Concept explainers
Interpretation:
The given compound has to be named.
Concept Introduction:
All the molecules have their unique names. These names must be known in order to communicate. But remembering all the chemical names is impossible as there are many numbers of molecules. To avoid this, the
Stereoisomerism | Substituents | Parent | Unsaturation | Functional Group |
Stereoisomerism indicate if the considered molecule has any stereocenters are present (R,S) and if double bond is present are cis/trans. In order to name a double bond as cis/trans, the important condition is that, an identical group has to be present on either side of the double bond. If the identical groups are present on the same side of double bond, then it is known as cis. If the identical groups are present on the opposite side of the double bond then it is known as trans.
The groups that are connected to the main carbon chain are known as substituents. The substituents are identified after the carbon chain is identified and the functional group present in the given compound also identified. The substituents are named by adding “yl” to the end of the name which indicate that it is a substituent is an alkyl. The
The longest carbon chain is known as the parent. Parent carbon chain is the lengthiest carbon chain in the molecule that must include the functional group that is present in the compound. The longest carbon chain is identified and the parent name is given by the number of carbon atoms that is present in it. It must be remembered that even though functional group is not present in parent carbon chain, the double bond, triple bond if present has to be included. Numbering becomes a part of all the parts in IUPAC name. After identifying the parent chain, the numbering is done. If functional group is present in the given compound, the numbering is given in such a way that the functional group gets the least number. Then the double bond, triple bond.
Number of carbon atoms in chain | Parent |
1 | Meth |
2 | Eth |
3 | Prop |
4 | But |
5 | Pent |
6 | Hex |
Unsaturation indicates that if any triple or double bonds are present in the molecule. If a compound contains a double bond it is named as “-en-” and if a triple bond is present “-yn-” is used. If a compound contains two double bonds, then it is named as “-dien-”. If three double bonds are present in the given compound, then it is named as “-trien-”. This same rule applies for the compound that contains multiple triple bonds also.
Functional group is the one after which the considered compound is being named.
Functional Group | Class of compound | Suffix |
![]() | Ester | -oate |
![]() | Ketone | -one |
![]() | Aldehyde | -al |
![]() | -oic acid | |
![]() | Alcohol | -ol |
![]() | -amine |
For naming a given compound, the reverse order has to be followed. First the functional group is considered followed by unsaturation, parent chain, substituents, and then the stereoisomerism.

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Chapter 5 Solutions
ORGANIC CHEMISTRY-NEXTGEN+BOX (1 SEM.)
- 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
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