A Lewis structure with lowest formal charges for M n O 4 − has to be drawn. Concept introduction: Lewis structure otherwise known as Lewis dot diagrams or electron dot structures that show the bond between atoms and lone pairs of electrons that are present in the molecule. Lewis structure represents each atom and their position in structure using the chemical symbol. Excess electrons forms the lone pair are given by pair of dots, and are located next to the atom.

Question

Want to see the full answer?

Answer 1

Question

Chapter 23, Problem 23.108P

Interpretation Introduction

Interpretation:

A Lewis structure with lowest formal charges for MnO4− has to be drawn.

Concept introduction:

Lewis structure otherwise known as Lewis dot diagrams or electron dot structures that show the bond between atoms and lone pairs of electrons that are present in the molecule. Lewis structure represents each atom and their position in structure using the chemical symbol. Excess electrons forms the lone pair are given by pair of dots, and are located next to the atom.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.

Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.

Five isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compound