(a) Interpretation: The glycosidic linkage and acetal carbon in the lactose are to be identified. Concept introduction: An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Question

Want to see the full answer?

Answer 1

Question

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

A solution contains 0.097 M Ce3+, 1.55x10-3 M Ce4+, 1.55x10-3 M Mn2+, 0.097 M MnO4-, and 1.00 M HClO4 (F= 9.649 x 104 C/mol). a) Write a balanced net reaction that can occur between species in this solution. b) Calculate deltaG0 and K for the reaction. c) Calculate E and deltaG for the conditions given. Ce4+ + e- = Ce3+ E0= 1.70 V MnO4- + 8H+ + 5e- = Mn2+ + 4H2O E0= 1.507 V

2. Specify the solvent and reagent(s) required to carry out each of the following FGI. If two reagent sets must be used for the FGI, specify the solvent and reagent(s) for each reagent set. If a reaction cannot be carried out with reagents (sets) class, write NP (not possible) in the solvent box for reagent set #1. Use the letter abbreviation for each solvent; use a number abbreviation for reagent(s). Solvents: CH2Cl2 (A); H₂O (B); Reagents: HBr (1); R₂BH (6); H2SO4 (2); CH3OH (C); Br₂ (3); CH3CO₂H (D) NaHCO3 (4); Hg(OAc)2 (5); H₂O2/HO (7); NaBH4 (8) Reagent Set #1 Reagent Set #2 FGI + enant OH Solvent Reagent(s) Solvent Reagent(s)

Germanium (Ge) is a semiconductor with a bandgap of 2.2 eV. How could you dope Ge to make it a p-type semiconductor with a larger bandgap? Group of answer choices It is impossible to dope Ge and have this result in a larger bandgap. Dope the Ge with silicon (Si) Dope the Ge with gallium (Ga) Dope the Ge with phosphorus (P)

Answer 2

Question

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Answer 6

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Answer 7

Chapter 9, Problem 9.77P

Interpretation Introduction

(a)

Interpretation:

The glycosidic linkage and acetal carbon in the lactose are to be identified.

Concept introduction:

An acetal is characterized by a C atom bonded to two alkyl (or H) groups and two alkoxy (RO–) groups. The compound in which the OH from a sugar molecule has been replaced by OR is called glycoside. The C-O-C bond that connects the sugars is called glycosidic linkage.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Answer 9

Interpretation Introduction

(b)

Interpretation:

The type of glycosidic linkage that lactose has is to be determined.

Concept introduction:

If the alkoxy (RO-) group on the acetal carbon is in axial position, it is designated as α, and if it is in equatorial position, it is designated as β. If the ring on the other side of glycosidic linkage is bonded to C4, it is notified as 1, 4’, where the prime (’) indicates that C4 and C1 are on different sugar units.

Answer 10

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Answer 11

Interpretation Introduction

(c)

Interpretation:

The complete, detailed mechanism and the products of the acid catalyzed hydrolysis of lactose are to be drawn.

Concept introduction:

The acid catalyzed hydrolysis of lactose gives two monosaccharides, which are linked by glycoside linkage. In Step 1, protonation of electron rich O atom takes place. The bond of acetal carbon and electron poor O of the glycosidic linkage breaks in heterolysis step. In Step 3, the newly formed carbocation is attacked by water as a nucleophile. This can take place on either side of the plane of the carbocation C. Finally, in Step 4, deprotonation removes the positive charge from O and forms D-Galactose.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 9 - Prob. 9.1P Ch. 9 - Prob. 9.2P Ch. 9 - Prob. 9.3P Ch. 9 - Prob. 9.4P Ch. 9 - Prob. 9.5P Ch. 9 - Prob. 9.6P Ch. 9 - Prob. 9.7P Ch. 9 - Prob. 9.8P Ch. 9 - Prob. 9.9P Ch. 9 - Prob. 9.10P

Solution Summary: The author explains the glycosidic linkage and acetal carbon in the lactose.

Concept explainers

Want to see the full answer?

Chapter 9 Solutions