Question

1. Circle the two alpha carbons and box the two leaving groups in methamphetamine. methamphetamine 2. Methamphetamine reacts with Br. Draw structure of the two possible products. (Hint: nucleophile reacts at each alpha carbon) 3. Assume the secondary alpha carbon in methamphetamine undergoes loss of leaving group (first step of a SN1 mechanism). Draw the structure of the carbocation.

0 0
Add a comment Improve this question Transcribed image text
Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
1. Circle the two alpha carbons and box the two leaving groups in methamphetamine. methamphetamine 2....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • 1. (a) In the first step of an SN1 reaction, the carbon-leaving group bond breaks and...

    1. (a) In the first step of an SN1 reaction, the carbon-leaving group bond breaks and a carbocation forms. In the second step, the carbocation undergoes a nucleophilic attack. The rate limiting step in an SN1 reaction is: The nucleophilic attack. The formation of the product. The formation of the carbocation. Submit Answer Tries 0/1 (b) In an SN2 reaction, the nucleophile attacks while the leaving group leaves in a concerted mechanism. The rate limiting step in an SN2 reaction...

  • a. First, circle the leaving group and label the alpha carbon as a. b. Then, label...

    a. First, circle the leaving group and label the alpha carbon as a. b. Then, label the adjacent sp3 beta carbons as B. c. Finally, draw in the H's that can get eliminated along with the leaving group in each of these compounds. The first compound has been completed for you. HH می ) و CH3 2. Now, let's perform the elimination to make the Zaitsev (most substituted) product. Draw in the H from before and the curved arrows to...

  • 2. (1.5 Credits) a. One carbon in Compound A is an alpha carbon and beta carbon....

    2. (1.5 Credits) a. One carbon in Compound A is an alpha carbon and beta carbon. Circle this carbon. b. Box the leaving group(s) in Compound A. If the leaving group is poor, how can it be made better? Br c. Compound A reacts with H SO4. Draw two possible intermediates of this reaction. Which intermediate most likely forms? Give reasons. d. One of your intermediates from part c forms Compound B. Use curved arrows to show how Compound B...

  • Please, make the answer clear and reable. BrEtOH a OEt Hace H2C Alkyl halides undergo nucleophilic...

    Please, make the answer clear and reable. BrEtOH a OEt Hace H2C Alkyl halides undergo nucleophilic substitution and elimination reactions. When the kinetics of the reaction are measured, if the rate of the reaction is found to be dependent only upon the concentration of the alkyl halide the reaction is first order. The substitution reaction is thus termed ction is termed E1. These reactions are unimolecular and occur in two steps. The first step is rate-limiting and involves the loss...

  • 1. Draw the structure of any secondary alkyl halide containing at least six carbons. 2. Name...

    1. Draw the structure of any secondary alkyl halide containing at least six carbons. 2. Name it. 3. Circle the alpha carbon. 4. Draw box around all the beta carbons. 5. Draw the product(s) for the elimination reaction product for your choosen alkyl halide.

  • 1) Draw the arrows to show bond formation and breaking for the following mechanisms and complete...

    1) Draw the arrows to show bond formation and breaking for the following mechanisms and complete questions b, c, and d. Marking scheme is also provided. (9 marks total) a) All tails of arrows must start at electrons. The arrows must start in the middle of a sigma bond or at a lone pair (2 marks) Take % mark for each mistake until you reach zero. All heads of arrows must end at an atom. (2 marks) Take X mark...

  • 5) Iodine is a better leaving group than bromine. But iodine is a better nucleophile than bromine. Why is that? Introdu...

    5) Iodine is a better leaving group than bromine. But iodine is a better nucleophile than bromine. Why is that? Introduction Alkyl halides can be prepared from alcohols by reactions with hydrogen halides (HCI, HBr, or HI) via nucleophilic substitution. In this type of reaction, the nucleophile displaces a leaving group from a carbon atom of an organic substrate (here the alcohol once protonated). Both electrons of the new bond to the carbon are provided by the nucleophile while the...

  • 1.   In both the sodium iodide test and the silver nitrate test, why does 2-bromobutane react faster...

    1.   In both the sodium iodide test and the silver nitrate test, why does 2-bromobutane react faster than 2-chlorobutane? Bromine is a better leaving group since it is a weaker base than chlorine is. 2.   a. Why does benzyl chloride react under both SN1 and SN2 conditions?             Benzyl chloride is a primary alkyl halide, hence reactive under SN2 conditions.             The primary carbocation formed due to the departure of Cl- is stabilized by the pi electrons in the benzene ring.         b. Why is...

  • 1) State which of the electrophiles given below will react preferentially by i) SN1, ii) by...

    1) State which of the electrophiles given below will react preferentially by i) SN1, ii) by SN2, or iii) capable of reacting by either of the two mechanisms depending on the given conditions. How can you affect those conditions to favour SN1 or SN2? Reason your predictions based on the structures of the compounds. Br-CH3, Br-CH2CH3, Br-CH(CH3)2, Br-C(CH3)3, Br-CH2-C5H6; C5H6 = phenyl Introduction Alkyl halides can be prepared from alcohols by reactions with hydrogen halides (HCI, HBr, or HI) via...

  • Multistep reaction mechanisms Group Activity Treating (S)-2-butanol with HBr results in the following two step reaction: Step 1: CH3CH(OH)CH2CH3 + HBr ⇌ CH3CH(OH2)CH2CH3+ + Br- Step 2: CH3CH(OH2)CH2CH3+ + Br- ⇌ CH3CH(Br)CH2CH3 + H2O Draw a mechanism for e

    Multistep reaction mechanisms Group ActivityTreating (S)-2-butanol with HBr results in the following two step reaction: Step 1: CH3CH(OH)CH2CH3 + HBr ⇌ CH3CH(OH2)CH2CH3+ + Br-Step 2: CH3CH(OH2)CH2CH3+ + Br- ⇌ CH3CH(Br)CH2CH3 + H2ODraw a mechanism for each step. Draw structures for all reactants and products, including correct stereochemistry (wedges and dashes). Show all the curved arrows in each step, and label each curved arrow according to whether it represents proton transfer, loss of a leaving group, nucleophilic attack, or carbocation rearrangement.

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT