(a)
Interpretation: The synthetic routes for the given transformations have to be proposed
Concept Introduction:
Friedel-Crafts Alkylation: The Friedel-Crafts alkylation involves the electrophilic substitution of alkyl groups on
An azide synthesis: An azide synthesis involves the reaction between alkyl halide and sodium azide. Alkyl azide is formed in the first step. Alkyl azide on reduction with hydrogen in the presence of platinum produces primary amine. In this way, primary
Reductive amination: Sodium cyanoborohydride is a strong reducing agent than sodium borohydride. It reduces the carbonyl group into amine group in a rapid way. So, it is called as reductive amination reactions.
To find: Propose the synthetic routes for the given transformation (a)
Apply a retrosynthetic analysis
(b)
Interpretation: The synthetic routes for the given transformations have to be proposed
Concept Introduction:
Friedel-Crafts Alkylation: The Friedel-Crafts alkylation involves the electrophilic substitution of alkyl groups on aromatic rings when arenes are treated with alkyl halides in presence of Lewis acids. This reaction is catalyzed by Lewis acids like anhydrous AlCl3, FeX3, ZnCl2, BF3 etc.
Sandmeyer reaction: Sandmeyer reactions use copper salts as the reagents. Here, aryldiazonium salt is converted into aryl halides or aryl cyanides by using copper halides or copper cyanides.
A compound containing an amino group is treated with sodium nitrite and HCl leading to the formation of diazonium salt.
To find: Propose the synthetic routes for the given transformation (b)
Apply a retrosynthetic analysis
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Chapter 22 Solutions
ORGANIC CHEMISTRY-NEXTGEN+BOX (2 SEM.)
- Nonearrow_forwardNonearrow_forwardman 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_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(9 Pts) In one of the two Rare Earth element rows of the periodic table, identify an exception to the general ionization energy (IE) trend. For the two elements involved, answer the following questions. Be sure to cite sources for all physical data that you use. a. (2 pts) Identify the two elements and write their electronic configurations. b. (2 pts) Based on their configurations, propose a reason for the IE trend exception. c. (5 pts) Calculate effective nuclear charges for the last electron in each element and the Allred-Rochow electronegativity values for the two elements. Can any of these values explain the IE trend exception? Explain how (not) - include a description of how IE relates to electronegativity.arrow_forward
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- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
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