Interpretation:
Given that the isobutyl cation spontaneously rearranges to the tert-butyl cation by a hydride shit. The nature of the rearrangement exergonic or endergonic is to be stated. Applying Hammond postulate, how the transition state for the hydride shift might look like is to be drawn.
Concept introduction:
The stabilities of the carbocations are in the order tertiary > secondary > primary. The conversion of a less stable system in to a more stable system will take place readily and it will be an exergonic process. According to Hammond postulate, the transition state of an exergonic reaction will resemble the structure of the reactants.
To state:
Whether the spontaneous rearrangement of isobutyl cation to the tert-butyl cation by a hydride shit is exergonic or endergonic.Further to draw, by applying Hammond postulate, the transition state for the hydride shift.
Trending nowThis is a popular solution!
Chapter 7 Solutions
Organic Chemistry
- Draw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3 CH3 CH3COCI/ AICI3 H₂C. You do not have to consider stereochemistry. Include all valence lone pairs in your answer. • In cases where there is more than one answer, just draw one.arrow_forwardMISTRY 2400 EXPERIMENT 7: BROMINATION OF CINNAMIC ACID WINTER 2022 For the trans isomer of cinnamic acid: Draw the structure of the trans isomer of cinnamic acid and the structure of the intermediate bromonium ion. Note that the bromonium ion can exist as two enantiomeric forms. You should draw both of these isomeric forms. Draw the structures of the two dibrominated products which can form when a bromide ion adds to the bromonium ion intermediate. These must be drawn to show the correct stereochemistry. Assign the chiral carbons as R or S. L. 20 Based on the information in the introduction, what is the expected melting point of these + enantiomeric products? yONng pont, which set of stereoisomersarrow_forwardChapter 6 Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt [References] Draw the structures of the two carbocation intermediates that might form during the reaction of propene (above) with HBr. • You do not have to consider stereochemistry. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. • Separate structures with + signs from the drop-down menu. 00-Sn [F ****arrow_forward
- 6. (Chapter 15-Q37) Indole is an aromatic heterocyclic that has a benzene ring fused to a pyrrole ring. Answer the following questions. Indole 6(a) What is the hybridization of N in this molecule? = 6(b) How many pi electrons N contributes to the ring? = 6() Which orbitals contribute to form a sigma bond between N and H in this molecule? = 6(c) What is the electronic relationship of Indole to naphthalene? Give the answer by comparing number of rings and number of pi electrons in both compounds, write x rings, y pi electrons=|arrow_forwardDraw the most stable resonance form for the intermediate in the following electrophilic substitution reaction. CH3 CH3 CH3CI / AICI3 • H3C • You do not have to consider stereochemistry. Include all valence lone pairs in your answer. In cases where there is more than one answer, just draw one.arrow_forwardDraw the energy diagram of the following reaction. Label and draw the reactants, intermediates, and products. Label the transition state of the rate limiting step and draw the transition state of the rate determining step. Is it a concerted or stepwise reaction? Write the rate law for this reaction. How do we increase the reaction rate? If we use deutrium-tert-butyl bromide such as (CD3)3CBr instead of (CH3)3CBr, will it increase or decrease the reaction rate, or is there no meaningful rate change? Write reasons for your answer. Draw mechanisms for reactions by drawing arrows and intermediates.arrow_forward
- The compound below is treated with chlorine in the presence of light. CH3 CH3CHCH₂CH3 Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. n [ ]# ?arrow_forward4. Carbocations can undergo rearrangements to form more stable carbocations. A fourmembered ring can expand to a five-membered ring and a five-membered ring can expand to a six-membered ring, but a six-membered ring will not expand to a seven-membered ring. Provide an explanation for why the expansion of four- and five-membered rings is favorable, but the expansion of six-membered rings is not. O ** Carrow_forwardThe addition of water to aldehydes and ketones occurs rapidly, although it is not thermodynamically favored. What would be the product for the reaction above? Hint: Think of the self-ionization of water and the polarity of the carbonyl group.arrow_forward
- For the following SN2 reaction, draw the major organic product and select the correct (R) or (S) designation around the stereocenter carbon in the organic substrate and organic product. Include wedge/dash bonds and H on a stereocenter -Ц CI The correct (R) or (S) designations for the stereocenter carbons are The organic substrate is (S) and the organic product is (R) The organic substrate is (R) and the organic product is (S). The organic substrate and organic product are both (S) The organic substrate and organic product are both (R).arrow_forward- For the dehydration shown, use curved arrows to show the formation of the carbocation intermediate in the presence of sulfuric acid H2SO4H2SO4, then draw the structure of the major product of the elimination. (Picture 1): Step 1: Use curved arrows to complete the protonation mechanism of the alcohol. Step 2: Use a curved arrow to show the formation of the carbocation intermediate. Note: HSO−4HSO4− is formed from step 1, but not shown. (Picture 2): Step 3: Use a curved arrow to show the carbocation rearrangement. Step 4: Draw curved arrows to show the elimination to form the major product. Water is added as a base. - Draw the major product of the elimination. H3O+H3O+ has been pre‑drawn for your convenience.arrow_forwardZaitsev's rule is useful in selecting which carbon adjacent to a carbocation will form the double bond in the alkene product. True or Falsearrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning