Concept explainers
(a)
Interpretation:
Whether the given
Concept introduction:
The rule for aromaticity applies only if the system is planar, cyclic, and has a continuous overlap of p-orbitals.
As per Huckel’s rules for aromaticity, the molecule must possess a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring. The number of electrons in the
If the number of electrons in the
All other compounds are nonaromatic.
(b)
Interpretation:
Whether the given compound is aromatic, antiaromatic, or nonaromatic is to be determined.
Concept introduction:
The rule for aromaticity applies only if the system is planar, cyclic, and has a continuous overlap of p-orbitals.
As per Huckel’s rule for aromaticity, The molecule must possess a pi system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring. The number of electrons in the
If the number of electrons in the
All other compounds are nonaromatic.
(c)
Interpretation:
Whether the given compound is aromatic, antiaromatic, or nonaromatic is to be determined.
Concept introduction:
The rule for aromaticity applies only if the system is planar, cyclic, and has a continuous overlap of p-orbitals.
As per Huckel’s rule for aromaticity, the molecule must possess a
If the number of electrons in the
All other compounds are nonaromatic.
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Chapter 14 Solutions
Organic Chemistry: Principles And Mechanisms (second Edition)
- Calculate E° for Ni(glycine)2 + 2e– D Ni + 2 glycine– given Ni2+ + 2 glycine– D Ni(glycine)2 K = 1.2×1011 Ni2+ + 2 e– D Ni E° = -0.236 Varrow_forwardOne method for the analysis of Fe3+, which is used with a variety of sample matrices, is to form the highly colored Fe3+–thioglycolic acid complex. The complex absorbs strongly at 535 nm. Standardizing the method is accomplished using external standards. A 10.00-ppm Fe3+ working standard is prepared by transferring a 10-mL aliquot of a 100.0 ppm stock solution of Fe3+ to a 100-mL volumetric flask and diluting to volume. Calibration standards of 1.00, 2.00, 3.00, 4.00, and 5.00 ppm are prepared by transferring appropriate amounts of the 10.0 ppm working solution into separate 50-mL volumetric flasks, each of which contains 5 mL of thioglycolic acid, 2 mL of 20% w/v ammonium citrate, and 5 mL of 0.22 M NH3. After diluting to volume and mixing, the absorbances of the external standards are measured against an appropriate blank. Samples are prepared for analysis by taking a portion known to contain approximately 0.1 g of Fe3+, dissolving it in a minimum amount of HNO3, and diluting to…arrow_forwardAbsorbance and transmittance are related by: A = -log(T) A solution has a transmittance of 35% in a 1-cm-pathlength cell at a certain wavelength. Calculate the transmittance if you dilute 25.0 mL of the solution to 50.0 mL? (A = εbc) What is the transmittance of the original solution if the pathlength is increased to 10 cm?arrow_forward
- Under what conditions will Beer’s Law most likely NO LONGER be linear? When the absorbing species is very dilute. When the absorbing species participates in a concentration-dependent equilibrium. When the solution being studied contains a mixture of ions.arrow_forwardCompared to incident (exciting) radiation, fluorescence emission will have a: Higher energy Higher frequency Longer wavelengtharrow_forwardLin and Brown described a quantitative method for methanol based on its effect on the visible spectrum of methylene blue. In the absence of methanol, methylene blue has two prominent absorption bands at 610 nm and 663 nm, which correspond to the monomer and the dimer, respectively. In the presence of methanol, the intensity of the dimer’s absorption band decreases, while that for the monomer increases. For concentrations of methanol between 0 and 30% v/v, the ratio of the two absorbance, A663/ A610, is a linear function of the amount of methanol. Use the following standardization data to determine the %v/v methanol in a sample if A610 is 0.75 and A663 is 1.07.arrow_forward
- The crystal field splitting energy, Δ, of a complex is determined to be 2.9 × 10-19 What wavelength of light would this complex absorb? What color of light is this? What color would the compound be in solution?arrow_forwardA key component of a monochromator is the exit slit. As the exit slit is narrowed, the bandwidth of light (i.e., the range of wavelengths) exiting the slit gets smaller, leading to higher resolution. What is a possible disadvantage of narrowing the exit slit? (Hint: why might a narrower slit lower the sensitivity of the measurement?).arrow_forwardAn x-ray has a frequency of 3.33 × 1018 What is the wavelength of this light?arrow_forward
- Choose the Lewis structure for the compound below: H2CCHOCH2CH(CH3)2 HH H :d H H H C. Η H H HH H H H H. H H H HH H H H H H- H H H C-H H H HHHHarrow_forwardEach of the highlighted carbon atoms is connected to hydrogen atoms.arrow_forwardく Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. More... No reaction. Explanation Check O + G 1. Na O Me Click and drag to start drawing a structure. 2. H + 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility 000 Ar Parrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning