If the natural abundance of 13 C is 1.10%, the relative heights of M + and M+1peaks in the mass spectrum of benzene are to be given. Concept introduction: Mass spectrum of compounds contains in addition to the molecular ion peak (M + ) a small peak at M+1. This peak arises due to the different isotopes of the atoms present in the molecule. The heights of M + and M+1peaks in the mass spectrum of a compound will be in proportion to the percentage of the different isotopes of atoms present in the molecule. To give: The relative heights of M + and M+1peaks in the mass spectrum of benzene if the natural abundance of 13 C is 1.10%.

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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 + 2C3H8O3

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Chapter 12.SE, Problem 22AP

Interpretation Introduction

Interpretation:

If the natural abundance of ¹³C is 1.10%, the relative heights of M⁺ and M+1peaks in the mass spectrum of benzene are to be given.

Concept introduction:

Mass spectrum of compounds contains in addition to the molecular ion peak (M⁺) a small peak at M+1. This peak arises due to the different isotopes of the atoms present in the molecule. The heights of M⁺ and M+1peaks in the mass spectrum of a compound will be in proportion to the percentage of the different isotopes of atoms present in the molecule.

To give:

The relative heights of M⁺ and M+1peaks in the mass spectrum of benzene if the natural abundance of ¹³C is 1.10%.

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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 + 2C3H8O3

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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.…