Hydrocarbons: Class 11 NCERT Chemistry Guide

Classification of Hydrocarbons

Hydrocarbons are broadly classified based on the types of carbon-carbon bonds present in their structure. This classification forms the basis for understanding their properties and reactivity.

1. Saturated Hydrocarbons (Alkanes):
These are the simplest hydrocarbons, containing only carbon-carbon single bonds (C-C). Each carbon atom is sp³ hybridized and bonded to four other atoms (either carbon or hydrogen). They are called 'saturated' because they have the maximum possible number of hydrogen atoms for a given number of carbon atoms and cannot undergo addition reactions. Their general formula is CₙH₂ₙ₊₂ (for open-chain alkanes). Because of their low reactivity, they are also known as paraffins (from Latin: parum = little, affinis = affinity). Example: Ethane (C₂H₆).

2. Unsaturated Hydrocarbons:
These hydrocarbons contain at least one carbon-carbon double bond (C=C) or triple bond (C≡C). They are 'unsaturated' because they can add more hydrogen atoms across the multiple bonds.

• Alkenes: Contain at least one C=C double bond. The carbon atoms involved in the double bond are sp² hybridized. Their general formula is CₙH₂ₙ (for one double bond). They are more reactive than alkanes. Example: Ethene (C₂H₄).
• Alkynes: Contain at least one C≡C triple bond. The carbon atoms in the triple bond are sp hybridized. Their general formula is CₙH₂ₙ₋₂ (for one triple bond). They are even more reactive than alkenes. Example: Ethyne (C₂H₂).

3. Aromatic Hydrocarbons (Arenes):
These are a special class of cyclic compounds. The most common parent compound is benzene (C₆H₆). They have a characteristic aroma (hence the name 'aromatic'). They exhibit unique stability due to delocalization of π-electrons across the ring, a property known as aromaticity. They undergo substitution reactions rather than the addition reactions typical of unsaturated compounds.

How to Name Hydrocarbons (IUPAC System)

1. Step 1: Identify the Parent Chain — Find the longest continuous chain of carbon atoms. This chain determines the 'parent' name (e.g., hexane for 6 carbons, heptane for 7). If there's a double or triple bond, the parent chain MUST include it, even if it's not the absolute longest chain.
2. Step 2: Number the Parent Chain — Number the carbon atoms in the parent chain starting from the end that gives the substituents (attached groups) the lowest possible numbers ('locants'). For alkenes and alkynes, start numbering from the end closer to the multiple bond.
3. Step 3: Identify and Name Substituents — Any group attached to the parent chain is a substituent. Name them based on the number of carbons (e.g., -CH₃ is methyl, -C₂H₅ is ethyl). Use prefixes like 'di-', 'tri-', 'tetra-' if the same substituent appears multiple times.
4. Step 4: Assemble the Full Name — Combine the parts in the order: [Locant]-[Substituent] [Parent Chain Name]. If there are multiple substituents, list them alphabetically (ignoring di-, tri- etc.). For alkenes/alkynes, the position of the multiple bond is indicated by a number before the suffix (e.g., Hex-2-ene).

Worked Examples of Key Reactions

• Example 1: Markovnikov's Addition to an Alkene Question: What is the major product when propene (CH₃-CH=CH₂) reacts with HBr? Explanation: This is an electrophilic addition reaction. The H⁺ from HBr acts as the electrophile. Step 1: Attack of the π bond on H⁺. The double bond attacks the H⁺, forming a carbocation intermediate. Step 2: Determine the most stable carbocation. The H⁺ can add to C1 or C2. Adding H⁺ to C1 gives a secondary carbocation (CH₃-⁺CH-CH₃). This is more stable. Adding H⁺ to C2 gives a primary carbocation (CH₃-CH₂-⁺CH₂). This is less stable. Step 3: Nucleophilic attack. The reaction proceeds via the more stable secondary carbocation. The bromide ion (Br⁻) then attacks this carbocation. Final Product: The major product is 2-Bromopropane (CH₃-CHBr-CH₃). Markovnikov's rule states that the negative part of the addendum (Br⁻) attaches to the carbon atom with fewer hydrogen atoms.
• Example 2: Free Radical Halogenation of an Alkane Question: Explain the mechanism for the chlorination of methane (CH₄) in the presence of UV light. Explanation: This reaction occurs via a free radical chain mechanism. Step 1: Initiation. UV light provides energy to break the Cl-Cl bond homolytically, forming two highly reactive chlorine free radicals (Cl•). Cl₂ → (UV light) → 2Cl• Step 2: Propagation. The chlorine radical attacks a methane molecule, abstracting a hydrogen atom to form HCl and a methyl radical (•CH₃). This methyl radical then attacks another Cl₂ molecule, forming chloromethane (CH₃Cl) and another chlorine radical, which continues the chain. Cl• + CH₄ → HCl + •CH₃ •CH₃ + Cl₂ → CH₃Cl + Cl• * Step 3: Termination. The reaction stops when two free radicals combine. This can happen in several ways: Cl• + Cl• → Cl₂ •CH₃ + •CH₃ → C₂H₆ (Ethane) Cl• + •CH₃ → CH₃Cl

Exam Traps and Common Mistakes

Markovnikov vs. Anti-Markovnikov: This is a classic point of confusion. Remember:

• Markovnikov's Rule (standard conditions, e.g., HBr alone): The negative part of the adding molecule goes to the carbon with fewer hydrogens. The mechanism involves the more stable carbocation.
• Anti-Markovnikov's Rule (Peroxide Effect, e.g., HBr + R₂O₂): The negative part goes to the carbon with more hydrogens. This applies only to HBr and proceeds via a free radical mechanism, not a carbocation.

Wurtz Reaction: Students often forget that the Wurtz reaction (treating alkyl halides with sodium in dry ether) is best for preparing symmetrical alkanes (R-R). Using two different alkyl halides (R-X + R'-X) leads to a mixture of products (R-R, R'-R', and R-R'), which is difficult to separate and gives a low yield of the desired unsymmetrical alkane.

Isomer Counting: When asked to draw isomers of a formula like C₅H₁₂, be systematic. First, draw the straight chain (n-pentane). Then, shorten the chain by one carbon and use it as a methyl group (isopentane/2-methylbutane). Finally, shorten the main chain again and see if you can arrange the carbons differently (neopentane/2,2-dimethylpropane). Don't double-count structures that are just rotated versions of each other.

Practice Questions with Solutions

• Q: Write the IUPAC name for the following compound: CH₃-CH(C₂H₅)-CH=C(CH₃)₂ A: Step 1: Identify the longest carbon chain containing the double bond. The longest chain has 6 carbons. We must include the C=C double bond. The chain is -CH(C₂H₅)-CH=C(CH₃)₂. Expanding C₂H₅ to -CH₂-CH₃ shows the longest chain is actually 6 carbons. Step 2: Number the chain to give the double bond the lowest possible number. Numbering from right to left gives the double bond position 2. Numbering from left to right gives it position 4. So, we number from the right. (6)CH₃-(5)CH₂-(4)CH(CH₃)-(3)CH=C(2)(CH₃)-(1)CH₃ - This is incorrect. Let's re-evaluate. Correct Chain: The chain starts from the left. CH₃-CH(CH₂CH₃)-CH=C(CH₃)₂. The main chain is Heptane if we go straight. But we must include the double bond. So, CH₃(6)-C(5)H(C₂H₅)-C(4)H=C(3)(CH₃)-C(2)H(CH₃). No, this is also wrong. Let's redraw and find the longest chain including the double bond: It is a 6-carbon chain (hexane). C(6)H₃-C(5)H₂-C(4)H(CH₃)-C(3)H=C(2)(CH₃)-C(1)H₃ is wrong. Let's reconsider: CH₃-CH(C₂H₅)-CH=C(CH₃)₂. The double bond is between C3 and C4 from left, or C2 and C3 from right. Oh, the C(CH₃)₂ means two methyls on one carbon. The structure is CH₃–CH(C₂H₅)–CH=C(CH₃)₂. Longest chain containing the double bond is 6 carbons. Numbering from the right gives the double bond position 2. Numbering from the left gives it position 4. So, we number from the right: C6-C5-C4=C3-C2(CH3)-C1. This structure is incorrect. Let's try again. Structure: CH₃(1)-CH₂(2)-CH(3)(CH₃)-CH(4)=C(5)(CH₃)-C(6)H₂-C(7)H₃. Nope. Let's write it out clearly: Carbon 1 is part of a CH₃ group. It's connected to Carbon 2 (CH). This C2 is connected to an ethyl group (C₂H₅) and Carbon 3 (CH). C3 is double bonded to Carbon 4 (C). C4 is connected to two methyl groups. The longest chain containing the double bond is: CH₃-CH₂-CH(from ethyl)-CH-CH=C(CH₃)₂. This chain has 6 carbons. Let's number from the end closer to the double bond (the right end). The structure is 4-ethyl-2-methylhex-2-ene. Step 1: Longest chain with double bond is 6 carbons (hex-). Number from right to give C=C locant 2. Step 2: Substituents are a methyl group at C2 and an ethyl group at C4. Step 3: Alphabetize substituents: ethyl before methyl. Final answer: 4-Ethyl-2-methylhex-2-ene
• Q: What happens when 1-Butyne is treated with dilute H₂SO₄ and HgSO₄ at 333 K? Write the reaction and name the product. A: Step 1: Identify the reaction. This is the hydration of an alkyne, specifically Kucherov's reaction. It involves the addition of a water molecule across the triple bond. Step 2: Apply Markovnikov's rule. The addition of H₂O (as H⁺ and OH⁻) follows Markovnikov's rule. The negative part (OH⁻) will add to the carbon atom of the triple bond that has fewer hydrogen atoms. In 1-butyne (CH₃-CH₂-C≡CH), the terminal carbon (C1) has one H, and C2 has zero H. So, OH⁻ adds to C2. Step 3: Form the intermediate enol. The initial product is an enol (a compound with a C=C double bond and an -OH group on one of the double-bonded carbons): CH₃-CH₂-C(OH)=CH₂. Step 4: Tautomerism. Enols are generally unstable and rapidly rearrange (tautomerize) into their more stable keto form. The hydrogen from the -OH group migrates to the adjacent carbon (CH₂), and the C=C double bond becomes a C=O double bond. Step 5: Write the final product. The final product is CH₃-CH₂-CO-CH₃. This is a ketone. Final answer: The product is Butan-2-one. Reaction: CH₃CH₂C≡CH + H₂O --(Hg²⁺/H⁺)--> [CH₃CH₂C(OH)=CH₂] --(Tautomerism)--> CH₃CH₂COCH₃.
• Q: Draw all the structural isomers of the alkane with the molecular formula C₅H₁₂ and give their IUPAC names. A: Step 1: Start with the straight-chain isomer. The simplest arrangement is a continuous chain of all five carbon atoms. This is called n-pentane. Structure: CH₃-CH₂-CH₂-CH₂-CH₃ IUPAC Name: Pentane Step 2: Shorten the parent chain by one carbon and use that carbon as a substituent. The parent chain is now 4 carbons (butane). Place the one-carbon (methyl) group as a branch. It can be placed on C2. Placing it on C3 is the same as C2 due to symmetry. Structure: CH₃-CH(CH₃)-CH₂-CH₃ IUPAC Name: 2-Methylbutane (commonly known as isopentane). Step 3: Shorten the parent chain again. The parent chain is now 3 carbons (propane). We have two carbons left to place as substituents. They must both be placed on the central carbon (C2), as placing them on the ends would just extend the chain. Structure: CH₃-C(CH₃)₂-CH₃ IUPAC Name: 2,2-Dimethylpropane (commonly known as neopentane). Step 4: Check for any other possibilities. If we shorten the chain to 2 carbons, we cannot arrange the remaining 3 carbons as branches without recreating one of the previous structures. Thus, there are only three isomers. Final answer: There are 3 structural isomers: Pentane, 2-Methylbutane, and 2,2-Dimethylpropane.
• Q: How will you convert Ethyne to Benzene? A: Step 1: Identify the starting material and product. Starting material is ethyne (C₂H₂), an alkyne. The product is benzene (C₆H₆), an aromatic compound. Step 2: Recognize the required transformation. This conversion involves the joining of three molecules of ethyne to form one molecule of benzene. This is a polymerization reaction, specifically a cyclic polymerization. Step 3: State the reaction conditions. This reaction occurs when ethyne gas is passed through a red-hot iron tube at a high temperature (around 873 K). Step 4: Write the balanced chemical equation. Three molecules of ethyne combine in a cyclic manner to form one molecule of benzene. Reaction: 3 CH≡CH --(Red hot Fe tube, 873 K)--> C₆H₆ Final answer: Ethyne can be converted to benzene by passing it through a red-hot iron tube at 873 K. This process is called cyclic polymerization.

Frequently Asked Questions