic character of nitrogen atom in the given molecule and the non-overlapping of lone pair of nitrogen with the double bond to give a total of six electrons in the pi system is to be stated.

Concept introduction:

According to Lewis concept, the species that can easily accept the electrons is known as an acid and the species that donates the electrons is known as a base.

(d)

Interpretation Introduction

Interpretation:

The information regarding the NMR data corresponding to the given structure is given. An explanation corresponding to the bonding in given compound on the basis of NMR data and similarity between the conclusion based on the NMR data and prediction done in part (a) is to be stated.

Concept introduction:

NMR data reveals the number and type of hydrogen atoms that are present in a given compound.

The number of NMR signal in a compound is equal to number of chemically non-equivalent protons present in that compound. The splitting of NMR signals take place according to N+1 rule, where N is the number of adjacent protons. Protons which are present in same chemical environment, that is between the same group of atoms are known as chemically equivalent protons and in 1HNMR all chemically equivalent proton generates one signal or one peak, whereas non-equivalent proton generates different signals.

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Chapter 16 Solutions

EP ORGANIC CHEMISTRY -MOD.MASTERING 18W

Ch. 16.8A - Repeat Problem16-10 for the cyclopentadienyl ions....Ch. 16.8C - Explain why each compound or ion should be...Ch. 16.8C - The following hydrocarbon has an unusually large...Ch. 16.8C - Prob. 16.14PCh. 16.8C - Prob. 16.15PCh. 16.9B - Prob. 16.16PCh. 16.9C - Show which of the nitrogen atoms in purine are...Ch. 16.9C - The proton NMR spectrum of 2-pyridone gives the...Ch. 16.9D - Prob. 16.19PCh. 16.9D - Prob. 16.20PCh. 16.10 - Prob. 16.21PCh. 16.12 - Ciprofloxacin is a member of the fluoroquinolone...Ch. 16.13 - Draw and name all the chlorinated benzenes having...Ch. 16.13 - Name the following compounds:Ch. 16.15 - The UV spectrum of 1-phenylprop-2-en-1-ol shows an...Ch. 16 - Prob. 16.26SPCh. 16 - Name the following compounds:Ch. 16 - Draw and name all the methyl, dimethyl, and...Ch. 16 - Four pairs of compounds are shown. In each pair,...Ch. 16 - One of the following hydrocarbons is much more...Ch. 16 - In Kekuls time cyclohexane was unknown, and there...Ch. 16 - Prob. 16.32SPCh. 16 - Azulene is a deep-blue hydrocarbon with resonance...Ch. 16 - Prob. 16.34SPCh. 16 - Prob. 16.35SPCh. 16 - Prob. 16.36SPCh. 16 - Prob. 16.37SPCh. 16 - Prob. 16.38SPCh. 16 - Prob. 16.39SPCh. 16 - Biphenyl has the following structure. a. Is...Ch. 16 - Anions of hydrocarbons are rare, and dianions of...Ch. 16 - How would you convert the following compounds to...Ch. 16 - Prob. 16.43SPCh. 16 - Prob. 16.44SPCh. 16 - A student found an old bottle labeled thymol on...Ch. 16 - Prob. 16.46SPCh. 16 - Prob. 16.47SPCh. 16 - Prob. 16.48SPCh. 16 - The proton NMR chemical shifts of the hydrogens in...Ch. 16 - Prob. 16.50SPCh. 16 - NMR has been used to probe many molecular...
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