(a)
Interpretation: The more basic
Concept Introduction:
If the lone pair of electrons on nitrogen atom in amine compound is delocalized, that compound is resonance stabilized. It is also involved in
To find: Decide the more basic amine compound in the given pairs of compounds (a)
Get a compound where lone pairs on nitrogen atoms are delocalized
(b)
Interpretation: The more basic amine compound has to be found for the given pairs of compounds
Concept Introduction:
If the lone pair of electrons on nitrogen atom in amine compound is delocalized, that compound is resonance stabilized. It is also involved in aromatic stabilization. As a result, a number of resonance structures are drawn for the delocalization of electrons in the structures. So, the compound has less basic character. If the lone pair of electrons on nitrogen atom in amine compound is localized, that compound doesn’t involve in resonance stabilization. So, that compound has more basic character.
To find: Decide the more basic amine compound in the given pairs of compounds (b)
Get a compound where lone pairs on nitrogen atoms are delocalized
(c)
Interpretation: The more basic amine compound has to be found for the given pairs of compounds
Concept Introduction:
If the lone pair of electrons on nitrogen atom in amine compound is delocalized, that compound is resonance stabilized. It is also involved in aromatic stabilization. As a result, a number of resonance structures are drawn for the delocalization of electrons in the structures. So, the compound has less basic character. If the lone pair of electrons on nitrogen atom in amine compound is localized, that compound doesn’t involve in resonance stabilization. So, that compound has more basic character.
To find: Decide the more basic amine compound in the given pairs of compounds (c)
Get a compound where lone pairs on nitrogen atoms are delocalized
(d)
Interpretation: The more basic amine compound has to be found for the given pairs of compounds
Concept Introduction:
If the lone pair of electrons on nitrogen atom in amine compound is delocalized, that compound is resonance stabilized. It is also involved in aromatic stabilization. As a result, a number of resonance structures are drawn for the delocalization of electrons in the structures. So, the compound has less basic character. If the lone pair of electrons on nitrogen atom in amine compound is localized, that compound doesn’t involve in resonance stabilization. So, that compound has more basic character.
To find: Decide the more basic amine compound in the given pairs of compounds (d)
Get a compound where lone pairs on nitrogen atoms are delocalized
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Chapter 22 Solutions
Organic Chemistry
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- Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the intermediates and product of the following reaction or mechanistic step(s).arrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the arrows to draw the intermediate and the product in this reaction or mechanistic step(s).arrow_forwardLook at the following pairs of structures carefully to identify them as representing a) completely different compounds, b) compounds that are structural isomers of each other, c) compounds that are geometric isomers of each other, d) conformers of the same compound (part of structure rotated around a single bond) or e) the same structure.arrow_forward
- Given 10.0 g of NaOH, what volume of a 0.100 M solution of H2SO4 would be required to exactly react all the NaOH?arrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward3.50 g of Li are combined with 3.50 g of N2. What is the maximum mass of Li3N that can be produced? 6 Li + N2 ---> 2 Li3Narrow_forward
- Concentration Trial1 Concentration of iodide solution (mA) 255.8 Concentration of thiosulfate solution (mM) 47.0 Concentration of hydrogen peroxide solution (mM) 110.1 Temperature of iodide solution ('C) 25.0 Volume of iodide solution (1) used (mL) 10.0 Volume of thiosulfate solution (5:03) used (mL) Volume of DI water used (mL) Volume of hydrogen peroxide solution (H₂O₂) used (mL) 1.0 2.5 7.5 Time (s) 16.9 Dark blue Observations Initial concentration of iodide in reaction (mA) Initial concentration of thiosulfate in reaction (mA) Initial concentration of hydrogen peroxide in reaction (mA) Initial Rate (mA's)arrow_forwardDraw the condensed or line-angle structure for an alkene with the formula C5H10. Note: Avoid selecting cis-/trans- isomers in this exercise. Draw two additional condensed or line-angle structures for alkenes with the formula C5H10. Record the name of the isomers in Data Table 1. Repeat steps for 2 cyclic isomers of C5H10arrow_forwardExplain why the following names of the structures are incorrect. CH2CH3 CH3-C=CH-CH2-CH3 a. 2-ethyl-2-pentene CH3 | CH3-CH-CH2-CH=CH2 b. 2-methyl-4-pentenearrow_forward
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