Organic Chemistry
Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 25, Problem 14P

The example of a silyl–Hilbert–Johnson nucleosidation reaction in Section 25.3 is presumed to involve an intermediate ribosyl cation that is stabilized by intramolecular interactions involving the C2 benzoyl group. This intermediate blocks attack by the heterocyclic base from the α face of the ribose ring but allows attack on the β face, as required for formation of the desired product. Propose a struc-ture for the ribosyl cation intermediate that explains the stereoselective bonding of the base

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When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.
When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.   Which experimental number must be initialled by the Lab TA for the first run of Part 1 of the experiment? a) the heat capacity of the calorimeter b) Mass of sample c) Ti d) The molarity of the HCl e) Tf
Predict products for the Following organic rxn/s by writing the structurels of the correct products. Write above the line provided" your answer D2 ①CH3(CH2) 5 CH3 + D₂ (adequate)" + 2 mited) 19 Spark Spark por every item. 4 CH 3 11 3 CH 3 (CH2) 4 C-H + CH3OH CH2 CH3 + CH3 CH2OH 0 CH3 fou + KMnDy→ C43 + 2 KMn Dy→→ C-OH ") 0 C-OH 1110 (4.) 9+3 =C CH3 + HNO 3 0 + Heat> + CH3 C-OH + Heat CH2CH3 - 3 2 + D Heat H 3 CH 3 CH₂ CH₂ C = CH + 2 H₂ → 2 2
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