EBK ORGANIC CHEMISTRY AS A SECOND LANGU
EBK ORGANIC CHEMISTRY AS A SECOND LANGU
4th Edition
ISBN: 9781119234715
Author: Klein
Publisher: VST
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Chapter 27, Problem 26PP
Interpretation Introduction

Interpretation:

Draw structure of alternating copolymer of vinyl chloride and ethylene.

Concept introduction:

Polymer constructed from a single type of monomer is known as homopoylmers.  If the polymer is constructed from two or more type of monomer then they are called as copolymers.  Copolymers are usually classified based on the distribution of monomer units.  If the monomer unit is present in an alternating fashion then it is known as alternating copolymer while if it is in random fashion then it is known as random copolymer.  If the copolymer consists of homopolymer subunits connected in a block fashion then it is known as block copolymer.  If the homopolymer subunit is grafted onto another homopolymer subunit then it is known as grafter copolymer. These four types of copolymer can be represented as,

EBK ORGANIC CHEMISTRY AS A SECOND LANGU, Chapter 27, Problem 26PP

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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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