Concept explainers
(a)
Interpretation:
The detailed mechanism for the production of given compounds from the respective alkyne is to be drawn.
Concept introduction:
The
Answer to Problem 11.39P
The detailed mechanism for the given reaction with a major product is:
Explanation of Solution
The structure for the given compound
The given compound is germinal dichloride, and thus can be produced by electrophilic addition of an excess of
In the given reaction the alkyne is the electron rich site and the hydrogen from the
In the second step, the chloride ion acts as a nucleophile and attacks at vinylic carbocation forming vinylic chloride product.
The vinylic chloride again undergoes the addition of
This carbocation further reacts with nucleophilic chloride ion to form a germinal dichloride product.
The detailed mechanism is drawn for the given reaction with showing the formation of stable carbocations and major product.
(b)
Interpretation:
The detailed mechanism for the production of given compounds from respective alkyne is to be drawn.
Concept introduction:
The alkynes are electron rich system like alkenes and can undergo an electrophilic addition reaction with strong Bronsted acids just like the alkenes do. The reaction proceeds with proton transfer reaction to form a stable carbocation followed by the action of water as a nucleophile. In excess of reagent, the reaction occurs twice forming a geminal dihalide compound as a major product.
Answer to Problem 11.39P
The detailed mechanism for the given reaction with the major product is:
Explanation of Solution
The structure for the given compound
The given compound is germinal dibromide, thus can be produced by electrophilic addition of an excess of
In the given reaction the alkyne is the electron rich site and the hydrogen from the
In the second step, the bromide ion acts as a nucleophile and attacks at vinylic carbocation forming vinylic bromide product.
The vinylic bromide again undergoes the addition of
This carbocation further reacts with nucleophilic bromide ion to form germinal dibromide product.
The detailed mechanism is drawn for the given reaction with showing the formation of stable carbocations and major product.
(c)
Interpretation:
The detailed mechanism for the production of given compounds from respective alkyne is to be drawn.
Concept introduction:
The alkynes are electron rich system like alkenes and can undergo an electrophilic addition reaction with strong Bronsted acids just like the alkenes do. The reaction proceeds with proton transfer reaction to form a stable carbocation followed by the action of water as a nucleophile. In a single addition reaction, the reaction occurs only once forming a vibylic halide compound as a major product. The deuterium is an isotope of a hydrogen atom and reacts the same as hydrogen.
Answer to Problem 11.39P
The detailed mechanism for the given reaction with a major product is:
Explanation of Solution
The structure for the given compound is:
The given compound is vinylic bromide having deuterium at adjacent carbon, thus can be produced by single electrophilic addition of
In the given reaction the alkyne is the electron rich site and the hydrogen from the
In the second step the bromide ion acts as a nucleophile and attacks at vinylic carbocation forming vinylic bromide product.
The detailed mechanism is drawn for the given reaction with showing the formation of stable carbocations and major product.
(d)
Interpretation:
The detailed mechanism for the production of given compounds from respective alkyne is to be drawn.
Concept introduction:
The alkynes are electron rich system like alkenes and can undergo an electrophilic addition reaction with strong Bronsted acids just like the alkenes do. The reaction proceeds with proton transfer reaction to form a stable carbocation followed by the action of water as a nucleophile. In excess of reagent, the reaction occurs twice forming a geminal dihalide compound as a major product.
Answer to Problem 11.39P
The detailed mechanism for the given reaction with the major product is:
Explanation of Solution
The structure for the given compound is:
The given compound is germinal dichloride, thus can be produced by electrophilic addition of an excess of
In the given reaction the alkyne is the electron rich site and the hydrogen from the
In the second step, the chloride ion acts as a nucleophile and attacks at vinylic carbocation forming vinylic chloride products.
The vinylic chloride again undergoes the addition of
This carbocation further reacts with nucleophilic chloride ion to form a germinal dichloride product.
The detailed mechanism is drawn for the given reaction with showing the formation of stable carbocations and major product.
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Chapter 11 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Complete boxes in the flow chart. Draw the structure of the organic compound foundin each layer after adding 3M NaOH and extraction. Make sure to include any charges. Provide explanation on answers.arrow_forward== Vid4Q2 Unanswered ☑ Provide IUPAC name of product in the reaction below A 3,4-dimethylcyclohexene B 1,2-dimethylcyclohexane C 1,2-dimethylcyclohexene D 3,4-dimethylcyclohexane H₂ Pdarrow_forward5. Use the MS data to answer the questions on the next page. 14.0 1.4 15.0 8.1 100- MS-IW-5644 26.0 2.8 27.0 6.7 28.0 1.8 29.0 80 4.4 38.0 1.0 39.0 1.5 41.0 1.2 42.0 11.2 43.0 100.0 44.0 4.3 79.0 1.9 80.0 2.6 Relative Intensity 40 81.0 1.9 82.0 2.5 93.0 8.7 20- 95.0 8.2 121.0 2.0 123.0 2.0 136.0 11.8 0 138.0 11.5 20 40 8. 60 a. Br - 0 80 100 120 140 160 180 200 220 m/z Identify the m/z of the base peak and molecular ion. 2 b. Draw structures for each of the following fragments (include electrons and charges): 43.0, 93.0, 95.0, 136.0, and 138.0 m/z. C. Draw a reasonable a-fragmentation mechanism for the fragmentation of the molecular ion to fragment 43.0 m/z. Be sure to include all electrons and formal charges. 6. Using the values provided in Appendix E of your lab manual, calculate the monoisotopic mass for the pyridinium ion (CsH6N) and show your work.arrow_forward
- Nonearrow_forwardStereochemistry: Three possible answers- diastereomers, enantiomers OH CH₂OH I -c=0 21108 1101 41745 HOR CH₂OH IL Но CH₂OH TIL a. Compounds I and III have this relationship with each other: enantiomers b. Compounds II and IV have this relationship with each other: c. Compounds I and II have this relationship with each other: d. *Draw one structure that is a stereoisomer of II, but neither a diastereomer nor an enantiomer. (more than one correct answer)arrow_forwardNonearrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning