CBSE Class 11 Chemistry Chapter 12: Organic Chemistry - Some Basic Principles and Techniques NCERT Solutions

To determine the correct IUPAC name, we first identify the longest continuous carbon chain, which is heptane (7 carbons). Next, we number the carbon atoms in this chain to give the substituents the lowest possible locant numbers. In this case, the ethyl group is attached to the third carbon, and two methyl groups are attached to the fourth carbon.

The substituents are an ethyl group and two methyl groups. According to IUPAC nomenclature rules, substituents are written in alphabetical order in the name. Therefore, 'ethyl' comes before 'dimethyl'. The locants assigned are 3 for the ethyl group and 4, 4 for the two methyl groups. Thus, the correct name is 3-ethyl-4, 4-dimethylheptane. The prefix 'bis' is used for identical substituents when they are already complex or have prefixes themselves, but here 'dimethyl' is sufficient.

In this molecule, there are two functional groups: a ketone (>C=O) and a carboxylic acid (—COOH). According to IUPAC rules, the carboxylic acid group has higher priority than the ketone group. Therefore, the parent chain is named as a carboxylic acid.

The longest carbon chain containing the carboxylic acid group has five carbons. The carboxylic acid group is at position 1. The ketone group is located at the fourth carbon atom of this chain. Thus, the name is derived by considering the five-carbon chain as pentanoic acid, with an 'oxo' group (ketone) at position 4. Hence, the IUPAC name is 4-oxopentanoic acid.

This is a tri-substituted benzene derivative. To determine the IUPAC name, we need to assign locants (numbers) to the substituents on the benzene ring. The rule is to assign the lowest possible set of locants. We can consider any of the substituents as position 1 and then number the ring to give the lowest numbers to the other substituents.

Let's consider the possibilities:

1. If we assign position 1 to Chlorine (Cl), then numbering clockwise gives Nitro (NO2) at position 2 and Methyl (CH3) at position 4. The locant set is 1, 2, 4.

2. If we assign position 1 to Nitro (NO2), then numbering clockwise gives Chlorine (Cl) at position 3 and Methyl (CH3) at position 6. The locant set is 1, 3, 6.

3. If we assign position 1 to Methyl (CH3), then numbering clockwise gives Nitro (NO2) at position 3 and Chlorine (Cl) at position 5. The locant set is 1, 3, 5.

Comparing the locant sets (1, 2, 4), (1, 3, 6), and (1, 3, 5), the set (1, 2, 4) is the lowest. Therefore, Chlorine is at position 1, Nitro is at position 2, and Methyl is at position 4.

When writing the name, the substituents are listed in alphabetical order: Chloro, Methyl, Nitro. Thus, the IUPAC name is 1-chloro-4-methyl-2-nitrobenzene.

The electronegativity of a carbon atom is directly related to its state of hybridization. The higher the percentage of s-character in the hybrid orbital, the greater is the electronegativity of the carbon atom.

• In option (a), $CH\_3\; -\; CH\_2\; -\; *CH\_2\; -\; CH\_3$, the asterisk marks a carbon atom with $sp^3$ hybridization. The s-character is 25%.
• In option (b), $CH\_3\; -\; *CH\; =\; CH\; -\; CH\_3$, the asterisk marks a carbon atom with $sp^2$ hybridization. The s-character is 33.3%.
• In option (c), $CH\_3\; -\; CH\_2\; -\; C\; \backslash equiv\; *CH$, the asterisk marks a carbon atom with $sp$ hybridization. The s-character is 50%.
• In option (d), $CH\_3\; -\; CH\_2\; -\; CH\; =\; *CH\_2$, the asterisk marks a carbon atom with $sp^2$ hybridization. The s-character is 33.3%.

Comparing the s-characters: $sp^3$ (25%) < $sp^2$ (33.3%) < $sp$ (50%). Therefore, the $sp$-hybridized carbon atom in option (c) is the most electronegative.

Functional group isomerism occurs when two or more compounds have the same molecular formula but different functional groups.

• Alcohols (R-OH) can show functional isomerism with ethers (R-O-R'). For example, ethanol ($C\_2H\_5OH$) and dimethyl ether ($CH\_3OCH\_3$) are functional isomers.
• Aldehydes (R-CHO) can show functional isomerism with ketones (R-CO-R'). For example, propanal ($CH\_3CH\_2CHO$) and propanone ($CH\_3COCH\_3$) are functional isomers.
• Cyanides (R-CN) can show functional isomerism with isocyanides (R-NC). For example, ethyl cyanide ($CH\_3CH\_2CN$) and ethyl isocyanide ($CH\_3CH\_2NC$) are functional isomers.
• Alkyl halides (R-X) do not have a common class of compounds with the same molecular formula but a different functional group. While positional and chain isomerism are possible within alkyl halides, functional group isomerism is not exhibited.

Therefore, alkyl halides do not show functional group isomerism.