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Directions: This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Which of the following isomeric alcohols has the highest solubility in water? (a)n-butyl alcohol (b) sec-butyl alcohol (c) isobutyl alcohol (d) tert-butyl alcohol

Short Answer

Expert verified
a. n-butyl alcohol b. sec-butyl alcohol c. isobutyl alcohol d. tert-butyl alcohol Answer: d. tert-butyl alcohol

Step by step solution

01

Understanding the structures of the given isomeric alcohols

The first step is to understand the structures of the given alcohols: 1. n-butyl alcohol (CH3CH2CH2CH2OH) 2. sec-butyl alcohol (CH3CH2CHOHCH3) 3. isobutyl alcohol (CH3CH2CH(CH3)OH) 4. tert-butyl alcohol (CH3C(CH3)2OH) Notice that all of them have a single OH group as a hydrogen bond donor.
02

Comparing the structures of the given alcohols

Now we will compare the structures of the alcohols. The hydrogen bonding capacity of each alcohol is identical, as they all have a single OH group. However, their solubility in water will be influenced by the size and branching of their alkyl chains. Alcohols with smaller and more branched alkyl chains will generally have a higher solubility in water due to the decrease in the hydrophobic (water-repelling) interactions with their alkyl groups.
03

Determining the alcohol with the highest solubility in water

Comparing the given alcohols, we can see that tert-butyl alcohol (d) has the shortest and most branched alkyl group and is likely to form the strongest hydrogen bonds with water molecules. Therefore, it has the highest solubility in water. So the correct answer is (d) tert-butyl alcohol.

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Key Concepts

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

Isomeric Alcohols
Isomeric alcohols are molecules that share the same molecular formula but differ in the arrangement of their atoms. Let's consider the four isomeric alcohols from the exercise: n-butyl, sec-butyl, isobutyl, and tert-butyl alcohols. Each of these alcohols contains the same number of carbon, hydrogen, and oxygen atoms, resulting in the same molecular formula: C4H10O. However, the way these atoms are structured within each molecule changes their properties, including solubility.

For instance, n-butyl alcohol has a straight chain structure. Sec-butyl alcohol has a secondary branch at the second carbon, causing a slight difference in the molecule. Isobutyl alcohol has a subtle branch near the end of the chain, and tert-butyl alcohol is the most branched with three methyl groups surrounding a central carbon.

These structural differences have significant impacts on how each alcohol interacts with water, as seen in their solubility levels, which is a core focus in understanding their chemical behaviors.
Hydrogen Bonding in Alcohols
Hydrogen bonds are a type of strong dipole-dipole attraction that occurs between molecules when a hydrogen atom is bonded to a highly electronegative element such as oxygen. In alcohols, the hydroxyl (OH) group is the key to hydrogen bonding. This group allows alcohols to form hydrogen bonds with water molecules, enhancing solubility.
  • Every isomeric alcohol mentioned in the exercise has a single hydroxyl group.
  • The number of hydroxyl groups is the same, meaning the potential for hydrogen bonding is identical among the isomers.


However, while the hydroxyl group is crucial, the overall structure of the alcohols affects how these groups can interact with water. More branching in the molecule can result in better packing with water, enhancing solubility despite each having one OH group.

Thus, even though all isomeric alcohols can form hydrogen bonds, the configuration of the rest of the molecule plays a significant role in determining the effectiveness of these interactions with water.
Branching in Organic Molecules
Branching refers to the structure of carbon atoms within a molecule. When an organic compound has branching, it features carbon atoms connected in a more complex, often three-dimensional arrangement as opposed to a straight line. This has a substantial impact on solubility.

In the context of the exercise, the more highly branched an alcohol like tert-butyl, the lower its hydrophobic interactions.
  • Branching generally decreases the surface area of the non-polar part of the molecule, which reduces its hydrophobic characteristics.
  • This allows for stronger and more frequent interactions with water molecules through hydrogen bonding.


For example, tert-butyl alcohol, with its three methyl groups branching off a central carbon, is far more soluble than its counterparts due to its compact structure. This unique configuration minimizes the non-polar surface, thus decreasing the overall hydrophobic interactions and increasing its affinity for water.

Therefore, branching directly influences the interaction of organic molecules with polar solvents like water, playing a crucial part in determining the physical properties such as solubility.

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Most popular questions from this chapter

The correct nucleophilicity order among the following is (a) \(\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{CH}_{3}^{-}\) (b) \(\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}\) (c) \(\mathrm{OH}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{CH}_{3}^{-}>\mathrm{F}^{-}\) (d) \(\mathrm{CH}_{3}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{NH}_{2}^{-}\)

Which of the following statements is correct? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~S}^{-}\) is a stronger base, but weaker nucleophile than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\). (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~S}^{-}\) is a weaker base, but more nucleophilic than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~S}^{-}\) is a stronger base and more nucleophilic than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\). (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{~S}^{-}\) is a weaker base and less nucleophilic than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\).

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene

The correct statements about the following compound are $$ \mathrm{CH}_{3}(\mathrm{OH}) \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} $$ (a) The total number of stereoisomers possible is 6 . (b) The total number of diastereomers possible is \(3 .\) (c) The number of enantiomers possible for the cis isomer is \(2 .\) (d) The number of enantiomers possible for the trans isomer is 4 .

The major product formed when 2 - bromopentane is treated with potassium ethoxide in ethanol is

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