a)
Interpretation:
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.
Concept introduction:
Acetoacetic ester synthesis converts an
Both reactions involve the same steps such as i) enolate ion formation ii) SN2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.
To show
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.
Answer to Problem 46AP
The compound shown can be prepared by using malonic ester synthesis.
Explanation of Solution
The compound shown is a derivative of carboxylic acid. Hence it can be prepared using malonic ester synthesis. The acid has two methyl groups attached to the carbon adjacent to ester groups. It can be prepared by replacing the two hydrogens on the active methylene group of malonic ester by two methyl groups. This is achieved by treating the ester with two equivalents of sodium ethoxide and two equivalents of methyl bromide.
The compound shown can be prepared by using malonic ester synthesis.
b)
Interpretation:
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.
Concept introduction:
Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.
Both reactions involve the same steps such as i) enolate ion formation ii) SN2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.
To show
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.
Answer to Problem 46AP
The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.
Explanation of Solution
: The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,6- dibromohexane displaces the bromide ion to produce a α- substituted acetoacetic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a β- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO2 molecule on heating to yield methyl cycloheptyl ketone.
The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.
c)
Interpretation:
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.
Concept introduction:
Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.
Both reactions involve the same steps such as i) enolate ion formation ii) SN2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.
To show
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.
Answer to Problem 46AP
The compound shown can be prepared by using malonic ester synthesis.
Explanation of Solution
The compound shown is a carboxylic acid. Hence it can be prepared using malonic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of malonic ester to yield the enolate anion. The nucleophilic attack of the anion on 1,3- dibromopropane displaces the bromide ion to produce a α- substituted malonic ester. The second acidic hydrogen of the ester is then removed by another ethoxide ion which is followed by the nucleophilic attack of the anion on the other carbon bearing bromine to produce a cyclic diester. Upon treating with aqueous acids the ester groups get hydrolyzed to give a dicarboxylic acid. The dicarboxylic acid eliminates a CO2 molecule on heating to yield cyclobutylcarboxylic acid.
The compound shown can be prepared by using malonic ester synthesis.
d)
Interpretation:
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis is to be shown.
Concept introduction:
Acetoacetic ester synthesis converts an alkyl halide in to a methyl ketone having three more carbons. The methyl ketone part comes from acetoacetic eater while the remaining carbon comes from the primary alkyl halide. Malonic ester synthesis converts an alkyl halide to a carboxylic acid having two more carbon atoms.
Both reactions involve the same steps such as i) enolate ion formation ii) SN2 attack of the enolate anion on the alkyl halide iii) hydrolysis and decarboxylation.
To show
How to prepare the compound shown using either an acetoacetic ester synthesis or malonic ester synthesis.
Answer to Problem 46AP
The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.
Explanation of Solution
The compound is a methyl ketone. Hence it can be prepared using aceto acetic ester synthesis. The base ethoxide ion abstracts a proton from the active methylene group of acetoacetic ester to yield the enolate anion. The nucleophilic attack of the anion on allyl bromide displaces the bromide ion to produce α- allylsubstituted acetoacetic ester. Upon treating with aqueous acids the ester group gets hydrolyzed to give a β- ketocarboxylic acid. The ketocarboxylic acid eliminates a CO2 molecule on heating to yield hex-5-ene-2-one.
The compound shown can be prepared by using an acetoacetic ester synthesis as shown below.
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Chapter 22 Solutions
Bundle: Organic Chemistry, 9th, Loose-Leaf + OWLv2, 4 terms (24 months) Printed Access Card
- For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. དྲ。 ✗MgBr ? O CI Will the first product that forms in this reaction create a new C-C bond? Yes No • ? Will the first product that forms in this reaction create a new CC bond? Yes No × : ☐ Xarrow_forwardPredict the major products of this organic reaction: OH NaBH4 H ? CH3OH Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. ☐ : Sarrow_forwardPredict the major products of this organic reaction: 1. LIAIHA 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. X : ☐arrow_forward
- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
- give example for the following(by equation) a. Converting a water insoluble compound to a soluble one. b. Diazotization reaction form diazonium salt c. coupling reaction of a diazonium salt d. indacator properties of MO e. Diazotization ( diazonium salt of bromobenzene)arrow_forward2-Propanone and ethyllithium are mixed and subsequently acid hydrolyzed. Draw and name the structures of the products.arrow_forward(Methanesulfinyl)methane is reacted with NaH, and then with acetophenone. Draw and name the structures of the products.arrow_forward
- 3-Oxo-butanenitrile and (E)-2-butenal are mixed with sodium ethoxide in ethanol. Draw and name the structures of the products.arrow_forwardWhat is the reason of the following(use equations if possible) a.) In MO preperation through diazotization: Addition of sodium nitrite in acidfied solution in order to form diazonium salt b.) in MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at low pH c.) In MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at pH 4.5 d.) Avoiding not cooling down the reaction mixture when preparing the diazonium salt e.) Cbvcarrow_forwardA 0.552-g sample of an unknown acid was dissolved in water to a total volume of 20.0 mL. This sample was titrated with 0.1103 M KOH. The equivalence point occurred at 29.42 mL base added. The pH of the solution at 10.0 mL base added was 3.72. Determine the molar mass of the acid. Determine the Ka of the acid.arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENT