Concept explainers
a)
Interpretation:
The complete mechanism for the conversion of 30 alcohol given to 30
Concept introduction:
Alcohols get protonated in the presence of strong acids. The protonated intermediate will lose a molecule of water to yield a carbocation. The nucleophilic attack by the halide ion will lead to the formation of the alkyl halide.
To propose:
The complete mechanism for the conversion of 30 alcohol given to 30 alkyl halide by treating with HCl involving two cationic intermediates.
b)
Interpretation:
The complete mechanism for the conversion of 30 alcohol given to 30 alkyl halide by treating with HBr involving two cationic intermediates is to be proposed.
Concept introduction:
Alcohols get protonated in the presence of strong acids. The protonated intermediate will lose a molecule of water to yield a carbocation. The nucleophilic attack by the halide ion will lead to the formation of the alkyl halide.
To propose:
The complete mechanism for the conversion of 30 alcohol given to 30 alkyl halide by treating with HBr involving two cationic intermediates.
c)
Interpretation:
The complete mechanism for the conversion of 30 alcohol given to 30 alkyl halide by treating with HCl involving two cationic intermediates is to be proposed.
Concept introduction:
Alcohols get protonated in the presence of strong acids. The protonated intermediate will lose a molecule of water to yield a carbocation. The nucleophilic attack by the halide ion will lead to the formation of the alkyl halide.
To propose:
The complete mechanism for the conversion of 30 alcohol given to 30 alkyl halide by treating with HCl involving two cationic intermediates.
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Chapter 17 Solutions
Study Guide with Student Solutions Manual for McMurry's Organic Chemistry, 9th
- 5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3arrow_forwardShow work. don't give Ai generated solution. How many carbons and hydrogens are in the structure?arrow_forward13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. cleavage Bond A •CH3 + 26.← Cleavage 2°C. + Bond C +3°C• CH3 2C Cleavage E 2°C. 26. weakest bond Intact molecule Strongest 3°C 20. Gund Largest argest a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. C Weakest bond A Produces Most Bond Strongest Bond Strongest Gund produces least stable radicals Weakest Stable radical b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 13°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. • CH3 methyl radical Formed in Gund A Cleavage c.…arrow_forward