Generate Quiz 68E428BEB8E2E

A positive iodoform test is appropriately indicated by the formation of yellow crystals. The iodoform test produces iodoform (CHI3) as a yellow precipitate when a compound with a methyl ketone or secondary alcohol group reacts with iodine and a base. The presence of the functional groups that the iodoform test is looking for is confirmed by the yellow precipitate, which is frequently seen as yellow crystals.

Greek letters are used to indicate the position of the carbon atoms close to the carbonyl group after the two alkyl groups on either side of the carbonyl group have been identified. This is the standard procedure for naming ketone compounds.

Now let’s learn about the α, β, and γ carbons.

α carbon= the one attached to carbonyl group

β carbon= the one next to α carbon

γ carbon= the one next to β carbon

For the compound in question, the alkyl groups are methyl (1 carbon on the right side) and ethyl (2 carbons on the left), while the hydroxy group is joined to the α carbon. When writing alkyl groups, the alphabetical order is also adhered to. Thus, α-hydroxy-ethyl methyl ketone is the compound’s common name.

Disproportionation, also known as dismutation, is a redox reaction in chemistry where a compound in an intermediate oxidation state changes into two compounds, one of which is in a lower oxidation state and the other is in a higher one.

“Iodine tincture” is a weak iodine solution used as an antiseptic. It is made by combining potassium iodide and 2-7% elemental iodine in a solution of ethanol and water.

By using the following reaction equation, butanone is oxidized to yield two molecules of acetic acid:

A is accurate since an acetal is created when alcohol is added to aldehyde.

Benedict’s solution tests are used to identify reducing agents that convert copper citrate (blue solution) to “copper(I)oxide or cuprous oxide (red precipitate)”, so option C is accurate.

Cyanide and formaldehyde can condense to form hydroxyacetonitriles.

A nucleophile (NH3) is added to an aldehyde’s carbonyl group (CH3CHO) in the given reaction to create an imine. Nucleophilic addition is the term for this kind of reaction.

Hemiacetals are created when alcohols reversibly add to aldehydes and ketones. Another alcohol can be added to this reaction to create an acetal or ketal. Due to their presence in sugars, these functional groups are significant. Option D is therefore accurate.

Primary alcohol, which has as many carbon atoms as Grignard’s reagent plus one, is produced when Grignard’s reagent reacts with formaldehyde and is then hydrolyzed.

In the aldol reaction, an enolate ion (produced from an aldehyde or ketone) is added to another aldehyde or ketone to form a beta-hydroxy carbonyl compound. Thus, an aldehyde or ketone (functional groups A) and the resulting beta-hydroxy carbonyl compound (having an alcohol functional group – functional group B) are both involved in the aldol reaction. Option C is therefore the right choice.

The aldol condensation reaction occurs in aldehydes with alpha hydrogen atoms.

An organic molecule with two distinct oxygen atoms single-bonded to a central carbon atom is called an acetal. Acetals are geminal-diether derivatives of aldehydes or ketones that are created when two equivalents (or more) of alcohol react with them and water is removed.

Friedel Craft Acylation is the reaction that is being described. C2H5O is added to the benzene ring when C2H5OCl is present along with AlCl3. CH3 is an ortho-para directing group because of this. As a result, substances are added at ortho-para positions, or (2,4) positions. Therefore, C2H5O is added at either the ortho or para position.

A yellow iodoform precipitate is produced when methyl ketone and iodine are heated together in the presence of sodium hydroxide.

Thus, the iodoform test is used to characterize methyl ketones.

Methyl ketone is changed into the sodium salt of carboxylic acid in this reaction.

Sugars contain both aldehydes and ketones. Aldoses are monosaccharides that have an aldehyde group, and ketoses are those that have a ketone group. Aldoses are lowering sugars as a result.

The carbon in a carbonyl group, -CHO, pulls electron density and acquires a partial negative charge because it is double bonded to an electronegative oxygen. The carbon simultaneously gains a partial positive charge. Since nucleophiles donate electrons, they target and prefer the carbon center that lacks electrons. Additionally, because the double bond between carbon and oxygen can break, releasing an electron to form a bond with the nucleophile, another group, carbonyls are subject to addition reactions. Option B is correct since we can prove that carbonyls (ketone and aldehyde) experience nucleophilic addition reactions.

is accurate since symmetrical ketones always result in a mixture of two carboxylic acids and only one carbon atom next to the carbonyl group is oxidized. This is written in PTB as well.

The carbonyl carbon has two bulkier groups attached to it, which explains this. The carbonyl carbon’s reactivity is reduced by these groups.

The following is a representation of the reaction:
CH3CH2CH2COCH3 + O2 → CH3COOH + CH3CH2CH2COOH
Therefore, butanoic acid and ethanolic acid will be the end products. Therefore, ethanoic acid will be the right response.

Acetaldehyde is the byproduct of the breakdown of calcium acetate and calcium formate when they are distilled together. With the chemical formula C2H4O, acetaldehyde is an organic compound that is frequently utilized as a precursor to other chemicals. Option B is therefore accurate.

C is the right choice.

In order to prevent further oxidation to acetic acid, acetaldehyde is rapidly distilled off in the laboratory after formation. The molecule acetaldehyde is comparatively unstable and readily oxidizes to acetic acid. Acetic acid is less likely to break down and is a more stable molecule.

Acetaldehyde will oxidize to acetic acid if it is not rapidly distilled off. By employing a reducing agent, like sodium sulfite, to stop the oxidation, this can be avoided. This isn’t always feasible, though, so it’s usually simpler to just distill the acetaldehyde as soon as it forms.

Because of the stronger I-effect of the two alkyl groups and the increased Stearic hindrance provided to the attacking nucleophile, acetone is less reactive to nucleophilic addition reactions than acetaldehyde.

Paraldehyde can be created by polymerizing formaldehyde (CH2O).

Aldehyde compounds are the primary target of the silver mirror test. The alkaline solution of ammoniacal silver nitrate, known as Tollen’s reagent, is typically used to perform the aldehyde test. It involves the presence of hydroxide ions and silver ions.

Option C is accurate since acetal is a “tetrahedral structure” that is created when two alcohol molecules condense with an aldehyde.

A carbonyl group, a functional group made up of a carbon atom double-bonded to an oxygen atom (-C=O), is present in carbonyl compounds, which are organic compounds. Depending on where the carbonyl group is located within the molecule, carbonyl compounds can be either aldehydes or ketones.