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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 18 Solutions
Organic Chemistry
- 2. Use Hess's law to calculate the AH (in kJ) for: rxn CIF(g) + F2(g) → CIF 3 (1) using the following information: 2CIF(g) + O2(g) → Cl₂O(g) + OF 2(g) AH = 167.5 kJ ΔΗ 2F2 (g) + O2(g) → 2 OF 2(g) 2C1F3 (1) + 202(g) → Cl₂O(g) + 3 OF 2(g) о = = -43.5 kJ AH = 394.1kJarrow_forwardci Draw the major product(s) of the following reactions: (3 pts) CH3 HNO3/H2SO4 HNO3/ H2SO4 OCH3 (1 pts)arrow_forwardProvide the product for the reactionarrow_forward
- What is the net ionic equation for the reaction between tin(IV) sulfide and nitric acid?arrow_forwardThe combustion of 28.8 g of NH3 consumes exactly _____ g of O2. 4 NH3 + 7 O2 ----> 4 NO2 + 6 H2Oarrow_forwardWhat is the molecular formula of the bond-line structure shown below OH HO ○ C14H12O2 ○ C16H14O2 ○ C16H12O2 O C14H14O2arrow_forward
- Check all molecules that are acids on the list below. H2CO3 HC2H3O2 C6H5NH2 HNO3 NH3arrow_forwardFrom the given compound, choose the proton that best fits each given description. a CH2 CH 2 Cl b с CH2 F Most shielded: (Choose one) Least shielded: (Choose one) Highest chemical shift: (Choose one) Lowest chemical shift: (Choose one) ×arrow_forwardConsider this molecule: How many H atoms are in this molecule? How many different signals could be found in its 1H NMR spectrum? Note: A multiplet is considered one signal.arrow_forward
- For each of the given mass spectrum data, identify whether the compound contains chlorine, bromine, or neither. Compound m/z of M* peak m/z of M + 2 peak ratio of M+ : M + 2 peak Which element is present? A 122 no M + 2 peak not applicable (Choose one) B 78 80 3:1 (Choose one) C 227 229 1:1 (Choose one)arrow_forwardShow transformation from reactant to product, step by step. *see imagearrow_forwardCheck the box if the molecule contains the listed item. *See imagearrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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