CBSE Class 11 Chemistry Chapter 13: Hydrocarbons NCERT Solutions

The chlorination of methane occurs through a free radical chain mechanism, which involves three distinct steps: initiation, propagation, and termination. Although methyl chloride (CH₃Cl) and hydrogen chloride (HCl) are the primary products, ethane (C₂H₆) can be formed as a by-product during the termination phase.

Step 1: Initiation

The process begins with the homolytic cleavage of the chlorine molecule (Cl₂) into chlorine free radicals, typically initiated by UV light (hν) or heat.

$$Cl\; -\; Cl\; \backslash xrightarrow\{h\backslash nu\}\; Cl\backslash cdot\; +\; Cl\backslash cdot$$

This step generates the reactive chlorine free radicals.

Step 2: Propagation

In this phase, the chlorine free radicals react with methane (CH₄) molecules. A chlorine radical abstracts a hydrogen atom from methane, forming a methyl free radical ($\backslash dot\{C\}H\_3$) and hydrogen chloride (HCl).

$$CH\_4\; +\; Cl\backslash cdot\; \backslash xrightarrow\{h\backslash nu\}\; \backslash dot\{C\}H\_3\; +\; HCl$$

The generated methyl radical can then react with another chlorine molecule:

$$\backslash dot\{C\}H\_3\; +\; Cl\; -\; Cl\; \backslash longrightarrow\; CH\_3\; -\; Cl\; +\; Cl\backslash cdot$$

This step produces methyl chloride (CH₃Cl) and regenerates a chlorine free radical, allowing the chain reaction to continue. Further reactions can lead to higher chlorinated products like CH₂Cl₂, CHCl₃, and CCl₄.

Step 3: Termination

The chain reaction ceases when free radicals combine with each other. The formation of ethane occurs when two methyl free radicals encounter each other and combine:

$$\backslash dot\{C\}H\_3\; +\; \backslash dot\{C\}H\_3\; \backslash longrightarrow\; CH\_3\; -\; CH\_3$$

(Ethane)

Other termination steps include the combination of two chlorine radicals ($Cl\backslash cdot\; +\; Cl\backslash cdot\; \backslash longrightarrow\; Cl\_2$) or a methyl radical and a chlorine radical ($CH\_3\backslash cdot\; +\; Cl\backslash cdot\; \backslash longrightarrow\; CH\_3Cl$). Thus, ethane is an incidental product arising from the termination of the radical chain process.

The IUPAC names are determined by identifying the longest carbon chain containing the principal functional group (double or triple bond) and numbering it to give the functional group the lowest possible number.

(a) $CH\_3CH\; =\; C(CH\_3)\_2$

The structure can be written as: $H\_3\backslash overset\{4\}\{C\}\; -\; \backslash overset\{3\}\{C\}H\; =\; \backslash overset\{2\}\{C\}(\backslash overset\{5\}\{C\}H\_3)\; -\; \backslash overset\{1\}\{C\}H\_3$. The longest carbon chain containing the double bond has 4 carbons. The double bond is between C2 and C3. There is a methyl group attached to C2. Numbering from the right gives the double bond position 2. Numbering from the left gives the double bond position 2. However, to give the substituent the lowest number, we number from the right. The parent chain is but-2-ene. The methyl group is at position 2.

IUPAC name: 2-Methylbut-2-ene

(b) $CH\_2\; =\; CH\; -\; C\; \backslash equiv\; C\; -\; CH\_3$

The structure is: $\backslash overset\{1\}\{C\}H\_2\; =\; \backslash overset\{2\}\{C\}H\; -\; \backslash overset\{3\}\{C\}\; \backslash equiv\; \backslash overset\{4\}\{C\}\; -\; \backslash overset\{5\}\{C\}H\_3$. The longest chain has 5 carbons. The double bond is at position 1 and the triple bond is at position 3. According to IUPAC rules, when both double and triple bonds are present, the suffix indicating the double bond (-ene) comes first, and the suffix indicating the triple bond (-yne) comes second. The chain is numbered to give the lowest locant to the double bond. The parent name is pent-1-ene-3-yne.

IUPAC name: Pent-1-ene-3-yne

(c) $CH\_2\; =\; CH\; -\; CH\; =\; CH\_2$

The structure is: $\backslash overset\{1\}\{C\}H\_2\; =\; \backslash overset\{2\}\{C\}H\; -\; \backslash overset\{3\}\{C\}H\; =\; \backslash overset\{4\}\{C\}H\_2$. This is a 4-carbon chain with two double bonds at positions 1 and 3. The prefix 'di-' is used before the suffix '-ene'.

IUPAC name: Buta-1,3-diene (or 1,3-Butadiene)

(d) $CH\_3(CH\_2)\_4CH(CH\_2)\_3CH\_3$

This represents a branched alkane. Let's expand it: $CH\_3\; -\; CH\_2\; -\; CH\_2\; -\; CH\_2\; -\; CH\_2\; -\; CH(CH\_3)\; -\; CH\_2\; -\; CH\_2\; -\; CH\_3$. The longest continuous carbon chain has 9 carbons. The chain is numbered from the end closer to the branch. Numbering from the left gives the methyl group at position 6. Numbering from the right gives the methyl group at position 4. Therefore, we number from the right.

The parent alkane is nonane. The methyl group is at position 4.

IUPAC name: 4-Methylnonane

(e) $CH\_3\; -\; CH\; =\; CH\; -\; CH\_2\; -\; CH\; =\; CH\; -\; CH\_2\; -\; CH\; =\; CH\_2$ with C₂H₅ attached to the second carbon of the main chain.

The structure is: $CH\_3\; -\; \backslash overset\{2\}\{C\}H(C\_2H\_5)\; -\; \backslash overset\{3\}\{C\}H\; =\; \backslash overset\{4\}\{C\}H\; -\; \backslash overset\{5\}\{C\}H\; =\; \backslash overset\{6\}\{C\}H\; -\; \backslash overset\{7\}\{C\}H\_2\; -\; \backslash overset\{8\}\{C\}H\; =\; \backslash overset\{9\}\{C\}H\_2$. The longest carbon chain containing the maximum number of double bonds is 9 carbons long. The double bonds are at positions 2, 4, and 7. However, the question implies the ethyl group is attached to the second carbon of a chain that starts with CH3. Let's re-evaluate the structure based on the provided linear formula and the ethyl group attachment. The linear formula has 9 carbons and 3 double bonds. If the ethyl group (C2H5) is attached to the second carbon, the longest chain needs to be identified carefully. Let's assume the structure is: $CH\_3\; -\; \backslash overset\{2\}\{C\}H(C\_2H\_5)\; -\; \backslash overset\{3\}\{C\}H\; =\; \backslash overset\{4\}\{C\}H\; -\; \backslash overset\{5\}\{C\}H\; =\; \backslash overset\{6\}\{C\}H\; -\; \backslash overset\{7\}\{C\}H\_2\; -\; \backslash overset\{8\}\{C\}H\; =\; \backslash overset\{9\}\{C\}H\_2$. The longest chain containing the double bonds would be 9 carbons. Numbering from the right gives the double bonds at 2, 5, and 8. Numbering from the left gives the double bonds at 2, 4, and 7. The ethyl group is at position 2. The numbering from the left is preferred as it gives lower numbers to the double bonds (2, 4, 7 vs 2, 5, 8). The parent chain is nona-2,4,7-triene. The ethyl group is at position 2.

IUPAC name: 2-Ethylnona-2,4,7-triene