ORGANIC CHEMISTRY MASTERINGCHEM ACCESS
ORGANIC CHEMISTRY MASTERINGCHEM ACCESS
9th Edition
ISBN: 9780137249442
Author: Wade
Publisher: INTER PEAR
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Chapter 24.9C, Problem 24.19P

(a)

Interpretation Introduction

To determine: The structure of the phenylthiohydantoin derivative of alanine.

Interpretation: The structure of the phenylthiohydantoin derivative of alanine is to be determined.

Concept introduction: The process in which a proper series of amino acids is maintained in the  peptide chain is known as edmann degradation. this process includes the labelling of terminal amino residue followed by the cleavage of this terminal amino residue from the peptide chain. the peptide bonds between other amino acids do not get disturb during this process.

(b)

Interpretation Introduction

To determine: The structure of the phenylthiohydantoin derivative of tryptophan.

Interpretation: The structure of the phenylthiohydantoin derivative of tryptophan is to be determined.

Concept introduction: The process in which a proper series of amino acids is maintained in the  peptide chain is known as edmann degradation. this process includes the labelling of terminal amino residue followed by the cleavage of this terminal amino residue from the peptide chain. the peptide bonds between other amino acids do not get disturb during this process.

(c)

Interpretation Introduction

To determine: The structure of the phenylthiohydantoin derivative of lysine.

Interpretation: The structure of the phenylthiohydantoin derivative of lysine is to be determined.

Concept introduction: The process in which a proper series of amino acids is maintained in the  peptide chain is known as edmann degradation. this process includes the labelling of terminal amino residue followed by the cleavage of this terminal amino residue from the peptide chain. the peptide bonds between other amino acids do not get disturb during this process.

(d)

Interpretation Introduction

To determine: The structure of the phenylthiohydantoin derivative of proline.

Interpretation: The structure of the phenylthiohydantoin derivative of proline is to be determined.

Concept introduction: The process in which a proper series of amino acids is maintained in the  peptide chain is known as edmann degradation. this process includes the labelling of terminal amino residue followed by the cleavage of this terminal amino residue from the peptide chain. the peptide bonds between other amino acids do not get disturb during this process.

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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

Chapter 24 Solutions

ORGANIC CHEMISTRY MASTERINGCHEM ACCESS

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