The enantiomers of the given compounds using perspective formula has to be drawn. Concept introduction: Enantiomers are stereoisomers of compounds having one asymmetric center, in which these isomers are non-superimposable mirror images of each other. Enantiomers can be drawn using perceptive formals or Fischer projections. Perceptive formula: draw two bonds of asymmetric carbon is in plane, one bond as a solid wedge that coming out of plane and the fourth bond as hatched wedge that pointing away from the plane. T he solid wedge is always below to the hatched wedge. Fischer projections: the asymmetric center (carbon) represents as the point of intersection of two perpendicular lines, in which the horizontal lines represent the bonds that project out of the plane of the paper and the vertical lines represent the bonds that away from the plane of the paper.

Question

Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m

Answer 1

Question

Chapter 4.7, Problem 18P

(a)

Interpretation Introduction

Interpretation:

The enantiomers of the given compounds using perspective formula has to be drawn.

Concept introduction:

Enantiomers are stereoisomers of compounds having one asymmetric center, in which these isomers are non-superimposable mirror images of each other.

Enantiomers can be drawn using perceptive formals or Fischer projections.

Perceptive formula: draw two bonds of asymmetric carbon is in plane, one bond as a solid wedge that coming out of plane and the fourth bond as hatched wedge that pointing away from the plane. T he solid wedge is always below to the hatched wedge.

Fischer projections: the asymmetric center (carbon) represents as the point of intersection of two perpendicular lines, in which the horizontal lines represent the bonds that project out of the plane of the paper and the vertical lines represent the bonds that away from the plane of the paper.

(b)

Interpretation Introduction

Interpretation:

The enantiomers of the given compounds using fisher projection has to be drawn.

Concept introduction:

Enantiomers are stereoisomers of compounds having one asymmetric center, in which these isomers are non-superimposable mirror images of each other.

Enantiomers can be drawn using perceptive formals or Fischer projections.

Perceptive formula: draw two bonds of asymmetric carbon is in plane, one bond as a solid wedge that coming out of plane and the fourth bond as hatched wedge that pointing away from the plane. The solid wedge is always below to the hatched wedge.

Fischer projections: the asymmetric center (carbon) represents as the point of intersection of two perpendicular lines, in which the horizontal lines represent the bonds that project out of the plane of the paper and the vertical lines represent the bonds that away from the plane of the paper.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ m

Determine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +

Please help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWL