Correct match of the {C}- {H} bonds (shown in bold) in Column {J} with their BDE in Column {K} is(S) {H}- {C} \equiv {CH} Column J Molecule Column K BDE (kcal mol ^{-1} ) (P) {H}- {CH}( {CH}_{3})_{2} (i) 132 (Q) {H}- {CH}_{2} {Ph} (ii) 110 (R) {H}- {CH}= {CH}_{2} (iii) 95 (S) {H}- {C} \equiv {CH} (iv) 88 [JEE Adv 2021 P2]

Correct Answer is : P−iii,Q−iv,R−ii,S−i

{P}- {iii}, {Q}- {iv}, {R}- {ii}, {S}- {i}

{P}- {i}, {Q}- {i}, {R}- {iii}, {S}- {iv}

{P}- {iii}, {Q}- {ii}, {R}- {i}, {S}- {iv}

{P}- {ii}, {Q}- {i}, {R}- {iv}, {S}- {iii}

Basics of Organic Chemistry

Paragraph for next 2 questions
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 :

Cl−Cl(g)⟶Cl∙(g)+Cl∙(g)‌∆H∘=58kcalmol−1
H3C−Cl(g)⟶H3C∙(g)+Cl∙(g)‌∆H∘=85kcalmol−1
H−Cl(g)⟶H∙(g)+Cl*(g)‌∆H∘=103kcalmol−1

Section: Chemistry

Question : 2 of 39

Marks: +1, -0

Correct match of the C−H bonds (shown in bold) in Column J with their BDE in Column K is(S) H−C≡CH

Column J Molecule	Column K BDE (kcal mol ‌−1 )
(P) H−CH(CH3)2	(i) 132
(Q) H−CH2Ph	(ii) 110
(R) H−CH=CH2	(iii) 95
(S) H−C≡CH	(iv) 88

[JEE Adv 2021 P2]

P−iii,Q−iv,R−ii,S−i
P−i,Q−i,R−iii,S−iv
P−iii,Q−ii,R−i,S−iv
P−ii,Q−i,R−iv,S−iii

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