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Match list I (organic compounds oxidized by \(\left.\mathrm{HIO}_{4}\right)\) with list II (products of HIO \(_{4}\) oxidation) and select the correct answer: List I 1\. \(\mathrm{CH}_{3} \mathrm{COCHO}\) 2\. 1, 2-cyclohexanedione 3\. \(\mathrm{PhCH}(\mathrm{OH}) \mathrm{CHO}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{COCH}_{3}\) List II (i) \(\mathrm{PhCH}=\mathrm{O}+\mathrm{HCOOH}\) (ii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}+\mathrm{HOOCCH}_{3}\) (iii) \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{COOH}\) (iv) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HCOOH}\) The correct matching is: 1 2 3 4 (a) (iii) (iv) (i) (iii) (b) (iii) (ii) (i) (iv) (c) (iv) (iii) (i) (ii) (d) (iv) (iii) (ii) (i)

Short Answer

Expert verified
The correct match is (c) (iv) (iii) (i) (ii).

Step by step solution

01

Understanding the Reaction with HIO4

Periodic acid (HIO_4) oxidizes 1,2-diols and 1,2-dicarbonyl compounds to form carboxylic acids or aldehydes. We need to identify possible cleavage products from compounds in List I.
02

Analyzing Compound 1: CH3COCHO

Compound 1, CH_3COCHO, is a 1,2-dicarbonyl compound. HIO_4 will cleave it into acetic acid (CH_3COOH) and formic acid (HCOOH), corresponding to List II option (iv).
03

Analyzing Compound 2: 1,2-Cyclohexanedione

Compound 2 is a cyclic 1,2-dicarbonyl compound. HIO_4 will break the carbon-carbon bond between the ketones, forming a dicarboxylic acid: adipic acid (HOOC(CH_2)_4COOH), which matches List II option (iii).
04

Analyzing Compound 3: PhCH(OH)CHO

Compound 3 is a 1,2-diol structure. HIO_4 will cleave between the alcohol and aldehyde groups, producing benzaldehyde (PhCHO) and formic acid (HCOOH), corresponding to List II option (i).
05

Analyzing Compound 4: CH3CH2CH(OH)COCH3

Compound 4 is a 1,2-diol structure. HIO_4 will cleave to produce propionaldehyde (CH_3CH_2CHO) and acetic acid (HOOCCH_3), corresponding to List II option (ii).
06

Selecting the Correct Answer

Based on the analysis: - Compound 1 matches option (iv) - Compound 2 matches option (iii) - Compound 3 matches option (i) - Compound 4 matches option (ii) The correct answer is (c) (iv) (iii) (i) (ii).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Reactions
In organic chemistry, oxidation reactions play a crucial role in transforming organic molecules. These reactions involve the increase in oxidation state, typically through the loss of electrons. For organic compounds, oxidation often involves the addition of oxygen or the removal of hydrogen atoms.

Common oxidizing agents include Periodic Acid (\( \mathrm{HIO}_4 \)), which is particularly effective at cleaving 1,2-diols and certain dicarbonyl compounds. These reactions are essential for converting specific functional groups in organic molecules into more oxidized forms such as aldehydes, ketones, or carboxylic acids.

Understanding how different oxidizing agents interact with various organic compounds is vital for predicting reaction products in synthesis and analysis.
Periodic Acid (HIO4)
Periodic acid is a strong oxidizing agent used specifically to cleave 1,2-diols and 1,2-dicarbonyl compounds. It acts by breaking carbon-carbon bonds between adjacent hydroxyl groups or carbonyl groups.

When periodic acid reacts with 1,2-diols, it converts these groups into aldehydes or ketones, depending on the surrounding structure of the molecule. The reaction is selective for the 1,2-positions due to the unique structure of \( \mathrm{HIO}_4 \), which allows it to form a cyclic intermediate with diol groups before cleavage.

This specificity makes periodic acid an invaluable tool for oxidative cleavage in organic synthesis, allowing chemists to precisely break bonds in multistep chemical reactions.
Dicarbonyl Compounds
Dicarbonyl compounds contain two carbonyl groups (C=O) either adjacent to one another or separated by another atom. These compounds can be highly reactive, especially when influenced by strong oxidizing agents like periodic acid.

The oxidative cleavage of dicarbonyl compounds involves the breaking of carbon-carbon bonds between the carbonyl groups. This is facilitated by \( \mathrm{HIO}_4 \), which effectively "slices" the molecule at the dicarbonyl site and converts the resulting fragments into carboxylic acids or other oxidized products.

Through this process, periodic acid reveals its power in organic transformations, breaking down complex dicarbonyl structures into simpler, functional derivatives.
1,2-Diol Cleavage
1,2-Diol cleavage is a specific reaction used to split diols, compounds with two hydroxyl groups on adjacent carbons, into smaller carbonyl compounds. This is most efficiently done using periodic acid due to its selective and powerful oxidative properties.

During the cleavage, periodic acid transforms the 1,2-diols into aldehydes or ketones. The mechanism typically involves the formation of a cyclic intermediate, which helps to break the carbon-carbon bond, releasing the oxidized products.

This type of reaction is important for various organic synthesis pathways because it allows chemists to modify complex molecules precisely, enabling the construction or deconstruction of specific chemical architectures. Such techniques are fundamental in pharmaceutical chemistry and the development of novel materials.

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Most popular questions from this chapter

Phenyl magnesium bromide reacts with methanol to give (a) a mixture of anisole and \(\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}\) (b) a mixture of benzene and \(\mathrm{Mg}(\mathrm{OMe}) \mathrm{Br}\) (c) a mixture of toluene and \(\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}\) (d) a mixture of phenol and \(\mathrm{Mg}(\mathrm{Me}) \mathrm{Br}\)

Which of the following reactions, yield a product with a three membered ring? (a) \(\mathrm{CH}_{3}-\mathrm{C}(\mathrm{O})-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} \stackrel{\mathrm{KOH}, \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\) (b) \(\mathrm{PhCHO}+\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{C}(\mathrm{O})-\mathrm{OEt}\) \(\mathrm{t}-\mathrm{BuO}^{-/} \mathrm{t}-\mathrm{BuOH}\) (c) \(\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\mathrm{mCPBA}}{\longrightarrow}\) (d) 3- bromobutan \(-2\) - ol \({ }^{-\mathrm{OH} / \mathrm{H}_{2} \mathrm{O}}\)

\(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\) can be converted into \(\mathrm{RCH}_{2} \mathrm{CH}_{2}\) COOH. The correct sequence of reagent is (a) \(\mathrm{PBr}_{3}, \mathrm{KCN}, \mathrm{H}^{+}\) (b) \(\mathrm{HCN}, \mathrm{PBr}_{3}, \mathrm{H}^{+}\) (c) \(\mathrm{KCN}, \mathrm{H}^{+}\) (d) \(\mathrm{PBr}_{3}, \mathrm{KCN}, \mathrm{H}_{2}\)

Which of the following will give yellow precipitate with \(\mathrm{I}_{2} / \mathrm{NaOH} ?\) 1\. \(\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3} \quad 2 . \mathrm{CH}_{3} \mathrm{COOCOCH}_{3}\) 3\. \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (a) 1,3 (b) 3,4 (c) 1,2 (d) 1,4

The cyanohydrin of a compound \(\mathrm{X}\) on hydrolysis gives lactic acid; the \(\mathrm{X}\) is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{HCHO}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CHO}\)

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