GSEB Class 11 Chemistry Solutions Chapter 13 Hydrocarbons

 

Question 1. How do you account for the formation of ethane during chlorination of methane?
Answer: One termination step during methane chlorination involves the side reaction of combining two methyl free radicals (\( \cdot \text{CH}_{3} \)) as shown below.
\( \cdot \text{CH}_{3} + \cdot \text{CH}_{3} \rightarrow \text{H}_{3}\text{C:CH}_{3} \) or \( \text{C}_{2}\text{H}_{6} \) (ethane)
In simple words: When methane is chlorinated, some methyl radicals are formed. These radicals can join together, creating ethane.

Exam Tip: Remember to identify the type of reaction (termination step in this case) and clearly show the reactants and products, including the intermediate free radicals.

 

Question 2. Write IUPAC names of the following compounds:
(a) \( \text{CH}_{3}\text{CH} = \text{C}(\text{CH}_{3})_{2} \)
(b) \( \text{CH}_{2} = \text{CH} – \text{C} = \text{C} – \text{CH}_{3} \)
(c) \( \text{CH}_{2}=\text{CH}-\text{CH}=\text{CH}_{2} \)
(d) \( \text{C}_{6}\text{H}_{5}-\text{CH}_{2}-\text{CH}_{2}-\text{CH}=\text{CH}_{2} \)
(e) \( \text{C}_{6}\text{H}_{4}(\text{OH})(\text{CH}_{3}) \) (ortho-methylphenol isomer)
(f) \( \text{CH}_{3}(\text{CH}_{2})_{4}\text{CH}(\text{CH}_{2})_{3}\text{CH}_{3} \)
\( \quad \quad \quad \quad |\)
\( \quad \quad \quad \quad \text{CH}_{2}-\text{CH}(\text{CH}_{3})_{2} \)
(g) \( \text{CH}_{3}-\text{CH}=\text{CH}-\text{CH}_{2}-\text{CH}=\text{CH}-\text{CH}(\text{C}_{2}\text{H}_{5})-\text{CH}_{2}-\text{CH}=\text{CH}_{2} \)
Answer: The IUPAC names for these compounds are:
(a) 2-Methylbut-2-ene
(b) Pent-1-en-3-yne
(c) Buta-1,3-diene
(d) 4-Phenylbut-1-ene
(e) 2-Methylphenol
(f) 5-(2-Methylpropyl)decane
(g) 4-Ethyldeca-1,5,8-triene
In simple words: For each chemical formula or structure, we give its official IUPAC name, which uniquely identifies the compound.

Exam Tip: When naming complex structures, always identify the longest continuous carbon chain, assign numbers to give substituents the lowest possible numbers, and correctly identify and alphabetize all branches and functional groups.

 

Question 3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having number of double or triple bonds as indicated:
(a) \( \text{C}_{4}\text{H}_{8} \) (one double bond)
(b) \( \text{C}_{5}\text{H}_{8} \) (one triple bond)
Answer: The structural formulas and IUPAC names for all possible isomers for the given compounds are as follows:
(a) For \( \text{C}_{4}\text{H}_{8} \) (one double bond):
(i) \( \text{CH}_{2} = \text{CH} – \text{CH}_{2} – \text{CH}_{3} \) (But-1-ene)
(ii) \( \text{CH}_{3} – \text{CH} = \text{CH} – \text{CH}_{3} \) (But-2-ene)
(iii) \( \text{CH}_{2} = \text{C}(\text{CH}_{3}) – \text{CH}_{3} \) (2-Methylpropene)
(b) For \( \text{C}_{5}\text{H}_{8} \) (one triple bond):
(i) \( \text{HC} \equiv \text{C} – \text{CH}_{2} – \text{CH}_{2} – \text{CH}_{3} \) (Pent-1-yne)
(ii) \( \text{CH}_{3} – \text{C} \equiv \text{C} – \text{CH}_{2} – \text{CH}_{3} \) (Pent-2-yne)
(iii) \( \text{CH}_{3} – \text{CH}(\text{CH}_{3}) – \text{C} \equiv \text{CH} \) (3-Methylbut-1-yne)
In simple words: We find all the different ways to arrange the atoms for each compound while keeping the specified number of double or triple bonds, then draw their structures and give their unique names.

Exam Tip: To find all possible isomers, systematically move the double/triple bond position and then move alkyl substituents along the chain, ensuring you don't repeat structures.

 

Question 4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
(i) Pent-2-ene
(ii) 3, 4-Dimethylhept-3-ene
(iii) 2-Ethylbut -1-ene
(iv) 1-Phenylbut-1-ene
Answer: The IUPAC names of the products resulting from the ozonolysis of these compounds are provided below:
(i) Pent-2-ene is \( \text{CH}_{3} – \text{CH} = \text{CH} – \text{CH}_{2} – \text{CH}_{3} \)
The products are \( \text{CH}_{3}\text{CHO} \) (Ethanal) and \( \text{CH}_{3}\text{CH}_{2}\text{CHO} \) (Propanal).
(ii) 3,4-Dimethylhept-3-ene is \( \text{CH}_{3}-\text{CH}_{2}-\text{C}(\text{CH}_{3})=\text{C}(\text{CH}_{3})-\text{CH}_{2}-\text{CH}_{2}-\text{CH}_{3} \)
The products are \( \text{CH}_{3}-\text{CH}_{2}-\text{C}(=\text{O})-\text{CH}_{3} \) (Butanone) and \( \text{CH}_{3}\text{C}(=\text{O})\text{CH}_{2}\text{CH}_{2}\text{CH}_{3} \) (Pentan-2-one).
(iii) 2-Ethylbut-1-ene is \( \text{CH}_{2}=\text{C}(\text{C}_{2}\text{H}_{5})-\text{CH}_{2}-\text{CH}_{3} \)
The products are \( \text{HCHO} \) (Methanal) and \( \text{CH}_{3}\text{CH}_{2}-\text{C}(=\text{O})-\text{CH}_{2}-\text{CH}_{3} \) (Pentan-3-one).
(iv) 1-Phenylbut-1-ene is \( \text{C}_{6}\text{H}_{5}\text{CH=CHCH}_{2}\text{CH}_{3} \). On ozonolysis, it produces \( \text{C}_{6}\text{H}_{5}\text{CHO} \) (Benzaldehyde) and \( \text{CH}_{3}\text{CH}_{2}\text{CHO} \) (Propanal).
In simple words: Ozonolysis breaks the double bond in each alkene and replaces it with two carbonyl groups, forming aldehydes or ketones. We then identify these products by their IUPAC names.

Exam Tip: To predict ozonolysis products, mentally "cut" the double bond and add an oxygen atom to each carbon of the original double bond. For unsymmetrical alkenes, you will get two different products.

 

Question 5. An alkene 'A' on ozonolysis gives a mixture of ethanal and 3-Pentanone. Write structure and IUPAC name of 'A'.
Answer: When an alkene 'A' undergoes ozonolysis, it gives a mixture of ethanal (\( \text{CH}_{3}\text{CHO} \)) and 3-Pentanone (\( \text{CH}_{3}\text{CH}_{2}\text{COCH}_{2}\text{CH}_{3} \)). To determine the structure of 'A', we connect the carbonyl carbons of the products with a double bond after removing the oxygen atoms. The structure of alkene 'A' is \( \text{CH}_{3}\text{CH=C(CH}_{2}\text{CH}_{3}\text{)CH}_{2}\text{CH}_{3} \). The IUPAC name of 'A' is 3-Ethylpent-2-ene.
In simple words: We take the two products from ozonolysis, ethanal and 3-pentanone, remove their oxygen atoms, and then connect the remaining carbon atoms with a double bond to find the original alkene's structure and name.

Exam Tip: To reverse ozonolysis, draw the products side-by-side, erase the oxygen atoms, and form a double bond between the carbons that originally held the carbonyl groups. Then, apply IUPAC naming rules to the resulting alkene.

 

Question 6. An alkene 'A' contains three C – C, eight C-H bonds and one C – C \( \pi \) bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'.
Answer: Alkene 'A' has three carbon-carbon single bonds, eight carbon-hydrogen sigma bonds, and one carbon-carbon pi bond. The ozonolysis of 'A' gives two moles of an aldehyde with a molar mass of 44 u. An aldehyde with a molar mass of 44 u is ethanal (\( \text{CH}_{3}\text{CHO} \)). Since two moles of ethanal are formed, alkene 'A' must be formed by joining two \( \text{CH}_{3}\text{CH} \) groups with a double bond. Therefore, alkene 'A' is \( \text{CH}_{3}\text{CH=CHCH}_{3} \). The IUPAC name for 'A' is But-2-ene.
In simple words: By looking at the number of bonds and the molar mass of the aldehyde produced by ozonolysis, we can work backward to figure out that the original alkene was But-2-ene.

Exam Tip: Remember that two moles of the same aldehyde often imply a symmetrical alkene, where the double bond is in the middle of a symmetrical chain.

 

Question 7. Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of alkene?
Answer: If propanal (\( \text{CH}_{3}\text{CH}_{2}\text{CHO} \)) and pentan-3-one (\( \text{CH}_{3}\text{CH}_{2}\text{COCH}_{2}\text{CH}_{3} \)) are the ozonolysis products of an alkene, then the structural formula of the original alkene can be found by reversing the ozonolysis process. This involves joining the carbonyl carbons of the products with a double bond after removing the oxygen atoms. The structural formula of the alkene is \( \text{CH}_{3}\text{CH}_{2}\text{CH=C(CH}_{2}\text{CH}_{3}\text{)CH}_{2}\text{CH}_{3} \). Its IUPAC name is 4-Ethylhex-3-ene.
In simple words: To find the original alkene, we combine the propanal and pentan-3-one molecules by connecting their carbon atoms (which were part of the double bond) with a double bond.

Exam Tip: Always be careful when recombining the fragments; ensure the double bond forms between the original carbonyl carbons, and then correctly name the resulting alkene.

 

Question 8. Write chemical equations of combustion reaction of the following by hydrocarbons.
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene
Answer: The chemical equations for the complete combustion reactions of the given hydrocarbons, producing carbon dioxide and water vapor, are presented below:
(i) Butane: \( \text{C}_{4}\text{H}_{10}\text{(g)} + \frac{13}{2} \text{O}_{2}\text{(g)} \xrightarrow{\Delta} 4\text{CO}_{2}\text{(g)} + 5\text{H}_{2}\text{O(g)} \)
(ii) Pentene: \( \text{C}_{5}\text{H}_{10}\text{(g)} + \frac{15}{2} \text{O}_{2}\text{(g)} \xrightarrow{\Delta} 5\text{CO}_{2}\text{(g)} + 5\text{H}_{2}\text{O(g)} \)
(iii) Hexyne: \( \text{C}_{6}\text{H}_{10}\text{(g)} + \frac{17}{2} \text{O}_{2}\text{(g)} \xrightarrow{\Delta} 6\text{CO}_{2}\text{(g)} + 5\text{H}_{2}\text{O(g)} \)
(iv) Toluene: \( \text{C}_{7}\text{H}_{8}\text{(g)} + 9\text{O}_{2}\text{(g)} \xrightarrow{\Delta} 7\text{CO}_{2}\text{(g)} + 4\text{H}_{2}\text{O(g)} \)
In simple words: We write a balanced chemical equation for each hydrocarbon burning in oxygen, always producing carbon dioxide and water.

Exam Tip: Ensure all equations are balanced for both atoms and charge. Remember that fractional coefficients for \( \text{O}_{2} \) are acceptable and can be converted to whole numbers by multiplying the entire equation by two.

 

Question 9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.pt. and why?
Answer: Hex-2-ene has the formula \( \text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH=CHCH}_{3} \). Its cis and trans forms are depicted below:
The cis isomer will exhibit a higher boiling point because its more polar nature results in stronger intermolecular dipole-dipole interactions. This requires a greater amount of heat energy to separate the molecules. In contrast, the trans form is non-polar (or weakly polar), possessing weak induced dipole interactions, and thus has a lower boiling point.
In simple words: Cis-hex-2-ene has a higher boiling point than trans-hex-2-ene because its structure is more polar, leading to stronger attractions between molecules that need more energy to break apart. Trans-hex-2-ene is less polar, so its molecules are easier to separate.

Exam Tip: Remember that cis isomers often have higher boiling points than trans isomers due to their greater polarity and more effective dipole-dipole interactions, while trans isomers tend to be more symmetrical and less polar.

 

Question 10. Why is benzene extra ordinarily stable though it contains three double bonds?
Answer: Benzene (\( \text{C}_{6}\text{H}_{6} \)) shows remarkable stability despite having three double bonds due to the phenomenon of resonance. Benzene possesses two resonating structures, also known as Kekule structures, which represent the delocalization of electrons within the ring.
This resonance between these two canonical structures provides benzene with enhanced stability, making it less reactive than a typical alkene.
In simple words: Benzene is very stable because its electrons are spread out (delocalized) across the whole ring, not just in fixed double bonds. This "electron sharing" makes it much stronger than expected.

Exam Tip: Always mention electron delocalization and resonance structures when explaining benzene's extraordinary stability. This is a key concept in aromaticity.

 

Question 11. What are the necessary conditions for any system to be aromatic?
Answer: For a system to be considered aromatic, it must satisfy the following essential conditions:

• The molecule needs to be planar.
• It must be cyclic, with alternating single and double bonds, and the cyclic pi-electron cloud should completely surround all carbon atoms within the ring.
• It must contain \( (4n + 2)\pi \) electrons, where \( n = 0, 1, 2, 3 \), and so on (Hückel's Rule).

A molecule that does not meet one or more of these conditions is considered non-aromatic.
In simple words: To be aromatic, a molecule must be flat, shaped like a ring with alternating single and double bonds, and have a special number of pi electrons following Hückel's Rule.

Exam Tip: When evaluating aromaticity, always check for planarity, cyclicity, continuous conjugation, and the \( (4n+2)\pi \) electron count. Missing even one condition means the compound is not aromatic.

 

Question 12. Explain why the following system are not aromatic?
(i) Answer: All three structures lack a continuous cyclic cloud of \( \pi \)-electrons that obeys Hückel's rule, which requires \( (4n+2)\pi \) electrons. This makes them non-aromatic.
In simple words: These molecules are not aromatic because their pi electrons aren't fully delocalized in a continuous ring and they don't follow Hückel's Rule for the correct number of pi electrons.

Exam Tip: For cyclopentadiene and cycloheptatriene, the \( \text{sp}^{3} \) hybridized \( \text{CH}_{2} \) group breaks the continuous overlap of p-orbitals. Cyclooctatetraene has 8 \( \pi \)-electrons, which is a \( 4n \) system, making it anti-aromatic if planar, but it avoids anti-aromaticity by adopting a non-planar (tub-like) conformation.

 

Question 13. How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m-nitrochlorobenzene
(iii) p-nitrotoluene
(iv) acetophenone?
Answer: The conversions of benzene into the specified products are performed through the following reaction pathways:
(i) Benzene into p-nitrobromobenzene:

(ii) Benzene into m-nitrochlorobenzene:

(iii) Benzene into p-nitrotoluene:

(iv) Benzene into acetophenone:
In simple words: These are step-by-step reaction pathways showing how benzene can be chemically changed into different products by adding specific reagents and conditions.

Exam Tip: For synthesis problems, always work backward from the target molecule to identify the necessary reactions. Pay attention to regioselectivity (ortho, meta, para) and the activating/deactivating effects of substituents.

 

Question 14. In the alkane \( \text{H}_{3}\text{C} – \text{CH}_{2} – \text{C} – (\text{CH}_{3})_{2} – \text{CH}_{2} – \text{CH} (\text{CH}_{3})_{2} \), identify \( 1^\circ, 2^\circ, 3^\circ \) carbon atoms and give the total number of H atoms bonded to each one of these.
Answer: To identify the primary (\(1^\circ\)), secondary (\(2^\circ\)), and tertiary (\(3^\circ\)) carbon atoms and count their bonded hydrogen atoms in the alkane \( \text{H}_{3}\text{C} – \text{CH}_{2} – \text{C} – (\text{CH}_{3})_{2} – \text{CH}_{2} – \text{CH} (\text{CH}_{3})_{2} \), we analyze its structure:

      CH3(1°)
      |
CH3(1°)-CH2(2°)-C(4°)-CH2(2°)-CH(3°)-CH3(1°)
                  |           |
                  CH3(1°)     CH3(1°)

Based on this analysis, the distribution of hydrogen atoms by carbon degree is as follows:

• 15 hydrogen atoms are bonded to 1° carbons (5 x \( \text{CH}_{3} \) groups).
• 4 hydrogen atoms are bonded to 2° carbons (2 x \( \text{CH}_{2} \) groups).
• 1 hydrogen atom is bonded to a 3° carbon (1 x \( \text{CH} \) group).

There are no hydrogen atoms bonded to the 4° carbon.
In simple words: We look at each carbon in the molecule and count how many other carbons it's attached to (its degree: primary, secondary, tertiary, or quaternary). Then we count how many hydrogen atoms are on each type of carbon.

Exam Tip: A carbon's degree is determined by the number of other carbon atoms it is directly bonded to. A \(1^\circ\) carbon is bonded to one other carbon, \(2^\circ\) to two, \(3^\circ\) to three, and \(4^\circ\) to four. Hydrogen count directly depends on the carbon's degree: \(1^\circ\) has 3H, \(2^\circ\) has 2H, \(3^\circ\) has 1H, and \(4^\circ\) has 0H.

 

Question 15. What effect does branching of an alkane chain has on its boiling point?
Answer: Increased branching in an alkane chain generally leads to a lower boiling point. This occurs because branching reduces the surface area available for effective van der Waals forces (intermolecular attractions) between molecules, making them easier to separate with less energy.
In simple words: When an alkane chain has more branches, its boiling point becomes lower. This is because branching makes the molecules more compact, reducing how well they can stick together, so they need less heat to boil.

Exam Tip: Remember that straight-chain alkanes have a larger surface area for intermolecular contact, leading to stronger van der Waals forces and higher boiling points compared to their branched isomers.

 

Question 16. Addition of HBr to propene yields 2-bromopropane, while in presence of benzoyl peroxide the same reaction yields 1-bromopropane. Explain and give mechanism.
Answer: Adding HBr to propene typically produces 2-bromopropane following Markovnikov's Rule. However, in the presence of benzoyl peroxide, the reaction shifts to yield 1-bromopropane, which follows the anti-Markovnikov rule. Let's explore why this happens and examine the mechanisms involved.
**Markovnikov Rule Explanation:** The addition of HBr to propene (an unsymmetrical alkene) normally follows Markovnikov's Rule. This rule states that the negative part of the adding molecule (the bromine atom from HBr) attaches to the carbon atom in the double bond that possesses fewer hydrogen atoms. Consequently, 2-bromopropane is the expected and major product in the absence of peroxides.
**Modern Mechanism: Electrophilic Addition.**
Step (i): It follows that proton \( \text{H}^{+} \) (an electrophile) from HBr gets attached to \( \text{C}_{1} \) [due to \( +1 \) effect exerted by \( –\text{CH}_{3} \) group, the double bond between \( \text{C}_{1} \) and \( \text{C}_{2} \) undergoes electromeric effect]
\( \text{H}_{3}\text{C} – \text{CH} = \text{CH}_{2} + \text{H}^{+}\text{Br}^{-} \rightarrow \text{H}_{3}\text{C} – \text{CH}^{+}-\text{CH}_{3} + \text{Br}^{-} \) (Carbocation)
Step (ii): The intermediate carbocation is attached by \( \text{Br}^{-} \) to form 2-bromopropane.
\( \text{CH}_{3} – \text{CH}^{+} – \text{CH}_{3} + \text{Br}^{-} \rightarrow \text{CH}_{3} – \text{CH}(\text{Br}) – \text{CH}_{3} \) (2-Bromopropane)
**Anti-Markovnikov Rule / Peroxide Effect (Kharash Effect) Explanation:** The addition of HBr to propene when peroxides are present, or under light, occurs contrary to Markovnikov's rule. This specific behavior, known as the peroxide effect or Kharash effect (or anti-Markovnikov addition), happens exclusively with HBr and not with HCl or HI. In this case, 1-bromopropane is formed as the major product.
**Mechanism:** Peroxide effect proceeds via free radical mechanism as given below:
\( (\text{C}_{6}\text{H}_{5}\text{COO})_{2} \xrightarrow{\text{Heat}} 2\text{C}_{6}\text{H}_{5}\text{COO}\cdot \)
\( \text{C}_{6}\text{H}_{5}\text{COO}\cdot \rightarrow \text{C}_{6}\text{H}_{5}\cdot + \text{CO}_{2} \)
\( \text{C}_{6}\text{H}_{5}\cdot + \text{HBr} \rightarrow \text{C}_{6}\text{H}_{6} + \text{Br}\cdot \)
\( \text{CH}_{3} – \text{CH} = \text{CH}_{2} + \text{Br}\cdot \rightarrow \text{CH}_{3} – \cdot \text{CH} – \text{CH}_{2}\text{Br} \) (More stable secondary radical)
\( \text{CH}_{3} – \cdot \text{CH} – \text{CH}_{2}\text{Br} + \text{HBr} \rightarrow \text{CH}_{3} – \text{CH}_{2} – \text{CH}_{2}\text{Br} + \text{Br}\cdot \) (1-Bromopropane)
In simple words: Usually, HBr adds to propene following Markovnikov's Rule, meaning the bromine goes to the carbon with fewer hydrogens. But if you add peroxide, it flips, and the bromine goes to the carbon with more hydrogens (anti-Markovnikov's Rule) because the reaction pathway changes to a free radical mechanism.

Exam Tip: Clearly distinguish between Markovnikov's (ionic mechanism, carbocation intermediate) and Anti-Markovnikov's (free radical mechanism, free radical intermediate) additions. The presence of peroxides is the key indicator for the latter, and it is specific to HBr.

 

Question 17. Write down the products of ozonolysis 1, 2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?
Answer: Ozonolysis of 1,2-dimethylbenzene (o-xylene) results in the formation of the following products, which helps to support the Kekule structure for benzene:
The formation of all three products cannot be explained by any single Kekule structure alone. This observation confirms that benzene is a resonance hybrid, oscillating between its two resonating structures, which accounts for the diverse ozonolysis products.
In simple words: When we break down o-xylene, we get three different products. This shows that benzene doesn't just have one fixed structure but is a mix of two different forms (resonance hybrids), which is what Kekule suggested.

Exam Tip: Remember that ozonolysis helps prove the delocalized nature of benzene. If benzene had fixed double bonds, ozonolysis of its substituted derivatives would yield fewer product variations than are actually observed.

 

Question 18. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for the behaviour.
Answer: When arranging benzene, n-hexane, and ethyne by their decreasing order of acidic behavior, we find the sequence is ethyne > benzene > n-hexane. This means ethyne is the most acidic, followed by benzene, and n-hexane is the least acidic. The primary reason for this behavior is the s-character of the carbon-hydrogen bond.
Ethyne's carbon atoms are sp hybridized, giving its C–H bonds 50% s-character. Benzene's carbon atoms are sp\(^{2}\) hybridized, resulting in 33% s-character. N-hexane's carbon atoms are sp\(^{3}\) hybridized, which provides the minimum s-character at 25%. A higher s-character implies that the s-orbital electrons are closer to the nucleus, leading to a more electronegative carbon and making the C–H bond more acidic as the hydrogen is more easily released as a proton. Consequently, ethyne is the most acidic due to its highest s-character.
In simple words: The acidity of these compounds depends on how much "s-character" their carbon-hydrogen bonds have. Ethyne has the most s-character (50%), making it the most acidic, then benzene (33%), and n-hexane has the least (25%), making it the least acidic.

Exam Tip: Correlate acidity with hybridization: sp hybridized carbons are more acidic than sp\(^{2}\), which are more acidic than sp\(^{3}\). This is a direct consequence of the increasing s-character pulling electrons closer to the nucleus, stabilizing the conjugate base.

 

Question 19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Answer: Benzene readily undergoes electrophilic substitution reactions but faces difficulty with nucleophilic substitutions. This is because benzene possesses a delocalized sextet of \( \pi \)-electrons (six \( \pi \)-electrons), making it an electron-rich species. Essentially, benzene acts as a Lewis base. Its electron-rich nature means it is easily attacked by electrophiles (reagents that are deficient in electrons). Conversely, benzene's electron density repels nucleophiles (electron-rich reagents), making nucleophilic attack much harder.
In simple words: Benzene is full of electrons, so it easily attracts electron-hungry things (electrophiles) for reactions. But because it has so many electrons, it pushes away other electron-rich things (nucleophiles), making those reactions harder.

Exam Tip: The electron-rich \( \pi \)-cloud of benzene is the key to understanding its reactivity. Electrophiles seek electron density, making electrophilic substitution characteristic of aromatic compounds.

 

Question 20. How would you convert the following compounds to benzene?
(i) Ethyne
(ii) Ethene
(iii) Hexane
Answer: Here's how you can convert the given compounds into benzene through various chemical reactions:
(i) Conversion of ethyne (\( \text{C}_{2}\text{H}_{2} \)) to benzene (\( \text{C}_{6}\text{H}_{6} \)):
\( 3 \text{HC} \equiv \text{CH} \xrightarrow[\text{873K}]{\text{Red hot Fe tube}} \text{Benzene} \)
(ii) Conversion of ethene to benzene:
\( \text{CH}_{2}=\text{CH}_{2} + \text{Br}_{2} \xrightarrow{} \text{CH}_{2}\text{Br}-\text{CH}_{2}\text{Br} \)
\( \text{CH}_{2}\text{Br}-\text{CH}_{2}\text{Br} + \text{NaNH}_{2} \xrightarrow{\text{Alc. KOH}} \text{HC}\equiv\text{CH} \)
\( \text{HC}\equiv\text{CH} \xrightarrow[\text{873K}]{\text{Red hot Fe tube}} \text{Benzene} \)
(iii) Conversion of hexane to benzene:
\( \text{C}_{6}\text{H}_{14} \xrightarrow[\text{773K. 10-20 atm.}]{\text{Cr}_{2}\text{O}_{3}/\text{V}_{2}\text{O}_{3}/\text{Mo}_{2}\text{O}_{3}} \text{C}_{6}\text{H}_{6} + 4\text{H}_{2} \)
In simple words: We can turn ethyne into benzene by heating it with an iron tube. Ethene needs more steps: first adding bromine, then removing it to make ethyne, and then using the iron tube method. Hexane can be directly converted to benzene by heating it with specific catalysts.

Exam Tip: These conversions involve specific reagents and conditions. Pay close attention to the catalysts, temperatures, and pressures required for each transformation to ensure you write accurate reaction schemes.

 

Question 21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Answer: The various possible alkenes that will yield 2-methylbutane upon hydrogenation are shown below with their corresponding names:
(i) \( \text{CH}_{2}=\text{C}(\text{CH}_{3})-\text{CH}_{2}-\text{CH}_{3} \) (2-Methylbut-1-ene)
(ii) \( \text{CH}_{3}-\text{C}(\text{CH}_{3})=\text{CH}-\text{CH}_{3} \) (2-Methylbut-2-ene)
(iii) \( \text{CH}_{3}-\text{CH}(\text{CH}_{3})-\text{CH}=\text{CH}_{2} \) (3-Methylbut-1-ene)
Therefore, the three possible alkenes are 2-Methylbut-1-ene, 2-Methylbut-2-ene, and 3-Methylbut-1-ene.
In simple words: We list all the different alkene molecules that, when hydrogen is added to them, will turn into 2-methylbutane.

Exam Tip: To find all possible alkenes that yield a specific alkane upon hydrogenation, identify all unique positions where a double bond could have existed in the alkane chain while maintaining the carbon skeleton. Remember to consider both chain and positional isomers.

 

Question 22. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, \( \text{E}^{+} \)
(a) Chlorobenzene, 2, 4 -dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, p – H\(_{3}\)C – C\(_{6}\)H\(_{4}\) – NO\(_{2}\), p- O\(_{2}\)N C\(_{6}\)H\(_{4}\) – NO\(_{2}\).
Answer: The compounds are arranged below in decreasing order of their relative reactivity towards an electrophile (\( \text{E}^{+} \)). Electron-donating groups activate the ring towards electrophilic attack, while electron-withdrawing groups deactivate it.
(a) Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene.
(b) Toluene > p-Nitrotoluene (\( \text{p-H}_{3}\text{C-C}_{6}\text{H}_{4}\text{-NO}_{2} \)) > p-Dinitrobenzene (\( \text{p-O}_{2}\text{N-C}_{6}\text{H}_{4}\text{-NO}_{2} \)).
In simple words: We sort these compounds by how easily they react with an electrophile. Compounds with electron-giving groups react faster, while those with electron-taking groups react slower.

Exam Tip: Remember that activating groups (like alkyls, -OH, -NH\(_{2}\)) increase electron density in the ring, enhancing reactivity towards electrophiles. Deactivating groups (like -NO\(_{2}\), -CHO, -COOH) decrease electron density, reducing reactivity. Halogens are deactivating but ortho-para directing.

 

Question 23. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration more easily and why?
Answer: Among benzene, m-dinitrobenzene, and toluene, toluene will undergo nitration most easily. Nitration is an electrophilic substitution reaction. The relative ease of nitration is determined by the electron density in the aromatic ring.
The nitronium ion (\( \text{NO}_{2}^{+} \)) acts as the attacking electrophile. In toluene, a hydrogen atom on the ring is substituted by a methyl (\( \text{-CH}_{3} \)) group, which is an electron-releasing group. This group increases the electron density in the benzene ring, particularly at the ortho and para positions, making electrons more readily available for the electrophile. Conversely, in m-dinitrobenzene, the two nitro (\( \text{-NO}_{2} \)) groups are strongly electron-withdrawing. These groups decrease the electron density in the ring, making it difficult for an electrophile to attack. Benzene falls in between these two extremes. Therefore, the order of reactivity for nitration is: Toluene > Benzene > m-Dinitrobenzene.
In simple words: Toluene is the easiest to nitrate because its methyl group pushes electrons into the ring, making it more attractive to the nitronium ion. M-dinitrobenzene is the hardest because its two nitro groups pull electrons away, making it less reactive. Benzene is in the middle.

Exam Tip: Remember that electron-donating groups (like alkyls) activate the benzene ring towards electrophilic attack, while electron-withdrawing groups (like nitro) deactivate it. The more activating groups, the faster the reaction; the more deactivating groups, the slower.

 

Question 24. Suggest name of another Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene,
Answer: Besides anhydrous aluminum chloride, another Lewis acid that can be utilized during the ethylation of benzene is ferric chloride (\( \text{FeCl}_{3} \)).
In simple words: Ferric chloride is another chemical that can act like anhydrous aluminum chloride to help with the ethylation of benzene.

Exam Tip: Recall common Lewis acids (electron pair acceptors) used in Friedel-Crafts alkylation and acylation reactions, such as \( \text{FeCl}_{3}, \text{BF}_{3}, \) and \( \text{ZnCl}_{2} \), in addition to \( \text{AlCl}_{3} \).

 

Question 25. Why is Wurtz reaction not preferred for preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Answer: The Wurtz reaction is generally not preferred for preparing alkanes that have an odd number of carbon atoms. This is because when you attempt to create such alkanes, the reaction often produces a mixture of various alkanes as side products, making it difficult to isolate the desired compound in pure form. This happens because cross-coupling can occur between different alkyl halides.
For instance, when trying to prepare propane (\( \text{C}_{3}\text{H}_{8} \)) or pentane (\( \text{C}_{5}\text{H}_{12} \)), several side products can form. If you start with a mix of 1-bromopropane and 1-bromobutane, you would get heptane, hexane, and octane in addition to the desired cross-coupled product.
\( \text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Br} + \text{BrCH}_{2}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3} \xrightarrow{\text{Dry ether}} \text{C}_{7}\text{H}_{16} + \text{C}_{6}\text{H}_{14} + \text{C}_{8}\text{H}_{18} \)
(1-Bromopropane + 1-Bromobutane \( \rightarrow \) Heptane + Hexane + Octane Side products)
Such a reaction results in a complex mixture of alkanes, making the isolation of a single, desired odd-numbered alkane difficult and impractical.
In simple words: The Wurtz reaction isn't good for making alkanes with an odd number of carbons because it creates a messy mix of products, making it hard to get just the alkane you want. For example, trying to make an odd-numbered alkane often yields three different alkanes instead of just one.

Exam Tip: When faced with a Wurtz reaction involving two different alkyl halides, remember that three possible alkane products can form: two symmetrical products (from each halide coupling with itself) and one unsymmetrical product (from cross-coupling). This complexity makes purification difficult for odd-carbon alkanes.

 

Question 24. Suggest name of another Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene,
Answer: Ferric chloride (\( \text{FeCl}_3 \)) can be used.

 

Question 25. Why is Wurtz reaction not preferred for preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Answer: When making larger alkanes using the Wurtz reaction, it is better to prepare only those alkanes that have an even number of carbon atoms, such as ethane, butane, and hexane, which contain 2, 4, and 6 carbon atoms, respectively.
\( \text{CH}_3\text{Br} + 2\text{Na} + \text{BrCH}_3 \xrightarrow{\text{Dry ether}} \text{CH}_3-\text{CH}_3 + 2\text{NaBr} \)
\( \text{C}_2\text{H}_5\text{Br} + 2\text{Na} + \text{BrC}_2\text{H}_5 \xrightarrow{\text{Dry ether}} \text{C}_4\text{H}_{10} + 2\text{NaBr} \)
If we try to prepare propane (\( \text{C}_3\text{H}_8 \)) or pentane (\( \text{C}_5\text{H}_{12} \)), there is a likelihood of forming unwanted side products. For instance, by beginning with 1-bromopropane and 1-bromobutane, other compounds like heptane, hexane, and octane can form in addition to the desired product.
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\text{Dry ether}} \text{C}_7\text{H}_{16} + \text{C}_6\text{H}_{14} + \text{C}_8\text{H}_{18} \)
1-Bromopropane \(\qquad\) 1-Bromobutane \(\qquad\) Heptane \(\qquad\) Hexane+Octane Side products