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Concept explainers
Interpretation:
The alkene should be identified from the given alcohol undergoes dehydration. Cis and trans alkene should be mentioned
Concept introduction:
Dehydration reaction:
Removal of water molecule from the reaction when the alcohol is treated with strong acid like sulfuric acid.
Alcohol is reaction with concentrated sulfuric acid, first alcohol gets protonated forms carbocation (more stable carbocation) followed by elimination of proton (
Tertiary carbocation is more stable than the secondary, secondary carbocation is more stable than primary.
In dehydration reaction, sulfuric acid is act as a proton donor, and which is used to protonate the alcohol and makes carbocation therefore sulfuric acid is the driving force of the reaction. Dehydration reaction will not go without acid (sulfuric acid).
Cis and Trans alkene:
In 1,2-disubstituted
In 1,2-disubstituted alkenes, if the two alkyl groups (
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Chapter 14 Solutions
Fundamentals of General, Organic, and Biological Chemistry, Books a la Carte Plus Mastering Chemistry with Pearson eText -- Access Card Package (8th Edition)
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- Problem 15 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 4 of 15 4G 54% Done On the following Lineweaver-Burk -1 plot, identify the by dragging the Km point to the appropriate value. 1/V 40 35- 30- 25 20 15 10- T Км -15 10 -5 0 5 ||| 10 15 №20 25 25 30 1/[S] Г powered by desmosarrow_forward1:30 5G 47% Problem 10 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a competitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' s mM¹ with 10 mg pe 20 V' 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
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