Interpretation:
The structure of compounds H and I is to be proposed.
Concept Introduction:
Isomers are the molecule that have the same number of atoms but have a different arrangement of the atoms in the space.
Stereoisomers have the same molecular formula, but the arrangement of atoms in the three-dimensional orientation is different.
Enantiomers are the stereoisomers whose molecules have chiral center and are mirrors image to 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
The equimolar mixture of two enantiomers is called a racemic mixture.
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
- How 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_forwardCompound A, C₁1 H160, 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.arrow_forwardCompound A has the formula C9H19Cl. B is a C9H19Br compound.A and B undergo base-promoted E2 elimination to give the same alkene C as the major product as well as different minor products.C reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form 2,6-dimethylheptane.Addition of HCl to C yields A as the major product.Propose structures for A and B.arrow_forward
- Compound 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_forwardSharpless epoxidation of allylic alcohol X forms compound Y. Treatment of Y with NaOH and C6H5SH in an alcohol–water mixture forms Z. Identify the structure of Y and draw a mechanism for the conversion of Y to Z. Account for the stereochemistry of the stereogenic centers in Z. Z has been used as an intermediate in the synthesis of chiral carbohydrates.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, C8H8O, was shown to be a methyl ketone while product D, C3H6O, was shown to be an aldehyde.arrow_forward
- Compound J, C16H16Br2, is optically active. On treatment with strong base, compounds K and L (each C16H14) are formed; K and L each absorb only 2 equivalents of hydrogen when reduced over a Pd/C catalyst. Compound K reacts with ozone to give phenylacetic acid (C6H5CH2COOH), while similar treatment of L gives 2 products. One product, M, is an aldehyde with formula C7H6O; the other product is glyoxal (CHO)2. Draw the structure of compound L.arrow_forwardA, a compound with molecular formula C6H10, contains three methylene units. A reacts with one equivalent of H2 over Pd/C to yield B. A reacts with aqueous acid to form a single product, C, and undergoes hydroboration/oxidation to form a pair of enantiomers, D and E. Ozonolysis of A followed by reaction with dimethyl sulfide forms F with molecular formula C6H10O2. Provide structures for A–F.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, CH10, 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 H,SO, dehydration occurred and an optically inactive alkene B, CH4 was produced as the major product. Alkene B, on ozonolysis, gave two products. Product C, C,H,0, was shown to be an aldehyde while product D, CHg0, 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, jtust draw one.arrow_forwardCompound A (C7H11Br) is treated with magnesium in ether to give B (C7H11MgBr), which reacts violently with D2O to give 1-methylcyclohexene with a deuterium atom on the methyl group (C). Reaction of B with acetone (CH3COCH3) followed by hydrolysis gives D (C10H18O). Heating D with concentrated H2 SO4 gives E (C10 H16), which decolorizes two equivalents of Br2 to give F (C10H16 Br4). E undergoes hydrogenation with excess H2 and a Pt catalyst to give isobutylcyclohexane. Determine the structures of compounds A through F, and show your reasoning throughout.arrow_forwardCompounds A and B are isomers having molecular formula C5H12. Heating A with Cl2 gives a single product of monohalogenation, whereas heating B under the same conditions forms three constitutional isomers. What are thestructures of A and B?arrow_forward