The internal oxidation of methanol yields methanoic acid, a...

Name: Problem 89, Chapter 13: Chemistry for Today: General, Organic, and Biochemistry
Uploaded: 2020-12-16T22:42:43-08:00
Duration: 11 min 36 s
Channel: Dr. Satish Ingale
Description: The internal oxidation of methanol yields methanoic acid, a highly toxic compound which can result in blindness, coma, and death. Determine the oxidation numbers on the carbon atom in the following reaction.

The internal oxidation of methanol yields methanoic acid, a highly toxic compound which can result in blindness, coma, and death. Determine the oxidation numbers on the carbon atom in the following reaction.

Key Concepts

Oxidation Numbers

Oxidation numbers are assigned values that represent the hypothetical charge an atom would have if all bonds were completely ionic. This concept is useful for understanding electron transfer in chemical reactions, balancing redox reactions, and tracking changes in oxidation state from reactants to products.

Redox Reactions

Redox reactions involve the transfer of electrons between species, leading to changes in oxidation states. In a redox process, one substance undergoes oxidation (loss of electrons) while another undergoes reduction (gain of electrons), and the relative change in oxidation numbers of elements signals these processes.

Organic Oxidation States

In organic chemistry, assigning oxidation states to carbon atoms can be complex due to covalent bonding and varying electronegativities. Determining the oxidation state in different functional groups helps in understanding reaction mechanisms, particularly in transformations like oxidation of alcohols to acids or aldehydes.

Internal Oxidation

Internal oxidation refers to a process where oxidation occurs within a single molecule, often leading to a rearrangement of electrons and conversion of functional groups. This concept is important in understanding how certain compounds, such as methanol under specific conditions, can be transformed into more oxidized, sometimes hazardous derivatives like methanoic acid.

Transcript

0:00 Hi everyone.

00:01 First of all, write the equation whatever they have given.

00:04 That is cs3 o 'h plus o2 gears co2 plus water.

00:20 So step one, assign oxidation number to each element.

00:27 And sign oxidation number to each element.

00:39 So what is that? that is cs3 -oh -plus -2 gives co2 plus water.

00:53 Now, you know, carbon oxidation state in this case, it is minus 2.

01:02 That is x plus 4 minus 2 so x plus 2 is equal 0 so x is equal minus 2 option is in the free state so it is a zero oxidation state then carbon in 02 oxidation state is plus 4 and this option is minus 2 now you know next one is you have to identify which undergo oxidation and which undergo reduction so you know carbon oxidation state changes from minus 2 to plus 4 so there is increase of plus 6 oxidation state and that's why it undergo oxidation it undergo oxidation similarly oxygen changes from that is 0 to minus 2 there is a decrease in the oxidation state and that's why this is a reduction what is this reduction.

02:05 So then this is that therefore cs3 is oxidized and o2 is reduced.

02:33 Then two separate half equation, one for oxidation, oxidation, cs3o8 gas co2.

02:53 Then another for reduction o2 gives water.

03:07 Now next step is you know balance the equation.

03:13 Now in order to balance these equations you know we will write oxidation as it is in the form of o.

03:21 Now in oxidation you know actually in both this first balance the number of first balance the number of atoms other than hydrogen and oxygen.

03:39 So you know in first equation carbon is balanced and in second oxygen is there so no need to balance.

03:46 So we will go for next that is balancing of oxygen.

03:51 Now in first equation we find there is only one option at lhs and two option at rhs and therefore you have to add one water molecule.

04:02 So one water molecule to lhs of the equation...

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