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Chapter 27: Problem 3

Which of the following is the most soluble in water? (a) isobutyl alcohol (b) n-butyl alcohol (c) sec-butyl alcohol (d) tert-butyl alcohol

Short Answer

Expert verified
Tert-butyl alcohol is the most soluble in water due to its high branching.

Step by step solution

01

Understanding Solubility

Alcohol solubility in water is influenced by the ability of the hydroxyl group (OH) to form hydrogen bonds with water. However, the solubility decreases with the increase in the carbon chain length due to the increased hydrophobic character.
02

Analyzing the Alcohols

The four butyl alcohols have the same carbon chain length, but their branching affects solubility. The more branched the alcohol, the more soluble it is in water because branching reduces the hydrophobic surface area.
03

Examining Structural Branching

Isobutyl alcohol (a) and n-butyl alcohol (b) have straight or slightly straight chains, whereas sec-butyl alcohol (c) and tert-butyl alcohol (d) are more branched.
04

Identifying the Most Branched Alcohol

Among the given alcohols, tert-butyl alcohol (d) is the most branched, with three methyl groups attached to the central carbon. This branching increases its solubility in water compared to the others.
05

Finding the Solution

Since tert-butyl alcohol (d) is the most branched, it has the least hydrophobic surface area and the highest ability to form hydrogen bonds with water, making it the most soluble in water.

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Key Concepts

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

Hydrogen Bonding in Alcohols
Alcohols are organic compounds that contain an -OH (hydroxyl) group. This hydroxyl group significantly influences their properties, especially solubility. One pivotal aspect of alcohols is their ability to form hydrogen bonds. Hydrogen bonding occurs when the hydrogen from the alcohol's -OH group forms a weak bond with an electronegative atom, like the oxygen in water.
  • Hydrogen bonds are strong dipole-dipole attractions but weaker than covalent bonds.
  • This bonding is crucial for solubility as it allows alcohol molecules to mix well with water.
In the context of solubility, the more hydrogen bonds an alcohol can form with water, the more soluble it becomes. However, other factors, such as the size and shape of the alcohol molecule, also play roles.
Effect of Carbon Chain Length on Solubility
The carbon chain length of an alcohol affects its solubility in water mostly due to its influence on the molecule's hydrophobic nature. As the carbon chain length increases, the hydrophobic (water-repelling) portion of the alcohol molecule also increases.

Here's what to keep in mind:
  • Shorter carbon chains have a higher ratio of hydrophilic (water-attracting) to hydrophobic regions, leading to better solubility.
  • Long carbon chains minimize the ability to form hydrogen bonds due to increased hydrophobicity.
In the exercise's context, since all options have the same carbon chain length, solubility differences arise primarily because of the molecular structure or branching.
Branching in Organic Compounds
Branching in alcohols is a fascinating aspect that influences their physical properties, particularly solubility. A branched alcohol has a more complex structure compared to its straight-chain counterparts. The degree of branching affects how the molecule interacts with water.

Here's how branching affects solubility:
  • Increased branching leads to reduced hydrophobic surface area, making the alcohol more water-soluble.
  • Branched alcohols have a more compact structure, facilitating better hydrogen bonding with water molecules.
In the given exercise, tert-butyl alcohol is the most branched and therefore the most soluble. Its increased branched structure decreases the overall surface area exposed to water, thereby enhancing solubility.

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Most popular questions from this chapter

Which one of the following undergoes reaction with \(50 \%\) sodium hydroxide solution to give the corresponding alcohol and acid? (a) phenol (b) benzaldehyde (c) butanal (d) benzoic acid

Ethylene reacts with \(1 \%\) cold alkaline \(\mathrm{KMnO}_{4}\) to form (a) ethylene glycol (b) oxalic acid (c) HCHO (d) ethyl alcohol

An organic compound ' \(X\) ' on treatment with acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) gives a compound ' \(\mathrm{Y}^{\prime}\) which reacts with \(\mathrm{I}_{2}\) and sodium carbonate to form triodomethane. The compound ' \(\mathrm{X}^{\prime}\) is (a) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{OH}\)

The compound 'A' when treated with methyl alcohol and few drops of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gave smell of winter green. The compound 'A' is (a) oxalic acid (b) tartaric acid (c) salicylic acid (d) succinic acid

Bakelite is obtained from phenol by reacting with (a) \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{CH}_{4} \mathrm{COCH}_{3}\) (d) \(\mathrm{HCHO}\)
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