Interpretation:
It is to be explained why the equilibrium percentage of the first given molecule in its keto form is lower than that of the second given molecule, referring to Table 7-1.
Concept introduction:
In aqueous basic or acidic conditions,
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Chapter 7 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Use the SN1 reaction shown below to answer questions 4-7. H2S -S-CH3 acetone Which option gives the correct order for bonds broken and bonds formed during the arrow-pushing mechanism for this reaction? Step 1: C-O bond breaks Step 2: S-C bond forms O A. Step 3: S-H bond breaks at the same time as an O-H bond forms Step 1: S-H bond breaks at the same time as an O-H bond forms O B. Step 2: C-O bond breaks at the same time as an S-C bond forms Step 1: S-C bond breaks OC. Step 2: C-O bond forms Step 1: S-H bond breaks at the same time as a C-O bond breaks O D. Step 2: S-C bond forms Step 3: O-H bond formsarrow_forwardFor each organic reaction, say whether at equilibrium there will be more reactants than products (so equilibrium lies to the left), or more products than reactants (so equilibrium lies to the right). You'll probably find some useful information in the ALEKS Data resource. OH + + H3O left + H₂O left + H₂O left + H₂O НО. right right right OH + H₂O + OH + OH + + H₂O* X Ś 10: 000 18 Ararrow_forwardAccording to Table 7-1, the equilibrium percentage of the first molecule in its keto form is lower than that of the second molecule. Explain why. Hint: What do you know about the stability of C=C double bonds?arrow_forward
- Draw an energy diagram for a reaction with keq = 1. What is the value of ∆G° in this reaction?arrow_forwardHO .OH cat. H2SO4 H₂O cat. H2SO4 Complete both mechanisms above. BRIEFLY explain how the different reaction conditions (in the presence of acid) result in different products being formed. How does equilibrium play a role?arrow_forward298) Select all of the following changes that would result in the equilibrium shown below being hifted toward the right (i.e., favoring products). If a change has no effect, then do not select it. C6H1206 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (g) Note that C6H1206 (s) refers to "glucose". O Increasing the amount of glucose O Increasing the concentration of oxygen gas Increasing the temperature O Decreasing the partial pressure of water vapor O Adding liquid water O Increasing the volume in which the reaction occurs O Decreasing the partial pressure of oxgyen O Adding a suitable catalyst to increase the reaction rate O Decreasing the volume in which the reaction occurs O Decreasing the temperaturearrow_forward
- Compound A is converted to compound B via a one-step reaction with an activation energy of +3 kcal/mol. The energy difference between Compound B and the transition state of the reaction is 7 kcal/mol. What is the AH° for the reaction? O -10 kcal/mol -4 kcal/mol +10 kcal/mol -3 kcal/mol +3 kcal/mol +4 kcal/molarrow_forwardor each organic reaction, say whether at equilibrium there will be more reactants than products (so equilibrium lies to the left), or more products than reactants (so equilibrium lies to the right). You'll probably find some useful information in the ALEKS Data resource. OH + + H30 t OH left + H₂O left + H₂O right -4. OH + left + H₂O HO right right ► -4 OH + H₂O + он + OH + + H30* X Ś 09: 00. 18 Ararrow_forwardH-C- OH Does the equilibrium lie to the left or to the right in the following reaction: Н-С—ОН + CH,-Ö:- ċ-0 + CH,-ö-H ??arrow_forward
- || C-H CH,OH Н-С-ОН Н-С -ОН Hydrogen catalyst НО -С —Н НО-С—Н H-C-OH Н-С—ОН H-C–OH Н-С-ОН CH,OH CH,OH Sorbitol Glucosearrow_forwardThe following reaction is reversible under acidic conditions: + H₂O f What step can be done to shift the equilibrium towards formation of more products? O Use excess amount of the alcohol O Use excess amount of the carboxylic acid O Use excess amount of water O Continuously remove water from the mixture O All of these O OH + CH3OHarrow_forwardDetermine the Kc of the following reaction: AB4(g) + B2C(g) ⇌ AC2(g) + B2(g) Kp = 21/62500arrow_forward