
(a)
Interpretation:
The given amine has to be classified as primary, secondary, or tertiary amine.
Concept Introduction:
Amine is an organic derivative. If in ammonia one or more alkyl, cycloalkyl, or aryl groups are substituted instead of hydrogen atom then it is known as amine. Depending on the number of substitution the
Amides are also organic derivative. In an amide, the nitrogen atom is bonded to a carbonyl group. The general structural formula of amide can be given as shown below,
The difference between amine and amide is that in amine, the nitrogen atom is bonded to a hydrocarbon chain. In case of amides, the nitrogen atom is bonded to a carbonyl group.
(b)
Interpretation:
The given amine has to be classified as primary, secondary, or tertiary amine.
Concept Introduction:
Amine is an organic derivative. If in ammonia one or more alkyl, cycloalkyl, or aryl groups are substituted instead of hydrogen atom then it is known as amine. Depending on the number of substitution the amines are classified as primary, secondary or tertiary amine. Primary amine is the one in which only one hydrogen atom in ammonia is replaced by a hydrocarbon group. Secondary amine is the one in which only two hydrogen atoms in ammonia is replaced by a hydrocarbon group. Tertiary amine is the one in which all three hydrogen atoms in ammonia is replaced by a hydrocarbon group. The generalized structural formula for all the amines is,
Amides are also organic derivative. In an amide, the nitrogen atom is bonded to a carbonyl group. The general structural formula of amide can be given as shown below,
The difference between amine and amide is that in amine, the nitrogen atom is bonded to a hydrocarbon chain. In case of amides, the nitrogen atom is bonded to a carbonyl group.
(c)
Interpretation:
The given amine has to be classified as primary, secondary, or tertiary amine.
Concept Introduction:
Amine is an organic derivative. If in ammonia one or more alkyl, cycloalkyl, or aryl groups are substituted instead of hydrogen atom then it is known as amine. Depending on the number of substitution the amines are classified as primary, secondary or tertiary amine. Primary amine is the one in which only one hydrogen atom in ammonia is replaced by a hydrocarbon group. Secondary amine is the one in which only two hydrogen atoms in ammonia is replaced by a hydrocarbon group. Tertiary amine is the one in which all three hydrogen atoms in ammonia is replaced by a hydrocarbon group. The generalized structural formula for all the amines is,
Amides are also organic derivative. In an amide, the nitrogen atom is bonded to a carbonyl group. The general structural formula of amide can be given as shown below,
The difference between amine and amide is that in amine, the nitrogen atom is bonded to a hydrocarbon chain. In case of amides, the nitrogen atom is bonded to a carbonyl group.
(d)
Interpretation:
The given amine has to be classified as primary, secondary, or tertiary amine.
Concept Introduction:
Amine is an organic derivative. If in ammonia one or more alkyl, cycloalkyl, or aryl groups are substituted instead of hydrogen atom then it is known as amine. Depending on the number of substitution the amines are classified as primary, secondary or tertiary amine. Primary amine is the one in which only one hydrogen atom in ammonia is replaced by a hydrocarbon group. Secondary amine is the one in which only two hydrogen atoms in ammonia is replaced by a hydrocarbon group. Tertiary amine is the one in which all three hydrogen atoms in ammonia is replaced by a hydrocarbon group. The generalized structural formula for all the amines is,
Amides are also organic derivative. In an amide, the nitrogen atom is bonded to a carbonyl group. The general structural formula of amide can be given as shown below,
The difference between amine and amide is that in amine, the nitrogen atom is bonded to a hydrocarbon chain. In case of amides, the nitrogen atom is bonded to a carbonyl group.

Trending nowThis is a popular solution!

Chapter 17 Solutions
General, Organic, and Biological Chemistry Seventh Edition
- Please help me calculate the undiluted samples ppm concentration. My calculations were 280.11 ppm. Please see if I did my math correctly using the following standard curve. Link: https://mnscu-my.sharepoint.com/:x:/g/personal/vi2163ss_go_minnstate_edu/EVSJL_W0qrxMkUjK2J3xMUEBHDu0UM1vPKQ-bc9HTcYXDQ?e=hVuPC4arrow_forwardProvide an IUPAC name for each of the compounds shown. (Specify (E)/(Z) stereochemistry, if relevant, for straight chain alkenes only. Pay attention to commas, dashes, etc.) H₁₂C C(CH3)3 C=C H3C CH3 CH3CH2CH CI CH3 Submit Answer Retry Entire Group 2 more group attempts remaining Previous Nextarrow_forwardArrange the following compounds / ions in increasing nucleophilicity (least to most nucleophilic) CH3NH2 CH3C=C: CH3COO 1 2 3 5 Multiple Choice 1 point 1, 2, 3 2, 1, 3 3, 1, 2 2, 3, 1 The other answers are not correct 0000arrow_forward
- curved arrows are used to illustrate the flow of electrons. using the provided starting and product structures, draw the cured electron-pushing arrows for thw following reaction or mechanistic steps. be sure to account for all bond-breaking and bond making stepsarrow_forwardUsing the graphs could you help me explain the answers. I assumed that both graphs are proportional to the inverse of time, I think. Could you please help me.arrow_forwardSynthesis of Dibenzalacetone [References] Draw structures for the carbonyl electrophile and enolate nucleophile that react to give the enone below. Question 1 1 pt Question 2 1 pt Question 3 1 pt H Question 4 1 pt Question 5 1 pt Question 6 1 pt Question 7 1pt Question 8 1 pt Progress: 7/8 items Que Feb 24 at You do not have to consider stereochemistry. . Draw the enolate ion in its carbanion form. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. ⚫ Separate multiple reactants using the + sign from the drop-down menu. ? 4arrow_forward
- Shown below is the mechanism presented for the formation of biasplatin in reference 1 from the Background and Experiment document. The amounts used of each reactant are shown. Either draw or describe a better alternative to this mechanism. (Note that the first step represents two steps combined and the proton loss is not even shown; fixing these is not the desired improvement.) (Hints: The first step is correct, the second step is not; and the amount of the anhydride is in large excess to serve a purpose.)arrow_forwardHi I need help on the question provided in the image.arrow_forwardDraw a reasonable mechanism for the following reaction:arrow_forward
- Draw the mechanism for the following reaction: CH3 CH3 Et-OH Et Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where needed. Electron-flow arrows should start on the electron(s) of an atom or a bond and should end on an atom, bond, or location where a new bond should be created. H± EXP. L CONT. י Α [1] осн CH3 а CH3 :Ö Et H 0 N о S 0 Br Et-ÖH | P LL Farrow_forward20.00 mL of 0.150 M NaOH is titrated with 37.75 mL of HCl. What is the molarity of the HCl?arrow_forward20.00 mL of 0.025 M HCl is titrated with 0.035 M KOH. What volume of KOH is needed?arrow_forward
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningGeneral, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
- Organic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,



