
(a)
Interpretation: Hexylamines have to be synthesized from various starting compounds.
Concept Introduction: The general formula for hexylamine is C6H13NH2. There are several methods available to prepare primary
Step-1: Formation of potassium phthalimide (deprotonation)
Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide.
Step-2: Formation of R−N bond by SN2 nucleophilic substitution
The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary
Step-3: Formation of primary amine by hydrolysis
The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative.
(b)
Interpretation: Hexylamines have to be synthesized from various starting compounds.
Concept Introduction: The general formula for hexylamine is C6H13NH2. There are several methods available to prepare primary amines. Among them, Gabriel synthesis plays a very important role for preparing it. In this method, secondary and tertiary amines are not formed as side products. It involves in three steps.
Step-1: Formation of potassium phthalimide (deprotonation)
Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide.
Step-2: Formation of R−N bond by SN2 nucleophilic substitution
The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize. As a result, a bond between nitrogen of phthalimide and carbon of R is formed. This is SN2 nucleophilic substitution reaction. Halogen atom is going away as halide anion.
Step-3: Formation of primary amine by hydrolysis
The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative.
(c)
Interpretation: Hexylamines have to be synthesized from various starting compounds.
Concept Introduction: The general formula for hexylamine is C6H13NH2. There are several methods available to prepare primary amines. Among them, Gabriel synthesis plays a very important role for preparing it. In this method, secondary and tertiary amines are not formed as side products. It involves in three steps.
Step-1: Formation of potassium phthalimide (deprotonation)
Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide.
Step-2: Formation of R−N bond by SN2 nucleophilic substitution
The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize. As a result, a bond between nitrogen of phthalimide and carbon of R is formed. This is SN2 nucleophilic substitution reaction. Halogen atom is going away as halide anion.
Step-3: Formation of primary amine by hydrolysis
The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative.
(d)
Interpretation: Hexylamines have to be synthesized from various starting compounds.
Concept Introduction: The general formula for hexylamine is C6H13NH2. There are several methods available to prepare primary amines. Among them, Gabriel synthesis plays a very important role for preparing it. In this method, secondary and tertiary amines are not formed as side products. It involves in three steps.
Step-1: Formation of potassium phthalimide (deprotonation)
Potassium phthalimide in alkaline KOH acts as the reagent which has negatively charged phthalimide. It is formed by the reaction between phthalimide and potassium hydroxide.
Step-2: Formation of R−N bond by SN2 nucleophilic substitution
The negative charged nitrogen atom in phthalimide can easily attract the positive side of R−X. In primary alkyl halides (R−X), R and X get positive and negative charges, respectively when they ionize. As a result, a bond between nitrogen of phthalimide and carbon of R is formed. This is SN2 nucleophilic substitution reaction. Halogen atom is going away as halide anion.
Step-3: Formation of primary amine by hydrolysis
The resultant product further goes for hydrolysis using hydrazine as the reagent. This reaction also follows nucleophilic substitution reaction. Finally, primary amine is formed with a side product of hydrazine derivative.

Want to see the full answer?
Check out a sample textbook solution
Chapter 23 Solutions
Organic Chemistry, Binder Ready Version
- please explain this in simple termsarrow_forwardK Most Reactive Na (3 pts) Can the metal activity series (shown on the right) or a standard reduction potential table explain why potassium metal can be prepared from the reaction of molten KCI and Na metal but sodium metal is not prepared from the reaction of molten NaCl and K metal? Show how (not). Ca Mg Al с Zn Fe Sn Pb H Cu Ag Au Least Reactivearrow_forward(2 pts) Why is O2 more stable as a diatomic molecule than S2?arrow_forward
- Draw the Lewis structure for the polyatomic phosphite (PO¾³¯) a anion. Be sure to include all resonance structures that satisfy the octet rule. C I A [ ]¯arrow_forwardDecide whether these proposed Lewis structures are reasonable. proposed Lewis structure Is the proposed Lewis structure reasonable? Yes. :0: Cl C C1: 0=0: : 0 : : 0 : H C N No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* Yes. ☐ No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | * If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0".arrow_forwardDraw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule. с [ ] - Garrow_forward
- 1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…arrow_forwardPlease draw, not just describe!arrow_forwardcan you draw each step on a piece of a paper please this is very confusing to mearrow_forward
- > Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? esc ? A O O •If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. olo 18 Ar Explanation Check BB Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibilityarrow_forwardName the structurearrow_forward> For each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that: 1. the rate of substitution doesn't depend on nucleophile concentration and 2. the products are a roughly 50/50 mixture of enantiomers. Substrate A Substrate B Faster Rate X CI (Choose one) (Choose one) CI Br Explanation Check Br (Choose one) C 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy A F10arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- 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





