Concept explainers
(a)
Interpretation:
Given amide has to be classified as primary, secondary, or tertiary amide.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Organic compounds contain heteroatom also. Some of them are nitrogen, sulfur, oxygen etc. Nitrogen containing organic compounds are of two important types and they are
One of the
Amides are also classified as primary, secondary, and tertiary amide.
Primary amide is the one that has a nitrogen atom that is bonded to two hydrogen atoms. Primary amides are also known as unsubstituted amides.
Secondary amide is the one that has a nitrogen atom that is bonded to one hydrogen atom and one alkyl (or aryl) group. Secondary amides are also known as monosubstituted amides.
Tertiary amide is the one that has a nitrogen atom that is bonded to two alkyl (or aryl) groups. Tertiary amides are also known as disubstituted amides.
Apart from linear amides, there are also cyclic amides. They are formed by intramolecular condensation. Cyclic amides are also known as lactams.
(b)
Interpretation:
Given amide has to be classified as primary, secondary, or tertiary amide.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Organic compounds contain heteroatom also. Some of them are nitrogen, sulfur, oxygen etc. Nitrogen containing organic compounds are of two important types and they are amines, amides.
One of the carboxylic acid derivatives is amide. In this the carboxyl
Amides are also classified as primary, secondary, and tertiary amide.
Primary amide is the one that has a nitrogen atom that is bonded to two hydrogen atoms. Primary amides are also known as unsubstituted amides.
Secondary amide is the one that has a nitrogen atom that is bonded to one hydrogen atom and one alkyl (or aryl) group. Secondary amides are also known as monosubstituted amides.
Tertiary amide is the one that has a nitrogen atom that is bonded to two alkyl (or aryl) groups. Tertiary amides are also known as disubstituted amides.
Apart from linear amides, there are also cyclic amides. They are formed by intramolecular condensation. Cyclic amides are also known as lactams.
(c)
Interpretation:
Given amide has to be classified as primary, secondary, or tertiary amide.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Organic compounds contain heteroatom also. Some of them are nitrogen, sulfur, oxygen etc. Nitrogen containing organic compounds are of two important types and they are amines, amides.
One of the carboxylic acid derivatives is amide. In this the carboxyl
Amides are also classified as primary, secondary, and tertiary amide.
Primary amide is the one that has a nitrogen atom that is bonded to two hydrogen atoms. Primary amides are also known as unsubstituted amides.
Secondary amide is the one that has a nitrogen atom that is bonded to one hydrogen atom and one alkyl (or aryl) group. Secondary amides are also known as monosubstituted amides.
Tertiary amide is the one that has a nitrogen atom that is bonded to two alkyl (or aryl) groups. Tertiary amides are also known as disubstituted amides.
Apart from linear amides, there are also cyclic amides. They are formed by intramolecular condensation. Cyclic amides are also known as lactams.
(d)
Interpretation:
Given amide has to be classified as primary, secondary, or tertiary amide.
Concept Introduction:
Organic compounds are the important basis of life. They include gasoline, coal, dyes, and clothing fibers etc. The compounds that are obtained from living organisms are termed as organic compounds and those obtained from the earth are known as inorganic compounds. Organic compounds are found in earth also apart from living organisms. All the organic compounds contain the element carbon. Urea was synthesized in the laboratory which is an organic compound.
Organic compounds contain heteroatom also. Some of them are nitrogen, sulfur, oxygen etc. Nitrogen containing organic compounds are of two important types and they are amines, amides.
One of the carboxylic acid derivatives is amide. In this the carboxyl
Amides are also classified as primary, secondary, and tertiary amide.
Primary amide is the one that has a nitrogen atom that is bonded to two hydrogen atoms. Primary amides are also known as unsubstituted amides.
Secondary amide is the one that has a nitrogen atom that is bonded to one hydrogen atom and one alkyl (or aryl) group. Secondary amides are also known as monosubstituted amides.
Tertiary amide is the one that has a nitrogen atom that is bonded to two alkyl (or aryl) groups. Tertiary amides are also known as disubstituted amides.
Apart from linear amides, there are also cyclic amides. They are formed by intramolecular condensation. Cyclic amides are also known as lactams.

Want to see the full answer?
Check out a sample textbook solution
Chapter 6 Solutions
EBK ORGANIC AND BIOLOGICAL CHEMISTRY
- Steps and explanation please. Add how to solve or target similar problems.arrow_forwardWould the following organic synthesis occur in one step? Add any missing products, required catalysts, inorganic reagents, and other important conditions. Please include a detailed explanation and drawings showing how the reaction may occur in one step.arrow_forwardThis organic molecule is dissolved in an acidic aqueous solution: OH OH A short time later sensitive infrared spectroscopy reveals the presence of a new C = O stretch absorption. That is, there must now be a new molecule present with at least one C = O bond. In the drawing area below, show the detailed mechanism that could convert the molecule above into the new molecule. Videos 849 Explanation Check C Click and drag to start dwing a structure. # 3 MAR 23 Add/Remove steparrow_forward||| 7:47 ull 57% ← Problem 19 of 48 Submit Curved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the product of this carbocation rearrangement. Include all lone pairs and charges as appropriate. H 1,2-alkyl shift +arrow_forwardWould the following organic synthesis occur in one step? Add any missing products, required catalysts, inorganic reagents, and other important conditions. Please include a detailed explanation and drawings showing how the reaction may occur in one step.arrow_forwardBelow is the SN1 reaction of (S)-3-chlorocyclohexene and hydroxide (OH). Draw the missing curved arrows, lone pairs of electrons, and nonzero formal charges. In the third box, draw the two enantiomeric products that will be produced. 5th attempt Please draw all four bonds at chiral centers. Draw the two enantiomeric products that will be produced. Draw in any hydrogen at chiral centers. 1000 4th attempt Feedback Please draw all four bonds at chiral centers. 8. R5 HO: See Periodic Table See Hint H Cl Br Jid See Periodic Table See Hintarrow_forwardShow that a molecule with configuration π4 has a cylindrically symmetric electron distribution. Hint: Let the π orbitals be equal to xf and yf, where f is a function that depends only on the distance from the internuclear axis.arrow_forward(a) Verify that the lattice energies of the alkali metal iodides are inversely proportional to the distances between the ions in MI (M = alkali metal) by plotting the lattice energies given below against the internuclear distances dMI. Is the correlation good? Would a better fit be obtained by plotting the lattice energies as a function of (1 — d*/d)/d, as theoretically suggested, with d* = 34.5 pm? You must use a standard graphing program to plot the graph. It generates an equation for the line and calculates a correlation coefficient. (b) From the graph obtained in (a), estimate the lattice energy of silver iodide. (c) Compare the results of (b) with the experimental value of 886 kJ/mol. If they do not agree, explain the deviation.arrow_forwardCan I please get help with #3 & 4? Thanks you so much!arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningOrganic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning


