Interpretation:
The structure of compound D and E is to be proposed with the help of given data.
Concept Introduction:
Isomers are molecules which have the same number of atoms but have different arrangements of the atoms in space.
Stereoisomers are of the same molecular formula, but the arrangement of atoms in the three-dimensional orientation is different.
Enantiomers are stereoisomers whose molecules have a chiral center and the molecules are mirror images of each other.
A pair of two mirror images that are non-identical is known as a pair of enantiomers.
The objects or molecules that are superimposable with their mirror images are achiral objects or molecules. These objects have a centre of symmetry or plane of symmetry.
The achiral compounds in which the plane of symmetry is present internally and consists of chiral centres are known as meso compounds, but they are optically inactive.
The stereoformula which is depicted in two dimensions, in which stereochemical information is not destroyed, is determined by the Fisher Projection formula.
The stereoisomers which are non-superimposable 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 which 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.
Plane of symmetry is the plane that bisects the molecule in two equal halves, such that they are mirror images of each other.
Compounds having plane of symmetry are usually achiral as they do not have different atoms around the central carbon atom.
A
Degree of unsaturation of compound can be calculated using expression as:
Here, C is the number of carbon atoms, N is the number of nitrogen atoms, X is the number of halogen and H is number of hydrogen atoms.
Want to see the full answer?
Check out a sample textbook solutionChapter 5 Solutions
Organic Chemistry
- (a) What product(s) are formed when the E isomer of C6H5CH = CHC6H5 is treated with Br2, followed by one equivalent of KOH? Label the resulting alkene(s) as E or Z. (b) What product(s) are formed when the Z isomer of C6H5CH = CHC6H5 is subjected to the same reaction sequence? (c) How are the compounds in parts (a) and (b) related to each other?arrow_forwardCompound X is treated with Br2 to yield meso-2,3-dibromobutane. Draw the structure of compound X.arrow_forwardTreatment of -D-glucose with methanol in the presence of an acid catalyst converts it into a mixture of two compounds called methyl glucosides (Section 25.3A). In these representations, the six-membered rings are drawn as planar hexagons. (a) Propose a mechanism for this conversion and account for the fact that only the OH on carbon 1 is transformed into an OCH3 group. (b) Draw the more stable chair conformation for each product. (c) Which of the two products has the chair conformation of greater stability? Explain.arrow_forward
- Draw and name the seven aldehydes and ketones with the formula C5H10O. Which are chiral?arrow_forwardThe first step in the metabolism of glycerol, formed by digestion of fats, is phosphorylation of the pro-R—CH2OH group by reaction with adenosine triphosphate (ATP) to give the corresponding glycerol phosphate plus adenosine diphosphate (ADP). Show the stereochemistry of the product.arrow_forwardCompound A, C11H16O, was found to be an optically active alcohol. Despite its apparent unsaturation, no hydrogen was absorbed on catalytic reduction over a Pd/C catalyst. On treatment of A with dilute H2SO4, dehydration occurred and an optically inactive alkene B, C11H14 was produced as the major product. Alkene B, on ozonolysis, gave two products. Product C, C7H6O, was shown to be an aldehyde while product D, C4H8O, was shown to be a ketone. Draw the structure of compound C. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. In cases where there is more than one answer, just draw one. HELP PLEASE I DONT UNDERSTAND THE PROCESSarrow_forward
- Compound A, C11H16O, was found to be an optically active alcohol. Despite its apparent unsaturation, no hydrogen was absorbed on catalytic reduction over a Pd/C catalyst. On treatment of A with dilute H2SO4, dehydration occurred and an optically inactive alkene B, C11H14 was produced as the major product. Alkene B, on ozonolysis, gave two products. Product C, C8H8O, was shown to be a methyl ketone while product D, C3H6O, was shown to be an aldehyde.arrow_forwardCompound X has molecular formula C5H10. In the presence of a metal catalyst, compound X reacts with one equivalent of molecular hydrogen to yield 2-methylbutane. (1) Suggest three possible structures for compound X. (2) Hydroboration-oxidation of compound X yields a product with no chirality centers. Identify the structure of compound X.arrow_forwardTreatment of (HOCH2CH2CH2CH2)2CO with acid forms a product of molecular formula C9H16O2 and a molecule of water. Draw the structure of the product and explain how it is formed.arrow_forward
- Vitamin D3, the most abundant of the D vitamins, is synthesized from 7-dehydrocholesterol, a compound found in milk and fatty fish such as salmon and mackerel. When the skin is exposed to sunlight, a photochemical electrocyclic ring opening forms provitamin D3, which is then converted to vitamin D3 by a sigmatropic rearrangement (Section 27.5). Draw the structure of provitamin D3.arrow_forwardHow many rings and π(pi) bonds are contained in compound A and draw one possible structure for this compound A. Compound A has molecular formula C6H10 and is hydrogenated to a compound having molecular formula C6H12arrow_forwardA chiral alkyne A with molecular formula C6H10 is reduced with H2 and Lindlar catalyst to B having the R configuration at its stereogenic center. What are the structures of A and B?arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning