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Concept explainers
Interpretation:
The parent chain has to be chosen and numbered correctly for the given compound.
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 is 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 |
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Chapter 5 Solutions
Organic Chemistry As a Second Language: First Semester Topics
- man Campus Depa (a) Draw the three products (constitutional isomers) obtained when 2-methyl-3-hexene reacts with water and a trace of H2SO4. Hint: one product forms as the result of a 1,2-hydride shift. (1.5 pts) This is the acid-catalyzed alkene hydration reaction.arrow_forwardNonearrow_forward. • • Use retrosynthesis to design a synthesis Br OHarrow_forward
- 12. Choose the best diene and dienophile pair that would react the fastest. CN CN CO₂Et -CO₂Et .CO₂Et H3CO CO₂Et A B C D E Farrow_forward(6 pts - 2 pts each part) Although we focused our discussion on hydrogen light emission, all elements have distinctive emission spectra. Sodium (Na) is famous for its spectrum being dominated by two yellow emission lines at 589.0 and 589.6 nm, respectively. These lines result from electrons relaxing to the 3s subshell. a. What is the photon energy (in J) for one of these emission lines? Show your work. b. To what electronic transition in hydrogen is this photon energy closest to? Justify your answer-you shouldn't need to do numerical calculations. c. Consider the 3s subshell energy for Na - use 0 eV as the reference point for n=∞. What is the energy of the subshell that the electron relaxes from? Choose the same emission line that you did for part (a) and show your work.arrow_forwardNonearrow_forward
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