Organic

u answered both wrongly :(

Why is smsm wrong? In Q1 D is indeed the most substituted alkene, it should have the least heat of hydrogenation...
And D is the least substituted in Q2, so it should have the greatest heat of combustion..

see answer is as B>E>C>A>D for 1 and
C>D>B>A for 2

don't know how isliye poocha :(

Q1]here carbon and hydrogen atoms are same in no

now ,
A=5 Hyperconjugating H
B = 2 Hyperconjugating H
C = 4 Hyperconjugating H
D =9 Hyperconjugating H
E = 3 Hyperconjugating H

more the unstability , more is the potential energy , and thus more is theenergy evolved on combustion

D> A >C >E >B ..stability

hence B > E > C > A >D ...heat of combustion

Q 2] same logic , more stable the double bond , less is its PE , so less is the heat given out on hydrogenation
now all have 4 pi bonds, and each have similar phenyl grps, so we can compare the remaining pi bond

double bond in A is most stable as it is in conjugation with the ring
also,
A = 8 HC + conjugation i.e resonance
B = 9 HC
C=4
D= 2

resonance dominates HC
so A > B > C > D stability ...
hence heat of hydrogenation is in the order D > C > B > A