Q: Q6. Draw the structures of the following compounds:\n(i) Hex-1,5-dien-3-yne\n(ii) 4-Ethyl-2,5-dimethylhex-2-ene\n(iii) Buta-1,3-diene\n(iv) 4-Phenylbut-1-ene\n(v) 2-Methylbutan-2-ol\n(

The structures for the given compounds are as follows:\n\n(i) **Hex-1,5-dien-3-yne**\n * Parent chain: Hexane (6 carbons)\n * Double bonds: at C1 and C5\n * Triple bond: at C3\n * Structure: CH2=CH—C≡C—CH2—CH=CH2\n\n(ii) **4-Ethyl-2,5-dimethylhex-2-ene**\n * Parent chain: Hexene (6 carbons) with a double bond at C2\n * Substituents: Ethyl group at C4, Methyl groups at C2 and C5\n * Structure: CH3—C(CH3)=CH—CH(CH2CH3)—CH(CH3)—CH3\n\n(iii) **Buta-1,3-diene**\n * Parent chain: Butane (4 carbons)\n * Double bonds: at C1 and C3\n * Structure: CH2=CH—CH=CH2\n\n(iv) **4-Phenylbut-1-ene**\n * Parent chain: Butene (4 carbons) with a double bond at C1\n * Substituent: Phenyl group (C6H5) at C4\n * Structure: CH2=CH—CH2—CH2—C6H5\n\n(v) **2-Methylbutan-2-ol**\n * Parent chain: Butane (4 carbons) with an -OH group at C2\n * Substituent: Methyl group at C2\n * Structure: CH3—C(CH3)(OH)—CH2—CH3\n\n(vi) **Pent-1-en-3-yne**\n * Parent chain: Pentane (5 carbons)\n * Double bond: at C1\n * Triple bond: at C3\n * Structure: CH2=CH—C≡C—CH3\n\n(vii) **Butane-2,3-diol**\n * Parent chain: Butane (4 carbons)\n * Hydroxyl groups: at C2 and C3\n * Structure: CH3—CH(OH)—CH(OH)—CH3\n\n(viii) **Octa-1,3,5,7-tetraene**\n * Parent chain: Octane (8 carbons)\n * Double bonds: at C1, C3, C5, and C7\n * Structure: CH2=CH—CH=CH—CH=CH—CH=CH2\n\n(ix) **3-Cyclopropylprop-1-ene**\n * Parent chain: Propene (3 carbons) with a double bond at C1\n * Substituent: Cyclopropyl group (a 3-membered carbon ring) at C3\n * Structure: CH2=CH—CH2—(cyclopropyl ring)\n * (Representing cyclopropyl as a triangle attached to CH2)\n\n(x) **4-Methylpent-2-ene**\n * Parent chain: Pentene (5 carbons) with a double bond at C2\n * Substituent: Methyl group at C4\n * Structure: CH3—CH=CH—CH(CH3)—CH3

Q: Q9. Write IUPAC names of the following compounds:\n(i) CH3 — CH2 — CH = CH — CH2 — CH3\n(ii) CH3 — CH2 — C ≡ C — CH2 — CH3\n(iii) CH2 = CH — CH2 — CH = CH2\n(iv) CH3 — CH — CH2 — CH =

The IUPAC names for the given compounds are:\n\n(i) **Hex-3-ene**\n * Longest chain has 6 carbons. Double bond is at position 3.\n\n(ii) **Hex-3-yne**\n * Longest chain has 6 carbons. Triple bond is at position 3.\n\n(iii) **Penta-1,4-diene**\n * Longest chain has 5 carbons. Double bonds are at positions 1 and 4.\n\n(iv) **3-Methylpent-1-ene**\n * Longest chain containing the double bond has 5 carbons. Double bond is at position 1. Methyl group is at position 3.\n\n(v) **3-Ethyl-5-methylheptane**\n * Longest chain has 7 carbons. Ethyl group at position 3, methyl group at position 5. Alphabetical order for substituents.\n\n(vi) **3-Ethylhex-3-ene**\n * Longest chain containing the double bond has 6 carbons. Double bond is at position 3. Ethyl group is also at position 3.\n\n(vii) **3-Propylhex-2-ene**\n * Longest chain containing the double bond has 6 carbons. Double bond is at position 2. Propyl group is at position 3.

Q: Q17. How will you convert the following compounds into benzene?\n(i) Ethyne\n(ii) n-Hexane\n(iii) Phenol

Benzene can be synthesized from the given compounds using the following reactions:\n\n**(i) Ethyne to Benzene**\nEthyne (acetylene) undergoes cyclic polymerization when passed through a red hot iron tube at 873 K. Three molecules of ethyne polymerize to form one molecule of benzene.\n\n**Reaction:**\n3 CH≡CH --(Red hot Fe tube, 873 K)--> C₆H₆\n(Ethyne) (Benzene)\n\n**(ii) n-Hexane to Benzene**\nn-Hexane can be converted to benzene by aromatization (catalytic reforming). This reaction involves heating n-hexane in the presence of catalysts like chromium oxide supported on alumina (Cr₂O₃/Al₂O₃) or molybdenum oxide supported on alumina (MoO₃/Al₂O₃) at high temperatures (773 K) and pressure (10-20 atm). This process involves dehydrogenation and cyclization.\n\n**Reaction:**\nCH₃(CH₂)₄CH₃ --(Cr₂O₃/Al₂O₃, 773 K, 10-20 atm)--> C₆H₆ + 4 H₂\n(n-Hexane) (Benzene)\n\n**(iii) Phenol to Benzene**\nPhenol can be converted to benzene by reduction with zinc dust. When phenol is distilled with zinc dust, the zinc removes the oxygen atom from the hydroxyl group, forming benzene and zinc oxide.\n\n**Reaction:**\nC₆H₅OH + Zn --(heat)--> C₆H₆ + ZnO\n(Phenol) (Benzene)

Question 1

Q1. How do you account for the formation of ethane during the chlorination of methane?

Accepted Answer

The chlorination of methane is a free radical substitution reaction that proceeds via a chain mechanism involving initiation, propagation, and termination steps. While the primary product is chloromethane (CH3Cl), followed by dichloromethane, trichloromethane, and tetrachloromethane, the formation of ethane (CH3-CH3) can be explained by a specific termination step in the free radical mechanism.

Question 2

Q2. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Accepted Answer

Hydrogenation is the addition of hydrogen (H2) across a double bond, converting an alkene into an alkane. To find the alkenes that give 2-methylbutane upon hydrogenation, we need to identify all possible positions where a double bond could have existed in the 2-methylbutane skeleton. The molecular formula of 2-methylbutane is C5H12. The corresponding alkenes will have the formula C5H10.

The structure of 2-methylbutane is:
CH3-CH(CH3)-CH2-CH3

Let's identify the possible alkenes:

1.  **2-Methylbut-1-ene**
    If the double bond is between C1 and C2 (of the main chain), the structure is:
    CH2=C(CH3)-CH2-CH3

2.  **2-Methylbut-2-ene**
    If the double bond is between C2 and C3 (of the main chain), the structure is:
    CH3-C(CH3)=CH-CH3

3.  **3-Methylbut-1-ene**
    If the double bond is between C1 and C2, and the methyl group is on C3, the structure is:
    CH2=CH-CH(CH3)-CH3

These are the three distinct alkene isomers that yield 2-methylbutane upon hydrogenation. Note that 2-methylbut-2-ene does not exhibit cis-trans isomerism because the carbon atoms of the double bond each have two identical groups attached (C2 has two methyl groups, C3 has H and CH3).

Question 3

Q3. Arrange the following set of compounds in order of their increasing boiling points:
(a) Pentan-1-ol, butan-1-ol, butanal, pentane, ethoxyethane
(b) Pentane, isopentane, neopentan

Accepted Answer

Boiling point is primarily determined by the strength of intermolecular forces (IMFs) and, to a lesser extent, by molecular weight and molecular shape (branching). Stronger IMFs require more energy to overcome, leading to higher boiling points. **General order of intermolecular forces strength:** Hydrogen bonding > Dipole-dipole interactions > London Dispersion Forces (LDFs) **Factors affecting LDFs:** 1. **Molecular Weight/Size:** Larger molecules have more electrons and larger surface areas, leading to stronger LDFs and higher boiling points. 2. **Molecular Shape/Branching:** More compact, branched molecules have smaller surface areas for intermolecular contact, resulting in weaker LDFs and lower boiling points compared to their straight-chain isomers. --- **(a) Pentan-1-ol, butan-1-ol, butanal, pentane, ethoxyethane** Let's analyze each compound: 1. **Pentane (C5H12):** An alkane. Only has weak London Dispersion Forces. Molecular weight = 72 g/mol. 2. **Ethoxyethane (Diethyl ether, CH3CH2OCH2CH3, C4H10O):** An ether. Has a polar C-O bond, leading to dipole-dipole interactions. No hydrogen bonding (no H directly bonded to O). Molecular weight = 74 g/mol. 3. **Butanal (CH3CH2CH2CHO, C4H8O):** An aldehyde. Has a polar carbonyl group (C=O), leading to strong dipole-dipole interactions. No hydrogen bonding. Molecular weight = 72 g/mol. 4. **Butan-1-ol (CH3CH2CH2CH2OH, C4H10O):** An alcohol. Can form strong hydrogen bonds due to the -OH group. Also has dipole-dipole interactions and LDFs. Molecular weight = 74 g/mol. 5. **Pentan-1-ol (CH3CH2CH2CH2CH2OH, C5H12O):** An alcohol. Can form strong hydrogen bonds. Also has dipole-dipole interactions and LDFs. Higher molecular weight than butan-1-ol, so stronger LDFs and hydrogen bonding (due to larger non-polar part). Molecular weight = 88 g/mol. **Comparison:** * **Pentane** has only LDFs, so it will have the lowest boiling point. * **Ethoxyethane** has dipole-dipole interactions, which are stronger than LDFs, so its boiling point will be higher than pentane. * **Butanal** also has dipole-dipole interactions, but the carbonyl group is more polar than the ether linkage, leading to stronger dipole-dipole forces than ethoxyethane. Its boiling point will be higher than ethoxyethane. * **Butan-1-ol** and **Pentan-1-ol** can form hydrogen bonds, which are the strongest intermolecular forces among these compounds. Therefore, they will have the highest boiling points. * Between butan-1-ol and pentan-1-ol, **Pentan-1-ol** has a higher molecular weight and a longer carbon chain, leading to stronger LDFs and slightly stronger hydrogen bonding, thus a higher boiling point than butan-1-ol. **Order of increasing boiling points:** Pentane < Ethoxyethane < Butanal < Butan-1-ol < Pentan-1-ol --- **(b) Pentane, isopentane, neopentane** These are all structural isomers of C5H12, meaning they have the same molecular formula and molecular weight (72 g/mol). They are all alkanes, so they only exhibit London Dispersion Forces. The difference in their boiling points arises from their molecular shapes (branching). 1. **Pentane (n-pentane):** CH3-CH2-CH2-CH2-CH3 (Straight chain) Has the largest surface area for intermolecular contact, leading to the strongest LDFs among the three. 2. **Isopentane (2-methylbutane):** CH3-CH(CH3)-CH2-CH3 (One branch) More compact than n-pentane, reducing the surface area for intermolecular contact, thus weaker LDFs than n-pentane. 3. **Neopentane (2,2-dimethylpropane):** C(CH3)4 (Highly branched, spherical shape) The most compact and spherical shape among the three, significantly reducing the surface area for intermolecular contact. This results in the weakest LDFs and the lowest boiling point. **Order of increasing boiling points (due to decreasing branching and surface area):** Neopentane < Isopentane < Pentane

Question 4

Q4. Write IUPAC names of the following compounds:
(a) CH3CH=C(CH3)CH2CH3
(b) (CH3)2C=CHCH2CH3
(c) CH2=CH-CH2-C(CH3)3
(d) CH3CH(CH3)CH2CH=CH2
(e) CH3(CH2)3CH(CH3)CH2CH=CH2
(f) CH3

Accepted Answer

To write the IUPAC names for alkenes, follow these rules:
1.  **Identify the longest continuous carbon chain** that contains the double bond(s). This chain is the parent chain.
2.  **Number the carbon atoms** of the parent chain starting from the end that gives the double bond(s) the lowest possible number(s).
3.  **Identify and name any substituents** attached to the parent chain.
4.  **Indicate the position of the double bond(s)** by using the number of the first carbon atom of the double bond. Use '-ene' for one double bond, '-diene' for two, '-triene' for three, etc.
5.  **Assemble the name** in the format: (substituent number)-(substituent name)-(parent chain number)-(double bond position)-(alkene suffix).

Let's apply these rules to each compound:

**(a) CH3CH=C(CH3)CH2CH3**
*   Longest chain containing the double bond: 5 carbons (pent-).
*   Numbering: Start from the left to give the double bond the lowest number (C2).
    CH3-CH=C(CH3)-CH2-CH3
    1   2   3    4    5
*   Substituent: Methyl group at C3.
*   Double bond at C2.
*   Name: **3-Methylpent-2-ene**

**(b) (CH3)2C=CHCH2CH3**
*   Longest chain containing the double bond: 5 carbons (pent-).
*   Numbering: Start from the right to give the double bond the lowest number (C2).
    CH3-CH2-CH=C(CH3)2
    5   4   3   2   1 (Incorrect numbering for lowest double bond number)
    Correct numbering: CH3-CH2-CH=C(CH3)2
                      4   3   2   1
    The longest chain containing the double bond is 4 carbons long (but-). The methyl groups are on C1. Let's redraw to be clear:
    CH3
    |  
    CH3-C=CH-CH2-CH3
    1  2  3  4  5 (Longest chain is 5 carbons, but double bond is at C2 from right)
    Correct numbering for longest chain containing double bond: 
    CH3-CH2-CH=C(CH3)2
    4   3   2   1
    The longest chain containing the double bond is 4 carbons (but-). The double bond is at C1. There are two methyl groups at C1.
    This is incorrect. Let's re-evaluate.
    The longest chain containing the double bond is 5 carbons. Numbering from the right gives the double bond at C2. Numbering from the left gives the double bond at C3. So, number from the right.
    CH3-CH2-CH=C(CH3)2
    5   4   3   2   1
    The double bond is between C2 and C3. The methyl groups are on C2.
*   Longest chain containing the double bond: 5 carbons (pent-).
*   Numbering: Start from the right to give the double bond the lowest number (C2).
    CH3-CH2-CH=C(CH3)2
    5   4   3   2   1
*   Substituents: Two methyl groups at C2.
*   Double bond at C2.
*   Name: **2,2-Dimethylpent-3-ene** (Wait, this is wrong. The double bond is at C2, not C3. Let's re-evaluate the numbering and parent chain.)

Let's draw the structure clearly:
      CH3
      |
CH3-CH2-CH=C-CH3
           |
           CH3

Longest chain containing the double bond: 5 carbons. Number from the end closer to the double bond. Right end is closer.
CH3-CH2-CH=C(CH3)2
5   4   3   2   1
Double bond is at C2. Methyl groups are at C2.
Name: **2-Methylpent-2-ene** (This is the correct interpretation)

**(c) CH2=CH-CH2-C(CH3)3**
*   Longest chain containing the double bond: 5 carbons (pent-).
*   Numbering: Start from the left to give the double bond the lowest number (C1).
    CH2=CH-CH2-C(CH3)3
    1   2   3   4   5
*   Substituents: Two methyl groups at C4.
*   Double bond at C1.
*   Name: **4,4-Dimethylpent-1-ene**

**(d) CH3CH(CH3)CH2CH=CH2**
*   Longest chain containing the double bond: 5 carbons (pent-).
*   Numbering: Start from the right to give the double bond the lowest number (C1).
    CH3-CH(CH3)-CH2-CH=CH2
    5   4    3    2   1
*   Substituent: Methyl group at C4.
*   Double bond at C1.
*   Name: **4-Methylpent-1-ene**

**(e) CH3(CH2)3CH(CH3)CH2CH=CH2**
*   Expand the condensed formula: CH3-CH2-CH2-CH2-CH(CH3)-CH2-CH=CH2
*   Longest chain containing the double bond: 8 carbons (oct-).
*   Numbering: Start from the right to give the double bond the lowest number (C1).
    CH3-CH2-CH2-CH2-CH(CH3)-CH2-CH=CH2
    8   7   6   5   4     3   2   1
*   Substituent: Methyl group at C4.
*   Double bond at C1.
*   Name: **4-Methyloct-1-ene**

**(f) CH3-CH=CH-CH2-CH=CH-CH2-CH3**
*   Longest chain containing both double bonds: 8 carbons (octa-).
*   Numbering: Start from either end, as both give double bonds at C2 and C6.
    CH3-CH=CH-CH2-CH=CH-CH2-CH3
    1   2   3   4   5   6   7   8
*   No substituents.
*   Double bonds at C2 and C6.
*   Name: **Octa-2,6-diene**

Question 5

Q5. Write IUPAC names of the following compounds:
(i) CH3—CH=C(CH3)2
(ii) CH2=CH—C≡C—CH3
(iii) CH2=CH—CH2—CH=CH2
(iv) CH3—CH(CH3)—CH2—CH=CH2
(v) CH3—CH(CH3)—CH2—C≡CH
(vi) CH3—CH2

Accepted Answer

The IUPAC names for the given compounds are as follows:

(i) **CH3—CH=C(CH3)2**
   *   Longest carbon chain containing the double bond has 4 carbons (but-). The double bond is between C1 and C2 if numbered from the right, or C2 and C3 if numbered from the left. Numbering from the right gives the double bond at C1 and the methyl group at C2. So, 2-methylbut-2-ene.
   *   **IUPAC Name: 2-Methylbut-2-ene**

(ii) **CH2=CH—C≡C—CH3**
   *   The longest chain has 5 carbons. It contains both a double bond and a triple bond. Numbering should give the lowest locants to the multiple bonds. Numbering from the left gives 1-ene and 3-yne. Numbering from the right gives 1-yne and 3-ene. When both are present, the double bond gets preference for numbering if there's a tie, or if the double bond is closer to an end. Here, both are at C1 from either end. So, the double bond gets preference for the suffix, and the triple bond is treated as a substituent. However, for numbering, the lowest locant rule for multiple bonds takes precedence. Here, 1-ene and 3-yne is correct.
   *   **IUPAC Name: Pent-1-en-3-yne**

(iii) **CH2=CH—CH2—CH=CH2**
   *   The longest carbon chain containing both double bonds has 5 carbons (penta-). Double bonds are at C1 and C4.
   *   **IUPAC Name: Penta-1,4-diene**

(iv) **CH3—CH(CH3)—CH2—CH=CH2**
   *   The longest carbon chain containing the double bond has 4 carbons (but-). Numbering starts from the end closer to the double bond. So, the double bond is at C1. The methyl group is at C3.
   *   **IUPAC Name: 3-Methylbut-1-ene**

(v) **CH3—CH(CH3)—CH2—C≡CH**
   *   The longest carbon chain containing the triple bond has 4 carbons (but-). Numbering starts from the end closer to the triple bond. So, the triple bond is at C1. The methyl group is at C3.
   *   **IUPAC Name: 3-Methylbut-1-yne**

(vi) **CH3—CH2—C(CH3)2—CH2—CH3**
   *   The longest carbon chain has 5 carbons (pentane). Numbering from either end gives the methyl groups at C3.
   *   **IUPAC Name: 3,3-Dimethylpentane**

(vii) **CH3—CH(CH3)—CH=CH—C(CH3)3**
   *   The longest carbon chain containing the double bond has 7 carbons (hept-). Numbering from the left gives the double bond at C3 and methyl groups at C2 and C6. Numbering from the right gives the double bond at C4 and methyl groups at C2 and C6. So, numbering from the left is preferred.
   *   Substituents: Methyl at C2, and a tert-butyl group at C6 (which is 2,2-dimethylpropyl, but here it's simpler to consider the methyls separately).
   *   Let's re-evaluate the longest chain. If we consider the tert-butyl group as a substituent, the main chain is 6 carbons long (hex-). CH3-CH(CH3)-CH=CH-C(CH3)3. Longest chain containing the double bond is 6 carbons. Numbering from left: 2-methyl, 3-ene, 5,5-dimethyl. So, 2,5,5-trimethylhex-3-ene. Let's try numbering from the right: 2,2-dimethyl, 3-ene, 5-methyl. So, 2,2,5-trimethylhex-3-ene. The lowest locant rule for substituents applies. 2,2,5 is lower than 2,5,5. So, 2,2,5-trimethylhex-3-ene.
   *   **IUPAC Name: 2,2,5-Trimethylhex-3-ene**

(viii) **CH3—CH2—CH=CH—CH2—CH=CH2**
   *   The longest carbon chain containing both double bonds has 7 carbons (hepta-). Numbering from the right gives double bonds at C1 and C4. Numbering from the left gives double bonds at C3 and C6. So, numbering from the right is preferred.
   *   **IUPAC Name: Hepta-1,4-diene**

Question 6

Q6. Draw the structures of the following compounds:
(i) Hex-1,5-dien-3-yne
(ii) 4-Ethyl-2,5-dimethylhex-2-ene
(iii) Buta-1,3-diene
(iv) 4-Phenylbut-1-ene
(v) 2-Methylbutan-2-ol
(

Accepted Answer

The structures for the given compounds are as follows:

(i) **Hex-1,5-dien-3-yne**
    *   Parent chain: Hexane (6 carbons)
    *   Double bonds: at C1 and C5
    *   Triple bond: at C3
    *   Structure: CH2=CH—C≡C—CH2—CH=CH2

(ii) **4-Ethyl-2,5-dimethylhex-2-ene**
    *   Parent chain: Hexene (6 carbons) with a double bond at C2
    *   Substituents: Ethyl group at C4, Methyl groups at C2 and C5
    *   Structure: CH3—C(CH3)=CH—CH(CH2CH3)—CH(CH3)—CH3

(iii) **Buta-1,3-diene**
    *   Parent chain: Butane (4 carbons)
    *   Double bonds: at C1 and C3
    *   Structure: CH2=CH—CH=CH2

(iv) **4-Phenylbut-1-ene**
    *   Parent chain: Butene (4 carbons) with a double bond at C1
    *   Substituent: Phenyl group (C6H5) at C4
    *   Structure: CH2=CH—CH2—CH2—C6H5

(v) **2-Methylbutan-2-ol**
    *   Parent chain: Butane (4 carbons) with an -OH group at C2
    *   Substituent: Methyl group at C2
    *   Structure: CH3—C(CH3)(OH)—CH2—CH3

(vi) **Pent-1-en-3-yne**
    *   Parent chain: Pentane (5 carbons)
    *   Double bond: at C1
    *   Triple bond: at C3
    *   Structure: CH2=CH—C≡C—CH3

(vii) **Butane-2,3-diol**
    *   Parent chain: Butane (4 carbons)
    *   Hydroxyl groups: at C2 and C3
    *   Structure: CH3—CH(OH)—CH(OH)—CH3

(viii) **Octa-1,3,5,7-tetraene**
    *   Parent chain: Octane (8 carbons)
    *   Double bonds: at C1, C3, C5, and C7
    *   Structure: CH2=CH—CH=CH—CH=CH—CH=CH2

(ix) **3-Cyclopropylprop-1-ene**
    *   Parent chain: Propene (3 carbons) with a double bond at C1
    *   Substituent: Cyclopropyl group (a 3-membered carbon ring) at C3
    *   Structure: CH2=CH—CH2—(cyclopropyl ring)
    *   (Representing cyclopropyl as a triangle attached to CH2)

(x) **4-Methylpent-2-ene**
    *   Parent chain: Pentene (5 carbons) with a double bond at C2
    *   Substituent: Methyl group at C4
    *   Structure: CH3—CH=CH—CH(CH3)—CH3

Question 7

Q7. Write the reactions involved in the preparation of the following hydrocarbons from an alkene.
(i) Ethane from Ethene
(ii) But-1-yne from But-1-ene
(iii) Hexane from Hex-1-ene

Accepted Answer

Here are the reactions involved in the preparation of the specified hydrocarbons from an alkene:

### (i) Ethane from Ethene

Ethane (an alkane) can be prepared from ethene (an alkene) by a hydrogenation reaction. This involves the addition of hydrogen gas across the double bond in the presence of a catalyst.

**Reaction:**
Ethene reacts with hydrogen gas in the presence of finely divided catalysts like Platinum (Pt), Palladium (Pd), or Nickel (Ni) to form ethane.

CH2=CH2 + H2 --(Pt/Pd/Ni, heat)--> CH3—CH3

**Explanation:**
This is a catalytic hydrogenation reaction. The hydrogen molecule adds across the double bond, converting the alkene into a saturated alkane. Nickel requires higher temperatures (200-300°C), while Platinum and Palladium work efficiently at room temperature.

### (ii) But-1-yne from But-1-ene

Converting an alkene (but-1-ene) to an alkyne (but-1-yne) typically involves a two-step process: first, halogenation of the alkene to form a dihaloalkane, followed by dehydrohalogenation using a strong base.

**Step 1: Halogenation of But-1-ene**
But-1-ene reacts with bromine (Br2) to form 1,2-dibromobutane. This is an addition reaction across the double bond.

CH3—CH2—CH=CH2 + Br2 --(CCl4)--> CH3—CH2—CH(Br)—CH2(Br)

**Step 2: Dehydrohalogenation of 1,2-Dibromobutane**
1,2-Dibromobutane is then treated with a strong base like alcoholic KOH or, more effectively, sodamide (NaNH2) in liquid ammonia to remove two molecules of HBr and form the alkyne.

CH3—CH2—CH(Br)—CH2(Br) + 2KOH (alc.) --(heat)--> CH3—CH2—C≡CH + 2KBr + 2H2O

*Alternatively, for terminal alkynes, a stronger base like NaNH2 is often preferred to prevent rearrangement and ensure the triple bond forms at the terminal position.*

CH3—CH2—CH(Br)—CH2(Br) + 2NaNH2 --(liq. NH3)--> CH3—CH2—C≡CH + 2NaBr + 2NH3

**Explanation:**
In the first step, bromine adds to the double bond via an electrophilic addition mechanism. In the second step, the strong base removes hydrogen and bromine atoms from adjacent carbons (β-elimination) to form the triple bond. Two such eliminations are required to form an alkyne from a dihaloalkane.

### (iii) Hexane from Hex-1-ene

Similar to the conversion of ethene to ethane, hexane (an alkane) can be prepared from hex-1-ene (an alkene) through catalytic hydrogenation.

**Reaction:**
Hex-1-ene reacts with hydrogen gas in the presence of a metal catalyst (Pt, Pd, or Ni) to yield hexane.

CH3—CH2—CH2—CH2—CH=CH2 + H2 --(Pt/Pd/Ni, heat)--> CH3—CH2—CH2—CH2—CH2—CH3

**Explanation:**
This is a straightforward catalytic hydrogenation. The double bond in hex-1-ene is saturated by the addition of hydrogen, resulting in the formation of the corresponding alkane, hexane.

Question 8

Q8. How will you convert the following compounds into benzene?
(i) Ethyne
(ii) Hexane
(iii) Hexene

Accepted Answer

Here are the methods to convert the given compounds into benzene:

### (i) Ethyne to Benzene

Benzene can be prepared from ethyne (acetylene) through a process called cyclic polymerization or trimerization. This reaction involves passing ethyne through a red hot iron tube.

**Reaction:**
When ethyne is passed through a red hot iron tube (at 873 K or 600°C), three molecules of ethyne polymerize to form one molecule of benzene.

3CH≡CH --(Red hot iron tube, 873 K)--> C6H6

**Explanation:**
This is a classic example of a cyclic polymerization reaction. The high temperature and the presence of the iron catalyst facilitate the formation of new C-C bonds and the rearrangement of electrons to form the stable aromatic ring structure of benzene.

### (ii) Hexane to Benzene

Hexane (an alkane) can be converted to benzene through a process called aromatization or catalytic reforming. This reaction involves heating hexane in the presence of catalysts like oxides of chromium, vanadium, or molybdenum supported on alumina.

**Reaction:**
Hexane undergoes dehydrocyclization (cyclization followed by dehydrogenation) when heated with Cr2O3/Al2O3 or Mo2O3/Al2O3 at high temperatures (773 K) and pressure.

CH3(CH2)4CH3 --(Cr2O3/Al2O3, 773 K, 10-20 atm)--> C6H6 + 4H2

**Explanation:**
In this reaction, hexane first cyclizes to form cyclohexane, which then undergoes dehydrogenation (removal of hydrogen) to form benzene. This process is crucial in the petroleum industry for increasing the octane number of gasoline and producing aromatic compounds.

### (iii) Hexene to Benzene

Hexene (an alkene) can also be converted to benzene through aromatization, similar to hexane. The presence of a double bond in hexene means it requires fewer dehydrogenation steps compared to hexane.

**Reaction:**
Hexene (e.g., hex-1-ene) is heated with catalysts like Cr2O3/Al2O3 at high temperatures and pressure. It first cyclizes and then undergoes dehydrogenation to form benzene.

CH3(CH2)3CH=CH2 --(Cr2O3/Al2O3, 773 K, 10-20 atm)--> C6H6 + 3H2

**Explanation:**
Similar to hexane, hexene undergoes dehydrocyclization. The six-carbon chain first forms a cyclohexane ring, and then hydrogen atoms are removed to establish the aromaticity of benzene. Since hexene already has a double bond, it requires the removal of fewer hydrogen molecules compared to hexane to achieve the aromatic structure.

Question 9

Q9. Write IUPAC names of the following compounds:
(i) CH3 — CH2 — CH = CH — CH2 — CH3
(ii) CH3 — CH2 — C ≡ C — CH2 — CH3
(iii) CH2 = CH — CH2 — CH = CH2
(iv) CH3 — CH — CH2 — CH =

Accepted Answer

The IUPAC names for the given compounds are:

(i) **Hex-3-ene**
    *   Longest chain has 6 carbons. Double bond is at position 3.

(ii) **Hex-3-yne**
    *   Longest chain has 6 carbons. Triple bond is at position 3.

(iii) **Penta-1,4-diene**
    *   Longest chain has 5 carbons. Double bonds are at positions 1 and 4.

(iv) **3-Methylpent-1-ene**
    *   Longest chain containing the double bond has 5 carbons. Double bond is at position 1. Methyl group is at position 3.

(v) **3-Ethyl-5-methylheptane**
    *   Longest chain has 7 carbons. Ethyl group at position 3, methyl group at position 5. Alphabetical order for substituents.

(vi) **3-Ethylhex-3-ene**
    *   Longest chain containing the double bond has 6 carbons. Double bond is at position 3. Ethyl group is also at position 3.

(vii) **3-Propylhex-2-ene**
    *   Longest chain containing the double bond has 6 carbons. Double bond is at position 2. Propyl group is at position 3.

Question 10

Q10. Suggest the free radical mechanism for the chlorination of methane.

Accepted Answer

The chlorination of methane is a free radical substitution reaction that occurs in the presence of UV light or high temperatures. It proceeds via a chain mechanism involving three main steps: Initiation, Propagation, and Termination.

### Free Radical Mechanism for Chlorination of Methane (CH4 + Cl2 → CH3Cl + HCl)

#### Step 1: Initiation
This step involves the homolytic cleavage of the chlorine molecule (Cl2) into two chlorine free radicals (Cl•) by absorbing energy from UV light or heat. This is an endothermic step and starts the chain reaction.

**Reaction:**
Cl — Cl  +  hν (UV light)  →  Cl•  +  Cl•
*(Chlorine molecule)*      *(Chlorine free radicals)*

#### Step 2: Propagation
This is the main step where the actual reaction takes place and the chain grows. It involves a series of two steps that regenerate a free radical, allowing the chain reaction to continue.

**Sub-step 2a: Hydrogen abstraction**
A chlorine free radical (Cl•) attacks a methane molecule (CH4), abstracts a hydrogen atom, forming hydrogen chloride (HCl) and a methyl free radical (CH3•).

**Reaction:**
Cl•  +  CH3 — H  →  H — Cl  +  CH3•
*(Chlorine radical)*  *(Methane)*    *(Hydrogen chloride)*  *(Methyl radical)*

**Sub-step 2b: Chlorine abstraction**
The methyl free radical (CH3•) then attacks another chlorine molecule (Cl2), abstracts a chlorine atom, forming chloromethane (CH3Cl) and regenerating a new chlorine free radical (Cl•). This regenerated chlorine radical can then continue the chain by reacting with another methane molecule (Sub-step 2a).

**Reaction:**
CH3•  +  Cl — Cl  →  CH3 — Cl  +  Cl•
*(Methyl radical)*  *(Chlorine molecule)*  *(Chloromethane)*  *(Chlorine radical)*

#### Step 3: Termination
This step involves the combination of two free radicals to form a stable molecule, thereby ending the chain reaction. This occurs when the concentration of reactants decreases or when radicals collide with each other.

**Possible Termination Reactions:**
1.  **Combination of two chlorine radicals:**
    Cl•  +  Cl•  →  Cl2

2.  **Combination of two methyl radicals:**
    CH3•  +  CH3•  →  CH3 — CH3 (Ethane)

3.  **Combination of a methyl radical and a chlorine radical:**
    CH3•  +  Cl•  →  CH3 — Cl (Chloromethane)

These termination steps lead to the formation of stable products and stop the propagation of the chain reaction.

Question 11

Q11. Write the reactions for the preparation of ethane from the following compounds:
(i) Sodium propanoate
(ii) Ethyl bromide
(iii) Iodoethane
(iv) Chloromethane

Accepted Answer

Here are the reactions for the preparation of ethane (CH3CH3) from the given compounds:

### (i) From Sodium propanoate (CH3CH2COONa)
Ethane can be prepared from sodium propanoate by decarboxylation. This can be achieved by two main methods:

**Method 1: Kolbe's Electrolytic Method**
Electrolysis of an aqueous solution of sodium propanoate yields ethane at the anode. This reaction involves free radical formation and coupling.

**Reaction:**
2CH3CH2COONa(aq) + 2H2O(l)  --Electrolysis-->  CH3CH3(g) + 2CO2(g) + 2NaOH(aq) + H2(g)
*(Sodium propanoate)*                                *(Ethane)*

**Mechanism (at anode):**
CH3CH2COO⁻  →  CH3CH2COO•  +  e⁻
CH3CH2COO•  →  CH3CH2•  +  CO2
2CH3CH2•  →  CH3CH3

**Method 2: Soda-lime Decarboxylation**
Heating sodium propanoate with soda-lime (a mixture of NaOH and CaO) results in the removal of a carboxyl group as CO2, forming ethane. This method typically produces an alkane with one carbon less than the carboxylic acid salt.

**Reaction:**
CH3CH2COONa  +  NaOH  --CaO, Δ-->  CH3CH3  +  Na2CO3
*(Sodium propanoate)*  *(Soda-lime)*       *(Ethane)*

### (ii) From Ethyl bromide (CH3CH2Br)
Ethane can be prepared from ethyl bromide through the Wurtz reaction or by reduction.

**Method 1: Wurtz Reaction**
Ethyl bromide reacts with sodium metal in dry ether to form ethane. This reaction involves the coupling of two alkyl halide molecules.

**Reaction:**
2CH3CH2Br  +  2Na  --Dry Ether-->  CH3CH3  +  2NaBr
*(Ethyl bromide)*                            *(Ethane)*

**Method 2: Reduction**
Ethyl bromide can be reduced to ethane using various reducing agents like Zn/HCl, LiAlH4, or H2/Pd.

**Reaction (using Zn/HCl):**
CH3CH2Br  +  Zn  +  2H⁺  →  CH3CH3  +  Zn²⁺  +  HBr
*(Ethyl bromide)*                      *(Ethane)*

### (iii) From Iodoethane (CH3CH2I)
Similar to ethyl bromide, iodoethane can be converted to ethane using the Wurtz reaction or by reduction.

**Method 1: Wurtz Reaction**
Iodoethane reacts with sodium metal in dry ether.

**Reaction:**
2CH3CH2I  +  2Na  --Dry Ether-->  CH3CH3  +  2NaI
*(Iodoethane)*                            *(Ethane)*

**Method 2: Reduction**
Iodoethane can be reduced to ethane using reducing agents.

**Reaction (using H2/Pd):**
CH3CH2I  +  H2  --Pd-->  CH3CH3  +  HI
*(Iodoethane)*                  *(Ethane)*

### (iv) From Chloromethane (CH3Cl)
Chloromethane can be converted to ethane using the Wurtz reaction. Since ethane has two carbon atoms, two molecules of chloromethane (each with one carbon) are required to couple.

**Method: Wurtz Reaction**
Chloromethane reacts with sodium metal in dry ether to form ethane.

**Reaction:**
2CH3Cl  +  2Na  --Dry Ether-->  CH3CH3  +  2NaCl
*(Chloromethane)*                        *(Ethane)*

Question 12

Q12. How would you convert the following compounds into benzene?
(i) Ethyne
(ii) Hexane
(iii) Phenol

Accepted Answer

Here are the conversions of the given compounds into benzene:

### (i) Conversion of Ethyne (Acetylene) to Benzene
Ethyne (C2H2) can be converted to benzene (C6H6) by cyclic polymerization (trimerization). When ethyne is passed through a red hot iron tube, three molecules of ethyne polymerize to form one molecule of benzene.

**Reaction:**
3CH≡CH  --Red hot iron tube, 873 K-->  C6H6
*(Ethyne)*                                    *(Benzene)*

### (ii) Conversion of Hexane to Benzene
n-Hexane (CH3(CH2)4CH3) can be converted to benzene by aromatization (catalytic reforming). This reaction involves cyclization and dehydrogenation. When n-hexane is heated to a high temperature (773 K) and pressure (10-20 atm) in the presence of catalysts like chromium oxide supported on alumina (Cr2O3/Al2O3) or molybdenum oxide supported on alumina (MoO2/Al2O3), it undergoes cyclization followed by dehydrogenation to form benzene.

**Reaction:**
CH3(CH2)4CH3  --Cr2O3/Al2O3, 773 K, 10-20 atm-->  C6H6  +  4H2
*(n-Hexane)*                                         *(Benzene)*

### (iii) Conversion of Phenol to Benzene
Phenol (C6H5OH) can be converted to benzene (C6H6) by reduction with zinc dust. When phenol is distilled with zinc dust, the zinc removes the oxygen atom from the hydroxyl group, leading to the formation of benzene.

**Reaction:**
C6H5OH  +  Zn  --Δ-->  C6H6  +  ZnO
*(Phenol)*                      *(Benzene)*

Question 13

Q13. How will you convert the following compounds into benzene? (i) Ethyne (ii) Hexane (iii) Phenol

Accepted Answer

Benzene is an aromatic hydrocarbon. The conversion of various compounds into benzene involves specific chemical reactions:

**(i) Ethyne to Benzene**
Ethyne (acetylene) can be converted to benzene by cyclic polymerization. When ethyne is passed through a red hot iron tube, three molecules of ethyne polymerize to form one molecule of benzene. This reaction is an example of trimerization.

**Reaction:**
3 CH≡CH  --[Red hot iron tube, 873 K]-->  C₆H₆
(Ethyne)                                         (Benzene)

**(ii) Hexane to Benzene**
n-Hexane can be converted to benzene by aromatization (also known as catalytic reforming). This process involves heating n-hexane in the presence of catalysts like chromium oxide (Cr₂O₃) supported on alumina (Al₂O₃) or molybdenum oxide (MoO₃) at high temperature (773 K) and pressure. During this reaction, n-hexane undergoes cyclization and dehydrogenation to form benzene.

**Reaction:**
CH₃(CH₂)₄CH₃  --[Cr₂O₃/Al₂O₃, 773 K, 10-20 atm]-->  C₆H₆ + 4 H₂
(n-Hexane)                                                  (Benzene)

**(iii) Phenol to Benzene**
Phenol can be converted to benzene by reduction using zinc dust. When phenol is distilled with zinc dust, the zinc removes the oxygen atom from the hydroxyl group, resulting in the formation of benzene.

**Reaction:**
C₆H₅OH + Zn  --[Heat]-->  C₆H₆ + ZnO
(Phenol)                                (Benzene)

Question 14

Q14. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Accepted Answer

2-Methylbutane has the following carbon skeleton:
CH₃-CH(CH₃)-CH₂-CH₃

To obtain 2-methylbutane upon hydrogenation, the original alkene must have the same carbon skeleton but with a double bond (C=C) at various possible positions. Hydrogenation adds hydrogen across the double bond, converting it to a single bond.

The possible positions for a double bond in the 2-methylbutane carbon skeleton are:

1.  **Between C1 and C2:**
    CH₂=C(CH₃)-CH₂-CH₃
    IUPAC Name: **2-Methylbut-1-ene**

2.  **Between C2 and C3:**
    CH₃-C(CH₃)=CH-CH₃
    IUPAC Name: **2-Methylbut-2-ene**
    (This compound does not exhibit cis-trans isomerism because the carbon atoms of the double bond are not each bonded to two different groups. The carbon with the methyl group and the double bond has two identical groups, CH₃, on one side of the double bond, and the other carbon has H and CH₃.)

3.  **Between C3 and C4 (of the main chain, but this is equivalent to C1-C2 if numbered from the other end):**
    If we consider the longest chain as butane, the methyl group is at position 2. So, the double bond can be at position 1 or 2.
    CH₃-CH(CH₃)-CH=CH₂ (This is 3-methylbut-1-ene, which is an isomer of 2-methylbut-1-ene, but the question asks for alkenes that give 2-methylbutane. Let's re-evaluate the carbon skeleton carefully.)

Let's draw the 2-methylbutane skeleton and then place the double bond:

C - C - C - C
        | 
        C

Possible double bond positions:

*   **Position 1: Between the primary carbon (C1) and the secondary carbon (C2) of the main chain.**
    CH₂=C(CH₃)-CH₂-CH₃
    This is **2-Methylbut-1-ene**.

*   **Position 2: Between the secondary carbon (C2) and the tertiary carbon (C3) of the main chain.**
    CH₃-C(CH₃)=CH-CH₃
    This is **2-Methylbut-2-ene**.

*   **Position 3: Between the primary carbon of the methyl group (C5) and the tertiary carbon (C2) of the main chain.**
    This would be the same as 2-Methylbut-1-ene, just viewed from a different angle or numbering. For example, if we consider the methyl group attached to C2 as part of the double bond, it would be CH₂=C(CH₂CH₃)-CH₃, which is 2-ethylprop-1-ene, but this is not 2-methylbutane skeleton. The carbon skeleton must remain 2-methylbutane after saturation.

Therefore, there are only two distinct alkenes that yield 2-methylbutane upon hydrogenation:

1.  **2-Methylbut-1-ene**
    Structure: CH₂=C(CH₃)CH₂CH₃

2.  **2-Methylbut-2-ene**
    Structure: CH₃C(CH₃)=CHCH₃

Question 15

Q15. Arrange the following set of compounds in order of their increasing boiling points: (a) n-pentane, isopentane, neopentane (b) n-butane, n-pentane, n-hexane

Accepted Answer

The boiling point of alkanes is primarily determined by the strength of intermolecular forces, specifically van der Waals forces. These forces depend on two main factors:

1.  **Molecular Mass:** As molecular mass increases, the number of electrons increases, leading to stronger London dispersion forces and thus higher boiling points.
2.  **Surface Area/Branching:** For isomers (compounds with the same molecular formula but different structures), increased branching leads to a more compact, spherical shape, which reduces the surface area available for intermolecular contact. This results in weaker van der Waals forces and lower boiling points.

**(a) n-pentane, isopentane, neopentane**
All three compounds are isomers of pentane (C₅H₁₂), meaning they have the same molecular mass. Therefore, branching is the deciding factor.

*   **n-pentane (n-C₅H₁₂):** Straight-chain structure, maximum surface area.
    CH₃-CH₂-CH₂-CH₂-CH₃
*   **Isopentane (2-methylbutane, C₅H₁₂):** One methyl branch, reduced surface area compared to n-pentane.
    CH₃-CH(CH₃)-CH₂-CH₃
*   **Neopentane (2,2-dimethylpropane, C₅H₁₂):** Two methyl branches, most compact (spherical) structure, minimum surface area.
    CH₃
    | 
    CH₃-C-CH₃
    | 
    CH₃

**Order of increasing boiling points:** Neopentane < Isopentane < n-pentane
(Due to decreasing branching and increasing surface area for intermolecular interactions)

**(b) n-butane, n-pentane, n-hexane**
These are straight-chain alkanes with increasing molecular mass.

*   **n-butane (C₄H₁₀):** Lowest molecular mass.
*   **n-pentane (C₅H₁₂):** Intermediate molecular mass.
*   **n-hexane (C₆H₁₄):** Highest molecular mass.

As the molecular mass increases, the number of electrons increases, leading to stronger van der Waals forces and higher boiling points.

**Order of increasing boiling points:** n-butane < n-pentane < n-hexane
(Due to increasing molecular mass and stronger van der Waals forces)

Question 16

Q16. Write the 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-en

Accepted Answer

Ozonolysis is a reaction used to locate the position of a double bond in an alkene. The double bond is cleaved, and the carbons involved in the double bond are converted into carbonyl compounds (aldehydes or ketones). The general reaction involves two steps: first, reaction with ozone (O₃) to form an ozonide, followed by reductive work-up with Zn/H₂O (or (CH₃)₂S) to cleave the ozonide into carbonyl compounds.

To determine the products, simply 'cut' the double bond and add an oxygen atom to each of the carbons that were part of the double bond.

**(i) Pent-2-ene**
Structure: CH₃-CH=CH-CH₂-CH₃

Cleaving the double bond between C2 and C3 and adding oxygen:
CH₃-CH=O  +  O=CH-CH₂-CH₃

Products:
1.  **Ethanal** (also known as Acetaldehyde)
    Structure: CH₃CHO
2.  **Propanal** (also known as Propionaldehyde)
    Structure: CH₃CH₂CHO

**(ii) 3,4-Dimethylhept-3-ene**
Structure: CH₃-CH₂-C(CH₃)=C(CH₃)-CH₂-CH₂-CH₃

Cleaving the double bond between C3 and C4 and adding oxygen:
CH₃-CH₂-C(CH₃)=O  +  O=C(CH₃)-CH₂-CH₂-CH₃

Products:
1.  **Butan-2-one** (also known as Methyl ethyl ketone)
    Structure: CH₃CH₂COCH₃
2.  **Pentan-2-one** (also known as Methyl propyl ketone)
    Structure: CH₃CH₂CH₂COCH₃

**(iii) 2-Ethylbut-1-ene**
Structure: CH₂=C(CH₂CH₃)-CH₂-CH₃
(Note: The longest chain is 4 carbons, with an ethyl group at C2. So, it's 2-ethylbut-1-ene)

Cleaving the double bond between C1 and C2 and adding oxygen:
O=CH₂  +  O=C(CH₂CH₃)-CH₂-CH₃

Products:
1.  **Methanal** (also known as Formaldehyde)
    Structure: HCHO
2.  **Pentan-3-one** (also known as Diethyl ketone)
    Structure: CH₃CH₂COCH₂CH₃

**(iv) 1-Phenylbut-1-ene**
Structure: C₆H₅-CH=CH-CH₂-CH₃

Cleaving the double bond between C1 and C2 and adding oxygen:
C₆H₅-CH=O  +  O=CH-CH₂-CH₃

Products:
1.  **Benzaldehyde**
    Structure: C₆H₅CHO
2.  **Propanal** (also known as Propionaldehyde)
    Structure: CH₃CH₂CHO

Question 17

Q17. How will you convert the following compounds into benzene?
(i) Ethyne
(ii) n-Hexane
(iii) Phenol

Accepted Answer

Benzene can be synthesized from the given compounds using the following reactions:

**(i) Ethyne to Benzene**
Ethyne (acetylene) undergoes cyclic polymerization when passed through a red hot iron tube at 873 K. Three molecules of ethyne polymerize to form one molecule of benzene.

**Reaction:**
3 CH≡CH  --(Red hot Fe tube, 873 K)-->  C₆H₆
(Ethyne)                                   (Benzene)

**(ii) n-Hexane to Benzene**
n-Hexane can be converted to benzene by aromatization (catalytic reforming). This reaction involves heating n-hexane in the presence of catalysts like chromium oxide supported on alumina (Cr₂O₃/Al₂O₃) or molybdenum oxide supported on alumina (MoO₃/Al₂O₃) at high temperatures (773 K) and pressure (10-20 atm). This process involves dehydrogenation and cyclization.

**Reaction:**
CH₃(CH₂)₄CH₃  --(Cr₂O₃/Al₂O₃, 773 K, 10-20 atm)-->  C₆H₆ + 4 H₂
(n-Hexane)                                            (Benzene)

**(iii) Phenol to Benzene**
Phenol can be converted to benzene by reduction with zinc dust. When phenol is distilled with zinc dust, the zinc removes the oxygen atom from the hydroxyl group, forming benzene and zinc oxide.

**Reaction:**
C₆H₅OH + Zn  --(heat)-->  C₆H₆ + ZnO
(Phenol)                               (Benzene)

Question 18

Q18. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Accepted Answer

Hydrogenation involves the addition of hydrogen (H₂) across a carbon-carbon double bond (C=C) to form a carbon-carbon single bond (C-C). To find the alkenes that give 2-methylbutane on hydrogenation, we need to identify all possible positions for a double bond in the carbon skeleton of 2-methylbutane.

First, let's draw the structure of 2-methylbutane:
```
    CH₃
    |
CH₃ - CH - CH₂ - CH₃
```
This is a 5-carbon chain with a methyl group at the second carbon. The carbon atoms can be numbered as follows:
```
      ⁴CH₃
      |
¹CH₃ - ²CH - ³CH₂ - ⁴CH₃
```
(Note: The IUPAC numbering for 2-methylbutane would be 1-2-3-4, with the methyl at C2. The numbering above is just to identify potential double bond positions relative to the branched structure.)

Now, let's place a double bond at all possible positions such that hydrogenation would yield 2-methylbutane:

1.  **Double bond between C1 and C2:**
    If the double bond is between the first and second carbon atoms (considering the longest chain starting from the double bond), the alkene would be 2-methylbut-1-ene.
    ```
          CH₃
          |
    CH₂ = C - CH₂ - CH₃
    (2-methylbut-1-ene)
    ```
    Hydrogenation: CH₂=C(CH₃)CH₂CH₃ + H₂ → CH₃CH(CH₃)CH₂CH₃ (2-methylbutane)

2.  **Double bond between C2 and C3:**
    If the double bond is between the second and third carbon atoms, the alkene would be 2-methylbut-2-ene.
    ```
          CH₃
          |
    CH₃ - C = CH - CH₃
    (2-methylbut-2-ene)
    ```
    Hydrogenation: CH₃C(CH₃)=CHCH₃ + H₂ → CH₃CH(CH₃)CH₂CH₃ (2-methylbutane)

3.  **Double bond between C3 and C4:**
    If the double bond is between the third and fourth carbon atoms, the alkene would be 3-methylbut-1-ene. However, 3-methylbut-1-ene is structurally identical to 2-methylbut-1-ene (just numbered from the other end to give the lowest possible number to the double bond). So, this is not a distinct alkene.
    ```
          CH₃
          |
    CH₃ - CH - CH = CH₂
    (3-methylbut-1-ene, same as 2-methylbut-1-ene)
    ```

Therefore, there are **two distinct alkenes** which on hydrogenation give 2-methylbutane:
1.  **2-methylbut-1-ene**
2.  **2-methylbut-2-ene**

Question 19

Q19. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile. Give reason.
Toluene, p-nitro toluene, chlorobenzene, benzene.

Accepted Answer

The reactivity of aromatic compounds towards electrophilic aromatic substitution (EAS) depends on the electron density of the benzene ring. Electron-donating groups (EDGs) increase electron density, activating the ring and increasing reactivity. Electron-withdrawing groups (EWGs) decrease electron density, deactivating the ring and decreasing reactivity.

Let's analyze each compound:

1.  **Toluene (C₆H₅-CH₃):**
    *   The methyl group (-CH₃) is an **electron-donating group** due to its **+I (inductive) effect** and **hyperconjugation**. It pushes electron density into the benzene ring.
    *   This increases the electron density of the ring, making it more susceptible to attack by an electrophile.
    *   Therefore, toluene is **activated** towards EAS.

2.  **p-Nitrotoluene (CH₃-C₆H₄-NO₂):**
    *   This compound has two substituents: a methyl group (-CH₃) and a nitro group (-NO₂).
    *   The methyl group is activating (as in toluene).
    *   The nitro group (-NO₂) is a **strong electron-withdrawing group** due to its **-M (mesomeric) effect** and **-I (inductive) effect**. It strongly pulls electron density out of the benzene ring.
    *   The strong deactivating effect of the nitro group far outweighs the weak activating effect of the methyl group. Overall, p-nitrotoluene is **strongly deactivated** towards EAS.

3.  **Chlorobenzene (C₆H₅-Cl):**
    *   The chlorine atom (-Cl) is an **electron-withdrawing group** due to its **-I (inductive) effect**. It pulls electron density away from the ring through the sigma bond.
    *   However, chlorine also has lone pairs of electrons that can be donated to the ring via **+M (mesomeric) effect**. This +M effect is weaker than the -I effect for reactivity, but it makes halogens ortho/para directing.
    *   Overall, the **-I effect dominates the reactivity**, making chlorobenzene **deactivated** compared to benzene, but less deactivated than compounds with strong EWGs like -NO₂.

4.  **Benzene (C₆H₆):**
    *   Benzene serves as the **reference point** as it has no substituents to activate or deactivate the ring.

**Order of Reactivity (Decreasing):**

**Toluene > Benzene > Chlorobenzene > p-Nitrotoluene**

**Reasoning for the order:**

*   **Toluene** is the most reactive because the methyl group is an electron-donating group (+I effect and hyperconjugation), which increases the electron density of the benzene ring, making it highly activated towards electrophilic attack.
*   **Benzene** is less reactive than toluene because it lacks any activating groups. It serves as the baseline for reactivity.
*   **Chlorobenzene** is less reactive than benzene because the chlorine atom is an electron-withdrawing group (-I effect is dominant over +M effect for reactivity), which decreases the overall electron density of the benzene ring, thus deactivating it.
*   **p-Nitrotoluene** is the least reactive because the nitro group is a strong electron-withdrawing group (-M and -I effects), which significantly reduces the electron density of the benzene ring. This strong deactivation makes it much less susceptible to electrophilic attack compared to the other compounds.

Question 20

Q20. Explain the directive influence of a methyl group in toluene.

Accepted Answer

The methyl group (-CH₃) in toluene is an **ortho-para directing** and **activating** group for electrophilic aromatic substitution (EAS) reactions. Its directive influence can be explained by a combination of the inductive effect and hyperconjugation.

### 1. Electronic Effects of the Methyl Group:

*   **Positive Inductive Effect (+I Effect):** The methyl group is an alkyl group, which is slightly electron-donating through the sigma bond. It pushes electron density towards the benzene ring, making the ring more nucleophilic.

*   **Hyperconjugation (No-bond Resonance):** This is the more significant effect. The C-H bonds of the methyl group are in conjugation with the π-electron system of the benzene ring. The electrons from the C-H sigma bonds can be delocalized into the ring, leading to the formation of temporary partial negative charges at the ortho and para positions. This can be represented by resonance structures:
    
    *   When one of the C-H bonds of the methyl group donates its electrons to the ring, a double bond forms between the methyl carbon and the ring carbon, and a negative charge appears at the ortho position.
    *   This negative charge can then delocalize to the para position, and then to the other ortho position.

### 2. Effect on Electron Density Distribution:

Due to both the +I effect and especially hyperconjugation, the electron density is increased more at the **ortho** and **para** positions of the benzene ring compared to the meta position. This makes the ortho and para positions more attractive to an incoming electrophile.

### 3. Stabilization of Intermediate Carbocation:

During electrophilic attack, a carbocation intermediate (sigma complex) is formed. The methyl group helps to stabilize this intermediate, particularly when the electrophile attacks at the ortho or para positions. This stabilization occurs because:

*   **Ortho attack:** The positive charge of the carbocation can be delocalized onto the carbon bearing the methyl group. The methyl group can then stabilize this positive charge through its +I effect and hyperconjugation.
*   **Para attack:** Similar to ortho attack, the positive charge can be delocalized to a position adjacent to the methyl group, which then stabilizes it through hyperconjugation and +I effect.
*   **Meta attack:** When the electrophile attacks at the meta position, the positive charge never appears on the carbon atom directly bonded to the methyl group, nor on an adjacent carbon that can be effectively stabilized by hyperconjugation. Thus, the meta intermediate is less stable.

### Conclusion:

Because the ortho and para positions are enriched with electron density and the carbocation intermediates formed by attack at these positions are more stable, the methyl group in toluene directs incoming electrophiles primarily to the **ortho and para positions**. It also **activates** the benzene ring towards electrophilic substitution by increasing the overall electron density.

Question 21

Q21. Write the structures of the main products of ozonolysis of the following compounds and also give their IUPAC names.
(i) 1-Methylcyclohexene
(ii) 3,4-Dimethyl-3-hexene
(iii) 2-Et

Accepted Answer

Ozonolysis is a reaction where an alkene is cleaved at the double bond by ozone, typically followed by a reductive workup (e.g., with Zn/H2O or dimethyl sulfide) to yield carbonyl compounds (aldehydes and ketones). The products depend on the substitution pattern of the double bond.

(i) **1-Methylcyclohexene**
   - The double bond is between C1 and C2 of the cyclohexene ring, with a methyl group at C1. Ozonolysis cleaves this bond, opening the ring and forming a keto-aldehyde.
   - **Products:** 6-Oxoheptanal
   - **IUPAC Name:** 6-Oxoheptanal

(ii) **3,4-Dimethyl-3-hexene**
   - The double bond is between C3 and C4. Both C3 and C4 are substituted with methyl and ethyl groups. Cleavage of this bond results in two identical ketone molecules.
   - **Products:** Butan-2-one (two molecules)
   - **IUPAC Name:** Butan-2-one

(iii) **2-Ethylbut-1-ene**
   - The double bond is between C1 and C2. C1 is a terminal carbon (CH2=), and C2 is substituted with an ethyl group. Cleavage yields an aldehyde (from CH2=) and a ketone (from the substituted carbon).
   - **Products:** Methanal (Formaldehyde) and Butan-2-one
   - **IUPAC Names:** Methanal, Butan-2-one

(iv) **1-Phenylbut-1-ene**
   - The double bond is between C1 and C2. C1 is attached to a phenyl group, and C2 is attached to an ethyl group. Cleavage yields two different aldehydes.
   - **Products:** Benzaldehyde and Propanal
   - **IUPAC Names:** Benzaldehyde, Propanal

Question 22

Q22. An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.

Accepted Answer

To determine the structure of the alkene 'A' from its ozonolysis products, we reverse the ozonolysis process. This involves removing the oxygen atoms from the carbonyl groups and joining the two carbon atoms that were originally part of the double bond.

**Given products:**
1.  Ethanal (Acetaldehyde): CH3CHO
2.  Pentan-3-one: CH3CH2COCH2CH3

**Steps to find Alkene 'A':**
1.  Write the structures of the products:
    CH3-CH=O
    O=C(CH2CH3)CH2CH3

2.  Remove the oxygen atoms from the carbonyl groups and connect the two carbon atoms with a double bond:
    CH3-CH=C(CH2CH3)CH2CH3

**Structure of Alkene 'A':**
CH3-CH=C(CH2CH3)CH2CH3

**IUPAC Name of 'A':**
To name the alkene, identify the longest carbon chain containing the double bond and the substituents.
-   Longest chain: 5 carbons (pentene)
-   Double bond position: between C2 and C3
-   Substituent: Ethyl group at C3

Therefore, the IUPAC name is **3-Ethylpent-2-ene**.

Question 23

Q23. An alkene 'A' with molecular formula C5H10 on hydrogenation gives 2-methylbutane. 'A' on ozonolysis gives propan-2-one and ethanal. Identify 'A' and write the chemical reactions in

Accepted Answer

This problem requires us to deduce the structure of alkene 'A' using two different reactions: hydrogenation and ozonolysis.

**Step 1: Analyze the hydrogenation product.**
-   Alkene 'A' has the molecular formula C5H10. This corresponds to the general formula CnH2n for alkenes, indicating one double bond or a cyclic structure. Since it undergoes hydrogenation, it must be an alkene.
-   Hydrogenation of 'A' gives 2-methylbutane. The structure of 2-methylbutane is:
    CH3-CH(CH3)-CH2-CH3
-   Hydrogenation removes the double bond and adds hydrogen atoms. This means the carbon skeleton of alkene 'A' must be the same as 2-methylbutane, but with a double bond somewhere in the chain. Possible positions for the double bond in a 2-methylbutane skeleton are:
    1.  Between C1 and C2: 2-methylbut-1-ene (CH2=C(CH3)-CH2-CH3)
    2.  Between C2 and C3: 2-methylbut-2-ene (CH3-C(CH3)=CH-CH3)
    3.  Between C3 and C4: 3-methylbut-1-ene (CH2=CH-CH(CH3)-CH3)

**Step 2: Analyze the ozonolysis products.**
-   Ozonolysis of 'A' gives propan-2-one and ethanal.
    -   Propan-2-one (Acetone): CH3COCH3
    -   Ethanal (Acetaldehyde): CH3CHO
-   To find the structure of 'A', we reverse the ozonolysis reaction. Remove the oxygen atoms from the carbonyl groups and join the carbon atoms with a double bond:
    CH3-C(CH3)=O  +  O=CH-CH3
    Combining these gives: CH3-C(CH3)=CH-CH3

**Step 3: Identify Alkene 'A'.**
-   The structure derived from ozonolysis is 2-methylbut-2-ene (CH3-C(CH3)=CH-CH3).
-   Let's verify if 2-methylbut-2-ene on hydrogenation gives 2-methylbutane:
    CH3-C(CH3)=CH-CH3 + H2 --(Ni/Pt/Pd)--> CH3-CH(CH3)-CH2-CH3 (2-methylbutane)
-   This matches the information given. Therefore, alkene 'A' is 2-methylbut-2-ene.

**Chemical Reactions Involved:**

1.  **Ozonolysis of 2-methylbut-2-ene:**
    CH3-C(CH3)=CH-CH3 + O3 --(followed by Zn/H2O)--> CH3COCH3 + CH3CHO
    (2-methylbut-2-ene)                                (Propan-2-one) (Ethanal)

2.  **Hydrogenation of 2-methylbut-2-ene:**
    CH3-C(CH3)=CH-CH3 + H2 --(Ni/Pt/Pd)--> CH3-CH(CH3)-CH2-CH3
    (2-methylbut-2-ene)                                (2-methylbutane)

Question 24

Q24. Propose a mechanism for the following reactions:
(a) Acid-catalyzed hydration of propene.
(b) Addition of HBr to ethene.
(c) Addition of HBr to propene.

Accepted Answer

All three reactions are examples of electrophilic addition reactions to alkenes, which proceed via carbocation intermediates. Markovnikov's rule is crucial for predicting the regioselectivity in unsymmetrical alkenes.

**(a) Acid-catalyzed hydration of propene**
This reaction involves the addition of water across the double bond of propene in the presence of an acid catalyst (typically H2SO4) to form an alcohol. It follows Markovnikov's rule, where the -OH group adds to the more substituted carbon of the double bond.

**Reaction:** CH3-CH=CH2 + H2O --(H+)--> CH3-CH(OH)-CH3 (Propan-2-ol)

**Mechanism:**
1.  **Protonation of Alkene:** The alkene (propene) acts as a nucleophile and attacks the electrophilic proton (H+) from the acid catalyst, forming a more stable secondary carbocation. The proton adds to the carbon with more hydrogen atoms (C1), leading to a secondary carbocation at C2, which is more stable than a primary carbocation.
    CH3-CH=CH2 + H+  →  CH3-C+H-CH3 (Secondary carbocation)

2.  **Nucleophilic Attack by Water:** The water molecule (H2O) acts as a nucleophile and attacks the positively charged carbon of the carbocation.
    CH3-C+H-CH3 + H2O  →  CH3-CH(O+H2)-CH3 (Protonated alcohol)

3.  **Deprotonation:** Another water molecule (or the conjugate base of the acid) acts as a base and removes a proton from the protonated alcohol, regenerating the acid catalyst and forming the final alcohol product.
    CH3-CH(O+H2)-CH3 + H2O  →  CH3-CH(OH)-CH3 + H3O+

**(b) Addition of HBr to ethene**
This is an electrophilic addition of hydrogen bromide to a symmetrical alkene (ethene). Since ethene is symmetrical, there is no regioselectivity issue, and only one product is formed.

**Reaction:** CH2=CH2 + HBr → CH3-CH2Br (Bromoethane)

**Mechanism:**
1.  **Protonation of Alkene:** The pi electrons of ethene attack the electrophilic proton (H+) from HBr, forming a primary carbocation. The bromide ion (Br-) is released.
    CH2=CH2 + H-Br  →  CH3-C+H2 + Br-

2.  **Nucleophilic Attack by Bromide Ion:** The bromide ion (Br-) acts as a nucleophile and attacks the positively charged carbon of the carbocation to form the final product.
    CH3-C+H2 + Br-  →  CH3-CH2Br

**(c) Addition of HBr to propene**
This is an electrophilic addition of hydrogen bromide to an unsymmetrical alkene (propene). This reaction follows Markovnikov's rule, where the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms, and the bromine atom adds to the carbon with fewer hydrogen atoms.

**Reaction:** CH3-CH=CH2 + HBr → CH3-CH(Br)-CH3 (2-Bromopropane)

**Mechanism:**
1.  **Protonation of Alkene:** The pi electrons of propene attack the electrophilic proton (H+) from HBr. Two possible carbocations can form:
    -   If H+ adds to C1: CH3-C+H-CH3 (Secondary carbocation, more stable)
    -   If H+ adds to C2: CH3-CH2-C+H2 (Primary carbocation, less stable)
    According to Markovnikov's rule, the more stable secondary carbocation is preferentially formed.
    CH3-CH=CH2 + H-Br  →  CH3-C+H-CH3 + Br-

2.  **Nucleophilic Attack by Bromide Ion:** The bromide ion (Br-) acts as a nucleophile and attacks the positively charged carbon of the more stable secondary carbocation to form the final product.
    CH3-C+H-CH3 + Br-  →  CH3-CH(Br)-CH3

Question 25

Q25. How will you convert the following compounds into benzene?
(i) Ethyne
(ii) Hexane
(iii) Phenol

Accepted Answer

Benzene can be synthesized from ethyne, hexane, and phenol through different chemical reactions:

**(i) From Ethyne (Acetylene)**
Ethyne undergoes cyclic polymerization when passed through a red hot iron tube at 873 K. Three molecules of ethyne polymerize to form one molecule of benzene.

**(ii) From Hexane (n-Hexane)**
n-Hexane can be converted to benzene by aromatization (reforming). When n-hexane is heated to 773 K at 10-20 atmospheric pressure in the presence of catalysts like Cr2O3 supported on Al2O3 or Mo2O3 supported on Al2O3 or V2O5 supported on Al2O3, it undergoes dehydrogenation and cyclization to form benzene.

**(iii) From Phenol**
Phenol can be converted to benzene by reduction. When phenol is distilled with zinc dust, the zinc removes the oxygen atom from the hydroxyl group, resulting in the formation of benzene.

Question 26

Q26. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Accepted Answer

2-Methylbutane has the following structure:
CH3-CH(CH3)-CH2-CH3

To obtain 2-methylbutane upon hydrogenation, the double bond must be present in a position that, when saturated, results in this carbon skeleton. The possible positions for the double bond are:

1.  **Between C1 and C2 of the main chain:** This gives 2-methylbut-1-ene.
    CH2=C(CH3)-CH2-CH3

2.  **Between C2 and C3 of the main chain:** This gives 2-methylbut-2-ene.
    CH3-C(CH3)=CH-CH3

Therefore, the alkenes which on hydrogenation give 2-methylbutane are:

(i) **2-Methylbut-1-ene**
    CH2=C(CH3)-CH2-CH3

(ii) **2-Methylbut-2-ene**
    CH3-C(CH3)=CH-CH3

Question 27

Q27. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+.
(a) Toluene, p-CH3-C6H4-NO2, p-O2N-C6H4-NO2
(b) C6H5-CH3, C6H5-N

Accepted Answer

The reactivity of aromatic compounds towards electrophilic substitution depends on the electron density of the benzene ring. Electron-donating groups (EDGs) activate the ring, increasing reactivity, while electron-withdrawing groups (EWGs) deactivate the ring, decreasing reactivity. The stronger the activating effect, the higher the reactivity, and vice-versa.

**(a) Toluene, p-CH3-C6H4-NO2, p-O2N-C6H4-NO2**
Let's analyze the substituents in each compound:
1.  **Toluene (C6H5-CH3):** The -CH3 group is an electron-donating group (+I effect and hyperconjugation). It activates the benzene ring.
2.  **p-CH3-C6H4-NO2 (p-Nitrotoluene):** This compound has two substituents: -CH3 (activating) and -NO2 (strongly deactivating, -M and -I effect). The overall effect on the ring's electron density will be a balance, but the strong deactivating effect of -NO2 will dominate compared to toluene.
3.  **p-O2N-C6H4-NO2 (1,4-Dinitrobenzene):** This compound has two strongly electron-withdrawing -NO2 groups. This will significantly deactivate the benzene ring, making it much less reactive than the other two.

**Order of Reactivity (Decreasing):**
*   **Toluene** (CH3 is activating) > **p-CH3-C6H4-NO2** (CH3 activates, but NO2 strongly deactivates, overall less reactive than toluene) > **p-O2N-C6H4-NO2** (two strong deactivating NO2 groups).

**(b) C6H5-CH3, C6H5-N(CH3)2, C6H5-NO2**
Let's analyze the substituents in each compound:
1.  **C6H5-CH3 (Toluene):** The -CH3 group is an electron-donating group (+I effect and hyperconjugation). It activates the benzene ring.
2.  **C6H5-N(CH3)2 (N,N-Dimethylaniline):** The -N(CH3)2 group is a very strong electron-donating group due to the lone pair on nitrogen which can be delocalized into the benzene ring (+M effect). This makes the ring highly activated.
3.  **C6H5-NO2 (Nitrobenzene):** The -NO2 group is a strong electron-withdrawing group (-M and -I effect). It strongly deactivates the benzene ring.

**Order of Reactivity (Decreasing):**
*   **C6H5-N(CH3)2** (Strongly activating due to +M effect) > **C6H5-CH3** (Moderately activating due to +I and hyperconjugation) > **C6H5-NO2** (Strongly deactivating due to -M and -I effect).

Question 28

Q28. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Accepted Answer

During the ethylation of benzene (a Friedel-Crafts alkylation reaction), a Lewis acid catalyst is required to generate the electrophile (ethyl carbocation). Anhydrous aluminium chloride (AlCl3) is commonly used, but other strong Lewis acids can also serve this purpose.

**Other Lewis acids that can be used:**
*   **Anhydrous Ferric Chloride (FeCl3)**
*   **Anhydrous Stannic Chloride (SnCl4)**
*   **Anhydrous Boron Trifluoride (BF3)**
*   **Anhydrous Zinc Chloride (ZnCl2)**

These Lewis acids can coordinate with the alkyl halide (e.g., ethyl chloride), facilitating the departure of the halide ion and generating the carbocation electrophile, which then attacks the benzene ring.

Question 29

Q29. Write the structures of all the alkenes which on hydrogenation give 2-methylbutane.

Accepted Answer

Hydrogenation is the addition of hydrogen (H2) across a carbon-carbon double bond, converting an alkene into an alkane. To find the alkenes that yield 2-methylbutane upon hydrogenation, we need to identify all possible positions where a double bond could have existed in the 2-methylbutane skeleton.

First, let's draw the structure of 2-methylbutane:
CH3 - CH(CH3) - CH2 - CH3

We can identify the carbon chain and the methyl substituent. Now, we consider placing a double bond such that its hydrogenation would result in this specific alkane structure. This means removing two hydrogen atoms from adjacent carbon atoms in 2-methylbutane.

Let's number the carbons of 2-methylbutane:
    1     2      3     4
CH3 - CH(CH3) - CH2 - CH3

Possible locations for a double bond:

1.  **Between C1 and C2:** Removing H from C1 and C2 leads to 2-methylbut-1-ene.
    CH2 = C(CH3) - CH2 - CH3
    (2-methylbut-1-ene)

2.  **Between C2 and C3:** Removing H from C2 and C3 leads to 2-methylbut-2-ene.
    CH3 - C(CH3) = CH - CH3
    (2-methylbut-2-ene)

3.  **Between C3 and C4:** Removing H from C3 and C4 leads to 3-methylbut-1-ene.
    CH3 - CH(CH3) - CH = CH2
    (3-methylbut-1-ene)

These are the three distinct alkene isomers that, upon hydrogenation, will yield 2-methylbutane.

Question 30

Q30. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile.
(a) Toluene, p-CH3-C6H4-NO2, p-CH3-C6H4-Cl
(b) C6H5-CH3, C6H5-H, C6H

Accepted Answer

The reactivity of aromatic compounds towards electrophilic aromatic substitution (EAS) depends on the electron density of the benzene ring. Electron-donating groups (EDGs) increase electron density, activating the ring and increasing reactivity. Electron-withdrawing groups (EWGs) decrease electron density, deactivating the ring and decreasing reactivity.

**General order of reactivity:** Activated benzene > Benzene > Deactivated benzene.

**(a) Toluene, p-CH3-C6H4-NO2, p-CH3-C6H4-Cl**

*   **Toluene (C6H5-CH3):** The -CH3 group is an electron-donating group (+I effect and hyperconjugation). It activates the benzene ring, making it more reactive than benzene itself.
*   **p-CH3-C6H4-NO2 (p-nitrotoluene):** This compound has two substituents. The -CH3 group is activating, but the -NO2 group is a very strong electron-withdrawing group (-M and -I effects). The strong deactivating effect of -NO2 dominates, making the ring significantly less reactive.
*   **p-CH3-C6H4-Cl (p-chlorotoluene):** This compound also has two substituents. The -CH3 group is activating. The -Cl group is an electron-withdrawing group (-I effect) but also has a weak electron-donating resonance effect (+M effect). Overall, halogens are deactivating due to the stronger inductive effect, but they are weaker deactivators compared to -NO2.

**Comparison:**
1.  **Toluene** is the most reactive as it has only an activating group.
2.  **p-chlorotoluene** is less reactive than toluene because of the deactivating -Cl group, but more reactive than p-nitrotoluene.
3.  **p-nitrotoluene** is the least reactive due to the strong deactivating effect of the -NO2 group.

**Decreasing order of reactivity:**
Toluene > p-CH3-C6H4-Cl > p-CH3-C6H4-NO2

**(b) C6H5-CH3, C6H5-H, C6H5-Cl, C6H5-NO2**

*   **C6H5-CH3 (Toluene):** -CH3 is an activating group (+I and hyperconjugation).
*   **C6H5-H (Benzene):** This is the reference compound, neither activated nor deactivated.
*   **C6H5-Cl (Chlorobenzene):** -Cl is a deactivating group (-I effect is stronger than +M effect).
*   **C6H5-NO2 (Nitrobenzene):** -NO2 is a strong deactivating group (-M and -I effects).

**Comparison:**
1.  **Toluene** is activated by the -CH3 group.
2.  **Benzene** is the reference.
3.  **Chlorobenzene** is deactivated by -Cl, but less so than nitrobenzene.
4.  **Nitrobenzene** is strongly deactivated by -NO2.

**Decreasing order of reactivity:**
C6H5-CH3 > C6H5-H > C6H5-Cl > C6H5-NO2

Question 31

Q31. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Accepted Answer

Ethylation of benzene is a Friedel-Crafts alkylation reaction. Friedel-Crafts reactions require a Lewis acid catalyst to generate the electrophile (in this case, the ethyl carbocation from ethyl halide).

Anhydrous aluminium chloride (AlCl3) is a commonly used Lewis acid. Other Lewis acids that can be used for this purpose include:

*   **Ferric chloride (FeCl3)**
*   **Stannic chloride (SnCl4)**
*   **Boron trifluoride (BF3)**
*   **Zinc chloride (ZnCl2)**

Any one of these can be suggested. For example, **anhydrous ferric chloride (FeCl3)**.

Question 32

Q32. An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.

Accepted Answer

To determine the structure of the alkene 'A', we need to reverse the ozonolysis reaction. Ozonolysis involves the cleavage of the carbon-carbon double bond (C=C) and the addition of oxygen atoms to form carbonyl compounds. Therefore, to find the original alkene, we remove the oxygen atoms from the carbonyl groups of the products and join the carbon atoms that were part of the carbonyl groups with a double bond.

Question 33

Q33. An alkene 'A' with molecular formula C₅H₁₀ on hydrogenation gives 2-methylbutane. 'A' on hydration in the presence of H₂SO₄ gives an optically active alcohol 'B'. Identify 'A' and

Accepted Answer

This problem requires us to deduce the structures of an alkene 'A' and an alcohol 'B' based on their reactions and properties. We will use the information from hydrogenation and hydration reactions, along with the concept of optical activity.

Question 34

Q34. Propanal and propanone are the ozonolysis products of an alkene. What is the structural formula of the alkene? Show the reactions involved.

Accepted Answer

To find the structural formula of the alkene, we need to reverse the ozonolysis reaction. Ozonolysis breaks the carbon-carbon double bond (C=C) and forms carbonyl compounds. By removing the oxygen atoms from the carbonyl groups of the products and joining the carbon atoms that were originally double-bonded to oxygen, we can reconstruct the original alkene.

Chapter 13: Hydrocarbons