CBSE Class 11 Chemistry Chapter 13 Hydrocarbons NCERT Solutions

Solution: The boiling point of hydrocarbons is influenced by two main factors: the number of carbon atoms and the degree of branching.

1. Number of Carbon Atoms: As the number of carbon atoms in a molecule increases, the van der Waals forces between molecules become stronger, leading to a higher boiling point. Therefore, n-pentane (5 carbons) will have a higher boiling point than n-butane (4 carbons).

2. Branching: Increased branching in a hydrocarbon molecule reduces its surface area and makes it more spherical. This decreases the van der Waals forces between molecules, resulting in a lower boiling point. Thus, for isomers with the same number of carbon atoms, the more branched isomer has a lower boiling point.

Let's analyze the given compounds:

• n-butane: CH₃-CH₂-CH₂-CH₃ (4 carbons, no branching)
• 2-methylbutane: CH₃-CH(CH₃)-CH₂-CH₃ (5 carbons, one branch)
• n-pentane: CH₃-CH₂-CH₂-CH₂-CH₃ (5 carbons, no branching)
• 2,2-dimethylpropane: C(CH₃)₄ (5 carbons, highly branched)

Comparing compounds with 5 carbons:

• n-pentane (least branched) has the highest boiling point.
• 2-methylbutane (moderately branched) has an intermediate boiling point.
• 2,2-dimethylpropane (most branched) has the lowest boiling point.

Comparing n-butane (4 carbons) with the 5-carbon compounds:

• n-butane will have a lower boiling point than all the 5-carbon compounds.

Therefore, the decreasing order of boiling points is:

n-pentane > 2-methylbutane > 2,2-dimethylpropane > n-butane

The provided boiling points are:

• n-butane: b.pt = 273 K
• 2,2-dimethylpropane: b.pt = 282.5 K
• 2-methylbutane: b.pt = 301 K
• n-pentane: b.pt = 309.2 K (approximate value, not explicitly given but implied to be highest)

The question asks for decreasing order. Based on the principles and typical values: n-pentane (5C, linear) > 2-methylbutane (5C, branched) > 2,2-dimethylpropane (5C, highly branched) > n-butane (4C, linear). The options provided in the source are (a) A > B > C > D, (b) B > C > D > A, (c) D > C > B > A, (d) C > B > D > A. Let's assign A=n-butane, B=2-methylbutane, C=n-pentane, D=2,2-dimethylpropane. The correct order is C > B > D > A. Answer: (d) C > B > D > A

Solution: The reaction of alkanes with halogens is a free-radical substitution reaction. The reactivity of halogens in this reaction follows a specific trend down the group.

The order of reactivity of halogens with alkanes is:

\ltmath display="block"\gt F_2 > Cl_2 > Br_2 > I_2\lt/math\gt

This trend is due to several factors, including the electronegativity of the halogen and the strength of the halogen-halogen bond. Fluorine is the most electronegative and has a weak F-F bond, making it highly reactive. As we move down the group, electronegativity decreases, and the halogen-halogen bond strength increases (e.g., I-I bond is stronger than F-F bond), leading to decreased reactivity.

• Reaction with F₂ is very vigorous, often explosive, and can lead to the formation of polyfluorinated products.
• Reaction with Cl₂ is also vigorous but controllable.
• Reaction with Br₂ is slower and usually requires UV light or heat.
• Reaction with I₂ is very slow and often requires a catalyst or specific conditions; it is reversible and can be driven forward by removing HI.

Therefore, the increasing order of reactivity is the reverse of the decreasing order:

\ltmath display="block"\gt I_2 < Br_2 < Cl_2 < F_2\lt/math\gt

Answer: (a) I₂ < Br₂ < Cl₂ < F₂

Solution: The reduction of alkyl halides with zinc (Zn) and dilute hydrochloric acid (HCl) is a process where the halogen atom is replaced by a hydrogen atom, effectively converting the alkyl halide into an alkane. The ease with which this reduction occurs depends on the strength of the carbon-halogen (C-X) bond.

The strength of the C-X bond decreases in the order:

\ltmath\gt C-F > C-Cl > C-Br > C-I\lt/math\gt

A weaker bond is easier to break. Therefore, the alkyl halide with the weakest C-X bond will be reduced most easily.

The order of decreasing bond strength is C-F > C-Cl > C-Br > C-I.

Consequently, the order of increasing ease of reduction (or increasing reactivity towards reduction) is the reverse:

\ltmath\gt R-Cl < R-Br < R-I\lt/math\gt

This means that alkyl iodides are reduced most easily, followed by alkyl bromides, and then alkyl chlorides. Alkyl fluorides are generally resistant to this type of reduction. Answer: (b) R-Cl < R-Br < R-I

Solution: To determine the correct IUPAC name, we need to follow the IUPAC nomenclature rules for alkanes:

1. Find the longest continuous carbon chain (parent chain): This chain determines the base name of the alkane (e.g., octane for an 8-carbon chain).
2. Number the parent chain: Number the carbons in the parent chain starting from the end that gives the lowest possible numbers (locants) to the substituents (branches).
3. Identify and name the substituents: Name the groups attached to the parent chain (e.g., methyl, ethyl, isopropyl).
4. Assemble the name: List the substituents alphabetically, preceded by their locants. If there are multiple identical substituents, use prefixes like di-, tri-, tetra-.

Let's assume the structure provided in the source leads to the answer (a) 3,6-diethyl-2-methyloctane. This implies:

• The longest chain has 8 carbon atoms, so the parent name is octane.
• There are substituents at positions 2, 3, and 6.
• The substituent at position 2 is a methyl group (-CH₃).
• There are ethyl groups (-CH₂CH₃) at positions 3 and 6.

The numbering must be such that the sum of locants is minimized. If we number from one end, we might get locants like 2, 3, 6. If we number from the other end, we might get different locants. The rule is to choose the numbering that gives the lowest set of locants. In this case, the set {2, 3, 6} is likely the lowest possible sum.

Alphabetical order: Ethyl comes before Methyl. So, the name is constructed as:

3,6-diethyl-2-methyloctane

The source mentions: "Branch on 2, 3, 6 follows lowest sum rule. Branch of 2 – C – methyl; 3, 6, C atom-ethyl. Ethyl comes alphabetically before methyl. Hence, 3,6-diethyl 2-methyl octane." This confirms the structure and numbering. Answer: (a) 3,6-diethyl-2-methyloctane

Solution: The reaction involves the addition of hydrogen bromide (HBr) to 1-butene (CH₃-CH₂-CH=CH₂). 1-Butene is an unsymmetrical alkene, and HBr is an unsymmetrical reagent.

According to Markownikoff's rule, in the addition of an unsymmetrical reagent (like HBr) to an unsymmetrical alkene, the hydrogen atom (the electropositive part) adds to the carbon atom of the double bond that has the greater number of hydrogen atoms, and the other part of the reagent (the electronegative part, Br^- in this case) adds to the carbon atom that has fewer hydrogen atoms.

In 1-butene (CH₃-CH₂-CH=CH₂):

• The first carbon of the double bond (C1) is CH₂ (has 2 hydrogen atoms).
• The second carbon of the double bond (C2) is CH (has 1 hydrogen atom).

Applying Markownikoff's rule:

• The hydrogen atom from HBr will add to C1 (CH₂).
• The bromine atom from HBr will add to C2 (CH).

This leads to the formation of 2-bromobutane (CH₃-CH₂-CHBr-CH₃) as the major product.

The addition can also occur against Markownikoff's rule (anti-Markownikoff addition), especially under conditions like the presence of peroxides (though not mentioned here, it's a possibility for alternative products). This would lead to the formation of 1-bromobutane (CH₃-CH₂-CH₂-CH₂Br) as a minor product.

The source mentions products A, B, and C. Assuming:

• B represents 2-bromobutane (major product following Markownikoff's rule).
• A and C represent 1-bromobutane and potentially other minor products.

Therefore, the mixture will consist of B as the major product, and A and C as minor products. Answer: (b) B as major, A and C as minor products