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(a)
Interpretation: The major product of the given reaction has to be found.
Concept Introduction:
Major product in the reaction of
General scheme:
It can be observed that the conjugate addition product has been formed by the addition of basic anion to the beta-carbon atom and that of hydrogen atom to the alpha-carbon carbon atom.
The examples for the weak bases are:
(b)
Interpretation: The major product of the given reaction has to be found.
Concept Introduction:
Major product in the reaction of
General scheme:
It can be observed that the conjugate addition product has been formed by the addition of basic anion to the beta-carbon atom and that of hydrogen atom to the alpha-carbon carbon atom.
The examples for the weak bases are:
(c)
Interpretation: The major product of the given reaction has to be found.
Concept Introduction:
Major product in the reaction of
General scheme:
It can be observed that the conjugate addition product has been formed by the addition of basic anion to the beta-carbon atom and that of hydrogen atom to the alpha-carbon carbon atom.
The examples for the weak bases are:
Major product in the reaction of
General scheme:
It can be observed that the direct addition product has been formed from the direct reaction of the strong base with the carbonyl
The examples for the strong bases are:
(d)
Interpretation: The major product of the given reaction has to be found.
Concept Introduction:
Major product in the reaction of
General scheme:
It can be observed that the conjugate addition product has been formed by the addition of basic anion to the beta-carbon atom and that of hydrogen atom to the alpha-carbon carbon atom.
The examples for the weak bases are:
Major product in the reaction of
General scheme:
It can be observed that the direct addition product has been formed from the direct reaction of the strong base with the carbonyl functional group of the
The examples for the strong bases are:
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Chapter 12 Solutions
Essential Organic Chemistry (3rd Edition)
- Don't used hand raiting and don't used Ai solutionarrow_forwardHighlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forwardQ4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forward
- Complete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
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