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Concept explainers
(a)
Interpretation: The given following compounds has to be prepared using an aldol addition in the first step.
Concept introduction: If two different carbonyl compound used in an aldol addition, known as a crossed-aldol reaction. If the two carbonyl compounds contain alpha carbon then four products are formed because two different enolate ion is formed.
If both carbonyl compounds have alpha hydrogens, primarily only one product is formed if LDA is used to remove the alpha hydrogen from carbonyl carbon that is needed for the enolate ion.
Because LDA is a strong base, all of the carbonyl carbon is converted to the enolate ion, so none of the carbonyl carbon is left behind to react with enolate ion. Therefore, the second carbonyl compound is added slowly.
(b)
Interpretation: The given following compounds has to be prepared using an aldol addition in the first step
Concept introduction: If two different carbonyl compound used in an aldol addition, known as a crossed-aldol reaction. If the two carbonyl compounds contain alpha carbon then four products are formed because two different enolate ion is formed.
If both carbonyl compounds have alpha hydrogens, primarily only one product is formed if LDA is used to remove the alpha hydrogen from carbonyl carbon that is needed for the enolate ion.
Because LDA is a strong base, all of the carbonyl carbon is converted to the enolate ion, so none of the carbonyl carbon is left behind to react with enolate ion. Therefore, the second carbonyl compound is added slowly.
(c)
Interpretation: The given following compounds has to be prepared using an aldol addition in the first step.
Concept introduction: If two different carbonyl compound used in an aldol addition, known as a crossed-aldol reaction. If the two carbonyl compounds contain alpha carbon then four products are formed because two different enolate ion is formed.
If both carbonyl compounds have alpha hydrogens, primarily only one product is formed if LDA is used to remove the alpha hydrogen from carbonyl carbon that is needed for the enolate ion.
Because LDA is a strong base, all of the carbonyl carbon is converted to the enolate ion, so none of the carbonyl carbon is left behind to react with enolate ion. Therefore, the second carbonyl compound is added slowly.
(d)
Interpretation: The given following compounds has to be prepared using an aldol addition in the first step.
Concept introduction: If two different carbonyl compound used in an aldol addition, known as a crossed-aldol reaction. If the two carbonyl compounds contain alpha carbon then four products are formed because two different enolate ion is formed.
If both carbonyl compounds have alpha hydrogens, primarily only one product is formed if LDA is used to remove the alpha hydrogen from carbonyl carbon that is needed for the enolate ion.
Because LDA is a strong base, all of the carbonyl carbon is converted to the enolate ion, so none of the carbonyl carbon is left behind to react with enolate ion. Therefore, the second carbonyl compound is added slowly.
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Chapter 17 Solutions
ORGANIC CHEMISTRY-W/S.G+SOLN.MANUAL
- Please answer the questions in the photos and please revise any wrong answers. Thank youarrow_forward(Please be sure that 7 carbons are available in the structure )Based on the 1H NMR, 13C NMR, DEPT 135 NMR and DEPT 90 NMR, provide a reasoning step and arrive at the final structure of an unknown organic compound containing 7 carbons. Dept 135 shows peak to be positive at 128.62 and 13.63 Dept 135 shows peak to be negative at 130.28, 64.32, 30.62 and 19.10.arrow_forward-lease help me answer the questions in the photo.arrow_forward
- For the reaction below, the concentrations at equilibrium are [SO₂] = 0.50 M, [0] = 0.45 M, and [SO3] = 1.7 M. What is the value of the equilibrium constant, K? 2SO2(g) + O2(g) 2SO3(g) Report your answer using two significant figures. Provide your answer below:arrow_forwardI need help with this question. Step by step solution, please!arrow_forwardZn(OH)2(s) Zn(OH)+ Ksp = 3 X 10-16 B₁ = 1 x 104 Zn(OH)2(aq) B₂ = 2 x 1010 Zn(OH)3 ẞ3-8 x 1013 Zn(OH) B4-3 x 1015arrow_forward
- Help me understand this by showing step by step solution.arrow_forwardscratch paper, and the integrated rate table provided in class. our scratch work for this test. Content attribution 3/40 FEEDBACK QUESTION 3 - 4 POINTS Complete the equation that relates the rate of consumption of H+ and the rate of formation of Br2 for the given reaction. 5Br (aq) + BrO3 (aq) + 6H (aq) →3Br2(aq) + 3H2O(l) • Your answers should be whole numbers or fractions without any decimal places. Provide your answer below: Search 尚 5 fn 40 * 00 99+ 2 9 144 a [arrow_forward(a) Write down the structure of EDTA molecule and show the complex structure with Pb2+ . (b) When do you need to perform back titration? (c) Ni2+ can be analyzed by a back titration using standard Zn2+ at pH 5.5 with xylenol orange indicator. A solution containing 25.00 mL of Ni2+ in dilute HCl is treated with 25.00 mL of 0.05283 M Na2EDTA. The solution is neutralized with NaOH, and the pH is adjusted to 5.5 with acetate buffer. The solution turns yellow when a few drops of indicator are added. Titration with 0.02299 M Zn2+ requires 17.61 mL to reach the red end point. What is the molarity of Ni2+ in the unknown?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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