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(a)
Interpretation:
The given salt has to be classified as an amine salt or quaternary ammonium salt.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As
(b)
Interpretation:
The given salt has to be classified as an amine salt or quaternary ammonium salt.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
(c)
Interpretation:
The given salt has to be classified as an amine salt or quaternary ammonium salt.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
(d)
Interpretation:
The given salt has to be classified as an amine salt or quaternary ammonium salt.
Concept Introduction:
Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom. This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.
Neutralization reaction is the one that takes place between an acid and a base to give salt as product. As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product. The salt formed is an amine salt. Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor. In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.
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Chapter 17 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- b) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forwardRedraw the flowchartarrow_forward
- redraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forward
- What is/are the product(s) of the following reaction? Select all that apply. * HI A B C OD OH A B OH D Carrow_forwardIn the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+. Assuming there is 2.00 L of solution, answer the following: The Ka of the left & right solution is? The pH of the left & right solution is? The acid on the left & right is what kind of acid?arrow_forwardNonearrow_forward
- 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
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