ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
ORGANIC CHEM. VOL.1+2-W/WILEYPLUS
12th Edition
ISBN: 9781119304241
Author: Solomons
Publisher: WILEY C
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Chapter 23, Problem 16P
Interpretation Introduction

Interpretation:

Using palmitoleic acid as an example, each of the given reaction of the double bond is to be illustrated.

Concept introduction:

Palmitoleic acid is a solid fat that is formed of a combination of 66% saturated fat (palmitic acid) and 34% unsaturated fat (oleic acid). The structural formula of palmitoleic acid can be given as follows:

ORGANIC CHEM. VOL.1+2-W/WILEYPLUS, Chapter 23, Problem 16P

Solid fats are the triacylglycerols and are solid at room temperature. Oil or liquid fat from plant source undergoes catalytic hydrogenation in order to get converted into solid fat. It needs extensive application to convert vegetable oil into solid commercial cooking fat for the longer shelf life of the fat/oil.

All fatty acids undergo reactions typical of carboxylic acids, such as with alcohols, mineral acids and bromine and phosphorus.

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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
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