Oxidation of one mole of glycerol with periodic acid gives (a) 1 mole each of methanol, methanal and methanoic acid. (b) 1 mole of methanal and 2 moles of methanoic acid. (c) 1 mole of methanoic acid and 2 moles of methanal. (d) three moles of methanal.

Short Answer

Expert verified
Answer: The products formed by the oxidation of glycerol with periodic acid are 1 mole of methanal and 2 moles of methanoic acid.

Step by step solution

01

Determine the structure of glycerol

Glycerol is a polyol with the molecular formula C3H8O3. Its structure consists of a three-carbon chain with each carbon atom bonded to a hydroxyl group (-OH). The IUPAC name for glycerol is 1,2,3-propanetriol.
02

Understand the oxidation reaction with periodic acid

Periodic acid (HIO4) is a strong oxidizing agent, which is capable of cleaving carbon-carbon bonds adjacent to alcohol groups in molecules like glycerol. Periodic acid essentially oxidizes the carbon atoms that are bonded to hydroxyl groups, leading to the formation of various oxidation products such as aldehydes (R-CHO), ketones (R-CO-R'), and carboxylic acids (R-COOH).
03

Find the products of the oxidation reaction

Given the structure of glycerol and the oxidation effect of periodic acid, let's analyze the possible products: 1. Oxidizing the hydroxyl group on C1 (first carbon) would lead to the formation of an aldehyde (methanal). 2. Oxidizing the hydroxyl group on C2 (middle carbon) would lead to the formation of both an aldehyde (from C1) and a carboxylic acid (from C3). 3. Oxidizing the hydroxyl group on C3 (last carbon) would lead to the formation of a carboxylic acid (methanoic acid).
04

Match the products with the given options

With the information gathered in Step 3, we can now identify the correct option from the given choices: (a) 1 mole each of methanol, methanal, and methanoic acid. This option is incorrect, as methanol isn't a product of the reaction. (b) 1 mole of methanal and 2 moles of methanoic acid. This option is correct, as the oxidation of glycerol leads to the formation of one mole of methanal and two moles of methanoic acid. (c) 1 mole of methanoic acid and 2 moles of methanal. This option is incorrect, as there is only one mole of methanal formed in the reaction. (d) three moles of methanal. This option is incorrect, as only one mole of methanal is formed in the reaction. Thus, the correct answer is option (b) 1 mole of methanal and 2 moles of methanoic acid.

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Most popular questions from this chapter

The haloalkane, 2 -iodo-2-methylpentane, can be prepared by treating an alkene (A) with hydroiodic acid. (A) can be (a) 2 -methylpent-2-ene (b) 2 -methylpent-1-ene (c) 2 -methyl-but-2-ene (d) both (a) and (b)

The correct decreasing order of reactivity of the halides for \(\mathrm{S}_{\mathrm{N}} 1\) reaction is (A) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) (B) \(\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}\) (d) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\)

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