Free radical substitution in alkanes


Free Radical Substitution in Alkanes

 1. Definition and Core Explanation

Free radical substitution is a reaction mechanism where a hydrogen atom in an alkane is replaced by a halogen atom, typically chlorine or bromine, resulting in the formation of alkyl halides. This process is initiated by the formation of highly reactive free radicals, which are species with unpaired electrons. Free radical substitution is a chain reaction, meaning it progresses through a series of steps that continually generate new reactive intermediates. This mechanism consists of three stages: initiation, propagation, and termination.

 

Initiation involves the formation of free radicals through the homolytic cleavage of a halogen molecule (e.g., \(\text{Cl}_2\)) upon exposure to ultraviolet (UV) light or heat. Propagation steps involve the newly formed radicals reacting with alkanes to generate alkyl radicals and hydrogen halide. These alkyl radicals then react with additional halogen molecules to form the alkyl halide product and another halogen radical, allowing the reaction to continue. Termination occurs when two free radicals combine, creating a stable molecule and ending the chain reaction.

 2. Fundamental Equations and Derivations

The general reaction for free radical substitution in alkanes can be represented as:

\[

\text{R-H} + \text{X}_2 \rightarrow \text{R-X} + \text{H-X}

\]

where \( R \) is the alkyl group, \( H \) is hydrogen, and \( X \) is the halogen (typically chlorine or bromine).

 

Mechanism Breakdown:

- Initiation Step:

  \[

  \text{Cl}_2 \xrightarrow{\text{UV}} 2 \cdot \text{Cl}

  \]

  The UV light splits a chlorine molecule into two chlorine radicals, each with an unpaired electron.

 

- Propagation Step 1:

  \[

  \text{CH}_4 + \cdot \text{Cl} \rightarrow \cdot \text{CH}_3 + \text{HCl}

  \]

  The chlorine radical abstracts a hydrogen atom from methane (or any other alkane), forming a methyl radical and hydrogen chloride.

 

- Propagation Step 2:

  \[

  \cdot \text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot \text{Cl}

  \]

  The methyl radical then reacts with a chlorine molecule, producing chloromethane and regenerating a chlorine radical. This allows the chain reaction to continue.

 

- Termination Steps: These steps occur when two radicals combine, effectively ending the chain reaction. For example:

  \[

  \cdot \text{CH}_3 + \cdot \text{Cl} \rightarrow \text{CH}_3\text{Cl}

  \]

  \[

  \cdot \text{CH}_3 + \cdot \text{CH}_3 \rightarrow \text{C}_2\text{H}_6

  \]

 

 3. Key Terms and Concepts

- Free Radicals: Atoms or molecules with unpaired electrons, making them highly reactive. Free radicals are denoted with a dot (\(\cdot\)) to indicate the unpaired electron.

- Homolytic Cleavage: This is the breaking of a covalent bond such that each of the bonded atoms retains one of the bonded electrons, forming two radicals.

- Chain Reaction: A reaction where intermediates generated in one step propagate further steps, leading to a self-sustaining sequence.

- Selectivity and Reactivity: Chlorination is less selective but more reactive, whereas bromination is more selective but less reactive. For example, bromination predominantly occurs at tertiary carbon positions due to its higher selectivity.

 4. Important Rules, Theorems, and Principles

- Order of Reactivity of Halogens: The reactivity of halogens in free radical substitution decreases as follows:

  \[

  \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2

  \]

  Fluorine is so reactive that it can cause explosions, while iodine generally does not participate in free radical substitution with alkanes.

 

- Stability of Free Radicals: Free radical stability follows the order: tertiary > secondary > primary > methyl. This order affects the likelihood of substitution at different carbon positions, especially during bromination, which is more selective than chlorination.

 

 5. Illustrative Diagrams and Visuals

 

Below is a diagram illustrating the free radical substitution mechanism:

 

Initiation Step:

\[

\text{Cl}_2 \xrightarrow{\text{UV}} \cdot \text{Cl} + \cdot \text{Cl}

\]

 

Propagation Step 1:

\[

\text{CH}_4 + \cdot \text{Cl} \rightarrow \cdot \text{CH}_3 + \text{HCl}

\]

 

Propagation Step 2:

\[

\cdot \text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot \text{Cl}

\]

 

Termination Step:

\[

\cdot \text{CH}_3 + \cdot \text{Cl} \rightarrow \text{CH}_3\text{Cl}

\]

[Include a detailed diagram showing each radical reaction step, including bond breakages and formations, with arrows pointing from the radicals to the products in each step.]

 

 6. Sample Problems and Step-by-Step Solutions

Example Problem 1: Predict the products of the reaction between ethane (\(\text{C}_2\text{H}_6\)) and chlorine (\(\text{Cl}_2\)) under UV light.

 

- Solution:

  - Step 1 (Initiation): \(\text{Cl}_2 \xrightarrow{\text{UV}} \cdot \text{Cl} + \cdot \text{Cl}\)

  - Step 2 (Propagation):

    - \(\text{C}_2\text{H}_6 + \cdot \text{Cl} \rightarrow \cdot \text{C}_2\text{H}_5 + \text{HCl}\)

    - \(\cdot \text{C}_2\text{H}_5 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \cdot \text{Cl}\)

  - Product: Ethyl chloride (\(\text{C}_2\text{H}_5\text{Cl}\)) and hydrogen chloride (HCl).

 

Example Problem 2: In the free radical chlorination of propane, identify the primary and secondary chlorination products and their relative amounts.

- Solution: Since secondary free radicals are more stable, secondary chlorination (at the middle carbon) will dominate, leading to 2-chloropropane as the major product, while 1-chloropropane will be the minor product.

 7. Common Tricks, Shortcuts, and Solving Techniques

- Predicting Products in Alkane Chlorination: For chlorination, all hydrogens are susceptible to substitution, but selectivity may lead to different amounts of products. Bromination, however, is highly selective and tends to occur at more stable (secondary or tertiary) carbon atoms.

- Recognizing Conditions for Free Radical Substitution: UV light or heat is a clear indicator of free radical substitution, as these conditions are necessary to initiate the chain reaction.

 8. Patterns in JEE Questions

In JEE, questions on free radical substitution frequently test understanding of the chain mechanism. Typical question types include:

- Predicting the products of halogenation of specific alkanes.

- Determining the major and minor products based on selectivity (especially in bromination).

- Explaining the initiation, propagation, and termination steps, especially how the reaction sustains itself.

 9. Tips to Avoid Common Mistakes

- Ignoring Reaction Conditions: Always check if UV light or heat is mentioned, as it is essential for initiating free radical substitution.

- Misidentifying Products: Students often confuse chlorination and bromination; remember, bromination is selective and predominantly targets tertiary carbons if available.

 10. Key Points to Remember for Revision

- Mechanism Steps: Familiarize yourself with each step (initiation, propagation, termination) and the role of each reactant.

- Halogen Reactivity and Selectivity: Chlorine is less selective, while bromine is highly selective.

- Free Radical Stability: Tertiary > Secondary > Primary. This affects product distribution in halogenation reactions.

 11. Real-World Applications and Cross-Chapter Links

- Industrial Uses: Chlorinated hydrocarbons are used in solvents, refrigerants, and pesticides (e.g., chloroform, CFCs).

- Cross-Concept Links: Free radical mechanisms are also relevant in polymerization processes and combustion reactions, connecting this concept to other topics in organic and environmental chemistry.

Questions

Q 1. In the free radical chlorination of methane, the chain initiating step involves the formation of;

(a) chlorine free radical;

(b) hydrogen chloride;

(c) methyl radical;

(d) chloromethyl radical.;

Monochlorination of alkanes


Monochlorination of Alkanes

 1. Definition and Core Explanation

Monochlorination is a specific type of halogenation reaction in which only one hydrogen atom in an alkane is replaced by a chlorine atom. This process typically occurs via a free radical mechanism and requires exposure to ultraviolet (UV) light or heat to proceed. Monochlorination is distinct from general chlorination in that it aims to substitute only one hydrogen atom, producing a single chlorinated product. For example, in the monochlorination of methane (\(\text{CH}_4\)), the reaction yields chloromethane (\(\text{CH}_3\text{Cl}\)) as the primary product.

 

The reaction mechanism is a chain reaction involving initiation, propagation, and termination steps, similar to other free radical substitution processes. The selective nature of monochlorination often depends on the structure of the alkane, as secondary and tertiary hydrogens are more reactive than primary hydrogens due to the stability of the resulting free radicals.

 

 2. Fundamental Equations and Derivations

The general equation for the monochlorination of an alkane can be represented as:

\[

\text{R-H} + \text{Cl}_2 \xrightarrow{\text{UV}} \text{R-Cl} + \text{HCl}

\]

where \( \text{R-H} \) is the alkane, \( \text{Cl}_2 \) is chlorine, \( \text{R-Cl} \) is the chlorinated product, and \( \text{HCl} \) is hydrogen chloride.

 

Mechanism of Monochlorination:

- Initiation Step: Chlorine molecules split into chlorine radicals under UV light:

  \[

  \text{Cl}_2 \xrightarrow{\text{UV}} 2 \cdot \text{Cl}

  \]

- Propagation Step 1: A chlorine radical abstracts a hydrogen atom from the alkane, forming an alkyl radical and hydrogen chloride:

  \[

  \text{R-H} + \cdot \text{Cl} \rightarrow \cdot \text{R} + \text{HCl}

  \]

- Propagation Step 2: The alkyl radical then reacts with another chlorine molecule, forming the chlorinated alkane (e.g., \(\text{R-Cl}\)) and another chlorine radical, which sustains the chain reaction:

  \[

  \cdot \text{R} + \text{Cl}_2 \rightarrow \text{R-Cl} + \cdot \text{Cl}

  \]

- Termination Steps: Free radicals combine to end the chain reaction. For example:

  \[

  \cdot \text{Cl} + \cdot \text{Cl} \rightarrow \text{Cl}_2

  \]

 

 3. Key Terms and Concepts

- Monochlorination: A halogenation process where only one hydrogen atom is replaced by a chlorine atom in an alkane.

- Selectivity in Chlorination: Chlorine is less selective than bromine and can chlorinate both primary and secondary carbons. However, secondary and tertiary carbons are more reactive due to the greater stability of the resulting radicals.

- Free Radical Mechanism: A chain reaction involving free radicals as reactive intermediates that perpetuate the reaction until termination.

 

 4. Important Rules, Theorems, and Principles

- Stability of Free Radicals: The order of stability for free radicals is tertiary > secondary > primary > methyl. This influences the likelihood of chlorination at various positions within the alkane structure.

- Selective Chlorination: In large alkanes, monochlorination tends to occur more readily at secondary or tertiary carbon atoms, as these positions yield more stable radicals. However, with smaller alkanes like methane, only one chlorination product is possible.

 

 5. Illustrative Diagrams and Visuals

 

Below is a step-by-step diagram illustrating the monochlorination of methane, showing each phase of the mechanism:

 

1. Initiation Step: \(\text{Cl}_2\) splits into two chlorine radicals under UV light.

   \[

   \text{Cl}_2 \xrightarrow{\text{UV}} \cdot \text{Cl} + \cdot \text{Cl}

   \]

 

2. Propagation Step 1: A chlorine radical abstracts a hydrogen from methane.

   \[

   \text{CH}_4 + \cdot \text{Cl} \rightarrow \cdot \text{CH}_3 + \text{HCl}

   \]

 

3. Propagation Step 2: The methyl radical (\(\cdot \text{CH}_3\)) reacts with another chlorine molecule to form chloromethane (\(\text{CH}_3\text{Cl}\)) and regenerates a chlorine radical.

   \[

   \cdot \text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot \text{Cl}

   \]

 

4. Termination Step: Free radicals combine to terminate the reaction.

   \[

   \cdot \text{Cl} + \cdot \text{Cl} \rightarrow \text{Cl}_2

   \]

 

[Diagrams should illustrate each of these steps clearly, showing radical movement and bond changes.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the major product when propane undergoes monochlorination in the presence of UV light.

 

- Solution:

  - Step 1 (Initiation): Chlorine molecules break into two chlorine radicals.

  - Step 2 (Propagation): The chlorine radicals react with propane (\(\text{C}_3\text{H}_8\)) at either a primary or secondary hydrogen.

  - Major Product: Since secondary radicals are more stable, the major product will be 2-chloropropane, though 1-chloropropane will also form in smaller quantities.

 

Example Problem 2: How would you carry out monochlorination of butane to get a single chlorinated product?

 

- Solution: The monochlorination of butane will yield a mixture of 1-chlorobutane and 2-chlorobutane. However, due to greater stability, 2-chlorobutane is the more abundant product in the reaction mixture.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Using Free Radical Stability to Predict Products: Secondary and tertiary carbons are more likely to be chlorinated due to the stability of the resulting radicals.

- Quick Identification of Monochlorination Products: If the alkane has symmetrical structures, monochlorination often produces a single product. For example, monochlorination of methane yields chloromethane, while ethane produces only one product, chloroethane.

 

 8. Patterns in JEE Questions

JEE questions on monochlorination often involve:

- Predicting the major product based on the structure of the alkane and radical stability.

- Identifying all possible monochlorinated products for a given alkane.

- Understanding the role of UV light and the free radical chain mechanism in halogenation.

 

 9. Tips to Avoid Common Mistakes

- Avoid Over-Chlorination: Monochlorination requires precise conditions to avoid further substitution. Ensure that only one hydrogen is replaced by chlorine.

- Confusion with Bromination: Chlorination is less selective than bromination, so it can produce multiple products. Remember that bromination selectively chlorinates more stable radicals, while chlorination can target multiple types of hydrogen.

 

---

 

 10. Key Points to Remember for Revision

- Mechanism Steps: Know each stage (initiation, propagation, and termination) and how they contribute to the chain reaction.

- UV Light Requirement: Remember that UV light is essential for initiating chlorination reactions.

- Radical Stability: Secondary and tertiary radicals are more stable and, thus, more likely to form in chlorination reactions.

 

 11. Real-World Applications and Cross-Chapter Links

- Industrial Uses of Chlorinated Alkanes: Chlorinated alkanes are widely used as solvents, refrigerants, and intermediates in chemical synthesis. For instance, chloromethane is a precursor in the production of silicone polymers.

- Cross-Concept Links: Monochlorination ties into broader organic chemistry concepts, such as reactivity in radical halogenation, effects of different substituents on stability, and the behavior of free radicals, which are foundational for understanding complex organic reactions.

Questions

Q 1. Which one of the following gives only one monochloro derivative?;

(a) n-hexane;

(b) 2-methylpentane;

(c) 2,3-dimethylpentane;

(d) neo-pentane;

Photochemical halogenation - example of free radical substitution


Photochemical Halogenation - Example of Free Radical Substitution

 1. Definition and Core Explanation

Photochemical halogenation is a reaction in which a halogen atom, typically chlorine or bromine, is introduced into an organic molecule under the influence of light, usually ultraviolet (UV) radiation. This reaction follows a free radical substitution mechanism and is a classic example of free radical halogenation. Photochemical halogenation specifically refers to halogenation initiated by light, where photons provide the energy needed to break the halogen-halogen bond (e.g., in \(\text{Cl}_2\) or \(\text{Br}_2\)), generating reactive free radicals.

 

In organic chemistry, photochemical halogenation is widely used to introduce halogen atoms into hydrocarbons, particularly alkanes. This process is essential for the synthesis of alkyl halides, which are useful intermediates in various organic reactions. The mechanism of photochemical halogenation involves three key steps: initiation, propagation, and termination.

 

2. Fundamental Equations and Derivations

The general form of a photochemical halogenation reaction is:

\[

\text{R-H} + \text{X}_2 \xrightarrow{\text{UV}} \text{R-X} + \text{H-X}

\]

where \( R \) is the alkyl group, \( H \) is hydrogen, and \( X \) is the halogen (usually chlorine or bromine).

 

Mechanism of Photochemical Halogenation:

- Initiation: UV light provides energy to split a halogen molecule into two halogen radicals.

  \[

  \text{Cl}_2 \xrightarrow{\text{UV}} 2 \cdot \text{Cl}

  \]

  This homolytic cleavage produces two highly reactive chlorine radicals, each with an unpaired electron.

 

- Propagation Step 1: A chlorine radical abstracts a hydrogen atom from the alkane (e.g., methane), producing a methyl radical and hydrogen chloride.

  \[

  \text{CH}_4 + \cdot \text{Cl} \rightarrow \cdot \text{CH}_3 + \text{HCl}

  \]

 

- Propagation Step 2: The methyl radical reacts with another chlorine molecule, forming chloromethane and another chlorine radical, which can initiate a new reaction cycle.

  \[

  \cdot \text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot \text{Cl}

  \]

 

- Termination: Free radicals combine, forming stable molecules and halting the chain reaction. For example:

  \[

  \cdot \text{CH}_3 + \cdot \text{Cl} \rightarrow \text{CH}_3\text{Cl}

  \]

  \[

  \cdot \text{CH}_3 + \cdot \text{CH}_3 \rightarrow \text{C}_2\text{H}_6

  \]

 

 3. Key Terms and Concepts

- Photochemical Reaction: A reaction driven by light energy, often requiring specific wavelengths, such as UV light, to initiate the reaction.

- Free Radical Substitution: A reaction mechanism involving free radicals, where one atom (e.g., hydrogen) is replaced by another (e.g., chlorine) in the presence of a halogen and UV light.

- Chain Reaction: A reaction sequence where reactive intermediates (free radicals) regenerate, sustaining the reaction through multiple cycles until termination occurs.

 

 4. Important Rules, Theorems, and Principles

- Reactivity Order of Halogens in Photochemical Halogenation: The reactivity order of halogens is:

  \[

  \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2

  \]

  Fluorine is too reactive and often results in uncontrolled reactions, while iodine is generally too unreactive. Therefore, chlorine and bromine are typically used in photochemical halogenation.

 

- Bond Dissociation Energy and Light Requirement: The energy required to break the \(\text{Cl}_2\) or \(\text{Br}_2\) bond determines the wavelength of light needed to initiate the reaction. For chlorine, UV light provides sufficient energy to dissociate the bond, making it suitable for photochemical halogenation.

 

 5. Illustrative Diagrams and Visuals

 

Below is a step-by-step diagram illustrating photochemical halogenation using methane as an example:

 

1. Initiation Step: \(\text{Cl}_2\) absorbs UV light and dissociates into two chlorine radicals.

   \[

   \text{Cl}_2 \xrightarrow{\text{UV}} \cdot \text{Cl} + \cdot \text{Cl}

   \]

 

2. Propagation Step 1: A chlorine radical abstracts a hydrogen from methane.

   \[

   \text{CH}_4 + \cdot \text{Cl} \rightarrow \cdot \text{CH}_3 + \text{HCl}

   \]

 

3. Propagation Step 2: The methyl radical reacts with a chlorine molecule to form chloromethane and regenerate a chlorine radical.

   \[

   \cdot \text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \cdot \text{Cl}

   \]

 

4. Termination Step: Free radicals combine to end the reaction cycle.

   \[

   \cdot \text{CH}_3 + \cdot \text{CH}_3 \rightarrow \text{C}_2\text{H}_6

   \]

 

[Include a series of detailed reaction mechanism diagrams here, illustrating radical formation, reaction progression, and termination.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the products of photochemical halogenation of ethane (\(\text{C}_2\text{H}_6\)) with chlorine under UV light.

 

- Solution:

  - Step 1 (Initiation): \(\text{Cl}_2 \xrightarrow{\text{UV}} 2 \cdot \text{Cl}\)

  - Step 2 (Propagation):

    - \(\text{C}_2\text{H}_6 + \cdot \text{Cl} \rightarrow \cdot \text{C}_2\text{H}_5 + \text{HCl}\)

    - \(\cdot \text{C}_2\text{H}_5 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \cdot \text{Cl}\)

  - Product: Chloroethane (\(\text{C}_2\text{H}_5\text{Cl}\)) and hydrogen chloride (HCl).

 

Example Problem 2: How would you achieve selective monochlorination of propane?

 

- Solution: In propane, chlorine radicals can abstract hydrogen from either a primary or secondary carbon. However, the secondary carbon will yield a more stable radical, making 2-chloropropane the major product, with 1-chloropropane as a minor product.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Predicting Reaction Outcomes Based on Radical Stability: Secondary and tertiary radicals are more stable and therefore more likely to form during photochemical halogenation.

- Distinguishing Between Monohalogenation and Polyhalogenation: Ensure that only one halogen atom is added per alkane molecule (monohalogenation) if that is the goal, by controlling reaction conditions and halogen concentration.

 

 8. Patterns in JEE Questions

JEE Advanced often includes questions that test:

- The identification of products formed in photochemical halogenation.

- Comparison of product distributions based on the stability of radicals formed during chlorination or bromination.

- Conditions required for the initiation of free radical halogenation.

 

 9. Tips to Avoid Common Mistakes

- Avoiding Over-Halogenation: If the goal is monochlorination or monobromination, be cautious about reaction conditions to prevent further substitution.

- Remembering the Role of UV Light: UV light is essential for initiating photochemical halogenation. If the light is absent, the reaction does not proceed.

 

 10. Key Points to Remember for Revision

- Mechanism Steps: Familiarize yourself with initiation, propagation, and termination, as well as their specific roles in sustaining or ending the reaction.

- Halogen Reactivity and Selectivity: Chlorine and bromine are preferred for photochemical halogenation. Fluorine is too reactive, and iodine is not reactive enough.

- Radical Stability: The stability of the radicals dictates the primary products formed in halogenation reactions.

 

 11. Real-World Applications and Cross-Chapter Links

- Applications in Organic Synthesis: Photochemical halogenation is a key step in the production of chlorinated and brominated hydrocarbons, used as solvents, intermediates, and pesticides.

- Cross-Concept Connections: The principles of photochemical halogenation tie into broader topics in organic chemistry, such as radical mechanisms in polymerization and the role of light in organic reactions.

Questions

Q 1. Photochemical halogenation of alkane is an example of;

(a) electrophilic substitution;

(b) electrophilic addition;

(c) nucleophilic substitution;

(d) free radical substitution;

Free radical bromination of alkanes


Free Radical Bromination of Alkanes

1. Definition and Core Explanation

Free radical bromination is a specific type of halogenation reaction where bromine atoms are introduced into an alkane through a free radical mechanism. Bromination, like chlorination, occurs through a chain reaction involving three main steps: initiation, propagation, and termination. However, bromine is more selective than chlorine, predominantly reacting at the position that forms the most stable radical. This selectivity makes bromination particularly useful in organic synthesis, as it often allows for more predictable and controlled product formation compared to chlorination.

 

Bromine's lower reactivity (compared to chlorine) results in a preference for tertiary carbon atoms, where the resulting free radical is most stable. For example, in the bromination of propane, bromine will predominantly attach to the secondary carbon rather than the primary carbon, forming 2-bromopropane as the major product. This selectivity arises because bromine radicals are less reactive, allowing for greater discrimination among potential reaction sites.

 

 2. Fundamental Equations and Derivations

The general reaction for free radical bromination can be represented as:

\[

\text{R-H} + \text{Br}_2 \xrightarrow{\text{UV}} \text{R-Br} + \text{HBr}

\]

where \( \text{R-H} \) is the alkane, \( \text{Br}_2 \) is bromine, \( \text{R-Br} \) is the brominated product, and \( \text{HBr} \) is hydrogen bromide.

 

Mechanism of Free Radical Bromination:

- Initiation: UV light or heat provides energy to break the bromine molecule into two bromine radicals.

  \[

  \text{Br}_2 \xrightarrow{\text{UV}} 2 \cdot \text{Br}

  \]

- Propagation Step 1: A bromine radical abstracts a hydrogen atom from the alkane, forming an alkyl radical and hydrogen bromide.

  \[

  \text{R-H} + \cdot \text{Br} \rightarrow \cdot \text{R} + \text{HBr}

  \]

- Propagation Step 2: The alkyl radical reacts with another bromine molecule, forming the brominated alkane and regenerating a bromine radical.

  \[

  \cdot \text{R} + \text{Br}_2 \rightarrow \text{R-Br} + \cdot \text{Br}

  \]

- Termination: Two free radicals combine to form a stable molecule, ending the chain reaction. For example:

  \[

  \cdot \text{Br} + \cdot \text{Br} \rightarrow \text{Br}_2

  \]

 

 3. Key Terms and Concepts

- Selective Halogenation: Bromine selectively halogenates tertiary carbons over secondary or primary carbons due to the stability of the resulting radicals.

- Free Radical Mechanism: A chain reaction involving reactive intermediates (free radicals) that propagate the reaction until termination.

- Radical Stability: The stability order for free radicals is tertiary > secondary > primary, which dictates the preferred site of bromination.

 

 4. Important Rules, Theorems, and Principles

- Reactivity and Selectivity of Bromine: Bromine is less reactive than chlorine but more selective, primarily reacting at the most stable carbon position. The selectivity factor in bromination allows for higher yields of specific products, especially when targeting tertiary carbons.

- Bond Dissociation Energy and Selectivity: Bromine radicals require less energy than chlorine radicals, resulting in higher selectivity. This property is particularly useful in complex organic molecules where controlled halogenation is necessary.

 

 5. Illustrative Diagrams and Visuals

 

Below is a step-by-step diagram of free radical bromination using propane as an example:

 

1. Initiation Step: \(\text{Br}_2\) absorbs UV light and dissociates into two bromine radicals.

   \[

   \text{Br}_2 \xrightarrow{\text{UV}} \cdot \text{Br} + \cdot \text{Br}

   \]

 

2. Propagation Step 1: A bromine radical abstracts a hydrogen atom from propane.

   \[

   \text{CH}_3\text{CH}_2\text{CH}_3 + \cdot \text{Br} \rightarrow \cdot \text{CH}_3\text{CHCH}_3 + \text{HBr}

   \]

 

3. Propagation Step 2: The secondary radical formed reacts with another bromine molecule, producing 2-bromopropane and another bromine radical.

   \[

   \cdot \text{CH}_3\text{CHCH}_3 + \text{Br}_2 \rightarrow \text{CH}_3\text{CHBrCH}_3 + \cdot \text{Br}

   \]

 

4. Termination Step: Free radicals combine, ending the chain reaction.

   \[

   \cdot \text{Br} + \cdot \text{Br} \rightarrow \text{Br}_2

   \]

 

[Include detailed diagrams for each step, showing radical formation, propagation, and termination, with bonds highlighted to indicate which are broken and formed.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the major product of free radical bromination of butane under UV light.

 

- Solution:

  - Step 1 (Initiation): Bromine molecules break into two bromine radicals.

  - Step 2 (Propagation):

    - \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + \cdot \text{Br} \rightarrow \text{CH}_3\text{CH}_2\cdot \text{CHCH}_3 + \text{HBr}\)

    - \(\text{CH}_3\text{CH}_2\cdot \text{CHCH}_3 + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CHBrCH}_3 + \cdot \text{Br}\)

  - Major Product: The major product will be 2-bromobutane, as bromine preferentially attaches to the secondary carbon, where the free radical is more stable.

 

Example Problem 2: Determine the product distribution when 2-methylpropane is subjected to free radical bromination.

 

- Solution: In 2-methylpropane, bromination will predominantly occur at the tertiary carbon due to the stability of the resulting tertiary radical. The primary product will be tert-butyl bromide (\(\text{(CH}_3\text{)}_3\text{CBr}\))

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Selectivity Based on Radical Stability: Since bromination is highly selective, always consider the most stable radical position for product prediction.

- Avoiding Multiple Products in Selective Bromination: If targeting specific sites in complex alkanes, bromination can be a more efficient choice over chlorination due to its selectivity.

 

 8. Patterns in JEE Questions

In JEE Advanced, questions on free radical bromination may include:

- Predicting the major products of bromination based on radical stability.

- Determining the product distribution in mixed alkanes with primary, secondary, and tertiary hydrogens.

- Understanding the influence of UV light in initiating free radical bromination.

 

 9. Tips to Avoid Common Mistakes

- Confusing Chlorination and Bromination Products: Remember that bromination is more selective than chlorination, which affects the product distribution.

- Incorrect Site Selection: Always choose the position that forms the most stable radical (usually tertiary) as the site for bromine attachment.

 10. Key Points to Remember for Revision

- Mechanism Steps: Be familiar with initiation, propagation, and termination in bromination, as each step plays a role in the reaction's progress and selectivity.

- Radical Stability and Selectivity: Tertiary > Secondary > Primary; bromine prefers more stable radicals, unlike chlorine, which is less selective.

- Role of UV Light: UV light is essential for initiating bromine radicals; without it, bromination does not proceed.

 

 11. Real-World Applications and Cross-Chapter Links

- Use in Organic Synthesis: Brominated compounds serve as intermediates in various organic syntheses, particularly in the pharmaceutical and agrochemical industries.

- Cross-Concept Connections: The concept of free radical stability is relevant across organic reactions, including polymerization, combustion, and other radical-driven processes.

Questions

Q 1. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly;

(a) 1-bromo-3-methylbutane;

(b) 2-bromo-3-methylbutane;

(c) 2-bromo-2-methylbutane;

(d) 1-bromo-2-methylbutane;

Complete combustion of hydrocarbons


Complete Combustion of Hydrocarbons

1. Definition and Core Explanation

Complete combustion of hydrocarbons refers to the reaction of a hydrocarbon compound with excess oxygen, resulting in the formation of carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) as the primary products. During this reaction, all the carbon and hydrogen atoms in the hydrocarbon are fully oxidized. Complete combustion is an exothermic process, meaning it releases a significant amount of energy, often in the form of heat and light, making it a common reaction in energy production, such as in fuel combustion for engines, power plants, and heating.

 

The general formula for the complete combustion of a hydrocarbon (\(\text{C}_x\text{H}_y\)) is:

\[

\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}

\]

This equation implies that complete combustion requires a sufficient supply of oxygen to ensure that all carbon atoms in the hydrocarbon convert to carbon dioxide, rather than forming partial oxidation products like carbon monoxide (\(\text{CO}\)) or elemental carbon (soot), which are common in incomplete combustion.

 

 2. Fundamental Equations and Derivations

The complete combustion reaction for various hydrocarbons can be represented as follows:

 

- For methane (\(\text{CH}_4\)):

  \[

  \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}

  \]

 

- For ethane (\(\text{C}_2\text{H}_6\)):

  \[

  \text{C}_2\text{H}_6 + 3.5\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}

  \]

 

- For propane (\(\text{C}_3\text{H}_8\)):

  \[

  \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}

  \]

 

Calculating the Amount of Oxygen Required:

For a hydrocarbon with the formula \(\text{C}_x\text{H}_y\), the oxygen requirement for complete combustion can be determined by balancing the combustion reaction:

\[

\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}

\]

3. Key Terms and Concepts

- Complete Combustion: A reaction where a hydrocarbon reacts with excess oxygen, producing carbon dioxide and water without any partial oxidation products.

- Stoichiometry of Combustion: The molar ratios of reactants and products in the combustion reaction. Accurate stoichiometry ensures complete combustion and maximizes energy output.

- Exothermic Reaction: A reaction that releases energy in the form of heat. Complete combustion is highly exothermic, which is why hydrocarbons are widely used as fuels.

 

 4. Important Rules, Theorems, and Principles

- Law of Conservation of Mass: In any combustion reaction, the total mass of reactants equals the total mass of products, ensuring that all atoms are accounted for in the balanced equation.

- Energy Release in Combustion: The enthalpy change (\(\Delta H\)) for complete combustion is negative, indicating that energy is released. This energy can be calculated using bond enthalpies of the reactants and products.

- Oxygen Supply Requirement: Complete combustion requires a sufficient amount of oxygen. If oxygen is limited, incomplete combustion occurs, producing carbon monoxide and soot, which are less efficient and more harmful byproducts.

 

 5. Illustrative Diagrams and Visuals

 

Below is a representation of the complete combustion reaction for methane, showing the reactants, products, and energy release.

 

1. Methane Molecule: A tetrahedral structure with one carbon atom bonded to four hydrogen atoms.

   - Reactants: \(\text{CH}_4\) and \(\text{O}_2\)

   - Products: \(\text{CO}_2\) (a linear molecule) and \(\text{H}_2\text{O}\)

 

2. Combustion Reaction Diagram: Illustrates the breaking of \(\text{C-H}\) bonds in methane and the formation of \(\text{C=O}\) bonds in carbon dioxide and \(\text{O-H}\) bonds in water, highlighting the energy release.

 

[Include detailed diagrams of the methane molecule, oxygen molecules, and the final products with bond changes and energy release marked.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Calculate the amount of oxygen required for the complete combustion of 10 grams of methane (\(\text{CH}_4\)).

 

- Solution:

  - Molar mass of \(\text{CH}_4 = 16 \, \text{g/mol}\).

  - Moles of \(\text{CH}_4\) = \(\frac{10 \, \text{g}}{16 \, \text{g/mol}} = 0.625 \, \text{mol}\).

  - From the balanced equation, 1 mole of \(\text{CH}_4\) requires 2 moles of \(\text{O}_2\).

  - Oxygen required = \(0.625 \times 2 = 1.25 \, \text{mol}\) of \(\text{O}_2\).

  - Molar mass of \(\text{O}_2 = 32 \, \text{g/mol}\).

  - Mass of \(\text{O}_2\) required = \(1.25 \times 32 = 40 \, \text{g}\).

 

Example Problem 2: How much \(\text{CO}_2\) is produced when 5 moles of propane (\(\text{C}_3\text{H}_8\)) undergo complete combustion?

 

- Solution:

  - From the balanced equation, 1 mole of \(\text{C}_3\text{H}_8\) produces 3 moles of \(\text{CO}_2\).

  - Therefore, 5 moles of \(\text{C}_3\text{H}_8\) will produce \(5 \times 3 = 15\) moles of \(\text{CO}_2\).

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Quick Molar Ratio Calculation: Use the balanced equation to determine the ratio of oxygen to hydrocarbon. For example, the complete combustion of \(\text{C}_n\text{H}_{2n+2}\) requires \(n + \frac{(2n+2)}{4}\) moles of oxygen.

- Estimating Energy Output: Calculate the enthalpy change of combustion using bond enthalpies for each type of hydrocarbon to approximate the energy released.

 

 8. Patterns in JEE Questions

Questions in JEE on complete combustion of hydrocarbons often include:

- Calculating the amount of oxygen needed for combustion.

- Determining the energy released based on bond enthalpies.

- Analyzing the products formed and their stoichiometric ratios.

 

 9. Tips to Avoid Common Mistakes

- Oxygen Requirements: Ensure the calculation accounts for complete combustion by using the correct oxygen-to-hydrocarbon ratio.

- Confusing Complete and Incomplete Combustion: Incomplete combustion produces carbon monoxide and soot, while complete combustion produces only \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

 

 10. Key Points to Remember for Revision

- Products of Complete Combustion: Always \(\text{CO}_2\) and \(\text{H}_2\text{O}\) with sufficient oxygen.

- Energy Release: Complete combustion is exothermic and releases significant energy.

- Balanced Equation for Combustion: Essential for calculating reactant and product amounts.

 

 11. Real-World Applications and Cross-Chapter Links

- Energy Production: Complete combustion of hydrocarbons is a fundamental process in engines, power plants, and household heating.

- Environmental Impact: Complete combustion emits carbon dioxide, a greenhouse gas. In contrast, incomplete combustion produces pollutants like carbon monoxide, impacting air quality.

- Cross-Concept Connections: This concept ties into thermodynamics (enthalpy of combustion), stoichiometry, and environmental science.

Questions

Q 1. Complete combustion of \(\mathrm{CH}_{4}\) gives :;

(a) \(\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\);

(b) \(\mathrm{CO}_{2}+\mathrm{H}_{2}\);

(c) \(\mathrm{COCl}_{2}\);

(d) \(\mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\);

IUPAC nomenclature of hydrocarbons


IUPAC Nomenclature of Hydrocarbons

 1. Definition and Core Explanation

IUPAC (International Union of Pure and Applied Chemistry) nomenclature is a standardized system for naming chemical compounds. In the case of hydrocarbons, IUPAC names help to systematically identify alkanes, alkenes, and alkynes based on their structure, carbon chain length, type of bonds (single, double, or triple), and functional groups. The IUPAC system enables chemists to communicate unambiguously about the specific structure and type of hydrocarbons.

 

Hydrocarbons are divided into three main classes:

- Alkanes: Saturated hydrocarbons with only single bonds.

- Alkenes: Unsaturated hydrocarbons with at least one double bond.

- Alkynes: Unsaturated hydrocarbons with at least one triple bond.

 

The naming of hydrocarbons follows specific rules that account for the longest carbon chain, position of double or triple bonds, substituents, and stereochemistry (if applicable).

 

 2. Fundamental Rules and Steps for Naming Hydrocarbons

The process of assigning an IUPAC name to a hydrocarbon follows these fundamental rules:

 

1. Identify the Longest Continuous Carbon Chain: This chain becomes the parent name and determines the base name of the hydrocarbon.

   - For example, a seven-carbon chain is called "heptane" for an alkane, "heptene" for an alkene, and "heptyne" for an alkyne.

 

2. Number the Carbon Chain: Number the chain from the end nearest to the first substituent or the first multiple bond (double or triple). For alkenes and alkynes, the location of the double or triple bond takes priority in numbering.

 

3. Identify and Name Substituents: Name any alkyl groups (e.g., methyl, ethyl) or other substituents attached to the main chain. Assign a number to each substituent based on its position on the carbon chain.

 

4. Assign Locants for Double/Triple Bonds (for alkenes and alkynes): The locant indicates the position of the multiple bond. Use the smallest possible number to locate the first carbon of the multiple bond.

 

5. Assemble the Name: Combine the elements in the following order: position of substituents, substituent names, root name of the longest chain, and locants for any double or triple bonds.

 

6. Use Prefixes for Multiple Identical Substituents: If the same substituent appears more than once, use prefixes such as di-, tri-, tetra-, etc., to indicate the quantity.

 

Examples of IUPAC Names:

- Methane (\(\text{CH}_4\)): The simplest alkane, with a single carbon atom.

- 2-Methylpropane (\(\text{C}_4\text{H}_{10}\)): A four-carbon chain with a methyl group on the second carbon.

- But-2-ene (\(\text{C}_4\text{H}_8\)): A four-carbon chain with a double bond starting at the second carbon.

- 2-Butyne (\(\text{C}_4\text{H}_6\)): A four-carbon chain with a triple bond starting at the second carbon.

 

 3. Key Terms and Concepts

- Parent Chain: The longest continuous chain of carbon atoms in the structure.

- Substituent: An atom or group attached to the parent chain, such as alkyl groups or halogens.

- Locant: A number that indicates the position of a substituent or a multiple bond.

- Prefix and Suffix: The prefix specifies substituents, while the suffix specifies the type of hydrocarbon (-ane, -ene, -yne for alkanes, alkenes, and alkynes, respectively).

 

 4. Important Rules, Theorems, and Principles

- Rule of Lowest Locants: Number the parent chain so that substituents and multiple bonds get the lowest possible numbers.

- Alphabetical Order for Substituents: Arrange substituents in alphabetical order when naming the compound, ignoring prefixes like di-, tri-, etc., for alphabetizing purposes.

- Functional Group Priority: For compounds with functional groups, the functional group takes precedence over multiple bonds in naming (e.g., alcohols, carboxylic acids).

 5. Illustrative Diagrams and Visuals

 

Here are a few examples of hydrocarbons with IUPAC names, showing the numbering, substituents, and bonds:

 

1. 2-Methylpentane:

   - Structure: A five-carbon chain with a methyl group on the second carbon.

   - Diagram: Show the carbon chain with a methyl group attached at the second carbon, with appropriate numbering.

 

2. 3,3-Dimethylhexane:

   - Structure: A six-carbon chain with two methyl groups attached to the third carbon.

   - Diagram: Display the six-carbon chain with two methyl groups on the third carbon.

 

3. But-1-ene:

   - Structure: A four-carbon chain with a double bond starting from the first carbon.

   - Diagram: Display the double bond between the first and second carbons, with numbering.

 

4. 3-Chloro-2-methylpentane:

   - Structure: A five-carbon chain with a chlorine substituent on the third carbon and a methyl group on the second carbon.

   - Diagram: Show the five-carbon chain with appropriate substituents and numbering.

 

[Include detailed diagrams with numbered carbons, substituent locations, and bond types clearly marked.]

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Name the following structure: \(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3\).

 

- Solution:

  1. Identify the longest chain: Five carbons – "pentane."

  2. Number the chain from the end nearest the substituent (methyl group).

  3. Position the substituent: Methyl group on the third carbon.

  4. Name: 3-Methylpentane.

 

Example Problem 2: Provide the IUPAC name for \(\text{CH}_3\text{CH}(\text{Cl})\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3\).

 

- Solution:

  1. Longest chain: Five carbons – "pentane."

  2. Number the chain from the left to give substituents the lowest numbers.

  3. Position the substituents: Chlorine on the second carbon and a methyl group on the third carbon.

  4. Arrange substituents alphabetically: 2-Chloro-3-methylpentane.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Identifying the Parent Chain: Always identify the longest continuous chain with the most double or triple bonds for alkenes and alkynes.

- Numbering Priority: For alkenes and alkynes, prioritize the position of double or triple bonds when numbering the chain.

- Combining Substituents: When the same substituent appears multiple times, combine them with prefixes like di-, tri-, and tetra-.

 

 8. Patterns in JEE Questions

JEE questions on IUPAC nomenclature may involve:

- Identifying the IUPAC names of complex structures with multiple substituents.

- Drawing structures from IUPAC names.

- Determining priority in numbering chains with both substituents and multiple bonds.

 

---

 

 9. Tips to Avoid Common Mistakes

- Misnumbering the Parent Chain: Start numbering from the end closest to the first substituent or double/triple bond.

- Ignoring Alphabetical Order: List substituents in alphabetical order, not in order of appearance.

- Incorrect Use of Prefixes: Remember to use prefixes (di-, tri-, etc.) for identical substituents and ignore them in alphabetical ordering.

 

 10. Key Points to Remember for Revision

- Types of Hydrocarbons: Alkanes (single bonds), Alkenes (double bonds), Alkynes (triple bonds).

- Substituent Naming and Positioning: Place substituents and multiple bonds at the lowest possible locants.

- Basic Nomenclature Structure: Position numbers – substituent names – root name – suffix for double or triple bonds.

 

 11. Real-World Applications and Cross-Chapter Links

- Chemical Databases: IUPAC names are essential for organizing and searching compounds in chemical databases.

- Communication in Chemistry: Standardized names allow chemists to identify and discuss compounds accurately.

- Cross-Concept Connections: The IUPAC system applies to various organic compounds, including alcohols, carboxylic acids, and amines, beyond hydrocarbons.

Questions

Q 1. Aromatisation of \(n\)-hexane gives :;

(a) cyclohexane;

(b) benzene;

(c) cycloheptane;

(d) toluene;

Classification of hydrocarbons - alkanes, alkenes, alkynes


Classification of Hydrocarbons - Alkanes, Alkenes, Alkynes

 

 1. Definition and Core Explanation

Hydrocarbons are organic compounds consisting solely of carbon and hydrogen atoms. They are classified based on the type of bonds between carbon atoms. The three primary types of hydrocarbons are:

 

- Alkanes: Saturated hydrocarbons containing only single bonds between carbon atoms. They are also known as paraffins.

- Alkenes: Unsaturated hydrocarbons containing at least one carbon-carbon double bond. They are also called olefins.

- Alkynes: Unsaturated hydrocarbons containing at least one carbon-carbon triple bond, known as acetylenes.

 

Each class has unique chemical properties due to the nature of bonding, which affects reactivity, stability, and types of reactions they undergo.

 

 2. Fundamental Properties and Structure of Hydrocarbons

- Alkanes (C\(_n\)H\(_{2n+2}\)):

  - Structure: Only single bonds (\(\text{C-C}\)) with a tetrahedral geometry around each carbon.

  - Saturation: Alkanes are fully saturated, meaning each carbon is bonded to as many hydrogen atoms as possible.

  - Reactivity: Alkanes are generally less reactive due to strong \(\text{C-C}\) and \(\text{C-H}\) bonds. They primarily undergo combustion and substitution reactions.

  - Example: Methane (\(\text{CH}_4\)), Ethane (\(\text{C}_2\text{H}_6\)).

 

- Alkenes (C\(_n\)H\(_{2n}\)):

  - Structure: At least one double bond (\(\text{C=C}\)) with trigonal planar geometry around the double-bonded carbons.

  - Unsaturation: Alkenes are unsaturated, meaning they have fewer hydrogen atoms than alkanes due to the presence of a double bond.

  - Reactivity: Alkenes are more reactive than alkanes, primarily undergoing addition reactions due to the double bond’s electron-rich nature.

  - Example: Ethene (\(\text{C}_2\text{H}_4\)), Propene (\(\text{C}_3\text{H}_6\)).

 

- Alkynes (C\(_n\)H\(_{2n-2}\)):

  - Structure: At least one triple bond (\(\text{CC}\)) with a linear geometry around the triple-bonded carbons.

  - Unsaturation: Alkynes are even more unsaturated than alkenes, with fewer hydrogen atoms due to the presence of a triple bond.

  - Reactivity: Alkynes are quite reactive, especially in reactions involving addition across the triple bond.

  - Example: Ethyne (acetylene) (\(\text{C}_2\text{H}_2\)), Propyne (\(\text{C}_3\text{H}_4\)).

 

 3. Key Terms and Concepts

- Saturation: A measure of hydrogen presence; alkanes are saturated, while alkenes and alkynes are unsaturated.

- Double Bond and Triple Bond: Functional groups within alkenes and alkynes, respectively, which make these compounds more reactive than alkanes.

- Hydrocarbon Series: Each class of hydrocarbon follows a general formula (alkanes: \(\text{C}_n\text{H}_{2n+2}\), alkenes: \(\text{C}_n\text{H}_{2n}\), alkynes: \(\text{C}_n\text{H}_{2n-2}\)).

 

 4. Important Rules, Theorems, and Principles

- Degree of Unsaturation: The number of hydrogen atoms decreases as the number of double or triple bonds increases. Alkanes are fully saturated, while alkenes and alkynes have degrees of unsaturation due to their multiple bonds.

- Bonding and Geometry:

  - Alkanes have sp\(^3\) hybridized carbons, leading to tetrahedral geometry.

  - Alkenes have sp\(^2\) hybridized carbons around the double bond, creating a planar geometry.

  - Alkynes have sp hybridized carbons around the triple bond, leading to linear geometry.

 

 5. Illustrative Diagrams and Visuals

 

1. Alkane Structure (Methane as Example):

   - Diagram showing methane’s tetrahedral geometry with single \(\text{C-H}\) bonds.

 

2. Alkene Structure (Ethene as Example):

   - Diagram showing ethene’s planar geometry with a double bond between carbons.

 

3. Alkyne Structure (Ethyne as Example):

   - Diagram showing ethyne’s linear geometry with a triple bond between carbons.

 

[Include illustrations for methane, ethene, and ethyne, marking bond types and hybridization at each carbon.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Identify the class of hydrocarbon and the number of pi bonds in \(\text{C}_5\text{H}_8\).

 

- Solution:

  1. Formula Analysis: \(\text{C}_n\text{H}_{2n-2}\), matching alkynes.

  2. Pi Bonds: Alkynes have one triple bond, which contains two pi bonds.

  - Answer: Alkyne with two pi bonds.

 

Example Problem 2: Calculate the degree of unsaturation for \(\text{C}_6\text{H}_{10}\) and identify the class of hydrocarbon.

 

- Solution:

  1. Degree of Unsaturation: \(\frac{2 \times 6 + 2 - 10}{2} = 2\).

  2. Interpretation: The structure could have one triple bond (alkyne) or two double bonds (alkene).

  - Answer: The compound could be either an alkyne or a diene (two double bonds).

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Using General Formulas: Recognize alkanes, alkenes, and alkynes by their general formulas (\(\text{C}_n\text{H}_{2n+2}\), \(\text{C}_n\text{H}_{2n}\), and \(\text{C}_n\text{H}_{2n-2}\)) to quickly identify the type of hydrocarbon.

- Degree of Unsaturation for Quick Identification: Use the degree of unsaturation formula \(\frac{2 \times \text{C} + 2 - \text{H}}{2}\) to determine if a compound is saturated, unsaturated, or aromatic.

 

 8. Patterns in JEE Questions

JEE Advanced often includes questions that test:

- Identifying hydrocarbons based on their formula.

- Calculating degrees of unsaturation.

- Determining the structural differences and properties among alkanes, alkenes, and alkynes.

- Recognizing reactions that are unique to each class of hydrocarbons (e.g., substitution for alkanes, addition for alkenes and alkynes).

 

 9. Tips to Avoid Common Mistakes

- Misidentifying Hydrocarbon Class: Use the general formula and degree of unsaturation to ensure correct classification.

- Counting Pi Bonds Incorrectly: Alkenes have one pi bond per double bond, while alkynes have two pi bonds per triple bond.

- Incorrect Geometry: Remember, alkanes are tetrahedral, alkenes are planar around the double bond, and alkynes are linear around the triple bond.

 

 10. Key Points to Remember for Revision

- General Formulas: Alkanes (\(\text{C}_n\text{H}_{2n+2}\)), Alkenes (\(\text{C}_n\text{H}_{2n}\)), Alkynes (\(\text{C}_n\text{H}_{2n-2}\)).

- Bonding and Geometry: Alkanes (sp\(^3\), tetrahedral), Alkenes (sp\(^2\), planar), Alkynes (sp, linear).

- Reactivity and Typical Reactions: Alkanes undergo substitution, while alkenes and alkynes are more reactive in addition reactions due to multiple bonds.

 

 11. Real-World Applications and Cross-Chapter Links

- Fuel and Energy Sources: Alkanes (natural gas, gasoline) are widely used as fuels, while ethene is a key industrial precursor.

- Industrial Synthesis: Ethyne (acetylene) is used in welding torches due to its high energy release on combustion.

- Cross-Concept Connections: Understanding hydrocarbon classification is fundamental for studying organic reactions, as reactivity varies across alkanes, alkenes, and alkynes.

Questions

Q 1. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by :;

(a) oxidation;

(b) cracking;

(c) distillation under reduced pressure;

(d) hydrolysis;

Preparation of alkanes by Wurtz reaction


Preparation of Alkanes by Wurtz Reaction

 

1. Definition and Core Explanation

The Wurtz reaction is a method used to synthesize alkanes by coupling two alkyl halides in the presence of metallic sodium in dry ether. It is primarily used to prepare alkanes with an even number of carbon atoms, as two identical alkyl groups combine to form a longer carbon chain. The reaction can be represented as:

 

\[

\text{R-X} + \text{R'-X} + 2\text{Na} \rightarrow \text{R-R'} + 2\text{NaX}

\]

 

where \( \text{R} \) and \( \text{R'} \) are alkyl groups and \( \text{X} \) is a halogen, typically bromine or iodine. The Wurtz reaction is especially useful in organic synthesis for forming larger alkanes from smaller alkyl halides. However, it is limited by side reactions and is more effective for symmetrical alkanes, where both alkyl groups are identical.

2. Fundamental Mechanism of the Wurtz Reaction

The Wurtz reaction proceeds through a radical mechanism, involving the following steps:

 

1. Single Electron Transfer (SET): The reaction begins when sodium donates an electron to an alkyl halide, producing a radical anion.

   \[

   \text{R-X} + \text{Na} \rightarrow \text{R} \cdot + \text{NaX}

   \]

 

2. Radical Formation: The radical anion splits into an alkyl radical and halide ion. Sodium donates another electron to the alkyl radical, forming an alkyl anion.

 

3. Coupling Step: The alkyl anions then combine to form a new carbon-carbon bond, producing the final alkane.

   \[

   \text{R} \cdot + \text{R} \cdot \rightarrow \text{R-R}

   \]

 

The reaction requires a nonpolar solvent, usually dry ether, to prevent side reactions and to stabilize the alkyl radicals and sodium ions.

3. Key Terms and Concepts

- Coupling Reaction: A reaction where two identical or similar molecules combine to form a single larger molecule.

- Radical Mechanism: A reaction mechanism involving free radicals, often requiring a single electron transfer (SET).

- Symmetrical Alkane: Alkanes with identical groups on either side of the newly formed carbon-carbon bond, which are easier to synthesize using the Wurtz reaction.

4. Important Rules, Theorems, and Principles

- Only Effective for Primary Alkyl Halides: The Wurtz reaction is best suited for primary alkyl halides, as secondary and tertiary halides are prone to elimination reactions, which can yield alkenes instead of alkanes.

- Sodium as a Reducing Agent: Sodium metal acts as a reducing agent, donating electrons to the alkyl halides and promoting the formation of radicals.

- Side Reactions: If dissimilar alkyl halides are used, mixtures of products are often formed, as cross-coupling can produce multiple alkanes.

5. Illustrative Diagrams and Visuals

 

1. Reaction Setup: Show an apparatus where sodium metal is added to a dry ether solution containing two alkyl halides.

2. Reaction Pathway: Illustrate the stepwise mechanism, including electron transfer from sodium to the alkyl halides, radical formation, and the coupling of radicals to form the final product.

 

[Include diagrams showing the electron transfer, radical formation, and combination steps, with each molecule and intermediate clearly labeled.]

6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the product of the Wurtz reaction between two molecules of chloromethane (\(\text{CH}_3\text{Cl}\)).

 

- Solution:

  1. Identify the Alkyl Groups: Both molecules are chloromethane, so they are identical.

  2. Reaction Setup: The Wurtz reaction between two \(\text{CH}_3\text{Cl}\) molecules in the presence of sodium will yield ethane.

  - Final Product: Ethane (\(\text{CH}_3\text{CH}_3\)).

 

Example Problem 2: Explain why the Wurtz reaction between bromopropane (\(\text{C}_3\text{H}_7\text{Br}\)) and bromoethane (\(\text{C}_2\text{H}_5\text{Br}\)) might lead to a mixture of products.

 

- Solution:

  1. Identify Possible Combinations: The reaction could produce:

     - Symmetrical alkanes: Hexane (\(\text{C}_3\text{H}_7\text{-}\text{C}_3\text{H}_7\)) and butane (\(\text{C}_2\text{H}_5\text{-}\text{C}_2\text{H}_5\)).

     - Cross-coupled alkane: Pentane (\(\text{C}_3\text{H}_7\text{-}\text{C}_2\text{H}_5\)).

  2. Conclusion: The reaction results in a mixture of hexane, butane, and pentane, making it less suitable for preparing unsymmetrical alkanes.

7. Common Tricks, Shortcuts, and Solving Techniques

- Using the Wurtz Reaction for Symmetrical Alkanes: The Wurtz reaction is most effective when synthesizing alkanes with an even number of carbon atoms, particularly symmetrical alkanes.

- Avoiding Side Products: For a single, pure product, use identical alkyl halides. Mixing different alkyl halides leads to multiple products due to possible cross-coupling.

8. Patterns in JEE Questions

Questions on the Wurtz reaction in JEE often involve:

- Predicting the products of the reaction when given one or two different alkyl halides.

- Identifying the reaction mechanism and understanding why sodium and dry ether are required.

- Determining limitations and applicability of the reaction, especially regarding side reactions and product mixtures.

9. Tips to Avoid Common Mistakes

- Incorrect Use of Secondary or Tertiary Halides: Secondary and tertiary alkyl halides can lead to elimination reactions, producing alkenes rather than alkanes.

- Misidentifying Products in Mixed Reactions: Be aware of the possibility of multiple products when different alkyl halides are used, due to cross-coupling.

10. Key Points to Remember for Revision

- Reaction Conditions: Sodium in dry ether is necessary for the Wurtz reaction.

- Reaction Mechanism: Radical mechanism involving electron transfer from sodium to the alkyl halide.

- Product Purity: The Wurtz reaction is most efficient for synthesizing symmetrical alkanes.

11. Real-World Applications and Cross-Chapter Links

- Industrial Relevance: The Wurtz reaction is primarily of academic interest, but it serves as a basis for understanding radical reactions and alkane synthesis.

- Cross-Concept Connections: The reaction mechanism introduces radical chemistry, applicable in understanding polymerization and certain organic transformations.

Questions

Q 1. n-Hexane isomerises in presence of anhydrous aluminium chloride and hydrogen chloride gas to give;

(a) 2-Methyl pentane;

(b) 3-Methyl pentane;

(c) Both (a) and (b);

(d) Neither (a) nor (b);

Markovnikov's rule in alkene addition reactions


Markovnikov's Rule in Alkene Addition Reactions

1. Definition and Core Explanation

Markovnikov's Rule is a guideline used in organic chemistry to predict the regioselectivity of addition reactions to unsymmetrical alkenes. Named after the Russian chemist Vladimir Markovnikov, this rule states that during the addition of a protic acid (like HCl, HBr, or H\(_2\)O) to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with the greater number of hydrogen atoms, while the other part of the reagent (such as Cl\(^-\) or OH\(^-\)) attaches to the carbon with fewer hydrogen atoms.

 

In simpler terms, Markovnikov's Rule helps predict the major product by indicating that the positive part of the adding molecule (often H\(^+\)) will bond to the carbon atom that already has more hydrogen atoms. This is because the formation of a more stable carbocation intermediate drives the reaction. As a result, the reaction pathway that creates the most stable carbocation is favored.

 

Example: For the addition of HBr to propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)), Markovnikov's Rule predicts that the hydrogen atom attaches to the terminal carbon (with more hydrogens), and the bromine attaches to the central carbon. The major product is therefore 2-bromopropane.

2. Fundamental Mechanism of Markovnikov's Rule

The addition reaction proceeds through a carbocation intermediate, and the stability of this intermediate determines the favored product:

 

1. Protonation of the Alkene: The double bond in the alkene acts as a nucleophile and reacts with the H\(^+\) ion from the reagent (e.g., HBr), leading to the formation of a carbocation.

   \[

   \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}^+\text{CH}_3

   \]

 

2. Carbocation Stability: The carbocation that forms will be the more stable carbocation possible. In this case, a secondary carbocation is more stable than a primary one due to electron-donating inductive effects from neighboring alkyl groups.

 

3. Nucleophilic Attack: The nucleophile (e.g., Br\(^-\)) then attaches to the carbocation, forming the final product.

   \[

   \text{CH}_3\text{CH}^+\text{CH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{CHBrCH}_3

   \]

 

The final product is 2-bromopropane, where the bromine attaches to the more substituted carbon.

 

 3. Key Terms and Concepts

- Regioselectivity: The preference of one direction of chemical bond formation over another, leading to the formation of a major product.

- Carbocation Stability: Stability order of carbocations is tertiary > secondary > primary. The stability influences the direction of addition.

- Electrophilic Addition: A type of reaction where an electrophile adds to a double bond, common in alkenes.

 

 4. Important Rules, Theorems, and Principles

- Carbocation Stability Principle: The more substituted the carbocation, the more stable it is, due to hyperconjugation and inductive effects. This stability drives the regioselective outcome predicted by Markovnikov's Rule.

- Anti-Markovnikov Addition: In certain cases (e.g., free radical addition of HBr in the presence of peroxides), the addition occurs in the opposite manner, with the hydrogen adding to the carbon with fewer hydrogen atoms.

 

 5. Illustrative Diagrams and Visuals

 

1. Addition of HBr to Propene:

   - Diagram showing propene with the double bond.

   - Step-by-step illustration of protonation leading to the secondary carbocation.

   - Final product: 2-bromopropane with bromine attached to the more substituted carbon.

 

2. Carbocation Stability Comparison:

   - Diagram comparing primary, secondary, and tertiary carbocations to illustrate why secondary and tertiary carbocations are more stable.

 

[Include clear diagrams showing the protonation step, carbocation formation, and final product with carbon-hydrogen and carbon-bromine bonds labeled.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the major product of HCl addition to 2-methylpropene (\(\text{CH}_2=\text{C}(\text{CH}_3)_2\)).

 

- Solution:

  1. Identify the Alkene Structure: 2-methylpropene is an unsymmetrical alkene.

  2. Apply Markovnikov's Rule: The hydrogen from HCl attaches to the carbon with more hydrogens (the terminal carbon).

  3. Carbocation Intermediate: The reaction forms a tertiary carbocation, which is highly stable.

  4. Final Product: 2-chloro-2-methylpropane, with Cl attached to the tertiary carbon.

 

Example Problem 2: For the addition of HBr to 1-butene, predict the major product.

 

- Solution:

  1. Identify the Structure of 1-Butene: \(\text{CH}_2=\text{CHCH}_2\text{CH}_3\).

  2. Markovnikov's Rule Application: H attaches to the carbon with more hydrogens (the terminal \(\text{CH}_2\) group).

  3. Carbocation Intermediate: A secondary carbocation forms, which is relatively stable.

  4. Final Product: 2-bromobutane, with Br attached to the second carbon.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Identifying Major Products: Always consider the carbocation stability; tertiary is more stable than secondary, which is more stable than primary.

- Recognizing Anti-Markovnikov Cases: When peroxides are involved in the addition of HBr, remember the reaction proceeds via a free radical mechanism, leading to anti-Markovnikov products.

 

 8. Patterns in JEE Questions

JEE Advanced questions on Markovnikov's Rule often involve:

- Predicting the products of addition reactions on unsymmetrical alkenes.

- Comparing Markovnikov and anti-Markovnikov products in reactions with or without peroxides.

- Determining carbocation stability and understanding its impact on the reaction pathway.

 

 9. Tips to Avoid Common Mistakes

- Ignoring Carbocation Stability: Ensure you always consider the stability of the intermediate carbocation in determining the major product.

- Misidentifying Anti-Markovnikov Products: If the problem states or implies the presence of peroxides, apply the anti-Markovnikov rule.

10. Key Points to Remember for Revision

- Markovnikov's Rule: "The rich get richer" – the carbon with more hydrogens receives the hydrogen atom in the addition.

- Carbocation Stability: Major product formation depends on forming the most stable carbocation.

- Anti-Markovnikov Exception: Free radical addition in the presence of peroxides reverses the regioselectivity, with hydrogen adding to the carbon with fewer hydrogens.

 11. Real-World Applications and Cross-Chapter Links

- Synthesis of Alkyl Halides: Markovnikov’s Rule is widely used in industrial processes for the synthesis of alkyl halides.

- Pharmaceuticals and Organic Synthesis: Selective addition reactions are essential in creating complex organic molecules with specific structures.

- Cross-Concept Connections: Understanding Markovnikov’s Rule is fundamental in predicting products in various organic reactions, including hydration, hydrohalogenation, and polymerization.

Questions

Q 1. Which of the following represents the correct reaction?;

(a) \(\mathrm{CH}_{4}+2 \mathrm{H}_{2} \mathrm{O} \xrightarrow{\mathrm{Ni}} \mathrm{CO}_{2}+4 \mathrm{H}_{2}\);

(b) \(\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \xrightarrow{\mathrm{Ni}} \mathrm{CO}+3 \mathrm{H}_{2}\);

(c) \(\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \xrightarrow{\mathrm{Ni}} \mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}_{2}\);

(d) \(\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \xrightarrow{\mathrm{Ni}} \mathrm{HCHO}+2 \mathrm{H}_{2}\);

Electrophilic addition in alkenes


Electrophilic Addition in Alkenes

1. Definition and Core Explanation

Electrophilic addition is a common reaction mechanism for alkenes, where an electrophile (electron-seeking species) adds to the double bond of an alkene. Alkenes, due to their electron-rich double bonds, readily undergo electrophilic addition reactions. The double bond provides a region of high electron density that can attract electrophiles, initiating the reaction.

 

The mechanism of electrophilic addition involves two main steps:

1. Attack by the Electrophile: The electrophile interacts with the alkene, leading to the formation of a carbocation intermediate.

2. Nucleophilic Attack: A nucleophile then attacks the carbocation, completing the addition process and forming the final product.

 

Example: In the addition of HBr to ethene (\(\text{CH}_2=\text{CH}_2\)), HBr adds across the double bond. The electrophilic hydrogen (\(\text{H}^+\)) bonds to one of the carbon atoms, while the bromine ion (\(\text{Br}^-\)) attaches to the other carbon, forming bromoethane (\(\text{CH}_3\text{CH}_2\text{Br}\)).

 

 2. Fundamental Mechanism of Electrophilic Addition

The electrophilic addition mechanism in alkenes can be broken down into the following steps:

 

1. Formation of a Carbocation Intermediate:

   - The \(\text{H}^+\) ion from HBr (or another electrophile) is attracted to the electron-rich double bond in the alkene.

   - This protonation breaks the double bond, resulting in a carbocation intermediate on the carbon that does not receive the hydrogen.

   \[

   \text{CH}_2=\text{CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}^+

   \]

 

2. Nucleophilic Attack:

   - The bromide ion (\(\text{Br}^-\)) from HBr then attacks the carbocation, bonding to the positively charged carbon.

   \[

   \text{CH}_3\text{CH}^+ + \text{Br}^- \rightarrow \text{CH}_3\text{CH}_2\text{Br}

   \]

 

The final product is bromoethane, with the H and Br atoms adding across the former double bond.

3. Key Terms and Concepts

- Electrophile: An electron-deficient species that seeks out electrons and reacts with electron-rich areas (e.g., \(\text{H}^+\) in HBr).

- Carbocation Intermediate: A positively charged intermediate formed during the reaction, with the positive charge localized on one of the carbons in the former double bond.

- Nucleophile: An electron-rich species that seeks a positively charged center, such as \(\text{Br}^-\) in HBr addition.

4. Important Rules, Theorems, and Principles

- Markovnikov's Rule: In the case of unsymmetrical alkenes, the proton from the electrophile will attach to the carbon with more hydrogen atoms, forming the most stable carbocation.

- Carbocation Stability: Carbocations are stabilized by alkyl groups through inductive and hyperconjugative effects. Therefore, tertiary carbocations are more stable than secondary, which are more stable than primary carbocations.

- Reaction Rates and Selectivity: The formation of a stable carbocation intermediate generally makes the reaction faster and more selective towards forming the major product.

5. Illustrative Diagrams and Visuals

 

1. Reaction Pathway for Ethene and HBr:

   - Show ethene reacting with HBr, where the double bond breaks to form a carbocation intermediate.

   - Illustrate the attack of \(\text{Br}^-\) on the carbocation to form the final product, bromoethane.

 

2. Carbocation Stability Chart:

   - Diagram showing primary, secondary, and tertiary carbocations, indicating the relative stability and why more stable carbocations are favored in electrophilic addition.

 

[Include diagrams illustrating each reaction step, highlighting the movement of electrons and showing the carbocation intermediate formation and final product.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the product of the electrophilic addition of HCl to propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)).

 

- Solution:

  1. Identify the Alkene Structure: Propene is an unsymmetrical alkene.

  2. Apply Markovnikov's Rule: The \(\text{H}^+\) from HCl will attach to the terminal carbon with more hydrogens, leading to a secondary carbocation at the central carbon.

  3. Carbocation Formation: The resulting carbocation forms on the second carbon.

  4. Nucleophilic Attack: \(\text{Cl}^-\) attacks the carbocation to form the product, 2-chloropropane.

 

Example Problem 2: For the addition of \(\text{H}_2\text{O}\) (hydration) to 1-butene, predict the major product.

 

- Solution:

  1. Identify the Structure of 1-Butene: \(\text{CH}_2=\text{CHCH}_2\text{CH}_3\).

  2. Electrophilic Attack: The \(\text{H}^+\) from water adds to the terminal carbon, forming a secondary carbocation at the second carbon.

  3. Nucleophilic Attack by OH: The \(\text{OH}^-\) (from water) then attaches to the carbocation, resulting in the formation of 2-butanol as the major product.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Using Markovnikov's Rule for Unsymmetrical Alkenes: Remember to add the hydrogen from the electrophile to the carbon with more hydrogen atoms for the formation of a more stable carbocation.

- Identifying Carbocation Stability for Prediction: Carbocation stability (tertiary > secondary > primary) can guide you in predicting the major product.

 

 8. Patterns in JEE Questions

JEE Advanced questions often test:

- Predicting the major product in electrophilic addition reactions.

- Applying Markovnikov's Rule to addition reactions of unsymmetrical alkenes.

- Analyzing reaction mechanisms by identifying intermediates, especially carbocations.

 

 9. Tips to Avoid Common Mistakes

- Ignoring Carbocation Stability: Carbocation stability plays a crucial role; be sure to predict intermediates based on the most stable carbocation.

- Incorrect Application of Markovnikov's Rule: Make sure to correctly apply Markovnikov’s Rule, especially for unsymmetrical alkenes, to avoid incorrect product prediction.

 

 10. Key Points to Remember for Revision

- Electrophilic Addition Mechanism: Consists of electrophile attack followed by nucleophilic addition to the carbocation.

- Carbocation Formation: Stability of carbocation determines the direction and outcome of addition.

- Markovnikov’s Rule Application: Essential for predicting products in unsymmetrical alkenes.

 

 11. Real-World Applications and Cross-Chapter Links

- Industrial Alkyl Halide Synthesis: Electrophilic addition reactions are widely used for synthesizing alkyl halides, alcohols, and other valuable compounds.

- Organic Synthesis Pathways: Understanding electrophilic addition is fundamental in designing multi-step organic synthesis processes.

- Cross-Concept Connections: This mechanism is foundational in organic chemistry, with applications in topics such as polymerization and reactivity of alkenes.

Questions

Q 1. How many conformations are possible for ethane?;

(a) 2;

(b) 3;

(c) infinite;

(d) one;

Aromaticity and stability of benzene


Aromaticity and Stability of Benzene

 

 1. Definition and Core Explanation

Aromaticity is a concept in organic chemistry that describes the unusual stability of certain cyclic, planar molecules with conjugated \(\pi\)-electron systems. Benzene (\(\text{C}_6\text{H}_6\)) is the simplest and most well-known aromatic compound, exhibiting remarkable stability due to its unique electronic structure.

 

Benzene's stability arises from a phenomenon called resonance and delocalization of electrons. In benzene, six carbon atoms form a planar, hexagonal ring with alternating single and double bonds. However, these bonds are not fixed; instead, the \(\pi\)-electrons are delocalized across the ring, creating a continuous ring of electron density above and below the plane of the molecule. This electron delocalization contributes to benzene's stability and lack of reactivity compared to typical alkenes.

 

The criteria for a molecule to be considered aromatic are defined by Hückel’s Rule, which states that a molecule must be cyclic, planar, fully conjugated, and contain \(4n + 2\) \(\pi\)-electrons (where \(n\) is a non-negative integer) to exhibit aromatic stability.

 

 2. Fundamental Properties of Aromatic Compounds

- Cyclic Structure: The molecule must form a closed ring, with atoms connected in a continuous loop.

- Planarity: All atoms in the ring must lie in the same plane to allow overlapping \(\pi\)-orbitals and delocalization of electrons.

- Conjugation: There must be an uninterrupted sequence of alternating single and double bonds around the ring, allowing for electron delocalization.

- Hückel's Rule: The molecule must have \(4n + 2\) \(\pi\)-electrons. For benzene, \(n = 1\), and the molecule has 6 \(\pi\)-electrons, which satisfies this rule.

 

Example of Aromaticity in Benzene:

Benzene (\(\text{C}_6\text{H}_6\)) has six \(\pi\)-electrons (one from each carbon’s p-orbital) that are fully conjugated and delocalized. This electron delocalization provides additional stability, making benzene less reactive than expected for a compound with alternating double bonds.

 

 

 3. Key Terms and Concepts

- Aromaticity: A property that imparts unusual stability to cyclic, planar, and conjugated molecules with \(4n + 2\) \(\pi\)-electrons.

- Hückel's Rule: A rule used to determine if a molecule is aromatic, requiring \(4n + 2\) \(\pi\)-electrons for stability.

- Resonance: The phenomenon of electron delocalization in a molecule, allowing electrons to be spread out over multiple bonds, reducing reactivity and increasing stability.

- Delocalization: The spreading of \(\pi\)-electrons across several atoms in a molecule, enhancing stability.

 

 4. Important Rules, Theorems, and Principles

- Hückel's Rule: A molecule is aromatic if it is cyclic, planar, conjugated, and contains \(4n + 2\) \(\pi\)-electrons.

- Stability of Aromatic Compounds: Aromatic compounds are more stable than expected due to electron delocalization. This stability results in reduced reactivity in electrophilic addition reactions and a preference for substitution reactions.

- Anti-Aromaticity: Molecules that are cyclic, planar, and conjugated but contain \(4n\) \(\pi\)-electrons (e.g., cyclobutadiene) are anti-aromatic and unusually unstable.

 

 5. Illustrative Diagrams and Visuals

 

1. Benzene Structure:

   - Show the benzene ring with alternating double bonds and a resonance hybrid structure representing electron delocalization across the ring.

 

2. Electron Cloud in Benzene:

   - Diagram of benzene with overlapping p-orbitals forming a continuous electron cloud above and below the ring, illustrating delocalization.

 

3. Application of Hückel's Rule:

   - Diagram showing examples of aromatic (benzene) and anti-aromatic (cyclobutadiene) molecules, with electron counting to verify Hückel’s Rule.

 

[Include clear diagrams for benzene’s resonance structures, the delocalized electron cloud, and examples demonstrating Hückel’s Rule.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Determine if cyclooctatetraene (\(\text{C}_8\text{H}_8\)) is aromatic, anti-aromatic, or non-aromatic.

 

- Solution:

  1. Structure and Planarity: Cyclooctatetraene is a cyclic molecule with conjugated double bonds, but it adopts a non-planar “tub” shape, preventing full conjugation across the ring.

  2. Hückel’s Rule Application: Although it has \(8\) \(\pi\)-electrons (\(4n\)), the non-planar structure means it cannot fully delocalize these electrons.

  3. Conclusion: Cyclooctatetraene is non-aromatic due to its lack of planarity.

 

Example Problem 2: Why is benzene more stable than 1,3,5-hexatriene, even though both have six \(\pi\)-electrons?

 

- Solution:

  1. Structure and Conjugation: Benzene’s six \(\pi\)-electrons are fully delocalized in a ring, contributing to resonance stability.

  2. Electron Delocalization: 1,3,5-hexatriene lacks the cyclic structure, so its electrons are not as effectively delocalized.

  3. Conclusion: Benzene is more stable due to aromaticity and resonance stabilization, which 1,3,5-hexatriene does not possess.

 

 7. Common Tricks, Shortcuts, and Solving Techniques

- Applying Hückel's Rule: For cyclic compounds, use \(4n + 2\) to quickly check for aromaticity. If the \(\pi\)-electron count fits this formula and the molecule is planar and conjugated, it is aromatic.

- Recognizing Stability through Resonance: When you see alternating double bonds in a ring, suspect aromaticity and check planarity and \(\pi\)-electron count.

 

 

 8. Patterns in JEE Questions

JEE questions on aromaticity and stability often include:

- Identifying aromatic, anti-aromatic, or non-aromatic compounds based on structure.

- Applying Hückel’s Rule to count \(\pi\)-electrons and determine aromaticity.

- Comparing stability between aromatic and non-aromatic compounds.

 

 9. Tips to Avoid Common Mistakes

- Forgetting Planarity Requirement: A molecule must be planar to be aromatic; non-planar conjugated rings cannot delocalize electrons fully.

- Incorrect Electron Counting: Ensure you count only \(\pi\)-electrons in conjugated systems for Hückel’s Rule.

 

 10. Key Points to Remember for Revision

- Aromaticity Requirements: Cyclic, planar, conjugated, and \(4n + 2\) \(\pi\)-electrons.

- Stability of Benzene: Benzene’s delocalized \(\pi\)-electrons create a stable, electron-rich structure.

- Anti-Aromaticity: Compounds with \(4n\) \(\pi\)-electrons are anti-aromatic and typically unstable.

 

 11. Real-World Applications and Cross-Chapter Links

- Pharmaceuticals and Aromatic Compounds: Aromatic rings are common in drug molecules due to their stability and ability to participate in specific interactions with biological targets.

- Polycyclic Aromatic Hydrocarbons (PAHs): Many stable molecules, such as naphthalene and anthracene, are based on aromatic rings, with applications in dyes, pigments, and organic electronics.

- Cross-Concept Connections: Aromaticity is foundational for understanding the chemistry of benzene derivatives, electrophilic substitution, and resonance effects.

Questions

Q 1. Spatial arrangements of atoms which can be converted into one another by rotation around a \(\mathrm{C}-\mathrm{C}\) single bond are called;

(a) Stereoisomers;

(b) Tautomers;

(c) Optical isomers;

(d) Conformers;

Structure and bonding in alkynes


Structure and Bonding in Alkynes

 

 1. Definition and Core Explanation

Alkynes are unsaturated hydrocarbons characterized by at least one carbon-carbon triple bond. The general formula for alkynes is \(\text{C}_n\text{H}_{2n-2}\), where \(n\) is the number of carbon atoms. The triple bond in alkynes consists of one sigma (\(\sigma\)) bond and two pi (\(\pi\)) bonds, making them more reactive than alkanes and alkenes due to the high electron density in the triple bond.

 

The structure of an alkyne is linear around the triple-bonded carbons because of the sp hybridization. The two sp-hybridized orbitals in each carbon form sigma bonds with the neighboring atoms (one sigma bond with the other carbon in the triple bond and one with a hydrogen or carbon atom in the chain). The remaining unhybridized p orbitals overlap to form two pi bonds, which lie perpendicular to each other. This unique bonding arrangement gives alkynes a linear geometry and affects their physical and chemical properties.

 

 2. Fundamental Structure and Hybridization in Alkynes

- Hybridization: The carbon atoms in the triple bond are sp hybridized, resulting in two sp orbitals and two unhybridized p orbitals. The sp orbitals form sigma bonds, while the unhybridized p orbitals form the pi bonds.

- Bond Formation:

  - Sigma Bond: Formed by the overlap of sp orbitals from each carbon atom in the triple bond.

  - Two Pi Bonds: Formed by the sideways overlap of the two unhybridized p orbitals on each carbon atom, creating electron density above and below the bond axis.

- Bond Length and Strength:

  - The carbon-carbon triple bond in alkynes is shorter (1.20 Å) than the double bond in alkenes and the single bond in alkanes due to stronger overlap in the triple bond.

  - The triple bond is stronger than double and single bonds, making alkynes more resistant to bond dissociation.

 

Example: In ethyne (acetylene, \(\text{C}_2\text{H}_2\)), each carbon is sp hybridized, and the molecule has a linear geometry with a bond angle of 180° around the triple bond.

 

 

 3. Key Terms and Concepts

- Triple Bond: A covalent bond formed by one sigma and two pi bonds between two carbon atoms.

- sp Hybridization: The type of hybridization where one s orbital and one p orbital mix to form two sp orbitals, leaving two p orbitals unhybridized for pi bonding.

- Linear Geometry: The structure around each sp-hybridized carbon in the triple bond, with a bond angle of 180°.

 

 4. Important Rules, Theorems, and Principles

- Bonding in Alkynes: The carbon atoms in alkynes form three bonds each – one sigma and two pi bonds – due to sp hybridization.

- Reactivity of Triple Bonds: The triple bond in alkynes is highly reactive in addition reactions, where electrophiles can add to the electron-rich pi bonds.

- Electronegativity and Acidity: The sp hybridized carbons in alkynes are more electronegative than sp\(^2\) or sp\(^3\)-hybridized carbons, giving terminal alkynes slight acidity, allowing them to react with strong bases to form acetylide ions.

5. Illustrative Diagrams and Visuals

1. Structure of Ethyne (Acetylene):

   - Show ethyne with two carbons linked by a triple bond and one hydrogen attached to each carbon.

   - Illustrate the linear geometry with 180° bond angles.

 

2. Hybridization in Alkynes:

   - Diagram showing sp hybridization with two sp orbitals forming sigma bonds and two unhybridized p orbitals forming pi bonds.

 

3. Electron Density in Pi Bonds:

   - Diagram of the electron clouds above and below the triple bond axis, representing the pi bonds formed by p orbital overlap.

 

[Include detailed diagrams showing the linear structure of ethyne, hybridization process, and orientation of pi bonds in alkynes.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Determine the bond angle and hybridization of carbons in propyne (\(\text{CH}_3\text{C}\equiv\text{CH}\)).

 

- Solution:

  1. Identify the Triple Bond: The carbons in the triple bond are sp hybridized.

  2. Determine Geometry: sp hybridization results in a linear structure with a bond angle of 180°.

  - Answer: The carbon atoms in the triple bond are sp hybridized with a bond angle of 180°.

 

Example Problem 2: Explain why terminal alkynes (\(\text{RC}\equiv\text{CH}\)) are more acidic than alkenes and alkanes.

 

- Solution:

  1. Hybridization Effect: The sp-hybridized carbon in a terminal alkyne is more electronegative than sp\(^2\) or sp\(^3\) hybridized carbons.

  2. Acidity: This electronegativity allows terminal alkynes to release \(\text{H}^+\) ions more readily, making them more acidic.

  - Conclusion: The greater electronegativity of sp-hybridized carbons makes terminal alkynes more acidic.

7. Common Tricks, Shortcuts, and Solving Techniques

- Recognizing Triple Bond Geometry: Any molecule with a \(\text{C}\equiv\text{C}\) bond will have a linear structure and sp hybridization around the triple bond.

- Using Acidity as an Indicator: Terminal alkynes are acidic and can form acetylide ions with strong bases, which is unique among hydrocarbons.

 

 8. Patterns in JEE Questions

JEE Advanced questions often test:

- Recognizing hybridization and bond angles in alkynes.

- Predicting reactivity based on the presence of a triple bond.

- Understanding acidity in terminal alkynes compared to alkenes and alkanes.

 

 9. Tips to Avoid Common Mistakes

- Incorrect Hybridization: Remember, carbons in a triple bond are sp hybridized, not sp\(^2\) or sp\(^3\).

- Ignoring Linear Geometry: Alkynes are linear around the triple bond. Ensure you draw bond angles as 180°.

10. Key Points to Remember for Revision

- Structure of Alkynes: Linear with sp hybridization and 180° bond angles.

- Bonding: Triple bond consists of one sigma and two pi bonds.

- Acidity of Terminal Alkynes: Terminal alkynes are acidic enough to react with strong bases to form acetylide ions.

 

11. Real-World Applications and Cross-Chapter Links

- Industrial Use of Acetylene: Ethyne (acetylene) is used in welding due to its high energy release upon combustion.

- Synthetic Applications: Alkynes are used as building blocks in organic synthesis, especially in reactions that exploit the triple bond’s reactivity.

- Cross-Concept Connections: Understanding alkynes is essential for studying addition reactions, acidity comparisons, and the concept of hybridization.

Questions

Q 1. General formula of alkenes and alkyl radicals are respectively:;

(a) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}\) and \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1}\);

(b) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}\) and \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}\);

(c) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-1}\) and \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}\);

(d) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+1}\) and \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}+2}\);

Addition reactions of alkynes


Addition Reactions of Alkynes

1. Definition and Core Explanation

Addition reactions in alkynes involve the addition of atoms or groups across the carbon-carbon triple bond, breaking the pi bonds and forming new sigma bonds. Due to the high electron density in the triple bond, alkynes are highly reactive towards electrophilic addition reactions. This reactivity enables them to undergo sequential addition, where one pi bond is broken at a time, leading first to an alkene and then to an alkane.

 

The general types of addition reactions in alkynes include hydrogenation, halogenation, hydrohalogenation, and hydration. Each type of reaction follows distinct mechanisms but ultimately reduces the unsaturation in the molecule by converting triple bonds into double or single bonds.

 

Example: In the hydrogenation of ethyne (acetylene), two equivalents of hydrogen add to the triple bond, converting it first to ethene and then to ethane if the reaction continues to completion.

2. Types of Addition Reactions in Alkynes

- Hydrogenation:

  - Partial Hydrogenation: One equivalent of hydrogen adds to the triple bond to form an alkene, typically in the presence of Lindlar’s catalyst for a cis-alkene or sodium in liquid ammonia for a trans-alkene.

  - Complete Hydrogenation: Two equivalents of hydrogen are added, converting the alkyne to an alkane.

  - Example: Ethyne (\(\text{C}_2\text{H}_2\)) undergoes complete hydrogenation to form ethane (\(\text{C}_2\text{H}_6\)).

 

- Halogenation:

  - Halogens such as \(\text{Cl}_2\) or \(\text{Br}_2\) add to alkynes, forming dihaloalkenes or tetrahaloalkanes depending on the amount of halogen added.

  - Example: Addition of \(\text{Br}_2\) to ethyne yields 1,2-dibromoethene (partial halogenation) or 1,1,2,2-tetrabromoethane (complete halogenation).

 

- Hydrohalogenation:

  - Hydrogen halides (\(\text{HCl}\), \(\text{HBr}\), \(\text{HI}\)) add across the triple bond, producing haloalkenes or dihaloalkanes. In unsymmetrical alkynes, Markovnikov’s Rule applies, with the hydrogen attaching to the carbon with more hydrogen atoms.

  - Example: Propyne (\(\text{CH}_3\text{C}\equiv\text{CH}\)) with HBr yields 2-bromopropene in the first addition, which can further react to form 2,2-dibromopropane.

 

- Hydration:

  - Water adds to alkynes in the presence of a catalyst (typically Hg\(^2+\) salts and \(\text{H}_2\text{SO}_4\)), forming an enol intermediate that rapidly tautomerizes to a ketone.

  - Example: Hydration of ethyne yields acetaldehyde through tautomerization.

3. Key Terms and Concepts

- Sequential Addition: In alkynes, addition occurs in two stages, as one pi bond breaks to form an alkene, followed by the breaking of the second pi bond to form an alkane.

- Markovnikov’s Rule: During hydrohalogenation, the hydrogen adds to the carbon with more hydrogens, producing a more stable carbocation intermediate.

- Tautomerization: A process in hydration where the enol intermediate rearranges to a more stable ketone form.

 

 4. Important Rules, Theorems, and Principles

- Markovnikov’s Rule in Unsymmetrical Alkynes: The rule applies to hydrohalogenation, where the hydrogen attaches to the carbon with more hydrogens, favoring the formation of the more stable carbocation intermediate.

- Regioselectivity and Stereoselectivity: Partial hydrogenation with Lindlar’s catalyst leads to cis-alkenes, while sodium in liquid ammonia produces trans-alkenes.

- Tautomerization in Hydration: The enol intermediate formed in hydration reactions is unstable and quickly tautomerizes to form a ketone, a more stable structure.

5. Illustrative Diagrams and Visuals

 

1. Hydrogenation Pathway:

   - Diagram showing ethyne converting to ethene (cis or trans) and finally to ethane with hydrogenation.

 

2. Halogenation Example:

   - Show the step-by-step addition of \(\text{Br}_2\) to ethyne, forming a dihaloalkene, and then the tetrahaloalkane with excess \(\text{Br}_2\).

 

3. Hydration and Tautomerization:

   - Diagram showing hydration of ethyne, formation of the enol, and tautomerization to acetaldehyde.

 

[Include diagrams that clearly show each step in the reaction pathways, indicating bond changes, intermediate formation, and the final products.]

6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the product(s) when 2-butyne (\(\text{CH}_3\text{C}\equiv\text{CCH}_3\)) is treated with excess \(\text{H}_2\) in the presence of a platinum catalyst.

 

- Solution:

  1. Complete Hydrogenation: Excess hydrogen and a platinum catalyst convert the triple bond completely.

  2. Final Product: 2-butane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)).

 

Example Problem 2: What is the major product of the addition of \(\text{HBr}\) to propyne (\(\text{CH}_3\text{C}\equiv\text{CH}\))?

 

- Solution:

  1. Markovnikov’s Rule: The hydrogen from HBr attaches to the terminal carbon (with more hydrogens).

  2. Carbocation Intermediate: A secondary carbocation forms on the middle carbon.

  3. Final Product: 2-bromopropene (\(\text{CH}_3\text{CBr}=\text{CH}_2\)).

7. Common Tricks, Shortcuts, and Solving Techniques

- Using Catalysts for Partial vs. Complete Hydrogenation: Lindlar’s catalyst yields cis-alkenes, while platinum or palladium leads to complete hydrogenation.

- Identifying Tautomerization: In hydration reactions, remember that enols formed as intermediates will quickly convert to more stable ketones or aldehydes.

8. Patterns in JEE Questions

JEE Advanced questions often include:

- Predicting the products of alkyne addition reactions.

- Applying Markovnikov’s Rule to determine product regioselectivity.

- Identifying the role of different catalysts in partial vs. complete hydrogenation.

9. Tips to Avoid Common Mistakes

- Misapplying Markovnikov’s Rule: Be careful to apply Markovnikov’s Rule correctly in hydrohalogenation reactions.

- Overlooking Tautomerization: Ensure you recognize that enols will tautomerize to ketones or aldehydes, particularly in hydration reactions.

10. Key Points to Remember for Revision

- Sequential Addition: Alkyne addition reactions often proceed in two steps, first forming an alkene.

- Hydrogenation Catalysts: Lindlar’s catalyst produces cis-alkenes; sodium in liquid ammonia produces trans-alkenes.

- Tautomerization in Hydration: Hydration of alkynes often results in ketones or aldehydes due to tautomerization of enol intermediates.

11. Real-World Applications and Cross-Chapter Links

- Industrial Synthesis of Alkanes and Alkenes: Hydrogenation of alkynes is commonly used in fuel processing and organic synthesis.

- Synthesis of Pharmaceuticals: Many complex molecules are synthesized through controlled addition reactions of alkynes.

- Cross-Concept Connections: Alkyne addition reactions are foundational for studying reactivity patterns in organic chemistry, relevant to synthetic pathways in biochemistry and pharmacology.

Questions

Q 1. The restricted rotation about carbon-carbon double bond in 2 - butene is due to;

(a) overlap of one s- and one \(\mathrm{sp}^{2}\)-hybridized orbitals;

(b) overlap of two \(\mathrm{sp}^{2}\)-hybridized orbitals;

(c) overlap of one \(\mathrm{p}\)-and one \(\mathrm{sp}^{2}\)-hybridized orbitals;

(d) sideways overlap of two p-orbitals;

Aromatic hydrocarbons and substitution reactions


Aromatic Hydrocarbons and Substitution Reactions

1. Definition and Core Explanation

Aromatic hydrocarbons, also known as arenes, are hydrocarbons containing one or more benzene rings. Benzene (\(\text{C}_6\text{H}_6\)) is the simplest aromatic hydrocarbon, characterized by a ring of six carbon atoms with delocalized \(\pi\)-electrons. Aromatic hydrocarbons are stable due to their resonance structure, making them less reactive in addition reactions but highly reactive in electrophilic substitution reactions.

 

Electrophilic aromatic substitution (EAS) is the primary type of reaction for aromatic hydrocarbons. In EAS, an electrophile replaces a hydrogen atom on the benzene ring. The high stability of the aromatic ring is maintained after substitution, as the conjugated \(\pi\)-system is preserved.

 

Example: In the nitration of benzene, a nitronium ion (\(\text{NO}_2^+\)) acts as the electrophile, substituting a hydrogen on the benzene ring to form nitrobenzene.

2. Types of Electrophilic Aromatic Substitution (EAS) Reactions

- Nitration:

  - A nitro group (\(\text{NO}_2\)) is introduced into the aromatic ring using concentrated nitric acid and sulfuric acid.

  - Example: Benzene reacts with \(\text{HNO}_3\) in the presence of \(\text{H}_2\text{SO}_4\) to form nitrobenzene.

 

- Halogenation:

  - A halogen atom (e.g., \(\text{Cl}\) or \(\text{Br}\)) replaces a hydrogen atom in the ring using a Lewis acid catalyst like \(\text{FeCl}_3\) or \(\text{AlCl}_3\).

  - Example: Chlorination of benzene with \(\text{Cl}_2\) in the presence of \(\text{FeCl}_3\) yields chlorobenzene.

 

- Sulfonation:

  - A sulfonic acid group (\(\text{SO}_3\text{H}\)) is introduced by reacting benzene with fuming sulfuric acid (\(\text{H}_2\text{SO}_4\) containing \(\text{SO}_3\)).

  - Example: Benzene reacts with fuming sulfuric acid to form benzenesulfonic acid.

 

- Friedel-Crafts Alkylation:

  - An alkyl group is added to the benzene ring using an alkyl halide and a Lewis acid catalyst.

  - Example: Benzene reacts with methyl chloride (\(\text{CH}_3\text{Cl}\)) in the presence of \(\text{AlCl}_3\) to form toluene.

 

- Friedel-Crafts Acylation:

  - An acyl group is introduced using an acyl chloride and a Lewis acid catalyst.

  - Example: Benzene reacts with acetyl chloride (\(\text{CH}_3\text{COCl}\)) in the presence of \(\text{AlCl}_3\) to form acetophenone.

3. Key Terms and Concepts

- Electrophilic Aromatic Substitution (EAS): A reaction in which an electrophile replaces a hydrogen on an aromatic ring.

- Electrophile: An electron-deficient species that seeks out electron-rich areas like the benzene ring.

- Resonance Stabilization: The delocalized electron system in benzene stabilizes the ring, allowing substitution without breaking aromaticity.

- Lewis Acid Catalysts: Compounds like \(\text{FeCl}_3\) or \(\text{AlCl}_3\) used to enhance electrophile formation in EAS reactions.

4. Important Rules, Theorems, and Principles

- Resonance Energy and Stability: The resonance stabilization of the benzene ring makes it less reactive toward addition reactions but more suited for substitution, as substitution preserves aromaticity.

- Directive Influence of Substituents: Groups already on the benzene ring influence the position of further substitutions. Electron-donating groups (activating) direct incoming substituents to the ortho and para positions, while electron-withdrawing groups (deactivating) favor the meta position.

- Reactivity of Substituted Benzenes: Activating substituents increase the reactivity of benzene, while deactivating substituents decrease it, affecting the speed and outcome of EAS.

5. Illustrative Diagrams and Visuals

 

1. Mechanism of Nitration:

   - Show the formation of the nitronium ion (\(\text{NO}_2^+\)) and its attack on the benzene ring, leading to the formation of nitrobenzene.

 

2. Directive Effects:

   - Diagram of benzene rings with activating and deactivating groups, showing preferred substitution positions (ortho, meta, para).

 

3. Friedel-Crafts Alkylation Mechanism:

   - Illustrate the generation of the carbocation (electrophile) and its attack on the benzene ring, forming an alkyl-substituted benzene.

 

[Include diagrams illustrating each mechanism with clear steps and the resulting products for each type of EAS reaction.]

 

 6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Predict the product of the nitration of toluene.

 

- Solution:

  1. Identify the Substituent on Benzene: Toluene has a methyl group, which is an activating, ortho/para-directing group.

  2. Apply Directive Influence: The nitro group (\(\text{NO}_2\)) will add to the ortho and para positions relative to the methyl group.

  3. Final Product: The major products are ortho-nitrotoluene and para-nitrotoluene.

 

Example Problem 2: Explain the product formed when benzene reacts with bromine in the presence of \(\text{FeBr}_3\).

 

- Solution:

  1. Reaction Type: This is an electrophilic aromatic substitution reaction (halogenation).

  2. Catalyst Role: \(\text{FeBr}_3\) acts as a Lewis acid, creating a bromonium ion (\(\text{Br}^+\)) that acts as the electrophile.

  3. Final Product: Bromobenzene is formed as the bromine substitutes for a hydrogen on the benzene ring.

7. Common Tricks, Shortcuts, and Solving Techniques

- Using Substituent Effects to Predict Positions: Remember that electron-donating groups favor ortho/para substitution, while electron-withdrawing groups favor meta substitution.

- Identifying the Type of EAS Reaction: Knowing the reagents and catalysts helps quickly identify the type of EAS reaction (e.g., \(\text{HNO}_3/\text{H}_2\text{SO}_4\) for nitration, \(\text{Cl}_2/\text{FeCl}_3\) for chlorination).

8. Patterns in JEE Questions

JEE Advanced questions on aromatic hydrocarbons and EAS may involve:

- Predicting the major product of an EAS reaction given a substituted benzene.

- Applying knowledge of directing groups to determine ortho, meta, or para positioning.

- Understanding the mechanism and role of catalysts in different substitution reactions.

9. Tips to Avoid Common Mistakes

- Ignoring Directive Influence of Substituents: Always consider the directive effects of any substituents on the benzene ring when predicting products.

- Incorrect Catalyst Use: Ensure that the correct catalyst (e.g., \(\text{FeCl}_3\), \(\text{AlCl}_3\)) is associated with each specific reaction type in EAS.

10. Key Points to Remember for Revision

- Types of EAS Reactions: Nitration, halogenation, sulfonation, Friedel-Crafts alkylation, and acylation.

- Directive Effects of Substituents: Activating groups direct to ortho/para positions, while deactivating groups favor meta substitution.

- Role of Catalysts: Lewis acids like \(\text{AlCl}_3\) and \(\text{FeCl}_3\) enhance electrophile formation in EAS reactions.

11. Real-World Applications and Cross-Chapter Links

- Synthesis of Dyes and Pharmaceuticals: EAS reactions are widely used in the synthesis of dyes, pharmaceuticals, and aromatic compounds in organic chemistry.

- Environmental Impact of Aromatics: Many aromatic compounds, such as polychlorinated benzenes, have environmental implications and are studied in environmental chemistry.

- Cross-Concept Connections: Understanding EAS is fundamental for advanced organic synthesis and is applied across topics like reactivity, resonance effects, and organic mechanisms.

Questions

Q 1. Bond angle in alkenes is equal to;

(a) \(120^{\circ}\);

(b) \(109^{\circ} 28\);

(c) \(180^{\circ}\);

(d) \(60^{\circ}\);

Combustion properties of hydrocarbons


Combustion Properties of Hydrocarbons

 1. Definition and Core Explanation

Combustion is a chemical reaction in which hydrocarbons react with oxygen to release energy in the form of heat and light. This exothermic reaction involves the breaking of carbon-hydrogen and carbon-carbon bonds in the hydrocarbon and the formation of new bonds in carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). The complete combustion of hydrocarbons produces carbon dioxide and water as the sole products, while incomplete combustion, due to limited oxygen, produces carbon monoxide (\(\text{CO}\)) or carbon (soot) alongside \(\text{H}_2\text{O}\).

 

The combustion of hydrocarbons is fundamental in energy generation, such as in engines, power plants, and household heating systems. The energy released during combustion depends on the molecular structure of the hydrocarbon, with larger hydrocarbons releasing more energy than smaller ones.

 

Example: The complete combustion of methane (\(\text{CH}_4\)) is represented as:

\[

\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + \text{Energy}

\]

2. Types of Combustion in Hydrocarbons

- Complete Combustion:

  - Occurs in the presence of sufficient oxygen, resulting in carbon dioxide and water.

  - Yields maximum energy and minimal pollutants.

  - Example: Ethane combusts completely as follows:

    \[

    \text{C}_2\text{H}_6 + 3.5\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{Energy}

    \]

 

- Incomplete Combustion:

  - Occurs when oxygen is limited, leading to carbon monoxide and/or elemental carbon (soot) as products.

  - Produces less energy and can release harmful byproducts.

  - Example: Incomplete combustion of propane can produce carbon monoxide:

    \[

    \text{C}_3\text{H}_8 + 2.5\text{O}_2 \rightarrow 3\text{CO} + 4\text{H}_2\text{O} + \text{Energy}

    \]

3. Key Terms and Concepts

- Complete Combustion: A combustion process with sufficient oxygen, producing \(\text{CO}_2\) and \(\text{H}_2\text{O}\) only.

- Incomplete Combustion: A combustion process with limited oxygen, producing \(\text{CO}\) or carbon as byproducts.

- Energy Release (Exothermic): Combustion reactions are exothermic, meaning they release heat.

- Air-to-Fuel Ratio: The amount of air needed to achieve complete combustion. An adequate air-to-fuel ratio ensures efficient combustion.

4. Important Rules, Theorems, and Principles

- Stoichiometry of Combustion: The amount of oxygen required for complete combustion can be calculated based on the molecular formula of the hydrocarbon. For \(\text{C}_x\text{H}_y\), the complete combustion reaction can be written as:

  \[

  \text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}

  \]

- Heat of Combustion: The energy released when a substance undergoes complete combustion. Larger hydrocarbons release more energy due to more C-H and C-C bonds being broken and reformed.

- Environmental Impact: Incomplete combustion releases pollutants like \(\text{CO}\) and soot, impacting air quality and health.

5. Illustrative Diagrams and Visuals

 

1. Complete Combustion of Methane:

   - Show methane and oxygen molecules as reactants and \(\text{CO}_2\) and \(\text{H}_2\text{O}\) as products, with energy release indicated.

 

2. Incomplete Combustion Pathways:

   - Diagram showing propane reacting with limited oxygen to form carbon monoxide and water.

 

3. Air-to-Fuel Ratio Chart:

   - Visual showing the required air-to-fuel ratio for complete combustion of various hydrocarbons.

 

[Include detailed diagrams to illustrate each pathway, the molecular breakdown, and the production of energy and byproducts.]

6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Calculate the amount of oxygen needed for the complete combustion of 10 grams of propane (\(\text{C}_3\text{H}_8\)).

 

- Solution:

  1. Molecular Weight of Propane: \(\text{C}_3\text{H}_8 = 44 \, \text{g/mol}\).

  2. Moles of Propane: \(10 \, \text{g} / 44 \, \text{g/mol} = 0.227 \, \text{mol}\).

  3. Stoichiometric Ratio: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\).

  4. Oxygen Required: \(0.227 \times 5 = 1.135 \, \text{mol of } \text{O}_2\).

  5. Mass of Oxygen: \(1.135 \times 32 = 36.32 \, \text{g}\).

 

Example Problem 2: What are the products of incomplete combustion of butane (\(\text{C}_4\text{H}_{10}\))?

 

- Solution:

  1. Identify Limited Oxygen Scenario: With limited oxygen, incomplete combustion can occur.

  2. Possible Products: Carbon monoxide (\(\text{CO}\)), carbon (soot), and water (\(\text{H}_2\text{O}\)).

  - Reaction: \(\text{C}_4\text{H}_{10} + 4.5\text{O}_2 \rightarrow 4\text{CO} + 5\text{H}_2\text{O}\).

7. Common Tricks, Shortcuts, and Solving Techniques

- Quick Oxygen Calculation: Use the general formula \(\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2\) for complete combustion to find oxygen requirements.

- Identifying Incomplete Combustion: Look for limited oxygen availability in the problem setup to predict incomplete combustion products like \(\text{CO}\) and soot.

8. Patterns in JEE Questions

JEE Advanced questions on combustion properties may include:

- Calculating oxygen required for combustion or energy released.

- Differentiating between complete and incomplete combustion products.

- Understanding environmental impacts and implications of combustion reactions.

9. Tips to Avoid Common Mistakes

- Inaccurate Oxygen Requirements: Ensure correct stoichiometry when calculating oxygen requirements for complete combustion.

- Misidentifying Incomplete Combustion Products: Incomplete combustion produces \(\text{CO}\) or soot, while complete combustion only yields \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

10. Key Points to Remember for Revision

- Complete vs. Incomplete Combustion: Complete combustion produces \(\text{CO}_2\) and \(\text{H}_2\text{O}\), while incomplete combustion produces \(\text{CO}\) and/or soot.

- Energy Release: Combustion reactions are exothermic, with energy output depending on hydrocarbon size.

- Environmental Considerations: Incomplete combustion emits pollutants, impacting air quality and health.

 

 11. Real-World Applications and Cross-Chapter Links

- Energy Production: Combustion of hydrocarbons is the basis for fuel usage in vehicles, heating, and power generation.

- Environmental Chemistry: Understanding the impact of pollutants from incomplete combustion is crucial for air quality and climate studies.

- Cross-Concept Connections: Combustion relates to thermochemistry, environmental science, and practical energy applications.

Questions

Q 1. The molecular formula of a compound in which double bond is present between \(\mathrm{C} \& \mathrm{C}\) :;

(a) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}+2\);

(b) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{\mathrm{n}}\);

(c) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}\);

(d) \(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}\);

Isomerism in hydrocarbons - structural and geometrical isomerism


Isomerism in Hydrocarbons - Structural and Geometrical Isomerism

1. Definition and Core Explanation

Isomerism is the phenomenon where compounds have the same molecular formula but different structures or spatial arrangements. In hydrocarbons, isomerism is primarily of two types: structural isomerism and geometrical (cis-trans) isomerism.

 

- Structural Isomerism: Hydrocarbons with the same molecular formula but different structural arrangements of atoms. This includes chain, position, functional, and tautomeric isomerism.

- Geometrical (Cis-Trans) Isomerism: A type of stereoisomerism observed in alkenes and cyclic compounds, where atoms or groups are arranged differently around a double bond or a ring structure. Cis isomers have groups on the same side, while trans isomers have groups on opposite sides.

 

Example:

- Structural Isomerism: Butane (\(\text{C}_4\text{H}_{10}\)) has two isomers: n-butane and isobutane.

- Geometrical Isomerism: In 2-butene, the methyl groups are on the same side in the cis isomer and on opposite sides in the trans isomer.

 

 2. Types of Structural Isomerism

- Chain Isomerism:

  - Different carbon skeletons (linear vs. branched chains) for the same molecular formula.

  - Example: Butane (\(\text{C}_4\text{H}_{10}\)) has n-butane (straight chain) and isobutane (branched chain) as chain isomers.

 

- Position Isomerism:

  - Same carbon skeleton and functional group but differing positions of the functional group.

  - Example: In 1-butene and 2-butene, the double bond is positioned differently within the carbon chain.

 

- Functional Isomerism:

  - Different functional groups for the same molecular formula.

  - Example: Ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and dimethyl ether (\(\text{CH}_3\text{OCH}_3\)) are functional isomers with the same molecular formula (\(\text{C}_2\text{H}_6\text{O}\)).

 

- Tautomeric Isomerism:

  - Isomers that can interconvert by the movement of a hydrogen atom and a double bond.

  - Example: Keto-enol tautomerism, such as acetone and its enol form.

3. Types of Geometrical (Cis-Trans) Isomerism

- Cis-Trans Isomerism in Alkenes:

  - Arises in alkenes where each carbon of the double bond has different groups attached. The double bond restricts rotation, leading to cis (same side) and trans (opposite side) forms.

  - Example: In 2-butene, the cis isomer has both methyl groups on the same side, while the trans isomer has them on opposite sides.

 

- Cis-Trans Isomerism in Cyclic Compounds:

  - Cyclic structures also restrict rotation, allowing cis-trans isomerism.

  - Example: In 1,2-dimethylcyclopropane, the cis isomer has both methyl groups on the same side of the ring, while the trans isomer has them on opposite sides.

4. Important Rules, Theorems, and Principles

- Double Bond Restriction: Geometrical isomerism in alkenes is possible because the double bond restricts rotation, making it possible for groups to be on the same (cis) or opposite (trans) sides.

- E/Z Nomenclature: For complex alkenes with different substituents, the E/Z system (based on Cahn-Ingold-Prelog priority rules) is used instead of cis/trans to denote the isomerism.

5. Illustrative Diagrams and Visuals

 

1. Structural Isomers of Butane:

   - Show n-butane and isobutane with different arrangements of carbon atoms.

 

2. Cis-Trans Isomerism in 2-Butene:

   - Illustrate the cis and trans forms of 2-butene, showing the placement of methyl groups.

 

3. Functional Isomers of \(\text{C}_2\text{H}_6\text{O}\):

   - Diagram showing ethanol and dimethyl ether, highlighting the difference in functional groups.

 

[Include diagrams that clearly illustrate chain, position, functional, and cis-trans isomerism in various hydrocarbon examples.]

6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: How many structural isomers are possible for \(\text{C}_5\text{H}_{12}\)?

 

- Solution:

  1. Identify Chain Isomers: \(\text{C}_5\text{H}_{12}\) has three chain isomers: n-pentane, isopentane, and neopentane.

  - Answer: There are three structural isomers of \(\text{C}_5\text{H}_{12}\).

 

Example Problem 2: Draw the cis and trans isomers of 2-pentene and indicate which isomer has a higher boiling point.

 

- Solution:

  1. Identify the Structure: 2-pentene has a double bond between the second and third carbons.

  2. Draw Isomers: In the cis isomer, the two larger alkyl groups are on the same side, while in the trans isomer, they are on opposite sides.

  3. Boiling Point Comparison: The cis isomer generally has a higher boiling point due to increased polarity.

  - Answer: The cis isomer of 2-pentene has a higher boiling point.

7. Common Tricks, Shortcuts, and Solving Techniques

- Counting Isomers: To count isomers, start with chain isomers, then look for position isomers, and finally check for functional group variations.

- Recognizing Cis-Trans Potential: Cis-trans isomerism is possible in alkenes where each carbon in the double bond has two different groups.

8. Patterns in JEE Questions

JEE questions on isomerism often include:

- Counting and identifying different types of isomers for a given molecular formula.

- Predicting and drawing the structures of cis and trans isomers.

- Comparing physical properties like boiling points between cis and trans isomers.

9. Tips to Avoid Common Mistakes

- Confusing Chain and Position Isomers: Remember that chain isomers have different carbon skeletons, while position isomers have the same skeleton but differ in the placement of functional groups.

- Ignoring Double Bond Restriction: Geometrical isomerism is only possible if the double bond carbons have two different substituents each.

10. Key Points to Remember for Revision

- Types of Structural Isomerism: Chain, position, functional, and tautomeric.

- Cis-Trans Isomerism: Restricted rotation around double bonds and rings leads to cis (same side) and trans (opposite side) isomers.

- E/Z Nomenclature: Used for complex alkenes when simple cis-trans notation is insufficient.

 11. Real-World Applications and Cross-Chapter Links

- Pharmaceuticals and Isomerism: Different isomers can have distinct biological activities, making isomerism crucial in drug design.

- Materials Science: Geometric isomers can influence properties such as melting and boiling points, impacting material applications.

- Cross-Concept Connections: Understanding isomerism aids in grasping stereochemistry, reactivity, and molecular structure in organic chemistry.

Questions

Q 1. IUPAC name of the following compound is ;

(a) 5-chloroheptene;

(b) 4-chloropent-1-ene;

(c) 5-chloropent-3-ene;

(d) 5-chlorohex-2-ene;

General concepts of hydrocarbons, their reactions, and properties


General Concepts of Hydrocarbons, Their Reactions, and Properties

1. Definition and Core Explanation

Hydrocarbons are organic compounds composed solely of carbon and hydrogen atoms. They are the fundamental building blocks of organic chemistry and can be classified into three main types: alkanes, alkenes, and alkynes, based on the types of bonds between the carbon atoms.

 

- Alkanes: Saturated hydrocarbons with only single bonds (\(\text{C-C}\)). They have the general formula \(\text{C}_n\text{H}_{2n+2}\).

- Alkenes: Unsaturated hydrocarbons with at least one carbon-carbon double bond (\(\text{C=C}\)), with the general formula \(\text{C}_n\text{H}_{2n}\).

- Alkynes: Unsaturated hydrocarbons with at least one carbon-carbon triple bond (\(\text{C}\equiv\text{C}\)), having the general formula \(\text{C}_n\text{H}_{2n-2}\).

 

Hydrocarbons are nonpolar and generally insoluble in water but soluble in nonpolar solvents. Their reactivity varies based on saturation; alkanes are relatively stable, while alkenes and alkynes are more reactive due to the presence of \(\pi\)-bonds that are more accessible to electrophiles.

2. Fundamental Properties of Hydrocarbons

- Physical Properties:

  - Boiling and Melting Points: Increase with molecular size and mass due to greater van der Waals forces.

  - Solubility: Insoluble in water but soluble in nonpolar solvents.

  - Density: Generally less dense than water; hydrocarbons float on the water surface.

 

- Chemical Properties:

  - Alkanes: Undergo combustion and substitution reactions, such as halogenation.

  - Alkenes: Undergo addition reactions like hydrogenation, halogenation, and hydrohalogenation.

  - Alkynes: Also undergo addition reactions and can form acetylides with strong bases due to the acidity of terminal alkynes.

3. Key Terms and Concepts

- Saturation: Refers to hydrocarbons with only single bonds, like alkanes. Unsaturated hydrocarbons contain double or triple bonds, like alkenes and alkynes.

- Hydrophobicity: Hydrocarbons are hydrophobic (water-repelling) due to their nonpolar nature.

- Electrophilic Addition: A reaction where an electrophile adds across the double or triple bonds in alkenes and alkynes.

- Substitution: A reaction where one atom or group in a molecule is replaced by another, common in alkanes.

4. Important Rules, Theorems, and Principles

- Markovnikov’s Rule: In unsymmetrical alkenes, the electrophile adds to the carbon with more hydrogen atoms, producing the most stable intermediate.

- Carbocation Stability: Plays a significant role in the reactivity and outcome of addition reactions, especially for alkenes and alkynes.

- Electronegativity and Reactivity: Alkynes (sp-hybridized carbons) are more electronegative than alkenes (sp\(^2\)-hybridized), which affects their reactivity and acidity.

5. Illustrative Diagrams and Visuals

 

1. Classification of Hydrocarbons:

   - Diagram showing the structures of alkanes, alkenes, and alkynes with examples (e.g., methane, ethene, ethyne).

 

2. General Reactions:

   - Show an alkane undergoing halogenation, an alkene undergoing hydrogenation, and an alkyne undergoing hydrohalogenation.

 

3. Markovnikov’s Rule:

   - Diagram of a hydrohalogenation reaction with an unsymmetrical alkene, illustrating the addition pattern according to Markovnikov’s Rule.

 

[Include diagrams to illustrate structural differences among alkanes, alkenes, and alkynes, along with sample reactions.]

6. Sample Problems and Step-by-Step Solutions

 

Example Problem 1: Identify the products of the complete combustion of hexane (\(\text{C}_6\text{H}_{14}\)).

 

- Solution:

  1. Write the Combustion Reaction:

     \[

     \text{C}_6\text{H}_{14} + 9.5\text{O}_2 \rightarrow 6\text{CO}_2 + 7\text{H}_2\text{O}

     \]

  2. Balance the Equation: Ensure all elements are balanced, yielding carbon dioxide and water.

  - Answer: Complete combustion produces \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

 

Example Problem 2: Predict the major product of the reaction between propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)) and HCl.

 

- Solution:

  1. Apply Markovnikov’s Rule: The H from HCl adds to the carbon with more hydrogens, and Cl attaches to the other carbon.

  2. Final Product: 2-chloropropane.

 

7. Common Tricks, Shortcuts, and Solving Techniques

- Using General Formulas: For quick identification, remember that alkanes follow \(\text{C}_n\text{H}_{2n+2}\), alkenes follow \(\text{C}_n\text{H}_{2n}\), and alkynes follow \(\text{C}_n\text{H}_{2n-2}\).

- Markovnikov’s Rule for Quick Prediction: When adding HX to an unsymmetrical alkene, apply Markovnikov’s Rule to find the major product.

 

 8. Patterns in JEE Questions

JEE questions on hydrocarbons often involve:

- Predicting products of combustion and substitution reactions.

- Applying Markovnikov’s Rule and understanding regioselectivity in addition reactions.

- Comparing physical properties and reactivity among alkanes, alkenes, and alkynes.

 9. Tips to Avoid Common Mistakes

- Incorrect Application of Markovnikov’s Rule: Ensure you apply the rule correctly in addition reactions involving unsymmetrical alkenes.

- Balancing Combustion Reactions: Remember to balance both reactants and products, especially when calculating oxygen requirements for combustion.

 

 10. Key Points to Remember for Revision

- Classifications of Hydrocarbons: Alkanes (saturated), Alkenes (unsaturated with double bonds), Alkynes (unsaturated with triple bonds).

- Reactivity Differences: Alkanes are stable, alkenes undergo addition reactions, and alkynes are reactive and can form acetylides.

- Key Reaction Types: Combustion (all hydrocarbons), substitution (alkanes), addition (alkenes and alkynes).

 

 11. Real-World Applications and Cross-Chapter Links

- Fuels and Energy Sources: Hydrocarbons are the primary components of fuels like gasoline, diesel, and natural gas, making their combustion properties essential for energy.

- Chemical Synthesis: Alkenes and alkynes are important feedstocks in chemical industries for synthesizing plastics, pharmaceuticals, and other organic compounds.

- Cross-Concept Connections: Understanding hydrocarbons is fundamental for organic synthesis, thermochemistry, and environmental studies.

Questions

Q 1. IUPAC name of the following compound will be ;

(a) 3-Ethyl-2-hexene;

(b) 3-Propyl-2-hexene;

(c) 3-Propyl-3-hexene;

(d) 4-Ethyl-4-hexene;

Q 2. Which of the following represents the correct IUPAC name of the compound \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Cl}\) ?;

(a) Allyl chloride;

(b) 1-chloro -3-propene;

(c) 3-chloro-1-propene;

(d) Vinyl chloride;

Q 3. Which of the following reactions of methane is incomplete combustion?;

(a) \(2 \mathrm{CH}_{4}+\mathrm{O}_{2} \xrightarrow{\mathrm{Cu} / 523 \mathrm{~K} / 100 \mathrm{~atm}} 2 \mathrm{CH}_{3} \mathrm{OH}\);

(b) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \xrightarrow{\mathrm{Mo}_{2} \mathrm{O}_{3}} \mathrm{HCHO}+\mathrm{H}_{2} \mathrm{O}\);

(c) \(\mathrm{CH}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\) l \()\);

(d) \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\);

Q 4. Which one of the following has highest boiling point?;

(a) n-Octane;

(b) 2,2 dimethyl pentane;

(c) Iso-octan;

(d) All have equal values;

Q 5. Which one of the following has the least boiling point?;

(a) 2,2-dimethylpropane;

(b) n-butane;

(c) 2-methylpropane;

(d) n-pentane;

Q 6. Sodium salts of carboxylic acids on heating with soda lime give alkanes containing \(\qquad\) than the carboxylic acid;

(a) one carbon more;

(b) one carbon less;

(c) two carbon less;

(d) Either (a) or (b);

Q 7. Pure methane can be produced by;

(a) Wurtz reaction;

(b) Kolbe's electrolytic method;

(c) Soda-lime decarboxylation;

(d) Reduction with \(\mathrm{H}_{2}\);

Q 8. The reaction, \(\mathrm{CH}_{3}-\mathrm{Br}+2 \mathrm{Na}+\mathrm{Br}-\mathrm{CH}_{3} \rightarrow\) the product, is called;

(a) Wurtz reaction;

(b) Perkin's reaction;

(c) Aldol condensation;

(d) Levit reaction;

Q 9. Which one of the following cannot be prepared by Wurtz reaction?;

(a) \(\mathrm{CH}_{4}\);

(b) \(\mathrm{C}_{2} \mathrm{H}_{6}\);

(c) \(\mathrm{C}_{3} \mathrm{H}_{8}\);

(d) \(\mathrm{C}_{4} \mathrm{H}_{10}\);

Q 10. The number of \(4^{\circ}\) carbon atoms in 2,2,4,4-tetramethyl pentane is -;

(a) 1;

(b) 2;

(c) 3;

(d) 4;

Topics

Free radical substitution in alkanes

Monochlorination of alkanes

Photochemical halogenation - example of free radical substitution

Free radical bromination of alkanes

Complete combustion of hydrocarbons

IUPAC nomenclature of hydrocarbons

Classification of hydrocarbons - alkanes, alkenes, alkynes

Preparation of alkanes by Wurtz reaction

Markovnikov's rule in alkene addition reactions

Electrophilic addition in alkenes

Aromaticity and stability of benzene

Structure and bonding in alkynes

Addition reactions of alkynes

Aromatic hydrocarbons and substitution reactions

Combustion properties of hydrocarbons

Isomerism in hydrocarbons - structural and geometrical isomerism

General concepts of hydrocarbons, their reactions, and properties

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