Interpretation:
The existence of enantiomeric forms of the given compound is to be explained.
Concept introduction:
The molecules that are nonsuperimposable or not identical with their mirror images are known as chiral molecules.
A pair of two mirror images that are nonidentical is known as enantiomers, which are optically active.
The objects or molecules that are superimposable with their mirror images are achiral objects or molecules and these objects have a centre of symmetry or plane of symmetry.
The achiral compounds in which plane of symmetry is present internally and consists of chiral centres are known as meso compounds but they are optically inactive.
The stereoisomers that are nonsuperimposable on each other and not mirror images of each other are known as diastereomers.
Chiral molecules are capable of rotating plane polarized light
The molecules that are superimposable or identical with their mirror images are known as achiral molecules, and achiral molecules are not capable of rotating the plane-polarised light.
Enantiomers are pairs of compounds that are nonsuperimposable mirror images of each other.
The compounds having a plane of symmetry do not exhibit enantiomeric forms.
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Organic Chemistry
- 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.arrow_forwardDraw the structures for (Z)-3-methylhex-3-ene and the two major organic products for its reaction with HBr. Be sure to show all stereoisomers using wedge and dash stereochemistry at any chirality center.arrow_forwardThe compound 3,4-dimethyl-hexan-3-ol of 3R, 4S configuration is treated with a concentrated HBr solution at room temperature. A mixture of two stereoisomers is obtained.If the reaction mixture above is heated, the appearance of several other compounds is observed. 1) Draw the different compounds obtained using the wedge-flywheel representation. 2) What is the majority product? Explain 3) Propose a modification of the experimental conditions in order to obtain the exclusive formation of these compounds obtained after heatingarrow_forward
- Trans-1-bromo-2-methylcyclohexane will yield a non-Zaitsev elimination product (3-methylcyclohexene) upon reaction with KOH. Show this reaction by drawing the chair conformations of the reactant and product. Include the curved arrows and explain why the product is not a non-Zaitsev product.arrow_forwardIn 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) Whichproduct predominates—the product of inversion or the product of retention of conguration? (c) Suggest an explanation for this phenomenon.arrow_forwardWhen (S)-2-bromobutane undergoes an SN2 reaction with CH3O-, the product is the compound shown below. What is/are the configuration(s) of the product(s) obtained from this reaction? O equal mixture of R and S H₂C a mixture of enantiomers with more R than S S only CH3 CH OR only O a mixture of enantiomers with more S than R H₂ CH₂arrow_forward
- (a) Show how you would synthesize the pure (R) enantiomer of 2-butyl methyl sulfide, starting with pure (R)-butan-2-oland any reagents you need.(b) Show how you would synthesize the pure (S) enantiomer of the product, still starting with (R)-butan-2-ol and anyreagents you need.arrow_forwardIn some nucleophilic substitutions under SN1 conditions, complete racemization does not occur and a small excess of one enantiomer ispresent. For example, treatment of optically pure 1-bromo-1phenylpropane 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.arrow_forwardbased on the structure of compound A and the conditions used, explain in one or two sentences whether the reaction will proceed by an Sn1, Sn2, E1 or E2 pathwayarrow_forward
- Explain why one diastereomer is thermodynamically more stable than the other. To do so,draw both chair conformations for cis-4-tert-butylcyclohexanol. Do the same for trans-4-tert-butylcyclohexanol. Label all 1,3-diaxial interactions for these four species. Indicate the most stable chair conformation of cis-4-tert-butylcyclohexanol. Do the same for trans-4-tert-butyl-cyclohexanol. Which of these latter two species is the most stable?arrow_forward5.) Given the following three reactions of enantiomerically pure starting materials, please identify products A and B for each line (given the product molecular formulae), using dashes and wedges where appropriate, the meso (optically inactive) product and the chiral (optically active product). NaBH4 CH3OH A CHo02 + B CSH1002 optically active HO optically inactive (R)-3-hydroxycyclopentanone OH Os04 (cat) (CH3)3COOH C CH1203 optically inactive + D CH1203 (2R, 3E)-2-hydroxy-3-pentene optically active OH Os04 (cat) (CH3)3COOH E CsH203 + F CSH1203 optically active optically inactive (2R, 3Z)-2-hydrox y-3-pentenearrow_forwardThe alkene shown undergoes bromination. (a) Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any). Use wedge‑and‑dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one. (b) Characterize the starting alkene as having the E or Z configuration. (c) characterize the product(s).arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning