(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
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Part C IN H N. Br₂ (2 equiv.) AlBr3 Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and + e (×) H± 12D T EXP. L CONT. דarrow_forward9. OA. Rank the expected boiling points of the compounds shown below from highest to lowest. Place your answer appropriately in the box. Only the answer in the box will be graded. (3) points) OH OH بر بد بدید 2 3arrow_forwardThere is an instrument in Johnson 334 that measures total-reflectance x-ray fluorescence (TXRF) to do elemental analysis (i.e., determine what elements are present in a sample). A researcher is preparing a to measure calcium content in a series of well water samples by TXRF with an internal standard of vanadium (atomic symbol: V). She has prepared a series of standard solutions to ensure a linear instrument response over the expected Ca concentration range of 40-80 ppm. The concentrations of Ca and V (ppm) and the instrument response (peak area, arbitrary units) are shown below. Also included is a sample spectrum. Equation 1 describes the response factor, K, relating the analyte signal (SA) and the standard signal (SIS) to their respective concentrations (CA and CIS). Ca, ppm V, ppm SCa, arb. units SV, arb. units 20.0 10.0 14375.11 14261.02 40.0 10.0 36182.15 17997.10 60.0 10.0 39275.74 12988.01 80.0 10.0 57530.75 14268.54 100.0…arrow_forward
- A mixture of 0.568 M H₂O, 0.438 M Cl₂O, and 0.710 M HClO are enclosed in a vessel at 25 °C. H₂O(g) + C₁₂O(g) = 2 HOCl(g) K = 0.0900 at 25°C с Calculate the equilibrium concentrations of each gas at 25 °C. [H₂O]= [C₁₂O]= [HOCI]= M Σ Marrow_forwardWhat units (if any) does the response factor (K) have? Does the response factor (K) depend upon how the concentration is expressed (e.g. molarity, ppm, ppb, etc.)?arrow_forwardProvide the structure, circle or draw, of the monomeric unit found in the biological polymeric materials given below. HO OH amylose OH OH 행 3 HO cellulose OH OH OH Ho HOarrow_forward
- OA. For the structure shown, rank the bond lengths (labeled a, b and c) from shortest to longest. Place your answer in the box. Only the answer in the box will be graded. (2 points) H -CH3 THe b Нarrow_forwardDon't used hand raitingarrow_forwardQuizzes - Gen Organic & Biological Che... ☆ myd21.lcc.edu + O G screenshot on mac - Google Search savings hulu youtube google disney+ HBO zlib Homework Hel...s | bartleby cell bio book Yuzu Reader: Chemistry G periodic table - Google Search b Home | bartleby 0:33:26 remaining CHEM 120 Chapter 5_Quiz 3 Page 1: 1 > 2 > 3 > 6 ¦ 5 > 4 > 7 ¦ 1 1 10 8 ¦ 9 a ¦ -- Quiz Information silicon-27 A doctor gives a patient 0.01 mC i of beta radiation. How many beta particles would the patient receive in I minute? (1 Ci = 3.7 x 10 10 d/s) Question 5 (1 point) Saved Listen 2.22 x 107 222 x 108 3.7 x 108 2.22 x 108 none of the above Question 6 (1 point) Listen The recommended dosage of 1-131 for a test is 4.2 μCi per kg of body mass. How many millicuries should be given to a 55 kg patient? (1 mCi = 1000 μСi)? 230 mCiarrow_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
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