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Concept explainers
(a)
Interpretation:
The structure of nitrogen-containing compound that is obtained when the given amide undergoes acidic hydrolysis has to be drawn.
Concept Introduction:
Amides are derivatives of
Acidic hydrolysis of amides gives the product as carboxylic acid and
Basic hydrolysis of amides gives the product as carboxylic acid salt and amine. Carboxylic acid salt is obtained because in basic conditions the carboxylic acid is converted into carboxylic acid salt.
(b)
Interpretation:
The structure of nitrogen-containing compound that is obtained when the given amide undergoes acidic hydrolysis has to be drawn.
Concept Introduction:
Amides are derivatives of carboxylic acid. Amides are not much reactive as of carboxylic acids. They are also stable relatively in aqueous solution. But under strenuous conditions amides undergo hydrolysis. The conditions are presence of acid, base or enzymes.
Acidic hydrolysis of amides gives the product as carboxylic acid and amine salt. Amine salt is obtained because in acidic conditions the amine is converted into amine salt.
Basic hydrolysis of amides gives the product as carboxylic acid salt and amine. Carboxylic acid salt is obtained because in basic conditions the carboxylic acid is converted into carboxylic acid salt.
(c)
Interpretation:
The structure of nitrogen-containing compound that is obtained when the given amide undergoes acidic hydrolysis has to be drawn.
Concept Introduction:
Amides are derivatives of carboxylic acid. Amides are not much reactive as of carboxylic acids. They are also stable relatively in aqueous solution. But under strenuous conditions amides undergo hydrolysis. The conditions are presence of acid, base or enzymes.
Acidic hydrolysis of amides gives the product as carboxylic acid and amine salt. Amine salt is obtained because in acidic conditions the amine is converted into amine salt.
Basic hydrolysis of amides gives the product as carboxylic acid salt and amine. Carboxylic acid salt is obtained because in basic conditions the carboxylic acid is converted into carboxylic acid salt.
(d)
Interpretation:
The structure of nitrogen-containing compound that is obtained when the given amide undergoes acidic hydrolysis has to be drawn.
Concept Introduction:
Amides are derivatives of carboxylic acid. Amides are not much reactive as of carboxylic acids. They are also stable relatively in aqueous solution. But under strenuous conditions amides undergo hydrolysis. The conditions are presence of acid, base or enzymes.
Acidic hydrolysis of amides gives the product as carboxylic acid and amine salt. Amine salt is obtained because in acidic conditions the amine is converted into amine salt.
Basic hydrolysis of amides gives the product as carboxylic acid salt and amine. Carboxylic acid salt is obtained because in basic conditions the carboxylic acid is converted into carboxylic acid salt.
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Chapter 17 Solutions
EBK GENERAL, ORGANIC, AND BIOLOGICAL CH
- Naturally occurring D-glucose is one of a pair of enantiomers. Its mirror image is L-glucose. Draw the two cyclic six-membered isomers of L-glucose that differ in the configuration around C1 and indicate which is a and which is B.arrow_forwardGive one specific sample structure (Fischer Projection or Haworth) of the following carbohydrate derivatives: - Amino-sugararrow_forwardIn the monosaccharide derivatives known as sugar alcohols, the carbonyl oxygen is reduced to a hydroxyl group. For example, D-glyceraldehyde can be reduced to glycerol. However, this sugar alcohol is no longer designated D or L. Why?arrow_forward
- Name the heterocyclic substituent found in the molecule shown according to the general type of nitrogen-containing ring to which it belongs.arrow_forwardFollowing are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers. (The sugars shown here are not all of the possible five-carbon sugars.) 1 СНО 2 СНО СНО Н-С—ОН Н-С—ОН Н—С—ОН Н-С—ОН НО —С—Н Н-С—ОН Н- С—ОН НО—С— Н НО—С—Н CH̟OH CH̟OH ČH̟OH в iоchemistry |61 STATE ATION Republic of the Philippines Romblon State University Romblan, Philippines 4 СНО 5 СНО 6 СНО НО—С—Н Н-С—ОН НО -С—Н Н—С—ОН НО -С—Н НО -С—Н H-C–OH Н-С—ОН НО—С—Н CH,OH ČH,OH CH,OHarrow_forwardOrnithine is an amino acid that is not used in the synthesis of proteins, but is an important intermediate in several metabloic pathways including the urea cycle and the synthesis of polyamines. It has a perfectly ordinary terminal amino group and terminal carboxyl group like any other amino acid and a side chain with a single ionizable side group with a pKa of 10.3. If ornithine is placed in solution at pH 7.0, it has a net charge of +1. What would the net charge on this amino acid be if the pH of the solution was raised to pH 12.0? Please explain your reasoning.arrow_forward
- Match the following structural composition of each polysaccharide with its identity Linear homoglycan of glucose connected by α1-4 linkages Linear sulfated chains of alternating β-D-galactopyranose and 3,6-anhydro- α-galactopyranosyl units Poly-β-Dmannopyranosyluronic acid and/or Poly-α-L-gulopyranosyluronic acid Repeating Poly-D-galacturonic acid residues deacetylated straight-chain amino-polysaccharide polymer linked in a β(1-4) type of linkage a. Alginate b. amylose c. chitosan d. carrageenan e. Pectinarrow_forwardDraw two different possible hydrogen-bonding interactions between two molecules of formamide (HCONH2). Clearly label the hydrogen-bond donor and acceptor atoms. Which of these two possible hydrogen-bonding interactions is more likely to occur? (Hint: Consider resonance structures for formamide.)arrow_forwardWith the aid of the simple generic diagram, identify and explain how the type of chemical bonding stabilizes a secondary structure present in 3GRS (glutathione reductase).arrow_forward
- The peptide aspartyl-glutamyl-leucyl-threonyl-alanine shown in the sketch drawing window, has several ionizable groups. Adjust the charges to show the molecule as it would exist at pH 5.00. (I'm lost, I appreciate any help with explanation. ?)arrow_forwardCarbohydrates Instructions: (A) Show the conversion from Fischer to Haworth projections. (B) Draw and give the systematic names of the two (2) possible Haworth structures for the following monosaccharides. 1.D- Ribose 2. D- Galactose 3. D- Fructosearrow_forwardFollowing are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers СНО СНО СНО H-C-OH H-C-OH H-C-OH H-C-OH Но-С—н H-C-OH H-C-OH Но-ҫ—н Но—с—н CHOH CHĻOH ČH,OH СНО СНО СНО Но —С— н H-C-OH HO-C-H Н—С—он Но -С—н HO-C-H H-c-OH H-C-OH HO-C-H CH̟OH ČHĻOH ČHĻOH Pairs of Enantiomers Pairs of Epimersarrow_forward
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