Question

Estimate the degree of reduction of methane, glucose and ethanol?

Answer 1

Answer:

Explanation:

First, calculate the number of electrons each atom of the molecule can donate to reach a full valence shell. For example, Hydrogen = 1, Carbon = 4, Oxygen = -2 (since it receives electrons instead of donating), Nitrogen = -3. ... For ethanol (2 Carbon atoms), Degree of reduction = 12 / 2 = 6.

Answer 2

The degree of reduction of methane, glucose, and ethanol is 8, 4, and 6 respectively.

Given:

The three compounds are methane, glucose, and ethanol.

To Find :

The degree of reduction of methane, glucose, and ethanol.

Solution:

To find the degree of reduction of methane, glucose, and ethanol we will follow. the following steps:

As we know,

The degree of reduction of a compound is calculated by taking the ratio of total valence electrons of all the atoms to the total carbon atoms.

The formula of methane, glucose, and ethanol are CH4, C6H12O6, and C2H5OH respectively.

Now,

The valence electrons of CH4 = 4 + 4×1 = 4 + 4 = 8

Now,

Degree of reduction =

$\frac{8}{1} = 8$

Now,

The valence electrons of C6H12O6 = 6 × 4 + 12×1 - 2×6 = 24 + 12 -12 = 24

So,

Degree of reduction =

$\frac{24}{6} = 4$

Now,

The valence electrons of C2H5OH = 4×2 + 5×1 -2 + 1= 8 + 5 -2 +1 = 12

So,

Degree of reduction =

$\frac{12}{2} = 6$

Henceforth, the degree of reduction of methane, glucose, and ethanol is 8, 4, and 6 respectively.

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