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Interpretation:
The syntheses of the two compounds, ethanolamine and diethanolamine, are to be proposed.
Concept Introduction:
They can act as nucleophiles and electrophiles in organic reactions.
Amines can be classified into three categories: primary amines, secondary amines, and tertiary amines.
Electrophiles are electron deficient species which has positive or partially positive charge. Lewis acids are electrophiles which accept electron pair.
Nucleophiles are electron rich species which has negative or partially negative charge. Lewis bases are nucleophiles which donate electron pair.
Free radical is an atom, molecule or ion that has unpaired electrons which makes it highly chemically reactive.
Substitution reaction: A reaction in which one of the hydrogen atoms of a hydrocarbon or a
Elimination reaction: A reaction in which two substituent groups are detached and a double bond is formed is called elimination reaction.
Addition reaction: It is the reaction in which unsaturated bonds are converted to saturated molecules by the addition of molecules.
Oxirane is a cyclic compound containing three atoms; it easily undergoes ring opening due to ring strain, and forms straight-chain compounds.
The reaction of oxirane with amine leads to the formation of straight-chain compounds containing both hydroxyl and amine groups, named ethanolamine.
The reaction of ethanolamine with another molecule of oxirane leads to the substitution of another alcohol group on the nitrogen atom, and forms the compound, diethanolamine.
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Chapter 20 Solutions
ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
- 18. Arrange the following carbocations in order of decreasing stability. 1 2 A 3124 B 4213 C 2431 D 1234 E 2134 SPL 3 4arrow_forwardAcetic acid is added to DI water at an initial concentration of 10 -6 M (Ka=1.8x10-5) A. Using the "ICE" Method, what would the pH be at equilibrium? State assumptions and show your work. B. Using the simultaneous equations method, what would the pH be at equilibrium? Show your workarrow_forward1. Show that the change in entropy for a fixed amount of ideal gas held at a constant temperature undergoing a volume change is given by the simple equation AS = NkB In Hint: Start with the equation M dS = du + (Œ) dv - Ž (#) an, dU du+av-dN; j=1 Why doesn't the equation for the entropy of an ideal gas depend on the strength of the intermolecular forces for the gas?arrow_forward
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- 2. Use Hess's law to calculate the AH (in kJ) for: rxn CIF(g) + F2(g) → CIF 3 (1) using the following information: 2CIF(g) + O2(g) → Cl₂O(g) + OF 2(g) AH = 167.5 kJ ΔΗ 2F2 (g) + O2(g) → 2 OF 2(g) 2C1F3 (1) + 202(g) → Cl₂O(g) + 3 OF 2(g) о = = -43.5 kJ AH = 394.1kJarrow_forwardci Draw the major product(s) of the following reactions: (3 pts) CH3 HNO3/H2SO4 HNO3/ H2SO4 OCH3 (1 pts)arrow_forwardProvide the product for the reactionarrow_forward
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