Homologous Series – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Define a homologous series. Explain its essential characteristics with suitable examples.

Answer: A homologous series is a group of organic compounds with the same functional group and similar chemical properties. Members have a general formula and each successive compound differs by a –CH₂– group. For example, alkanes have the general formula $C*nH*{2n+2}$ and include methane ( $CH_4$ ), ethane ( $C_2H_6$ ), propane ( $C_3H_8$ ). They show a regular gradation in physical properties like boiling point. Their chemical reactions are similar due to the common functional group. All these features make homologous series easy to identify and classify.

Q2. What is the significance of the –CH₂– (methylene) group in a homologous series? Illustrate using any one series.

Answer: The –CH₂– group is the unit that separates each member in a homologous series. This means that each successive compound in the series contains one more –CH₂– group than the previous one. For example, in the alkane series: methane ( $CH_4$ ), ethane ( $C_2H_6$ ), propane ( $C_3H_8$ ), each compound is exactly one –CH₂– group apart. This repeated difference leads to regular increases in molecular weight and a gradual change in physical properties. However, their chemical properties remain similar.

Q3. Why do compounds belonging to the same homologous series have similar chemical properties but show variation in physical properties?

Answer: Compounds in the same homologous series have the same functional group, which decides their chemical reactivity. Therefore, their chemical reactions are similar. On the other hand, physical properties like melting point, boiling point, and solubility depend on molecular size and mass. As each new member has one more –CH₂– group, their molecular size increases, leading to gradual changes in physical properties. Thus, chemical properties are similar but physical properties vary regularly.

Q4. Explain with examples how the general formula helps in recognizing members of a homologous series.

Answer: The general formula acts as a pattern for a homologous series. For example, for alkenes, the general formula is $C*nH*{2n}$ . By substituting values for $n$ , you can generate all members:

$n = 2$ gives $C_2H_4$ (ethene)
$n = 3$ gives $C_3H_6$ (propene)

Similarly, for alcohols with the formula $C*nH*{2n+1}OH$ :

$n = 1$ gives $CH_3OH$ (methanol)
$n = 2$ gives $C_2H_5OH$ (ethanol)

By checking a compound’s formula against the general formula, you can confirm if it is a member of the series.

Q5. What is the importance of homologous series in the study of organic chemistry?

Answer: Homologous series bring order and simplicity to studying organic compounds. Since all members have similar chemical properties, learning about one member helps understand others. The systematic variation in properties helps in predicting how compounds behave physically. Naming compounds using IUPAC rules is easier within a series. They allow for easy comparison and classification of organic substances, which is crucial for study and research. Therefore, they are a foundational concept in organic chemistry.

High Complexity (Analysis & Scenario-Based)

Q6. Imagine a compound has the formula $C_4H_9OH$ . Which homologous series does it belong to? Write the names and formulas of the next two members in the series.

Answer: The compound $C*4H_9OH$ matches the general formula $C_nH*{2n+1}OH$ , which represents alcohols (with the –OH group). The name is butanol. For the next two members:

Increasing $n$ by 1 gives $n = 5$ : $C*5H*{11}OH$ , which is pentanol.
$n = 6$ : $C*6H*{13}OH$ , which is hexanol.

All three are alcohols differing by a –CH₂– unit and share similar chemical properties.

Q7. If the boiling point of propane ( $C3H_8$ ) is –42°C and butane ( $C_4H{10}$ ) is –0.5°C, explain the reason for this trend and predict the boiling point of pentane ( $C5H{12}$ ) as higher or lower than these values.

Answer: As we move along a homologous series like alkanes, each member has a larger molecular mass than the previous. This increases intermolecular forces like van der Waals forces, raising the boiling point. Since propane’s boiling point is –42°C and butane’s is –0.5°C, pentane’s boiling point will be higher than both. The trend shows that with each additional –CH₂– group, the boiling point increases.

Q8. You have identified a group of organic compounds with the formulas: $CH_3COOH$ , $C_2H_5COOH$ , $C_3H_7COOH$ . How are these compounds related? Discuss their functional group and one chemical property they will share.

Answer: These compounds ( $CH_3COOH$ , $C_2H_5COOH$ , $C_3H_7COOH$ ) belong to the carboxylic acids homologous series. Each has the –COOH functional group. They increase by one –CH₂– group per step. Due to their common functional group, all will turn blue litmus red, showing they are acidic. They can also participate in esterification reactions with alcohols.

Q9. Describe a method to experimentally demonstrate the gradation in solubility in water among the first four members of the alcohol homologous series. What observation do you expect?

Answer: Take four test tubes and add equal volumes of water. Add equal amounts of methanol, ethanol, propanol, butanol (first four alcohols) into separate tubes. Shake well and observe the solubility. You will notice methanol and ethanol dissolve completely, propanol is slightly less soluble, and butanol shows the lowest solubility. This experiment shows solubility decreases with increasing chain length in the homologous series.

Q10. Why can predicting the chemical behavior of higher members in a homologous series be much easier if you know the properties of the lower members? Support with an example.

Answer: All members of a homologous series have the same functional group, so their chemical properties are almost the same. If you know that ethanol ( $C_2H_5OH$ ) reacts with sodium to produce hydrogen, you can be sure that propanol ( $C_3H_7OH$ ) and butanol ( $C_4H_9OH$ ) will react similarly. This predictability saves time and effort in experiments and helps in identifying unknown compounds with similar formulas.