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- In some nucleophilic substitutions under SN1 conditions, completeracemization does not occur and a small excess of one enantiomer ispresent. For example, treatment of optically pure 1-bromo-1-phenylpropane with water forms 1-phenylpropan-1-ol. (a) Calculate how much of each enantiomer is present using the given optical rotation data.(b) Which product predominates—the product of inversion or the product of retention of configuration? (c) Suggest an explanation for this phenomenon.In some nucleophilic substitutions under SN1 conditions, complete racemization does not occur and a small excess of one enantiomer is present. For example, treatment of optically pure 1-bromo-1-phenylpropane with water forms 1-phenylpropan-1-ol. (a) Calculate how much of each enantiomer is present using the given optical rotation data. (b) Which product predominates—the product of inversion or the product of retention of configuration? (c) Suggest an explanation for this phenomenon.By considering the stereochemical requirements for E2-elimination, and using an appropriate illustration need help indicating whether the more stable conformer (conformer B) of trans-1 can undergo E2 elimination. Thank you :)
- 1(a) Explain briefly what is meant by ONE of the following terms (use words and diagrams).(i) Markovnikov’s rule(ii) Transition state 1(b) Draw the mechanism (using structures, arrows and appropriate symbols) for the formation of thefollowing products:Draw the Newman projection of 3-ethyl-2,4-dimethylhexane looking through C3-C4 bond and perform a full conformation alanalysis by rotating along the C2-C3 bond. Make sure to clearly identify the structure of all minima (on the xaxis) and their relative Energies (on they axis).I am working on a practice assignment for my organic II course and am having difficulty with a question that asks to identify the reaction sequence used to synthesize isopropylcyclopentane. I would really appreciate the help!
- Friedel–Crafts alkylation of benzene with (R)-2-chlorobutane and AlCl3 affords sec-butylbenzene.a. How many stereogenic centers are present in the product?b. Would you expect the product to exhibit optical activity? Explain, with reference to the mechanism.To the right are two enantiomeric natural products with vastly different olfactory properties. I find this example of enantiomeric pairs particularly intriguing, as I love caraway and rye but loathe searment: Answer the questions below with the understanding that an optically pure solution of the (-)-enantiomer of carvone rotates light counterclockwise 60*. a. what would be the observed optical rotation of a solution comprised of a 3:1 ratio of the (+) to (-) enantiomers of carvone? b. a different solution of both carvone enantiomers exhibits an observed optical rotation of -10*. determine the enantiomeric excess for this solution. c. which enantiomer (R or S) is in excess in the solution from (b)? how do you know? d. using your answer (b) above, calculate the actual percentages of each enantiomer in the solution.(a) How many Stereoisomers are possible for 9,10- dibromoexadecanoic acid?(b) Addition of bromine to palmitoleic acid primarily yields a set of enantiomers, (±) -three-9,10-dibromohexadecanoic acid. The addition of bromine is an anti-addition to the double bond (ie, it appears to occur through the intermediary of the bromonium ion). Taking into account the stereochemistry of the cis double bond of palmitoleic acid and the stereochemistry of adding bromine,Write a three-dimensional structure for (±) -three-9,10-dibromohexadecanoic acid!
- Chemistry 1,3,5, -Cycloheptatriene (C) can undergo two different electrocyclic ring closures as givenbelow under thermal and photochemical conditions. Identify the mode of rotation (Conrotatoryor Dirotatory) in each product D and E. Using Aromatic transition state theory show that bothreactions are allowed.1. Give the major organic product(s) of the reaction and include all the stereochemistry as appropriate. Identify any meso compounds. 2. Indicate whether a solution of the product would be optically active or no not active and provide a simple reasoning to support your choice. Need the last 2Compound K, L and M are three isomers with the molecular formula C5H10O.Compound K cannot be oxidized, while compound M and L can be oxidized.Oxidation of compound L with hot acidified potassium permanganate,KMnO4 solution yields 2-methylbutanoic acid. When treated with Iodoformreagent, yellow precipitation only occurs in compound K. Fehling test onlyyielded positive results for compound L and negative for compound M andK. Compound M is then reacted with hydrogen chloride, HCl produceschlorocyclopentane