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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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Answer:

The IHD = 0

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Explanation:

IHD = Index for Hydrogen Deficiency .It determines total number of unsaturation or Pi- bond present in the molecule.It is calculated using the formula:

For Molecule:

C_{c}H_{h}O_{o}N_{n}X_{x}

Here . C = carbon

H = Hydrogen

O = Oxygen

N = Nitrogen

X = Halogen

The IHD is calculated as :

\frac{2c-h-x+n+2}{2}...............(1)

For example :

C4H8O2

Here c = 4 , h = 8 , o = 2 .

n = x=0

Put the value in equation (1) and find IHD

\frac{2(4)-8-0+0+2}{2}

\frac{8-8+2}{2}

\frac{2}{2}

IHD = 1

For C4H9Cl

c = 4 , h = 9 and  x = 1

IHD for this molecule can be calculated as :

\frac{2c-h-x+n+2}{2}

\frac{2(4)-9-1+2}{2}

\frac{8-9-1+2}{2}

= 0

This means there is no unsaturation in the molecule.

Each bond is sigma bond and properly saturated.

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Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.
goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

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Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}

In the first solution:

X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625

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340torr = 0.625P^0_{A}+0.375P^0_{B} <em>(1)</em>

For the second equation:

X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700

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370torr = 0.700P^0_{A}+0.300P^0_{B}<em>(2)</em>

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340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})

340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}

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P^0_{A} = 490 torr

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

370torr = 0.700*490torr+0.300P^0_{B}

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

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