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Dafna1 [17]
3 years ago
5

A is a d-aldohexose and B is an L-aldohexose. they afford the same optically active aldaric acid after dilute nitric acid oxidat

ion.
what are the structures and names for A and B respectively?

Chemistry
2 answers:
Alexeev081 [22]3 years ago
8 0

Answer:

D-Glucose and L-Glucose

Explanation:

Aldohexose are the sugars which have six number of carbons and ends up in having an aldehyde group at one end. When dilute nitric acid is treated with any of them, the molecule gets oxidized (gets oxygen) and therefore turns into carboxylic acid.

The name of A is D-Glucose, and B is L-Glucose. Please find the structural formula attached.

Gnoma [55]3 years ago
7 0

Answer: D-glucose reacts with dilute nitric acid to give an aldaric acid that was optically active, i.e. did not contain a plane of symmetry.

Explanation:Treatment of aldoses with dilute nitric acid converts them into aldaric acids. The aldehyde function at one end of the molecule and the primary alcohol at the other are both oxidized to carboxylic acids.

Fischer had previously developed a method to interchange the ends of a sugar (the

aldehyde is converted to a CH2OH and the CH2OH is converted to an aldehyde, Fischer reasoned that if ends of 2 were

interchanged, a new L-aldohexose would be obtained. On the other hand, if the

ends of 3 were interchanged, the product would be the same (structure 3). When

the ends of (+)-glucose were interchanged a new sugar was obtained, which Fischer

named L-gulose. When the ends of (+)-mannose were interchanged, the product

was (+)-mannose. Therefore the structure of (+) glucose is structure 2, and

strucutre 3 is (+)-mannose!!!

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Dmitriy789 [7]

Answer:

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Explanation:

Given

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According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

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T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

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∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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Answer:

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