(a)
Interpretation: The nucleophile, leaving group and product formed needs to be identify for the given substitution reaction.
Concept Introduction: In a nucleophilic substitution reaction, the electron rich nucleophile attacks on the electrophilic center of an electrophile (electron deficient) to form nucleophile substituted product. Here, nucleophile can be neutral or have a negative charge. The substitution of nucleophile on the electrophile causes leaving group to leave the electrophile.
(b)
Interpretation: The nucleophile, leaving group and product formed needs to be identify for the given substitution reaction.
Concept Introduction: In a nucleophilic substitution reaction, the electron rich nucleophile attacks on the electrophilic center of an electrophile (electron deficient) to form nucleophile substituted product. Here, nucleophile can be neutral or have a negative charge. The substitution of nucleophile on the electrophile causes leaving group to leave the electrophile.
(c)
Interpretation: The nucleophile, leaving group and product formed needs to be identify for the given substitution reaction.
Concept Introduction: In a nucleophilic substitution reaction, the electron rich nucleophile attacks on the electrophilic center of an electrophile (electron deficient) to form nucleophile substituted product. Here, nucleophile can be neutral or have a negative charge. The substitution of nucleophile on the electrophile causes leaving group to leave the electrophile.
(d)
Interpretation: The nucleophile, leaving group and product formed needs to be identify for the given substitution reaction.
Concept Introduction: In a nucleophilic substitution reaction, the electron rich nucleophile attacks on the electrophilic center of an electrophile (electron deficient) to form nucleophile substituted product. Here, nucleophile can be neutral or have a negative charge. The substitution of nucleophile on the electrophile causes leaving group to leave the electrophile.
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Chapter 7 Solutions
Organic Chemistry (6th Edition)
- > each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that: 1. the rate of substitution doesn't depend on nucleophile concentration and 2. the products are a roughly 50/50 mixture of enantiomers. Substrate A Substrate B Faster Rate X Ś CI (Choose one) (Choose one) CI Br Explanation Check Br (Choose one) © 2025 McGraw Hill LLC. All Rights Farrow_forwardNMR spectrum of ethyl acetate has signals whose chemical shifts are indicated below. Which hydrogen or set of hydrogens corresponds to the signal at 4.1 ppm? Select the single best answer. The H O HỌC—C—0—CH, CH, 2 A ethyl acetate H NMR: 1.3 ppm, 2.0 ppm, 4.1 ppm Check OA B OC ch B C Save For Later Submit Ass © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center |arrow_forwardHow many signals do you expect in the H NMR spectrum for this molecule? Br Br Write the answer below. Also, in each of the drawing areas below is a copy of the molecule, with Hs shown. In each copy, one of the H atoms is colored red. Highlight in red all other H atoms that would contribute to the same signal as the H already highlighted red Note for advanced students: In this question, any multiplet is counted as one signal. 1 Number of signals in the 'H NMR spectrum. For the molecule in the top drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. Check For the molecule in the bottom drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. O ✓ No additional Hs to color in top molecule ง No additional Hs to color in bottom…arrow_forward
- in the kinetics experiment, what were the values calculated? Select all that apply.a) equilibrium constantb) pHc) order of reactiond) rate contstantarrow_forwardtrue or false, given that a 20.00 mL sample of NaOH took 24.15 mL of 0.141 M HCI to reach the endpoint in a titration, the concentration of the NaOH is 1.17 M.arrow_forwardin the bromothymol blue experiment, pKa was measured. A closely related compound has a Ka of 2.10 x 10-5. What is the pKa?a) 7.1b) 4.7c) 2.0arrow_forward
- calculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forwardtrue or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forwardI2(g) + CI2(g) ↔ 2ICIK for this reaction is 81.9. Find the equilibrium concentration of I2 if the inital concentration of I2 and CI2 are 0.010 Marrow_forward
- true or false,the equilibrium constant for this reaction is 0.50.PCI5(g) ↔ PCI3(g) + CI2(g)Based on the above, the equilibrium constant for the following reaction is 0.25.2PCI5(g) ↔. 2PCI3(g) + 2CI2(g)arrow_forwardtrue or false, using the following equilibrium, if carbon dioxide is added the equilibrium will shift toward the productsC(s) + CO2(g) ↔ 2CO(g)arrow_forward2S2O2/3- (aq) + I2 (aq) ---> S4O2/6- (aq) +2I- (aq) Experiment I2 (M) S2O3- (M) Initital Rate (M/s) 1 0.01 0.01 0.0004 2 0.01 0.02 0.0004 3 0.02 0.01 0.0008 Calculate the overall order for this reaction using the table data a) 3b) 0c) 2d) 1arrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning