Generate Quiz 68EB569222BFD

The listed compounds—furan, thiophene, and pyrrole—are all benzene derivatives.

The chemical formula for benzoene is C6H6, which is a six-membered carbon ring with three alternating double bonds. Heterocyclic compounds with a five-membered ring and one or more carbon atoms swapped out for nitrogen (N), oxygen (O), and sulfur (S), respectively, are pyrrole, furan, and thiophene. Because of their similar aromatic structures and characteristics, these compounds are frequently regarded as benzene derivatives.

The reaction occurs in compliance with Markovnikov’s rule when an unsymmetrical alkene is involved. Step 1: An electrophile attacks an alkene, protonating it to create a carbocation. Step 2: The nucleophile attacks the carbocation. Step 3: Alcohol formation through deprotonation.

A procedure known as nitration can be used to create picric acid from phenols. The process of adding a nitro group (-NO2) to a molecule is known as nitration. When phenols react with a combination of sulfuric and nitric acids, the hydroxyl group (-OH) is changed to a nitro group. Picric acid, a yellow crystalline substance frequently found in dyes and explosives, is the product of this reaction.

Option A is a ketone that is reduced to a secondary alcohol by the reducing agent NaBH4. Due to the presence of the carbonyl functional group in the middle of the chain, a secondary alcohol is produced when the hydroxyl group is formed in the same location following reduction.

This choice implies that the iodoform precipitates are yellow in color. This assertion is accurate. Iodoform is a pale yellow solid that can precipitate as yellowish flakes or crystals during certain chemical reactions.

The development of yellow crystals indicates a positive iodoform test result. Iodoform (CHI₃), a pale yellow solid with a distinct medicinal smell, is what these crystals are.

This is a summary:

The iodoform test is used to find out whether an alcohol that can oxidize to a carbonyl group (CH₃CH(OH)-) or a carbonyl group (C=O) with a methyl group directly attached to it (CH₃CO-) is present.

Positive Outcome: Iodine (I₂) and a base (like NaOH) react to form iodoform (CHI₃) when they are added to a compound that contains these groups. This iodoform separates into distinct yellow crystals as it precipitates out of the solution.

The other choices are not accurate:

When carboxylic acids react with bases, a general property that is not unique to the iodoform test is the formation of carboxylate salt.

Fehling’s test, which is used to identify aldehydes, is characterized by the formation of brick-red precipitate.

Water formation is not a sign of a positive iodoform test, although it is frequently a byproduct of numerous organic reactions.

Thus, yellow crystals is the right response.

The stability of the anion that forms after the proton donation and the ease with which a proton can be removed determine an acid’s strength.

In essence, starch breaks down to produce glucose, which then goes through the glycolysis process to produce pyruvate, NADH, and ATP. Ethanal is produced when pyruvate is decarboxylated, and ethanol, an ethyl alcohol, is produced when NADH reduces ethanal. The answer is B because this is essentially a summary of the fermentation of alcohol.

The OH group of ethanol, a tiny polar molecule, forms a hydrogen bond with water.

Since hexanol is a long-chain alcohol, it is insoluble, just like the majority of higher alcohols.

Dimethyl ether and phenol are marginally soluble.

Intermolecular hydrogen bonding is made possible by 2-butanol’s larger molecular size and -OH group. Higher melting and boiling points are the result of stronger intermolecular forces caused by hydrogen bonding. 2. The melting and boiling points of butanol are -114.5°C and 99.5°C, respectively.