Organic Chemistry: Principles and Mechanisms (Second Edition)
Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Chapter C, Problem C.15P
Interpretation Introduction

(a)

Interpretation:

The complete IUPAC name for the given compound is to be written.

Concept introduction:

When assigning priorities to substituents, the atom having the greater atomic number has higher priority. In case of comparison between isotopes, the one having the greater atomic mass gets higher priority. If the substituents are attached by the same atom, then the set of atoms one bond away from the point of attachment are compared. In each set, a comparison is made between the highest priority atoms. If the sets of atoms one bond away from the point of attachment are identical, then the sets of atoms one additional bond away from the point of attachment are compared. If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration. When more than one double bond is present, each is assigned E or Z configuration, and the location of each double bond appears immediately before the E or Z designation.

Interpretation Introduction

(b)

Interpretation:

The complete IUPAC name for the given compound is to be written.

Concept introduction:

When assigning priorities to substituents, the atom having the greater atomic number has higher priority. In case of comparison between isotopes, the one having the greater atomic mass gets higher priority. If the substituents are attached by the same atom, then the set of atoms one bond away from the point of attachment are compared. In each set, a comparison is made between the highest priority atoms. If the sets of atoms one bond away from the point of attachment are identical, then the sets of atoms one additional bond away from the point of attachment are compared. If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration. When more than one double bond is present, each is assigned E or Z configuration, and the location of each double bond appears immediately before the E or Z designation.

Interpretation Introduction

(c)

Interpretation:

The complete IUPAC name for the given compound is to be written.

Concept introduction:

When assigning priorities to substituents, the atom having the greater atomic number has higher priority. In case of comparison between isotopes, the one having the greater atomic mass gets higher priority. If the substituents are attached by the same atom, then the set of atoms one bond away from the point of attachment are compared. In each set, a comparison is made between the highest priority atoms. If the sets of atoms one bond away from the point of attachment are identical, then the sets of atoms one additional bond away from the point of attachment are compared. If the higher priority groups, attached to the double bonded carbon atoms, are on the same side of the double bond, the alkene is assigned Z configuration. If the higher priority groups, attached to the double bonded carbon atoms, are on the opposite side of the double bond, the alkene is assigned E configuration. When more than one double bond is present, each is assigned E or Z configuration, and the location of each double bond appears immediately before the E or Z designation.

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1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)
19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+
Li+ is a hard acid.  With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl
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