
Concept explainers
Interpretation: The resonance structure of the radical that results from the abstraction of a hydrogen atom from the antioxidant BHA is to be drawn.
Concept introduction: Most of the organic structures cannot be represented using single Lewis structure. Therefore, there exists more than one Lewis structure for representing a molecule or ion. These structures are known as resonance structures. These are the hypothetical structures and do not specify the exact structure. These resonance structure combine together to give resonance hybrid that is lower in energy and is the most stable structure.
The delocalization of electrons results in the formation resonance structure.

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Chapter 15 Solutions
Package: Loose Leaf for Organic Chemistry with Biological Topics with Connect Access Card
- In the spectroscopic technique known as photoelectron spectroscopy (PES), ultraviolet radiation is directed at an atom or molecule. Electrons are ejected from the valence shell and their kinetic energies are measured. Since the energy of the incident ultraviolet photons is known and the kinetic energy of the ejected electron is measured, the ionization energy, I, can be deduced because total energy is conserved. (a) Show that the velocity, v, of the ejected electron and the frequency, n, of the incident radiation are related by hv = I + (1/2)mv^2? (b) Use this relation to calculate the ionization energy of a rubidium atom, knowing that light of wavelength 58.4 nm produces electrons with a velocity of 2,450 km/s Recall that 1 J = 1 kg.m^2/s^2arrow_forwardI) In Millikan's experiment, each droplet observed by the technicians contained an even number of electrons. If they had been unaware of this limitation, how would it have affected their report of an electron's charge?II) Millikan measured the charge of an electron in electrostatic units, esu. The data he collected included the following series of charges found on oil drops: 9.60 X 10^-10 esu, 1.92 X 10^-9 esu; 2.40 X 10^-9 esu; 2.88 X 10^-9 esu; and 4.80 X 10^-9 esu. (a) From this series, find the probable charge of the electron in electrostatic units. (b) Estimate the number of electrons in an oil drop with a charge of 6.72 X 10^-9 esu. The actual charge (in Coulombs) of an electron is 1.602 X 10^-19 C. What is the relationship between esu and Coulombs?arrow_forwardmy ccc edu - Search X Quick Access X D2L Homepage - Spring 2025 x N Netflix X Dimensional Analysis - A x+ pp.aktiv.com Q ☆ X Question 59 of 70 The volume of 1 unit of plasma is 200.0 mL If the recommended dosage for adult patients is 10.0 mL per kg of body mass, how many units are needed for a patient with a body mass of 80.0 kg ? 80.0 kg 10.0 DAL 1 units X X 4.00 units 1 1 Jeg 200.0 DAL L 1 units X 200.0 mL = 4.00 units ADD FACTOR *( ) DELETE ANSWER RESET D 200.0 2.00 1.60 × 10³ 80.0 4.00 0.0400 0.250 10.0 8.00 & mL mL/kg kg units/mL L unit Q Search delete prt sc 111 110 19arrow_forward
- Identify the starting material in the following reaction. Click the "draw structure" button to launch the drawing utility. draw structure ... [1] 0 3 C10H18 [2] CH3SCH3 Harrow_forwardIn an equilibrium mixture of the formation of ammonia from nitrogen and hydrogen, it is found that PNH3 = 0.147 atm, PN2 = 1.41 atm and Pн2 = 6.00 atm. Evaluate Kp and Kc at 500 °C. 2 NH3 (g) N2 (g) + 3 H₂ (g) K₂ = (PN2)(PH2)³ = (1.41) (6.00)³ = 1.41 x 104arrow_forwardWhat alkene or alkyne yields the following products after oxidative cleavage with ozone? Click the "draw structure" button to launch the drawing utility. and two equivalents of CH2=O draw structure ...arrow_forward
- H-Br Energy 1) Draw the step-by-step mechanism by which 3-methylbut-1-ene is converted into 2-bromo-2-methylbutane. 2) Sketch a reaction coordinate diagram that shows how the internal energy (Y- axis) of the reacting species change from reactants to intermediate(s) to product. Brarrow_forward2. Draw the missing structure(s) in each of the following reactions. The missing structure(s) can be a starting material or the major reaction product(s). C5H10 H-CI CH2Cl2 CIarrow_forwardDraw the products of the stronger acid protonating the other reactant. དའི་སྐད”“ H3C OH H3C CH CH3 KEq Product acid Product basearrow_forward
- Draw the products of the stronger acid protonating the other reactant. H3C NH2 NH2 KEq H3C-CH₂ 1. Product acid Product basearrow_forwardWhat alkene or alkyne yields the following products after oxidative cleavage with ozone? Click the "draw structure" button to launch the drawing utility. draw structure ... andarrow_forwardDraw the products of the stronger acid protonating the other reactant. H3C-C=C-4 NH2 KEq CH H3C `CH3 Product acid Product basearrow_forward
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