Although we learned in Section 5.4 that uncharged nitrogenatoms generally cannot be chiral centers (due to nitrogeninversion), an exception is Tröger's base. Tröger's base has twoenantiomers that can be separated from each other. They aredifferent in their configurations at the N atoms. One enantiomeris shown on the right, viewed from two different perspectives.(a) Draw the second enantiomer of Tröger's base. (b) Explainwhy the two enantiomers do not interconvert.Tröger's base

Troger's base is a white solid tetracyclic organic compound. Troger's base and it's analogs are…

Answered: Although we learned in Section 5.4 that…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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During your experimental investigation of the conversion of glutamine to proline, you have labeled the γ-carbonyl group of glutamine with 14C. What carbon in the proline product will be labeled?
The 6-member pyranose ring of glucose is formed through the interaction of the hydroxyl group on C5 with the aldehyde group What kind of ring is formed when the hydroxyl group on C4 attacks instead? Draw the Haworth projection of the products with all the carbons labelled accordingly and name them Why would ring formation through C6 or C3 be unfavorable?
Do reducing sugars have a reactive anomeric carbon? explain
Which bond(s) in α-D-glucose must be broken tochange its configuration to β-D-glucose? Which bond(s) to convert D-glucose to Dmannose? Which bond(s) to convert one “chair” form of D-glucose to the other?
why do so many pharmaceutical drugs contain nitrogen? How would an NH2 group benefit this dimer rather than the OH groups?
Sucrose, a major product of photosynthesis in green leaves, is synthesized by a battery of enzymes. The substrates for sucrose synthesis, D -glucose and D fructose, are a mixture of α and β anomers as well as acyclic compounds in solution. Nonetheless, sucrose consists of α - D -glucose linked by its carbon-1 atom to the carbon-2 atom of βD -fructose. How can the specificity of sucrose be explained in light of the potential substrates?
Why is it possible for electrons to flow from a redox center with a more positive ℰ°′ to one with a more negative ℰ°′ within an electrontransfer complex?
Gentiobiose has the molecular formula C12H22O11 and has been isolated from gentian root and by hydrolysis of amygdalin. Gentiobiose exists in two different forms, one melting at 86°C and the other at 190°C. The lower melting form is dextrorotatory ([α] 16°), the higher melting one is levorotatory ([α] -6°). The rotation of an aqueous solution of either form, however, gradually changes until a final value of [α] 9.6° is observed. Hydrolysis of gentiobiose is efficiently catalyzed by emulsin and produces two moles of D-glucose per mole of gentiobiose. Gentiobiose forms an octamethylether, which on hydrolysis in dilute acid yields 2,3,4,6-tetra-O-methyl-D-glucose and 2,3,4-tri-O-methyl-D-glucose. What is the structure of gentiobiose? EXPLAIN IN DETAIL.
In many biochemical reactions which involves the formation of an enolate intermediate, the carbonyl oxygen of the substrate is coordinated to a divalent metal ion (usually zinc or magnesium) in the active site. Explain with structural drawings, how this ion-dipole interactions affect the acidity of the a-protons?
Why can six-carbon sugars sometimes form furanose rings and other times pyranose rings?
What are the donor atoms involved in Aspartame-Cu(II) binding? A) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of alpha carboxyilic acid group of Aspartic acid B) Nitrogen atom of alpha amine group of Phenyl alanine methyl ester and an oxygen atom of alpha carboxyilic acid group of Aspartic acid C) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of alpha carboxyilic acid group of Phenyl alanine methyl ester D) Oxygen atoms of the dipeptide Aspartame E) Nitrogen atom of alpha amine group of Aspartic acid and an oxygen atom of carboxyilic acid group of the R group of Aspartic acid
Draw the (B1-->A1) linked dissacharide formed between the C3 epimer of B-D-glucose and the C2 epimer of a-D-galactose. Explain in 1-2 sentences why (or why not) a precipiate would be observed when mixing this dissacharide in an alkaline solution of copper.

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