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Basics of Organic Chemistry question

2021 · Shift 2 · Q14
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Basics of Organic Chemistry question

2021 · Shift 2 · Q14

JEE AdvancedChemistryBasics of Organic ChemistryMCQ+3 / −1
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below : JEE Advanced 2021 Paper 2 Online Chemistry - Basics of Organic Chemistry Question 26 English Comprehension ClClCl −-− ClClCl (g) →\to→ ClClCl (g)∙{}_{(g)}^ \bullet(g)∙​ + ClClCl (g)∙Δ{}_{(g)}^ \bullet \Delta(g)∙​Δ H ∘{}^\circ∘ = 58 kcal mol −-− 1 H3CH_3CH3​C −-− ClClCl (g) →\to→ H3CH_3CH3​C (g)∙{}_{(g)}^ \bullet(g)∙​ + ClClCl (g)∙Δ{}_{(g)}^ \bullet \Delta(g)∙​Δ H ∘{}^\circ∘ = 85 kcal mol −-− 1 HHH −-− ClClCl (g) →\to→ HHH (g)∙{}_{(g)}^ \bullet(g)∙​ + ClClCl (g)∙Δ{}_{(g)}^ \bullet \Delta(g)∙​Δ H ∘{}^\circ∘ = 103 kcal mol −-− 1For the following reaction CH4CH_4CH4​(g) + Cl2Cl_2Cl2​(g) →light\xrightarrow{light}light​ CH3ClCH_3ClCH3​Cl(g) + HClHClHCl (g) the correct statement is
  1. A
    Initiation step is exothermic with Δ\DeltaΔ H ∘{}^\circ∘=−-− 58 kcal mol −-− 1.
  2. B
    Propagation step involving ∙CH3{}^ \bullet C{H_3}∙CH3​ formation is exothermic with Δ\DeltaΔ H ∘{}^\circ∘=−-− 2 kcal mol −-− 1.
  3. C
    Propagation step involving CH3ClCH_3ClCH3​Cl formation is endothermic with Δ\DeltaΔ H ∘{}^\circ∘ = +27 kcal mol −-− 1.
  4. D
    The reaction is exothermic with Δ\DeltaΔ H ∘{}^\circ∘=−-− 25 kcal mol −-− 1.
View written solutionFree

Correct answer: D

  1. Identify the free-radical chlorination steps

For the reaction

CH4(g)+Cl2(g)→hνCH3Cl(g)+HCl(g)CH_4(g)+Cl_2(g)\xrightarrow{h\nu} CH_3Cl(g)+HCl(g)CH4​(g)+Cl2​(g)hν​CH3​Cl(g)+HCl(g)

The mechanism is:

  • Initiation:
Cl2→2Cl∙Cl_2 \rightarrow 2Cl^\bulletCl2​→2Cl∙
  • Propagation 1:
CH4+Cl∙→CH3∙+HClCH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HClCH4​+Cl∙→CH3∙​+HCl
  • Propagation 2:
CH3∙+Cl2→CH3Cl+Cl∙CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bulletCH3∙​+Cl2​→CH3​Cl+Cl∙
  1. Check option A: Initiation step

Initiation involves breaking the Cl−ClCl-ClCl−Cl bond:

Cl2→2Cl∙Cl_2 \rightarrow 2Cl^\bulletCl2​→2Cl∙

Given:

ΔH∘=+58 kcal mol−1\Delta H^\circ = +58\ \text{kcal mol}^{-1}ΔH∘=+58 kcal mol−1

Bond breaking is endothermic, not exothermic.

So option A is false.

  1. Check option B: Propagation step involving CH3∙CH_3^\bulletCH3∙​ formation

Reaction:

CH4+Cl∙→CH3∙+HClCH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HClCH4​+Cl∙→CH3∙​+HCl

Here:

  • one C−HC-HC−H bond is broken
  • one H−ClH-ClH−Cl bond is formed

Using bond energies:

  • C−HC-HC−H in methane is approximately 105 kcal mol−1105\ \text{kcal mol}^{-1}105 kcal mol−1
  • H−Cl=103 kcal mol−1H-Cl = 103\ \text{kcal mol}^{-1}H−Cl=103 kcal mol−1

Thus,

ΔH=BDE broken−BDE formed=105−103=+2 kcal mol−1\Delta H = \text{BDE broken} - \text{BDE formed} = 105 - 103 = +2\ \text{kcal mol}^{-1}ΔH=BDE broken−BDE formed=105−103=+2 kcal mol−1

So this step is endothermic by +2+2+2 kcal mol−1^{-1}−1, not exothermic.

Hence B is false.

  1. Check option C: Propagation step involving CH3ClCH_3ClCH3​Cl formation

Reaction:

CH3∙+Cl2→CH3Cl+Cl∙CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bulletCH3∙​+Cl2​→CH3​Cl+Cl∙

Here:

  • one Cl−ClCl-ClCl−Cl bond is broken: 58 kcal mol−158\ \text{kcal mol}^{-1}58 kcal mol−1
  • one C−ClC-ClC−Cl bond is formed: 85 kcal mol−185\ \text{kcal mol}^{-1}85 kcal mol−1

Therefore,

ΔH=58−85=−27 kcal mol−1\Delta H = 58 - 85 = -27\ \text{kcal mol}^{-1}ΔH=58−85=−27 kcal mol−1

So this step is exothermic by 272727 kcal mol−1^{-1}−1, not endothermic.

Hence C is false.

  1. Check overall reaction enthalpy

Overall reaction:

CH4+Cl2→CH3Cl+HClCH_4 + Cl_2 \rightarrow CH_3Cl + HClCH4​+Cl2​→CH3​Cl+HCl

Bonds broken:

  • one C−HC-HC−H bond in methane: 105 kcal mol−1105\ \text{kcal mol}^{-1}105 kcal mol−1
  • one Cl−ClCl-ClCl−Cl bond: 58 kcal mol−158\ \text{kcal mol}^{-1}58 kcal mol−1

Bonds formed:

  • one C−ClC-ClC−Cl bond: 85 kcal mol−185\ \text{kcal mol}^{-1}85 kcal mol−1
  • one H−ClH-ClH−Cl bond: 103 kcal mol−1103\ \text{kcal mol}^{-1}103 kcal mol−1

So,

ΔH∘=(105+58)−(85+103)\Delta H^\circ = (105+58) - (85+103)ΔH∘=(105+58)−(85+103) ΔH∘=163−188=−25 kcal mol−1\Delta H^\circ = 163 - 188 = -25\ \text{kcal mol}^{-1}ΔH∘=163−188=−25 kcal mol−1

Thus the overall reaction is exothermic.

So D is correct.

  1. Final conclusion
  • A: False
  • B: False
  • C: False
  • D: True

Therefore, the correct option is:

D\boxed{D}D​

This matches the stored correct answer.

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