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Basics of Organic Chemistry question

2021 · Shift 2 · Q13
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Basics of Organic Chemistry question

2021 · Shift 2 · Q13

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 25 English ComprehensionCorrect match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is

Column J
Molecule		 Column K
BDE (kcal mol−1mo{l^{ - 1}}mol−1)
(P) H-CH(CH3C{H_3}CH3​) 2 (i) 132
(Q) H-CH 2 Ph (ii) 110
(R) H-CH=CH 2 (iii) 95
(S) H-C ≡\equiv≡ CH (iv) 88
  1. A
    P - iii, Q - iv, R - ii, S - i
  2. B
    P - i, Q - ii, R - iii, S - iv
  3. C
    P - iii, Q - ii, R - i, S - iv
  4. D
    P - ii, Q - i, R - iv, S - ii
View written solutionFree

Correct answer: A

  1. Identify the types of C–H bonds

From Column J:

  • (P)(P)(P) H−CH(CH3)2\mathbf{H-CH(CH_3)_2}H−CH(CH3​)2​ : the marked C–H is on the central carbon of propane-like structure, i.e. a secondary alkyl C–H bond.
  • (Q)(Q)(Q) H−CH2Ph\mathbf{H-CH_2Ph}H−CH2​Ph : this is a benzylic C–H bond.
  • (R)(R)(R) H−CH=CH2\mathbf{H-CH=CH_2}H−CH=CH2​ : this is an allylic C–H bond.
  • (S)(S)(S) H−C≡CH\mathbf{H-C\equiv CH}H−C≡CH : this is an acetylenic (sp C–H) bond.
  1. Use BDE trends

Bond dissociation energy decreases when the radical formed is more stable.

Also, greater sss-character makes a C–H bond shorter and stronger:

sp>sp2>sp3sp > sp^2 > sp^3sp>sp2>sp3

So:

  • Acetylenic C–H (spspsp) should have the highest BDE.
  • Benzylic and allylic C–H bonds have lower BDE due to resonance stabilization of radicals.
  • Benzylic radical is more stabilized than allylic/secondary alkyl in typical comparison, so benzylic C–H has the lowest BDE among these.
  1. Match each with numerical values

Given BDE values:

  • (i) 132(i)\ 132(i) 132
  • (ii) 110(ii)\ 110(ii) 110
  • (iii) 95(iii)\ 95(iii) 95
  • (iv) 88(iv)\ 88(iv) 88

Now assign:

(S) H−C≡CH\mathbf{H-C\equiv CH}H−C≡CH

Acetylenic C–H is strongest due to high sss-character.

S→132⇒S−iS \to 132 \Rightarrow S - iS→132⇒S−i

(Q) H−CH2Ph\mathbf{H-CH_2Ph}H−CH2​Ph

Benzylic radical is strongly resonance stabilized, so lowest BDE.

Q→88⇒Q−ivQ \to 88 \Rightarrow Q - ivQ→88⇒Q−iv

(R) H−CH=CH2\mathbf{H-CH=CH_2}H−CH=CH2​

Allylic radical is resonance stabilized, so BDE is low, but higher than benzylic.

R→110⇒R−iiR \to 110 \Rightarrow R - iiR→110⇒R−ii

(P) H−CH(CH3)2\mathbf{H-CH(CH_3)_2}H−CH(CH3​)2​

Secondary alkyl C–H has BDE around mid-90s.

P→95⇒P−iiiP \to 95 \Rightarrow P - iiiP→95⇒P−iii

  1. Final matching

P−iii, Q−iv, R−ii, S−iP-iii,\ Q-iv,\ R-ii,\ S-iP−iii, Q−iv, R−ii, S−i

This corresponds to Option A.

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