Which best describes why carbocation formation is the rate determining step of a unimolecular substitution reaction? O The ABC step is slow. The nucelophile is too weak to attack without a carbocation present. The carbocation formed is unstable and high in energy. Carbocation formation is reversible so it doesn't readily occur. The solvent is acting as the nucleophile.
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- Explain why these alkyl halides form carbocations at different rates. Your answer should use some chemical representation, not only words. Br Br forms carbocation forms carbocation slower fasterWhat are the organic products of these reactions? Are they electrophiles or nucleophiles? What is their decreasing order of rate?Polar aprotic solvents enhance the rate of an SN2 reaction by stabilizing the cations and the anions. changing the polarizibility of the nucleophile. raising the energy of the nucleophile. O lowering the energy of the nucleophile. Save for Later
- 10. Draw an energy diagram for the addition of HBr to 1-pentene. Let one curve on your diagram show the formation of 1-bromopentane product and another curve on the same diagram show the formation of 2- bromopentane product. Draw and label the positions for all reactants, intermediates, and products. Which curve has the higher-energy carbocation intermediate? Which curve has the higher-energy first transition state?Use the dropdown menu to indicate whether the rate of the reaction shown below will increase, decrease, or remain the same when the reaction conditions are changed to X or to Y or to Z (see below). NaN3 'N3 Br CH3CN X: Change the leaving group from Br to CI Y: Increase the concentration of haloalkane Z: Increase the concentration of NaN3 X: choose your answer.. ^ Y: choose your answer... V Z: choose your answer... choose your answer... Remain the same Nex Decrease IncreaseUse the dropdown menu to indicate whether the rate of the reaction shown below will increase, decrease, or remain the same when the reaction conditions are changed to X or to Y or to Z (see below). NaCN Br CN CH3CN X: Change the leaving group from Br to CI Y: Increase the concentration of haloalkane Z: Increase the concentration of NaCN X: choose your answer... Y: choose your answer... Z: choose your answer... >
- Which statement is true concerning the two carbocations formed by the protonation (attachment of an H*) of 2-methylpropene? a) The activation energy associated with the primary carbocation intermediate is less than the activation energy associated with the tertiary carbocation. b) The rate of the formation of the tertiary carbocation is less (slower) than the rate of the formation of the primary carbocation. c) The activation energy and the rate of formation associated with the formation of the primary carbocation is greater. d) The activation energy is greater for the formation of the primary carbocation but its rate of formation is less. O a O bWhich is/are NOT TRUE about bimolecular nucleophilic substitution reactions? Select one or more: 1. A carbocation intermediate is formed. 2. A strong nucleophile displaces a halogen atom in a concerted mechanism. 3. Presence of polar aprotic solvents promotes this reaction. 4. Methyl halides react faster than secondary alkyl halides.For the given SN2 reaction, draw the organic and inorganic products of the reaction, and identify the nucleophile, substrate, and leaving group. Include wedge-and-dash bonds and draw hydrogen on a stereocenter. Hmm.. CEN + CI Organic product Inorganic product
- 5. Classify the following reagents as either nucleophiles or electrophiles: Zn2* , CH3NH2 , HS , OH2 , CH3COOH, H2SO42. Draw the structures and explain why CH3CH₂O and CH3CO₂ are good nucleophiles but CH3SO3, water, and alcohols (R-OH) are poor nucleophiles. Propose a 'cutoff' for the amount of negative charge needed to be a good nucleophile. CH3CH₂O CH3CO₂ CH3SO3 H₂O CH₂OHDraw the least stable resonance form for the intermediate in the following electrophilic substitution reaction. NO2 _NO2 Br2 / FeBr3 Br • 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. opy aste