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Chapter 19: Problem 19

Which of the following pairs of compounds can be separated from each other by an extraction procedure? a. CH\(_3\)(CH\(_2\))\(_6\)COOH and CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH=CH\(_2\) b. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH=CH\(_2\) and (CH\(_3\)CH\(_2\)CH\(_2\))\(_2\)O c. CH\(_3\)(CH\(_2\))\(_6\)COOH and NaCl d. NaCl and KCI

Short Answer

Expert verified
Pairs a and c can be separated by extraction.

Step by step solution

01

Identify Functional Groups in Compounds

First, identify the functional groups present in each compound of the pairs listed. For example, CH\(_3\)(CH\(_2\))\(_6\)COOH contains a carboxylic acid group, while CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH=CH\(_2\) is an alkene.
02

Understand Extraction Principles

Extraction is based on the solubility differences in different solvents. Compounds can be separated by extraction if they dissolve differently in polar and non-polar solvents. Polar compounds generally dissolve better in polar solvents and vice versa.
03

Analyze Pair a

CH\(_3\)(CH\(_2\))\(_6\)COOH (carboxylic acid) is polar and can be ionized with a base, making it soluble in water, a polar solvent. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH=CH\(_2\) (alkene) is non-polar, insoluble in water but soluble in organic solvents. These compounds can be separated by extraction using water and an organic solvent.
04

Analyze Pair b

Both CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)CH=CH\(_2\) and (CH\(_3\)CH\(_2\)CH\(_2\))\(_2\)O are non-polar and would dissolve in non-polar organic solvents similarly. They can't be separated by extraction easily using water and organic solvent.
05

Analyze Pair c

CH\(_3\)(CH\(_2\))\(_6\)COOH can be deprotonated with a base to dissolve in water, whereas NaCl, an ionic compound, remains in the aqueous phase. These can be separated by extraction using an organic solvent and an aqueous solution containing a base.
06

Analyze Pair d

NaCl and KCl are both ionic compounds and have similar properties, making them difficult to separate by extraction using water and organic solvents.
07

Conclusion

After analyzing, pairs a and c contain compounds that can be separated by extraction. In case of pair a, using an organic solvent and water separates the polar carboxylic acid from the non-polar alkene. For pair c, an aqueous base separates the carboxylic acid in its ionic form from NaCl.

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Key Concepts

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

Functional Groups Identification
When we talk about organic compounds, functional groups are key players. They are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. Understanding functional groups is essential for predicting how a compound will react or behave in a solution.

Here's what to look for:
  • Carboxylic acids, like the one in CH extsubscript{3}(CH extsubscript{2}) extsubscript{6}COOH, have a -COOH group. This group makes the molecule acidic and gives it the ability to donate a hydrogen ion (proton).
  • Alkenes, such as CH extsubscript{3}CH extsubscript{2}CH extsubscript{2}CH extsubscript{2}CH=CH extsubscript{2}, contain a carbon-carbon double bond, which makes them less reactive to polar solvents.
  • Ethers, like (CH extsubscript{3}CH extsubscript{2}CH extsubscript{2}) extsubscript{2}O, have an -O- linkage and are relatively non-polar.
Identifying these groups helps us understand why certain compounds can be separated through methods like extraction.
Polarity and Solubility
To separate compounds, we harness their differences in polarity and solubility. Polarity refers to the distribution of electrical charge over the atoms in a molecule. This property affects how a compound interacts with solvents.

Some quick points on polarity and solubility:
  • Polar compounds have uneven charge distribution and tend to dissolve in polar solvents like water.
  • Non-polar compounds, having even charge distribution, dissolve better in organic solvents such as oils or alcohols.
Understanding these properties helps in choosing the correct solvents to extract or separate compounds. For example, since carboxylic acids are polar, they can be extracted from non-polar environments using water or by creating ionic forms with bases.
Carboxylic Acid Separation
Carboxylic acids can be uniquely separated from mixtures by exploiting their acidic nature. They can react with bases to form water-soluble salts. Here's how it works:

Steps to separate carboxylic acids:
  • First, dissolve your mixture in an organic solvent.
  • Next, add an aqueous base; it converts the carboxylic acid into its ionized form, a carboxylate salt, making it soluble in the aqueous phase.
  • Finally, separate the two layers. The aqueous layer will contain the carboxylate salt and the organic layer will contain other non-polar compounds.
By understanding this process, we can efficiently separate and isolate carboxylic acids from compounds like alkenes or other non-polar substances.
Ionic Compound Separation
Ionic compounds, like NaCl and KCl, are generally soluble in water due to their charge. Separating ionic compounds can be challenging using standard extraction techniques due to their similar solubility profiles.

Here are some insights into ionic compound separation:
  • Ionic compounds dissociate into ions in polar solvents like water, making them difficult to extract into non-polar solvents.
  • Special techniques such as fractional crystallization or using specific ion-exchange resins are employed to separate mixtures of similar ionic compounds.
  • Because of their charged nature, changing the pH or using selective precipitation can also sometimes aid in separating different ionic species.
While simple extraction isn鈥檛 typically effective for these compounds, these alternative methods provide valuable solutions for separating closely related ionic substances.

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Most popular questions from this chapter

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