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
EBK ORGANIC CHEMISTRY
- Don't used hand raitingarrow_forwardCHEM2323 Problem 2-24 Tt O e: ל Predict the product(s) of the following acid/base reactions. Draw curved arrows to show the formation and breaking of bonds. If the bonds needed are not drawn out, you should redraw them. + BF3 (a) (b) HI + (c) OH -BF Problem 2-25 Use curved arrows and a proton (H+) to draw the protonated form of the following Lewis bases. Before starting, add all missing lone pairs. (a) (b) :0: (c) N 1 CHEM2323 PS CH02 Name:arrow_forwardCHEM2323 Problem 2-26 Tt O PS CH02 Name: Use the curved-arrow formalism to show how the electrons flow in the resonance form on the left to give the one on the right. (Draw all lone pairs first) (a) NH2 NH2 + (b) Problem 2-27 Double bonds can also act like Lewis bases, sharing their electrons with Lewis acids. Use curved arrows to show how each of the following double bonds will react with H-Cl and draw the resulting carbocation. (a) H2C=CH2 (b) (c) Problem 2-28 Identify the most electronegative element in each of the following molecules: (a) CH2FCI F Problem 2-29 (b) FCH2CH2CH2Br (c) HOCH2CH2NH2 (d) CH3OCH2Li F 0 0 Use the electronegativity table in Figure 2.3 to predict which bond in the following pairs is more polar and indicate the direction of bond polarity for each compound. (a) H3C-Cl or Cl-CI (b) H3C-H or H-CI (c) HO-CH3 or (CH3)3Si-CH3 (d) H3C-Li or Li-OHarrow_forward
- Reagan is doing an atomic absorption experiment that requires a set of zinc standards in the 0.4-1.6 ppm range. A 1000 ppm Zn solution was prepared by dissolving the necessary amount of solid Zn(NO3)2 in water. The standards can be prepared by diluting the 1000 ppm Zn solution. Table 1 shows one possible set of serial dilutions (stepwise dilution of a solution) that Reagan could perform to make the necessary standards. Solution A was prepared by diluting 5.00 ml of the 1000 ppm Zn standard to 50.00 ml. Solutions C-E are called "calibration standards" because they will be used to calibrate the atomic absorption spectrometer. a. Compare the solution concentrations expressed as ppm Zn and ppm Zn(NO3)2. Compare the concentrations expressed as M Zn and M Zn(NO3)2 - Which units allow easy conversion between chemical species (e.g. Zn and Zn(NO3)2)? - Which units express concentrations in numbers with easily expressed magnitudes? - Suppose you have an analyte for which you don't know the molar…arrow_forwardNonearrow_forwardHow will you prepare the following buffers? 2.5 L of 1.5M buffer, pH = 10.5 from NH4Cl and NH3arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning