Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Chapter 15, Problem 15.33P
Interpretation Introduction
Interpretation:
A compound, whose formula is
Concept introduction:
IHD accounts for the number of rings or multiple bonds the structure could possess.
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Which statement best describes the following pair of compounds?
C4H6 and C6H12
A.) Circulated functional groups do not show absorption in the infrared spectrum because they have the same mass
B.) The reduced mass of both compounds is the same so they do not absorb IR (infrared) energy.
C.) The functional group specified in both compounds does not undergo changes in the dipole moment due to symmetry
D.) Both represent an absorption band around 2000 cm ^ -1
Draw the structure for a compound that meets this description (can be any structure that meets this description):
C3H6O, with no IR absorption above 3000 cm^-1.
Draw and name each of the following:
a. An ether with molecular formula C3H8O
b. An alkene with the formula C5H10 that has stereoisomers.
c. A tertiary (30) alcohol with molecular formula C5H11OH that is chiral.
d. A primary (10) aliphatic amine with molecular formula C3H9
e. A thiol with molecular formula C6H14
Chapter 15 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
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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- Identify the structures of isomers H and I (molecular formula C8H11N).a.Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm−1b.Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm−1arrow_forwardIdentify the structures of A and B, isomers of molecular formula C10H12O2, from their IR data and 1H NMR spectra.a. IR absorption for A at 1718 cm−1b. IR absorption for B at 1740 cm−1arrow_forwardDraw the structure of the one tertiary (3°) amine with molecular formula C5H13N that does not contain a 3-carbon chain. You do not have to consider stereochemistry. You do not have to explicitly draw H atomsarrow_forward
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- An unknown compound has the formula C5H6. Its longest-wavelength UV–vis absorption is centered at 215 nm. Draw four isomers that are consistent with these results.arrow_forwardDraw the structure of the two tertiary (3°) amines with molecular formula C6H15N that contain a group with 3 carbon atoms in the largest group. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms.arrow_forwardwhich of the following cannot be distinguished by an infrared spectrum I) resonant hybrids II) tautomers (keto-enol) III) enantiomers IV) cis-trans isomers a) II,I,III b) II,III c)I,III d)IVarrow_forward
- which of the following cannot be distinguished by an infrared spectrum a) resonant hybrids b) tautomers (keto-enol) c) enantiomers d) cis-trans isomersarrow_forwardCompound B has molecular formula C9H10. The IR spectrum is shown below. The 1H-NMR spectrum shows a multiplet at 7.2 ppm integrating to 4H, a triplet at 2.9 ppm integrating to 4H, and a triplet at 2.1 ppm integrating to 2 H. Suggest a structure for B and explain your reasoningarrow_forwardHow many stereoisomers are possible for 1,4,7-octatriene? a) 2 b) 4 c) 8 d) 16arrow_forward
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