ORGANIC CHEMISTRY:PRIN...(PB)-W/ACCESS
2nd Edition
ISBN: 9780393666151
Author: KARTY
Publisher: NORTON
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Chapter 15, Problem 15.31P
Interpretation Introduction
Interpretation:
The longest-wavelength absorption of
Concept introduction:
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In the UV-vis spectrum of buta-1,2-diene (CH3CH=c=CH2), the longest-wavelength absorption appears at 178 nm.
Compare this to the longest-wavelength absorption in buta-1,3-diene (see Table 15-1) and explain the significant difference.
The H1H1 NMR spectrum shown corresponds to an unknown compound with the molecular formula C6H10C6H10. There are no strong IR bands between 2100 and 2300 or 3250 and 3350 cm−1. Deduce and draw the structure of the molecule that corresponds to the spectrum.
+
11
10
1H
9
A ¹H NMR spectrum is
shown for a molecule with
the molecular formula of
C8H8O2. Draw the structure
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1H 1H 1H 1H
ill
8
6
5
A
3H
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3
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2
1
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Chapter 15 Solutions
ORGANIC CHEMISTRY:PRIN...(PB)-W/ACCESS
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42PCh. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Prob. 15.46PCh. 15 - Prob. 15.47PCh. 15 - Prob. 15.48PCh. 15 - Prob. 15.49PCh. 15 - Prob. 15.50PCh. 15 - Prob. 15.51PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - Prob. 15.60PCh. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - Prob. 15.63PCh. 15 - Prob. 15.64PCh. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - Prob. 15.1YTCh. 15 - Prob. 15.2YTCh. 15 - Prob. 15.3YTCh. 15 - Prob. 15.4YTCh. 15 - Prob. 15.5YTCh. 15 - Prob. 15.6YTCh. 15 - Prob. 15.7YTCh. 15 - Prob. 15.8YTCh. 15 - Prob. 15.9YTCh. 15 - Prob. 15.10YTCh. 15 - Prob. 15.11YTCh. 15 - Prob. 15.12YTCh. 15 - Prob. 15.13YTCh. 15 - Prob. 15.14YTCh. 15 - Prob. 15.15YTCh. 15 - Prob. 15.16YTCh. 15 - Prob. 15.17YTCh. 15 - Prob. 15.18YTCh. 15 - Prob. 15.19YTCh. 15 - Prob. 15.20YTCh. 15 - Prob. 15.21YTCh. 15 - Prob. 15.22YTCh. 15 - Prob. 15.23YTCh. 15 - Prob. 15.24YTCh. 15 - Prob. 15.25YTCh. 15 - Prob. 15.26YTCh. 15 - Prob. 15.27YTCh. 15 - Prob. 15.28YTCh. 15 - Prob. 15.29YTCh. 15 - Prob. 15.30YT
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- The H NMR spectra corresponds to a hydrocarbon with the molecular formula C6H₁4. Deduce the structure from the spectra. Draw the unknown compound. 'H NMR 12 H Select Draw Rings More 1.5 1.4 2 H 1.5 1.0 0.5 0.0 300-MHz H NMR spectrum ppm (8) 0.9 0.8 (CH3)4Si G G Q2Earrow_forwardConsider a molecule with the molecular formula of C3H7NO2. The IR and ¹H NMR spectra are shown below. Draw the structure that best fits this data. 4000 3000 2H 2000 1500 Wavenumbers (cm-¹) + 2H 1000 3H 500 ppmarrow_forwardSaturated aldehyde Aromatic aldehyde At what position would you expect to observe IR absorption for the carbonyl groups in the following molecules? Infrared Absorptions of Some Aldehydes and Ketones. Carbonyl type Absorption cm-1 1730 1705 Example CH3CHO PhCHO a,ẞ-Unsaturated H2C=CHCHO 1705 aldehyde Saturated ketone CH3 COCH3 1715 Cyclohexanone 1715 Cyclopentanone 1750 Cyclobutanone Aromatic ketone 1785 PhCOCH3 a,ẞ-Unsaturated ketone H2C=CHCOCH3 1685 1690 Molecule #1: Molecule #2: CHO CHO CH3 cm-1 cm -1arrow_forward
- 10- The strongest absorption area of the benzene ring in visible and ultraviolet spectroscopy at 255 nm 11- A Calverti rearrangement occurs when the compound has a hydrogen atom in theẞ. Position 12-Carboxylic acids show an absorption peak of 44 m/e Q/ true or falsearrow_forwardThe IR spectrum of a sample contains absorptions at 1750 cm-1. To what class of organic compound does this sample most likely belong? Alkene Ketone Alkane Alcoholarrow_forwardCompound B has molecular formula C9H12. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.arrow_forward
- Molecular Formula: C7H10O IR: Only notable main absorptions is at 1710, 2945, and 3100 cm-¹ ¹H spectrum 1H, s 13C spectrum 200 180 160 140 3H, s 2H, t J = 7.1 Hz 2H, d J = 7.1 Hz dr 2H, t J = 7.1 Hz PPM 120 PPM 100 80 60 0 I T 40 20arrow_forwardDraw the structure of molecular formula C8H10O that produced the 1H NMR spectra shown below. The IR spectrum does not show a broad absorbance at 3300 cm–1 or a strong absorbance at 1710 cm–1.arrow_forwardThe 13C NMR spectrum of 1-bromo-3-chloropropane contains peaks at δ 30, δ 35, and δ 43. Assign these signals to the appropriate carbons.arrow_forward
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