Concept explainers
a)
Interpretation:
How to prepare tert-butyl chloride from tert-butylalcohol is to be stated.
Concept introduction:
tert-Alcohols when treated with HCl, HBr or HI in ether at 0oC give the corresponding tert-
To state:
How to prepare tert-butyl chloride from tert-butylalcohol.
b)
Interpretation:
How to .prepare 2-bromo-4-methylpentane from the corresponding alcohol is to be stated.
Concept introduction:
A secondary alkyl bromide is required. Secondary alkyl halides can be prepared by treating the alcohol required with PBr3 in ether solution.
To state:
How to prepare 2-bromo-4-methylpentane from the corresponding alcohol.
c)
Interpretation:
How to .prepare 1-bromo-5-methylhexane from the corresponding alcohol is to be stated.
Concept introduction:
A primary alkyl bromide is required. Primary alkyl halides can be prepared by treating the alcohol required with PBr3 in ether solution.
To state:
How to prepare1-bromo-5-methylhexane from the corresponding alcohol.
d)
Interpretation:
How to .prepare 1-fluoro-3,3-dimethylcyclopentane from the corresponding alcohol is to be stated.
Concept introduction:
A secondary alkyl fluoride is required. Secondary alkyl fluorides can be prepared by treating the alcohol required with diethylaminosulphur trifluoride and HF in pyridine.
To state:
How to prepare1-fluoro-3,3-dimethylcyclopentane from the corresponding alcohol.
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Chapter 10 Solutions
EP ORGANIC CHEMISTRY,24 MONTH-OWLV2
- Don't used Ai solutionarrow_forward5. 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_forward
- 13. (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_forwardBr. COOH Br, FCH COOH E FeBr ASOCI B NH (CH,CO),OD Br₂ 2 C alcKOHarrow_forwardFind A to F (all)arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning