Question

Which compound is soluble in aqueous \(\mathrm{NaOH}\) ? (A) Chlorobenzene (C) \(1,3,5\)-trichlorobenzene (B) 1,3 -dichlorobenzene (D) \(1,2,3,5\)-tetrachlorobenzene

Short answer

None of the given compounds (A, B, C, and D) are soluble in aqueous NaOH as they all lack acidic hydrogen atoms in their structures.

Step-by-step solution

Understanding the basicity of NaOH

Sodium hydroxide (NaOH) is a strong base that can dissolve acidic compounds. Therefore, we need to check which of these compounds has acidic protons that can be abstracted by the base.

Analyzing the acidity of each compound

Let's analyze the acidity of each compound: (A) Chlorobenzene is not acidic as there are no acidic hydrogens in its structure, so it will not dissolve in NaOH. (B) 1,3-dichlorobenzene also does not have any acidic hydrogens and will not dissolve in NaOH. (C) 1,3,5-trichlorobenzene has no acidic hydrogens due to its symmetrical structure and absence of acidic groups in its structure, so it will not dissolve in NaOH. (D) 1,2,3,5-tetrachlorobenzene: This compound does not contain any acidic hydrogen atoms either, and it won't dissolve in NaOH.

Conclusion

All of the compounds have no acidic hydrogen in their structure, so none of them will be soluble in aqueous NaOH.

Key concepts

Understanding Acidic Protons

In chemistry, an acidic proton is a hydrogen atom that can be released as a proton (H+) by a compound under certain conditions, often involving reaction with a base. Compounds with acidic protons typically have high polar bonds between hydrogen and a more electronegative element, making these hydrogens more prone to be abstracted by a base.

Hydrogen atoms attached to sp2 or sp hybridized carbons in benzene rings generally aren't acidic. However, when strong electron-withdrawing groups, such as nitro (-NO2) or carbonyl groups are present, they can increase the acidity of nearby hydrogens, as they draw electron density away from the hydrogen, making it more susceptible to removal. Halogens have an electronegative nature, but in the case of the halogenated benzene compounds in the exercise, they don't increase the acidity of any hydrogens to a point where the compounds would dissolve in NaOH.

The Basicity of Sodium Hydroxide (NaOH)

Sodium hydroxide, better known as lye or caustic soda, is a strong inorganic base used in various applications. When we say a base is 'strong,' we mean that it readily dissociates completely in water to give hydroxide ions (OH-). The basicity of NaOH is relevant to this exercise because only substances with acidic protons can react with the hydroxide ions to form water and the corresponding salt. This is essentially a neutralization reaction.

However, NaOH is not a universal solvent for all compounds. Its efficiency in dissolving a compound depends on the compound's ability to react with OH- ions. The halogenated benzene compounds listed in the original exercise lack significant acidity, meaning the hydroxide ions from NaOH have nothing to react with, resulting in low or no solubility in aqueous NaOH.

Acidity of Halogenated Benzene Compounds

Halogenated benzene compounds might intuitively seem acidic due to the presence of electronegative halogen atoms. However, their acidity is not pronounced enough for them to be considered acid-reactive with a strong base like NaOH.

The halogens attached to the benzene ring can, to some extent, withdraw electron density from the carbon atom to which they are bonded. But this effect isn't strong enough to significantly enhance the acidity of the hydrogen atoms on the carbon. In fact, due to the resonance stabilization of the benzene ring, any slight acidity that might be conferred by the halogens is further diminished.

Thus, despite their name, halogenated benzene compounds do not usually have sufficiently acidic protons that can react with NaOH. Therefore, none of the specific halogenated benzene compounds mentioned in the exercise (chlorobenzene, 1,3-dichlorobenzene, 1,3,5-trichlorobenzene, and 1,2,3,5-tetrachlorobenzene) will be soluble in aqueous NaOH.