Question

1. 36 г спирта «А» разделили в равных количествах на два сосуда. В первом сосуде при окислении получилась кислота «В». Определите спирт(ы), если при взаимодействии спирта «А» во втором сосуде и кислоту «В» получился сложный эфирм массой 34.8 г

Answer 1

Ответ:

Пропанол-1

Объяснение:

(18/X)=Y/(X+14) - спирт в кислоту

Отношение эфира, спирта и кислоты  

34,8/(X+X+14-18)=(18/X)=Y/(X+14) => X=60 г/моль

Спирт

Пропанол-1 (из пропанола-2 при окислении выходит кетон)

Эфир

Пропиловый эфир пропионовой кислоты

Answer 2

Problem Analysis

We are given that 36 g of alcohol "A" was divided equally into two vessels. When alcohol "A" was oxidized in the first vessel, it formed acid "B". In the second vessel, alcohol "A" reacted with acid "B" to form a complex ether with a mass of 34.8 g. We need to determine the alcohol(s) involved in the reaction.

Solution

Let's assume that the alcohol "A" is a primary alcohol with the general formula R-CH2-OH, where R represents an organic group. When alcohol "A" is oxidized, it forms an aldehyde or a carboxylic acid, depending on the oxidizing agent used. In this case, it forms acid "B".

The reaction between alcohol "A" in the second vessel and acid "B" to form a complex ether is an esterification reaction. The general equation for this reaction is:

Alcohol + Acid → Ester + Water

To determine the alcohol(s) involved, we need to consider the molecular weights and the mass balance of the reaction.

Let's assume the molecular weight of alcohol "A" is MA g/mol and the molecular weight of acid "B" is MB g/mol.

According to the mass balance equation:

Mass of alcohol "A" + Mass of acid "B" = Mass of complex ether + Mass of water

Using the given information, we have:

36 g + Mass of acid "B" = 34.8 g + Mass of water

Simplifying the equation, we get:

Mass of acid "B" = 34.8 g - 36 g + Mass of water

Now, let's consider the molecular weights of the substances involved:

Mass of acid "B" = (Mass of acid "B" / MB) × (Molecular weight of acid "B")

Mass of water = (Mass of water / MW) × (Molecular weight of water)

Substituting these values into the mass balance equation, we get:

(Mass of acid "B" / MB) × (Molecular weight of acid "B") = 34.8 g - 36 g + (Mass of water / MW) × (Molecular weight of water)

Simplifying further, we have:

(Mass of acid "B" / MB) × (Molecular weight of acid "B") = -1.2 g + (Mass of water / MW) × (Molecular weight of water)

Since the molecular weight of water is known (18 g/mol), we can rewrite the equation as:

(Mass of acid "B" / MB) × (Molecular weight of acid "B") = -1.2 g + (Mass of water / 18) × 18

Now, we need to consider the molecular weights of the substances involved in the esterification reaction:

Alcohol + Acid → Ester + Water

The molecular weight of the ester is the sum of the molecular weights of the alcohol and the acid. Let's assume the molecular weight of the ester is ME g/mol.

Using the molecular weights, we can rewrite the equation as:

(Mass of acid "B" / MB) × (Molecular weight of acid "B") = -1.2 g + (Mass of water / 18) × 18 = Mass of complex ether × (Molecular weight of ester)

Simplifying further, we have:

(Mass of acid "B" / MB) × (Molecular weight of acid "B") = -1.2 g + Mass of complex ether × (Molecular weight of ester)

Now, let's consider the molecular weights of the substances involved in the oxidation of alcohol "A":

Alcohol "A" + Oxidizing agent → Acid "B"

The molecular weight of alcohol "A" is MA g/mol, and the molecular weight of acid "B" is MB g/mol.

Using the molecular weights, we can rewrite the equation as:

(Mass of alcohol "A" / MA) × (Molecular weight of alcohol "A") = Mass of acid "B" × (Molecular weight of acid "B")

Substituting the value of the mass of acid "B" from the previous equation, we get:

(Mass of alcohol "A" / MA) × (Molecular weight of alcohol "A") = (-1.2 g + Mass of complex ether × (Molecular weight of ester)) × (Molecular weight of acid "B")

Simplifying further, we have:

(Mass of alcohol "A" / MA) × (Molecular weight of alcohol "A") = -1.2 g × (Molecular weight of acid "B") + Mass of complex ether × (Molecular weight of ester) × (Molecular weight of acid "B")

Now, we have two equations:

1. (Mass of acid "B" / MB) × (Molecular weight of acid "B") = -1.2 g + Mass of complex ether × (Molecular weight of ester) 2. (Mass of alcohol "A" / MA) × (Molecular weight of alcohol "A") = -1.2 g × (Molecular weight of acid "B") + Mass of complex ether × (Molecular weight of ester) × (Molecular weight of acid "B")

We can solve these equations simultaneously to determine the values of the mass of acid "B" and the mass of alcohol "A".

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