Question

Arrange the following compounds in the increasing order of the property indicated against each. Give reasons for your answer (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}:\) solubility in water (b) Propan- 1- ol, propan -2-ol, 2- methyl-2-propanol : Reaction with Lucas reagent (c) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}:\) Boiling point

Short answer

Question: Arrange the given compounds in increasing order of solubility in water, reactivity with Lucas reagent, and boiling point. Answer: 1. Increasing order of solubility in water: C6H5OH < C6H5CH2OH < CH3CH2OH. 2. Increasing order of reactivity with Lucas reagent: Propan-1-ol < Propan-2-ol < 2-methyl-2-propanol. 3. Increasing order of boiling point: (CH3)3CCH2OH < (CH3)2CHCH2CH2OH < CH3(CH2)3CH2OH.

Step-by-step solution

Arranging compounds in increasing order of solubility in water

For compounds (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$: The solubility of these compounds in water depends on their ability to form hydrogen bonds with water molecules. The greater the ability to form hydrogen bonds, the higher the solubility in water. 1. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) - This compound has a hydroxyl group allowing it to form hydrogen bonds with water molecules. 2. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) - This compound has a hydroxyl group and a larger benzene ring, reducing the effectiveness of hydrogen bonding with water molecules. 3. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) - This compound has a hydroxyl group, but the large benzene ring in the compound significantly reduces its ability to form hydrogen bonds with water molecules. Increasing order of solubility: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} < \mathrm{C}_{6}\mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH} < \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\).

Arranging compounds in increasing order of reactivity with Lucas reagent

For compounds (b) propan-1-ol, propan-2-ol, 2-methyl-2-propanol: The reaction with Lucas reagent (a solution of \(\mathrm{ZnCl}_{2}\) in concentrated \(\mathrm{HCl}\)) involves the formation of alkyl halides, and the rate of reaction depends on the stability of the carbocation formed as an intermediate. 1. Propan-1-ol: Forms a primary carbocation, which is the least stable. 2. Propan-2-ol: Forms a secondary carbocation, which is more stable than a primary carbocation. 3. 2-methyl-2-propanol: Forms a tertiary carbocation, which is the most stable among the three compounds. Increasing order of reaction with Lucas reagent: Propan-1-ol < Propan-2-ol < 2-methyl-2-propanol.

Arranging compounds in increasing order of boiling point

For compounds (c) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\): The boiling point of these compounds depends primarily on the strength of the intermolecular forces. Weak London dispersion forces are stronger in molecules with larger surface areas, and hydrogen bonding is stronger in molecules with higher electronegativity and polarity. 1. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH}\) - This compound has a straight chain, allowing for stronger London dispersion forces, and a hydroxyl group, forming hydrogen bonds with other molecules. 2. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) - This compound has a more branched structure, reducing London dispersion forces. The hydroxyl group still allows for hydrogen bonding. 3. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) - This compound has a less branched structure than the previous, resulting in stronger London dispersion forces. The hydroxyl group allows for hydrogen bonding. Increasing order of boiling point: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}< \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Key concepts

Solubility in Water

Understanding the solubility of organic compounds in water is critical for students preparing for examinations like IIT JEE, particularly in the context of organic chemistry. Solubility is largely determined by the capability of a compound to form hydrogen bonds with water molecules, a property conferred by hydrophilic (water-attracting) functional groups in the molecule.

For instance, a hydroxyl (-OH) group is highly hydrophilic due to its ability to engage in hydrogen bonding. In the case of alcohols, which have a hydroxyl group attached to a carbon chain, the length of the carbon chain and the presence of additional functional groups can greatly influence solubility. Short-chain alcohols with fewer carbons are generally more soluble in water than their long-chain counterparts because the hydrophobic (water-repellent) effect of the carbon chain is lesser, and the influence of the hydroxyl group is more pronounced.

Moreover, the presence of other hydrophilic groups or atoms, such as amines or carboxylates, can increase solubility, while large hydrophobic structures, like benzene rings, diminish it, as observed when comparing benzene-containing compounds against simple alcohols.

Reaction with Lucas Reagent

The reaction with Lucas reagent is a classic test for differentiating alcohols based on their reactivity, which plays an important role in organic synthesis discussions for IIT JEE students. The Lucas reagent, comprised of zinc chloride in concentrated hydrochloric acid, reacts with alcohols to form alkyl chlorides. The rate of this reaction is a reflection of carbocation stability, which serves as an intermediate in the reaction process.

From a student's perspective, knowing this concept helps to predict reaction outcomes for primary (1°), secondary (2°), and tertiary (3°) alcohols. Primary alcohols, which form the least stable carbocations, react slowly with the Lucas reagent. Secondary alcohols, which generate more stable carbocations, react at a moderate rate. And tertiary alcohols, with the most stable carbocations, react quickly. Thus, one can anticipate the speed of this transformation based on the alcohol's structure and use it as a reliable method for characterization in the lab. This understanding is pivotal for predicting reactions involving alcohols and for solving IIT JEE Organic Chemistry problems efficiently.

Boiling Point of Organic Compounds

When it comes to the boiling points of organic compounds, multiple factors are at play, making it an insightful topic for students targeting IIT JEE. The boiling point is fundamentally related to the strength and nature of intermolecular forces present in a substance. With organic compounds, two primary forces are London dispersion forces and hydrogen bonding.

London dispersion forces are weak intermolecular forces that increase in strength with the size of the molecule or molecular surface area. Therefore, straight-chain compounds generally have higher boiling points compared to their branched-chain isomers due to a greater surface area for the dispersion forces to act upon. On the other hand, hydrogen bonds are strong intermolecular forces that occur when hydrogen is bound to a highly electronegative atom such as nitrogen, oxygen, or fluorine. This is why compounds with hydroxyl or amine groups have higher boiling points, as these groups lead to the formation of hydrogen bonds.

The context of boiling points in organic chemistry goes beyond just memorizing facts; it encompasses understanding the effects of molecular structure on the intermolecular forces and subsequently predicting the physical properties of these molecules, which is key in various applications, from designing distillation processes to drug formulation in pharmaceuticals.