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Lemur [1.5K]
3 years ago
15

Which of the following statements does NOT describe the anomeric carbon? A. This carbon is attached to two oxygens. B. This carb

on is chiral. C. The hydroxyl group on this carbon can be above or below the ring. D. It is the carbon of the carbonyl in the open-chain form of the sugar. E. All of the statements describe the anomeric carbon

Chemistry
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

The answer is E. All of the statements describe the anomeric carbon.

Explanation:

When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.

As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).

It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).

The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)

It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.

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Two different compounds are obtained by combining nitrogen with oxygen. the first compound results from combining 46.7 g of n wi
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Second

17.9 g of N and 82.0 g of O.

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Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 6.80 g of lauryl alco
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Answer:

The approximate molar mass of lauryl alcohol is 174.08 g/m

Explanation:

An excersise to apply the colligative property of Freezing-point depression.

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First of all, think the T° of fusion of benzene → 5.5°C

ΔT = T° pure solvent - T° fusion solution

Kf for benzene: 5.12 °C/m

5.5°C - 4.5°C = 5.12 °C /m  . m

1°C /  5.12 m /°C = m

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This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.

I have to find out in 0.2 kg.

1 kg sv ____ 0.195 moles solute

0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute

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6.80 g/ 0.039 moles = 174.08 g/m

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