Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Question
Chapter 27, Problem 27.62AP
Interpretation Introduction
(a)
Interpretation:
The explanation for iodoacetic acid reaction with side chain thiol group of a cysteine residue is to be stated. The reason as to why this reaction is necessary before sequencing is to be stated.
Concept introduction:
The reagent DTT is known as dithiothreitol and is a common redox reagent. It is also known as Cleland’s reagent. It is relatively not a stable compound due to air oxidation. It is used as a deprotecting group and to break the disulfide bond in protein.
Interpretation Introduction
(b)
Interpretation:
The product obtained on oxidation of disulfide bond with
Concept introduction:
The compound hydrogen peroxide represented by
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(a) The two most acidic hydrogens of uracil have pKa’s of 9.5 and 14.2, respectively. Match these pKa’s with the hydrogens in the structural formula and provide structures for the most stable resonance contributors of the monoanion and the dianion.(b) The pKa of the conjugate acid of triethylamine is 10.4. Is triethylamine a strong enough base to convert uracil to its monoanion? To its dianion?
(b) Using alanine as an example, draw possible conformations of alanine in a peptide chain
and identify which conformer is favoured.
(c) Discuss how considering 'allylic strain' [see pages 33-39 of handout 1] can explain the
favoured conformation in part b above, and how this is relevant to the B-sheet
conformation in peptides/proteins.
A tripeptide contains lysine (R= -(CH₂)4NH₂), phenylalanine (R= -CH₂C6H5) and glycine (R= -H). It was found that the tripeptide reacts with 2 moles
of DNFB to give compound A. When compound A was treated with SOCI2, glycine was activated. When hydrolyzed, both NH₂ groups of lysine were
found to be DNP-labelled. What is the structure of the tripeptide?
Select one:
H
(CH₂)2NH2
CH
NH
CH
H₂N
O
O
H₂N
H₂N
H₂N
C
(CH2)4NH2
CH
010
(CH2)4NH2
CH
C=O
CH,Ph
CH
NH
NH
NH
CH
CH,Ph
CH
H
010
CH
CH,Ph
CH
(CH2)4NH2
ΝΗ
NH
NH
NH
CH,Ph
CH
H
CH
COOH
H
CH
COOH
COOH
COOH
Chapter 27 Solutions
Organic Chemistry
Ch. 27 - Prob. 27.1PCh. 27 - Prob. 27.2PCh. 27 - Prob. 27.3PCh. 27 - Prob. 27.4PCh. 27 - Prob. 27.5PCh. 27 - Prob. 27.6PCh. 27 - Prob. 27.8PCh. 27 - Prob. 27.9PCh. 27 - Prob. 27.10PCh. 27 - Prob. 27.11P
Ch. 27 - Prob. 27.12PCh. 27 - Prob. 27.13PCh. 27 - Prob. 27.14PCh. 27 - Prob. 27.15PCh. 27 - Prob. 27.16PCh. 27 - Prob. 27.17PCh. 27 - Prob. 27.18PCh. 27 - Prob. 27.19PCh. 27 - Prob. 27.20PCh. 27 - Prob. 27.21PCh. 27 - Prob. 27.22PCh. 27 - Prob. 27.23PCh. 27 - Prob. 27.24PCh. 27 - Prob. 27.25PCh. 27 - Prob. 27.26PCh. 27 - Prob. 27.27PCh. 27 - Prob. 27.28PCh. 27 - Prob. 27.29PCh. 27 - Prob. 27.30PCh. 27 - Prob. 27.31PCh. 27 - Prob. 27.32PCh. 27 - Prob. 27.33PCh. 27 - Prob. 27.34PCh. 27 - Prob. 27.35PCh. 27 - Prob. 27.36PCh. 27 - Prob. 27.37PCh. 27 - Prob. 27.38PCh. 27 - Prob. 27.39PCh. 27 - Prob. 27.40PCh. 27 - Prob. 27.41PCh. 27 - Prob. 27.42PCh. 27 - Prob. 27.43APCh. 27 - Prob. 27.44APCh. 27 - Prob. 27.45APCh. 27 - Prob. 27.46APCh. 27 - Prob. 27.47APCh. 27 - Prob. 27.48APCh. 27 - Prob. 27.49APCh. 27 - Prob. 27.50APCh. 27 - Prob. 27.51APCh. 27 - Prob. 27.52APCh. 27 - Prob. 27.53APCh. 27 - Prob. 27.54APCh. 27 - Prob. 27.55APCh. 27 - Prob. 27.56APCh. 27 - Prob. 27.57APCh. 27 - Prob. 27.58APCh. 27 - Prob. 27.59APCh. 27 - Prob. 27.60APCh. 27 - Prob. 27.61APCh. 27 - Prob. 27.62APCh. 27 - Prob. 27.63APCh. 27 - Prob. 27.64APCh. 27 - Prob. 27.65APCh. 27 - Prob. 27.66APCh. 27 - Prob. 27.67APCh. 27 - Prob. 27.68APCh. 27 - Prob. 27.69APCh. 27 - Prob. 27.70APCh. 27 - Prob. 27.71APCh. 27 - Prob. 27.72APCh. 27 - Prob. 27.73APCh. 27 - Prob. 27.74APCh. 27 - Prob. 27.75APCh. 27 - Prob. 27.76APCh. 27 - Prob. 27.77APCh. 27 - Prob. 27.78APCh. 27 - Prob. 27.79APCh. 27 - Prob. 27.80APCh. 27 - Prob. 27.81APCh. 27 - Prob. 27.82APCh. 27 - Prob. 27.83AP
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Identify monomeric and multimeric (di-,tri-meric, for example) saccharides from this group. If you come across multimeric polypeptides, state the mono- mers that form these multimers. State the nature of the bond that links two units together in each case. (a) Maltose (b) Trehalose (c) Lactose (d) Sucrose (e) Cellobiose (f) Agarosearrow_forward(b) Describe how the charge of some amino groups in a protein might differ at pH 9.0 and pH 5.0. the charge on the amino group will differ at pH 5 and pH 9 which will depend on the pKa of the amino acid. (c) Describe how the charge of some carboxyl groups in a protein might differ at pH 9.0 and pH 5.0. charge on carboxyl group will differ at pH 5 and pH 9 which will depend on the pKa of the amino acid. (d) Given your answers to parts (b) and (c), what kind of intramolecular interactions in beta-galactosidase are most likely to be affected by a change in pH from 9.0 to 5.0? (e) Could the interactions you mention in part (d) affect the catalytic activity of beta-galactosidase?arrow_forwardTaking the structure of the amino acids into account, give a comparative discussion of these pK values : Alanine : 2.34 : 6.1 : 9.70 Histidine: 1.82 : 9.17arrow_forward
- The peptide Proline-Serine-Alanine-Phenylalanine-Glutamine is present at pH 7. Draw the peptide and include stereochemistry.arrow_forward(a) An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed bydilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.NHHCH2RR2NH2 + sugarRHHOH OHOH Oan aliphatic ribosideH3O+ +(b) Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid.Using your mechanism for part (a), show why cytidine and adenosine (for example)are not so readily hydrolyzed. Explain why this stability is important for livingorganisms.arrow_forwardAn amino acid mixture of phenylalanine, glycine and glutamic acid is to be separated bypaper chromatography. The solvent is less polar than water. Which of these amino acids will have the highest Rf value and which the lowest? Explain. Treatment of a new protein with dansyl chloride reveals two (2) dansyl-labelled derivatives of amino acids, alanine and methionine. What can you deduce about the structure of the protein from these results? Name the amino acids that contribute atoms to both purine and pyrimidine rings.arrow_forward
- The anticlotting property of heparin is partly the result of the negative charges it carries. ( Q.) Identify the functional groups that provide the negative charges.arrow_forward2arrow_forwardA naturally occurring amino acid such as alanine has a group that is a carboxylic acid and a group that is a protonated amine. The pKa values of the two groups are shown.(a). If the pKa value of a carboxylic acid such as acetic acid is about 5, then why is the pKa value of the carboxylic acid group of alanine so much lower? (b). Draw the structure of alanine in a solution at pH = 0. (c). Draw the structure of alanine in a solution at physiological pH (pH 7.4).(d). Draw the structure of alanine in a solution at pH = 12. (e). Is there a pH at which alanine is uncharged (that is, neither group has a charge)? (f) At what pH does alanine have no net charge (that is, the amount of negative charge is the same as the amount of positive charge)?arrow_forward
- (a) Discuss the role of carbonic anhydrase in CO2 transport. (b) Discuss the cause of ‘cold denaturation’ of proteins.arrow_forwardNow let's consider histidine as a free amino acid in aqueous solution. a) Draw the most likely structure of histidine under biochemical standard state conditions. b) Given that free histidine has the following three pKa values, assign each to its corresponding acidic hydrogen or conjugate base in your structure from part a). pK1 = 1.7, pKa2 = 6.0, pK23 = 9.1 c) For each pKa, give the corresponding expression for the equilibrium constant. It helps to write out the chemical equation for each.arrow_forwardb) The methyl group at the C terminus of aspartame can be cleaved by concentrated sulfuric acid (conc. H₂SO4) to reveal the carboxylic acid (compound 1), which has a pK₂ of 2.43. OH NH₂ OCH3 conc. H₂SO4 OH NH₂ OH Compound 1 The other two pKa values in the molecule are unchanged. It is useful to calculate the isoelectric point (pl) when planning to separate charged species from neutral species with electrophoresis. pKal + PK₂2 2 pl = HINT: i) Calculate the isoelectric point of aspartame. ii) Calculate the isoelectric point of compound 1 and explain your reasoning behind the calculation.arrow_forward
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