7.7. Exercises

7.1 Write IUPAC names of the following compounds: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) C H –O–C H 6 5 2 5 (xi) C H –O–C H (n–) (xii) 6 5 7 15

In this exercise, students are tasked with assigning IUPAC names to various chemical compounds. This requires knowledge of organic nomenclature rules, such as identifying the longest carbon chain, substituents, and functional groups present in these compounds. For graduated naming, students should also revise how to indicate position and multiplicity of substituents appropriately.

Think of naming these compounds like giving each person a unique name based on their characteristics (like the number of carbon atoms, whether they have groups like ‘-OH’ for alcohols). For example, just as some friends might have nicknames based on their hobbies or traits, compounds are named based on their structural features.

7.2 Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane -1, 3, 5-triol (iv) 2,3 – Diethylphenol (v) 1 – Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 4-Chloro-3-ethylbutan-1-ol.

This exercise challenges students to visualize and sketch the structures based on the IUPAC names provided. This reinforces their understanding of molecular structure, functional groups, and how to correlate the written name to the corresponding structural formula. Students must pay attention to the placement of substituents as indicated in the names.

Consider this like drawing a detailed map for each of your friends’ houses based on a description. Each description contains important details about where things are located (like the garage, garden, etc. — or in chemical terms, where the functional groups are placed on the carbon chain).

7.3 (i) Draw the structures of all isomeric alcohols of molecular formula C H O 5 12 and give their IUPAC names. (ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.

In this part of the exercise, students are asked to draw and name all isomers of the given molecular formula. Understanding isomers is essential in organic chemistry because different arrangements of the same atoms can lead to very different properties. Students must also identify the type of alcohol (primary, secondary, tertiary) based upon the carbon atom that the -OH group is attached to.

This is similar to how different flavors of ice cream can be made using the same base mix but varying the ingredients — like adding chocolate, strawberry, or adding pieces of cookie. Each arrangement (or combination) leads to distinct flavors, just as different isomers lead to distinct chemical properties.

7.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Here, students are required to discuss the boiling point relationship between propanol and butane based on intermolecular forces. Propanol (an alcohol) exhibits hydrogen bonding due to the presence of an -OH group, which significantly increases its boiling point compared to butane, which is a hydrocarbon and only exhibits Van der Waals forces.

Think of it like holding hands with a friend (hydrogen bonding in propanol) versus just standing next to someone (Van der Waals forces in butane). If you're holding hands, it takes more effort (energy) to separate, similar to how it takes more energy to boil propanol because of its strong intermolecular forces.

7.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

This question asks students to explore the concept of solubility and the role of functional groups. Alcohols, due to their polar -OH groups, can form hydrogen bonds with water molecules, enhancing their solubility. On the other hand, hydrocarbons are non-polar and cannot interact with water in the same manner, making them less soluble.

Imagine trying to mix oil and water — they don’t blend well due to their differing properties. In contrast, think of sugar dissolving easily in water because it can interact positively with it. This is analogous to how alcohols dissolve better in water compared to hydrocarbons.

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