(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.
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Chapter 17 Solutions
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
- Use a textbook or other valid source to research the physical and chemical properties of each element listed in Data Table 1 using the following as a guideline: Ductile (able to be deformed without losing toughness) and malleable (able to be hammered or pressed permanently out of shape without breaking or cracking) or not ductile or malleable Good, semi, or poor conductors of electricity and heat High or low melting and boiling points Occur or do not occur uncombined/freely in nature High, intermediate, or low reactivity Loses or gains electrons during reactions or is not reactivearrow_forwardProvide the Physical and Chemical Properties of Elements of the following elements listedarrow_forwardQuestions 4 and 5arrow_forward
- For a titration of 40.00 mL of 0.0500 M oxalic acid H2C2O4 with 0.1000 M KOH, calculate the pH at each of the following volume of KOH used in the titration: 1) before the titration begin;2) 15 mL; 3) 20 mL; 4) 25 mL; 5) 40 mL; 6) 50 mL. Ka1 = 5.90×10^-2, Ka2 = 6.50×10^-5 for oxalic acid.arrow_forwardPredict the major organic product(s), if any, of the following reactions. Assume all reagents are in excess unless otherwise indicated.arrow_forwardPredict the major organic product(s), if any, of the following reactions. Assume all reagents are in excess unless otherwise indicated.arrow_forward
- How many signals would you expect to find in the 1 H NMR spectrum of each given compound? Part 1 of 2 2 Part 2 of 2 HO 5 ☑ Х IIIIII***** §arrow_forwardA carbonyl compound has a molecular ion with a m/z of 86. The mass spectra of this compound also has a base peak with a m/z of 57. Draw the correct structure of this molecule. Drawingarrow_forwardCan you draw this using Lewis dot structures and full structures in the same way they are so that I can better visualize them and then determine resonance?arrow_forward
- Synthesize the following compound from cyclohexanol, ethanol, and any other needed reagentsarrow_forwardFor a titration of 20.00 mL of 0.0500 M H2SO4 with 0.100 M KOH, calculate the pH at each of the following volume of KOH used in the titration: 1) before the titration begin; 2) 10.00 mL; 3) 20.00 mL; 4) 30.00 mL. Ka2 = 1.20×10-2 for H2SO4.arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s) Be sure to account for all bond-breaking and bond-making steps Problem 73 of 10 Drawing Amows ro HO Donearrow_forward
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