Concept explainers
Interpretation:
The complete IUPAC name of the given molecule is to be written.
Concept introduction:
When assigning priorities to substituents, the atom having the greater
If substituents are attached by the same atom, the tiebreaker is applied.
For each substituent, the set of three atoms one bond away from its point of attachment is to be identified.
In each set, arrange the three atoms from the highest to the lowest priority. Compare each set’s highest-priority atom.
If they are different, then the atom that has the higher priority corresponds to the higher-priority substituent.
If the highest-priority atoms from each set are identical, then compare each set’s second highest priority corresponding to the higher priority substituent. If the second-highest-priority atoms from each set are identical, then compare each set’s lowest-priority atom to break the tie.
When the fourth priority substituent is pointing away (it is attached by a dash bond) and the first, second, and third priority substituents are arranged clockwise, the configuration is R.
When the fourth priority substituent is pointing away (it is attached by a dash bond) and the first, second, and third priority substituents are arranged counterclockwise, the configuration is S.
If the fourth priority substituent is attached by a wedge bond, then the clockwise or counterclockwise arrangement of the first, second, and third priority substituents is determined, and that arrangement is reversed before assigning R or S.
If the fourth priority substituent is in the plane of the page, then it is switched with the substituent that points away. Then the clockwise or counterclockwise arrangement of the first, second, and third priority substituents is determined, and that arrangement is reversed before assigning R or S.
When writing the IPUAC name, the R or S designation is written in parenthesis for each asymmetric carbon atom, and hyphens are used to separate those designations from the rest of the IUPAC name.

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Chapter C Solutions
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
- Part I. Draw reaction mechanism for the transformations of benzophenone to benzopinacol to benzopinaco lone and answer the ff: Pinacol (2,3-dimethyl, 1-3-butanediol) on treatment w/ acid gives a mixture of pina colone and (3,3-dimethyl-2-butanone) 2,3-dimethyl-1,3-butadiene. Give reasonable mechanism the formation of the products Forarrow_forwardShow the mechanism for these reactionsarrow_forwardDraw the stepwise mechanismarrow_forward
- Draw a structural formula of the principal product formed when benzonitrile is treated with each reagent. (a) H₂O (one equivalent), H₂SO₄, heat (b) H₂O (excess), H₂SO₄, heat (c) NaOH, H₂O, heat (d) LiAlH4, then H₂Oarrow_forwardDraw the stepwise mechanism for the reactionsarrow_forwardDraw stepwise mechanismarrow_forward
- Part I. Draw reaction mechanism for the transformations of benzophenone to benzopinacol to benzopinaco lone and answer the ff: a) Give the major reason for the exposure of benzophenone al isopropyl alcohol (w/acid) to direct sunlight of pina colone Mechanism For b) Pinacol (2,3-dimethy 1, 1-3-butanediol) on treatment w/ acid gives a mixture (3,3-dimethyl-2-butanone) and 2, 3-dimethyl-1,3-butadiene. Give reasonable the formation of the productsarrow_forwardwhat are the Iupac names for each structurearrow_forwardWhat are the IUPAC Names of all the compounds in the picture?arrow_forward
- 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)arrow_forward19.78 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+arrow_forwardLi+ 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 LiClarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
