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(a)
Interpretation:
The given synthesis scheme is to be converted to word form that can be used as instructions in the laboratory.
Concept introduction:
The synthesis scheme is the balanced chemical equation written for carrying out a sequence of reactions with specified steps. The synthetic step displays that the reactants are converted to the product by reacting with reagents in the required conditions.
The structures on the left side of the reaction arrow () are the reactants. The reagents used are written above the arrow, and the reaction conditions including solvent, temperature, pH, time of reaction, etc. are written below the arrow. If more than one sequence is combined in one step, then the reagents are numbered according to their sequence and can be written above and below the reaction arrow. In such cases, the reaction conditions are separated from the reactant or reagent by either a comma or a slash. The numbers given to reagents represent that the reaction goes to completion before the next reagent is added. The structures on the right side of the reaction arrow
() are the products. The product of the previous step is the reactant of the next step. While writing the synthetic scheme in word form, the
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Answer to Problem 13.28P
The word form for the given synthesis scheme is:
The starting . Then, treat
with phosphoric acid at
to form
.
Explanation of Solution
The given synthesis scheme is:
In the given synthetic route, the reactant having functional group epoxide is converted to an alcohol named which is further converted to an
on reaction with appropriate reagents as mentioned. In the first step, the first reagent used is phenylmagnesium bromide in the solvent diethyl ether, and the second reagent is
which represents the aqueous acidic condition. In the second step, the reagent is phosphoric acid and
is the reaction temperature. Thus, the word form of the above synthetic scheme can be written as follows:
The starting epoxide reacts with phenylmagnesium bromide in the solvent diethyl ether followed by aqueous acid, to form . Then, treat
with phosphoric acid at
to form
.
The given synthesis scheme is converted to word form by identifying the names of reactants, reagents, and products.
(b)
Interpretation:
The given synthesis scheme is to be converted to word form that can be used as instructions in the laboratory.
Concept introduction:
The synthesis scheme is the balanced chemical equation written for carrying out a sequence of reactions with specified steps. The synthetic step displays that the reactants are converted to products by reacting with the reagents in the required conditions.
The structures on the left side of the reaction arrow () are the reactants. The reagents used are written above the arrow, and the reaction conditions including solvent, temperature, pH, time of reaction, etc. are written below the arrow. If more than one sequence is combined in one step, then the reagents are numbered according to their sequence and can be written above and below the reaction arrow. In such cases, the reaction conditions are separated from the reactant or reagent by either a comma or a slash. The numbers given to reagents represent that the reaction goes to completion before the next reagent is added. The structures on right side of the reaction arrow (
) are the products. The product of the previous step is the reactant of the next step. While writing the synthetic scheme in word form, the functional groups involved in the reactants and the functional groups produced in the product are to be identified.
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Answer to Problem 13.28P
The word form for the given synthesis scheme is:
Phenylehtanone reacts with lithium diisopropylamide followed by iodomethane, to produce phenylpropanone. Then, add the molecular bromine in the presence of acetic acid to form which further reacts with sodium acetate to yield the final product.
Explanation of Solution
The given synthesis scheme is:
The given synthetic route is of three steps. In the first step, the reactant having functional group alpha bromo ketone named
, which, in the third step, is further converted to the final product having the ester functional group. In the first step, the first reagent used is lithium diisopropylamide, and the second reagent is iodomethane. In the second step, the reagent is molecular bromine in acetic acid. The reagent for the third step is sodium acetate. Thus, the word form of the above synthetic scheme can be written as follows:
Phenylehtanone reacts with lithium diisopropylamide followed by iodomethane to produce phenylpropanone. Add the molecular bromine in presence of acetic acid to phenylpropanone to form which further reacts with sodium acetate to yield THE final product.
The given synthesis scheme was converted to word form by identifying the names of reactants, reagents, and products.
(c)
Interpretation:
The given synthesis scheme is to be converted to word form that can be used as instructions in the laboratory.
Concept introduction:
The synthesis scheme is the balanced chemical equation written for carrying out a sequence of reactions with specified steps. The synthetic step displays that the reactants are converted to the product by reacting with reagents in the required conditions.
The structures on the left side of the reaction arrow () are the reactants. The reagents used are written above the arrow, and the reaction conditions including solvent, temperature, pH, time of reaction, etc. are written below the arrow. If more than one sequence is combined in one step, then the reagents are numbered according to their sequence and can be written above and below the reaction arrow. In such cases, the reaction conditions are separated from the reactant or reagent by either a comma or a slash. The numbers given to reagents represent that the reaction goes to completion before the next reagent is added. The structures on right side of the reaction arrow (
) are the products. The product of the previous step is the reactant of the next step. While writing the synthetic scheme in word form, the functional groups involved in the reactants, and the functional groups produced in the product are to be identified.
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Answer to Problem 13.28P
The word form for the given synthesis scheme is:
Heat in aqueous hydrochloric acid to form
, add phosphorous tribromide to it to produce
. Then, react
with sodium azide to yield the final product.
Explanation of Solution
The given synthesis scheme is:
The given synthetic route is of three steps. The nitro functional group remains as it is throughout the reaction sequence, and so, is not considered. In the first step, ether functional group is converted to an alcohol. In the second step, group of alcohol is replaced by bromine, which in the third step is replaced by
. In the first step, the reagent used is
that represents acidic condition where
is the symbol used for heat. In the second step, the reagent is
, phosphorous tribromide, and in the third step, the reagent used is sodium azide,
. Thus, the word form of the above synthetic scheme can be written as follows:
Heat in aqueous hydrochloric acid to form
, and add phosphorous tribromide to it to produce
. Then, react
with sodium azide to yield the final product.
The given synthesis scheme was converted to word form by identifying the names of reactants, reagents, and products.
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Chapter 13 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- 1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward
- 6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forwardNonearrow_forwardQ1: Draw the most stable and the least stable Newman projections about the C2-C3 bond for each of the following isomers (A-C). Are the barriers to rotation identical for enantiomers A and B? How about the diastereomers (A versus C or B versus C)? H Br H Br (S) CH3 (R) CH3 H3C (S) H3C H Br Br H A C enantiomers H Br H Br (R) CH3 H3C (R) (S) CH3 H3C H Br Br H B D identicalarrow_forward
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- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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