All the constitutional isomers that have the molecular formula C 3 H 6 F 2 O , in which the oxygen is bonded to two carbon atoms, are to be drawn. Concept introduction: Constitutional isomers have the same molecular formula. They differ in the connectivity of the atoms. They do not interconvert and can be separated from each other. When drawing the constitutional isomers for a given molecular formula, the first step is to determine the IHD. This allows finding the possible number of double bonds, triple bonds, or rings in each isomer. Then all possible isomers are drawn omitting double bonds, triple bonds, and halogen atoms. Finally, double and triple bonds are added at various locations to generate as many structures as possible. Halogen atoms are added to form as many unique connectivities as possible.

Question

Want to see the full answer?

Answer 1

Question

Chapter 4, Problem 4.52P

Interpretation Introduction

Interpretation:

All the constitutional isomers that have the molecular formula C3H6F2O, in which the oxygen is bonded to two carbon atoms, are to be drawn.

Concept introduction:

Constitutional isomers have the same molecular formula. They differ in the connectivity of the atoms. They do not interconvert and can be separated from each other. When drawing the constitutional isomers for a given molecular formula, the first step is to determine the IHD. This allows finding the possible number of double bonds, triple bonds, or rings in each isomer. Then all possible isomers are drawn omitting double bonds, triple bonds, and halogen atoms. Finally, double and triple bonds are added at various locations to generate as many structures as possible. Halogen atoms are added to form as many unique connectivities as possible.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

8. Draw all the resonance forms for each of the fling molecules or ions, and indicate the major contributor in each case, or if they are equivalent (45) (2) -PH2 سمة مد

A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3

1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.