
Concept explainers
Interpretation:
Cyclic amides are designated as what has to be chosen from the given options.
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.

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Chapter 17 Solutions
General, Organic, and Biological Chemistry
- Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWLarrow_forwardDraw the skeletal ("line") structure of 2-hydroxy-4-methylpentanal. Click and drag to start drawing a structure. Xarrow_forwardDetermine whether the following molecule is a hemiacetal, acetal, or neither and select the appropriate box below. Also, highlight the hemiacetal or acetal carbon if there is one. hemiacetal acetal Oneither OHarrow_forward
- What is the missing reactant R in this organic reaction? ་ ་ ་ ་ ་ ་ ་ ་ ་ ་ +R H3O+ • Draw the structure of R in the drawing area below. N • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure.arrow_forwardWrite the systematic name of each organic molecule: H structure H OH OH H OH name ☐ OHarrow_forwardDetermine whether each of the following molecules is a hemiacetal, acetal, or neither and select the appropriate box in the table. CH3O OH OH OH hemiacetal acetal neither hemiacetal acetal neither Xarrow_forward
- What is the missing reactant R in this organic reaction? N N དལ་ད་་ + R • Draw the structure of R in the drawing area below. • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure. ㄖˋarrow_forwardDraw the condensed structure of 4-hydroxy-3-methylbutanal. Click anywhere to draw the first atom of your structure.arrow_forwardUsing the bond energy values, calculate the energy that must be supplied or is released upon the polymerization of 755 monomers. If energy must be supplied, provide a positive number; if energy is released, provide a negative number. Hint: Avogadro’s number is 6.02 × 1023.arrow_forward
- -AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningOrganic And Biological ChemistryChemistryISBN:9781305081079Author:STOKER, H. Stephen (howard Stephen)Publisher:Cengage Learning,

