What was the JEE Advanced 2021 matching question on C-H bond dissociation enthalpies?
The question required matching (P) H-CH(CH₃)₂, (Q) H-CH₂Ph, (R) H-CH=CH₂ and (S) H-C≡CH to BDE values 132, 110, 95 and 88 kcal mol^{-1}. The correct option is A (P-iii, Q-iv, R-ii, S-i). This tests one skill: rank the stability of the carbon radical formed on homolysis, then read the matching value.
The official logic treats the four molecules as a single stability ladder instead of four isolated facts. Once ranked, the values fall into place in under 90 seconds.
How does radical stability decide BDE in hydrocarbons?
BDE is inversely proportional to the stability of the carbon radical produced on homolytic cleavage. Homolytic cleavage gives C• + H•, so lower BDE corresponds to more stable C•. The stability order is resonance-stabilised benzyl > secondary alkyl > vinylic > sp-hybridised alkynyl.
These map to the anchor values benzyl ≈ 88, 2° alkyl ≈ 95, vinyl ≈ 110, terminal alkyne ≈ 132 kcal mol^{-1}. Fix this ladder and every future matching list becomes two steps: draw the radical, place it on the ladder, assign the value.
What is the official step-by-step solution to the 2021 JEE Advanced BDE matching question?
Bond dissociation enthalpy depends on the stability of the radical formed after homolytic cleavage of the C-H bond. More stable radical formation means lower BDE.
P: (CH₃)₂CH• (secondary alkyl) → 95 kcal mol^{-1} matches (iii).
Q: PhCH₂• (benzyl, resonance stabilised) → 88 kcal mol^{-1} matches (iv).
R: CH₂=CH• (vinylic) → 110 kcal mol^{-1} matches (ii).
S: HC≡C• (sp carbon, 50 % s-character) → 132 kcal mol^{-1} matches (i).
The matching is P-iii, Q-iv, R-ii, S-i. Therefore the correct option is A.
Search every JEE Main paper from 2002 and every Advanced paper from 2007 by chapter in the past-paper archive for similar questions with worked solutions.
How does confusing acidity with BDE produce the wrong option in this question?
One procedural error is to import the acidity (heterolytic) order instead of the radical-stability (homolytic) order. Terminal alkyne is most acidic because the sp carbanion is stable, yet it has the highest BDE because the sp radical is unstable.
This swaps the values for R and S or assigns 132 to the vinyl position, producing options such as C or D. The correct method is to always draw the carbon radical first, never the carbanion. Write the dot on carbon, count resonance or hybridisation, and the BDE falls into place.
Which practice questions test the same radical stability logic for BDE in hydrocarbons?
Apply the identical ladder to these three problems.
Question 1: Arrange BDE of allylic, vinylic and benzylic C-H bonds with justification.
Benzylic < allylic < vinylic. The benzyl radical enjoys resonance delocalisation into the aromatic ring. The allyl radical has resonance but lacks the extra aromatic stabilisation. The vinylic radical has no resonance and sits on an sp² carbon.
Question 2: Match BDE of CH₄, CH₃CH₃ and CH₂=CH₂ with values 105, 98 and 110 kcal mol^{-1}.
CH₃CH₃ (primary alkyl radical) matches 98. CH₄ (methyl radical) matches 105. CH₂=CH₂ (vinylic radical) matches 110. The stability sequence is primary alkyl > methyl > vinyl.
Question 3: Predict which C-H in Ph-CH₂-CH=CH₂ has the lowest BDE and why.
The benzylic C-H on the Ph-CH₂ group has the lowest BDE. Homolysis produces a radical stabilised by both phenyl resonance and allylic resonance. The vinylic positions produce far less stable radicals.
How do you solve BDE matching lists in under 90 seconds in JEE Advanced?
Write the radical for every bold C-H before looking at the numbers. Memorise only the four anchor values 88 (benzyl), 95 (2°), 110 (vinyl), 132 (terminal alkyne). In matching lists, eliminate options that violate the stability ladder benzyl > 2° alkyl > vinyl > sp-alkynyl.
This check takes 15 seconds and the remaining choice is correct. When a new functional group appears, reduce its radical to the nearest neighbour on the ladder and the value follows.
When you get stuck on a similar question, photograph the doubt for a step-by-step solution with a diagram when needed.
Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).
For a worked example of the same idea, see Atomic Structure JEE 2021: Helium Atom Recoil Velocity.
Frequently asked questions
What was the JEE Advanced 2021 matching question on hydrocarbons BDE?
It required matching (P) H-CH(CH₃)₂, (Q) H-CH₂Ph, (R) H-CH=CH₂ and (S) H-C≡CH to 132, 110, 95 and 88 kcal mol^{-1}. The correct option is P-iii, Q-iv, R-ii, S-i based on the stability of the carbon radical formed after homolysis.
How does radical stability decide BDE values in hydrocarbons JEE questions?
BDE is inversely proportional to the stability of the carbon radical produced on homolytic cleavage. More stable radicals correspond to lower BDE. The order is resonance-stabilised benzyl (88) > secondary alkyl (95) > vinylic (110) > sp-hybridised alkynyl (132).
Why is BDE of terminal alkyne highest even though it is most acidic?
Terminal alkyne has the highest BDE (132 kcal mol^{-1}) because the sp-hybridised radical is unstable due to high s-character. Acidity order uses carbanion stability after heterolytic cleavage, which is different. Always draw the radical, never the anion, for BDE questions.
How to solve BDE matching questions in under 90 seconds in JEE Advanced?
Write the radical for each C-H bond, rank it on the stability ladder using the four anchor values (88 benzyl, 95 2°, 110 vinyl, 132 alkyne), then match. Eliminate options violating benzyl > 2° alkyl > vinyl > alkynyl order. This method works for every similar hydrocarbons question.
What practice questions test BDE and radical stability for JEE hydrocarbons?
Arrange BDE of allylic, vinylic and benzylic C-H bonds; match BDE of CH₄, CH₃CH₃ and CH₂=CH₂ to 105, 98 and 110; predict the lowest BDE carbon in Ph-CH₂-CH=CH₂. All follow the same radical stability logic.