39Why 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..
1. Decreasing order of heat of combustion
2. Decreasing heat of hydrogenation
Please explain the reason also :)
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6 Answers
11. D<C<A<E<B
reason = number of hydrogen available for hyperconjugation...
this the increasing order
2. D>C>B>A
the same reason..
39
1see answer is as B>E>C>A>D for 1 and
C>D>B>A for 2
don't know how isliye poocha :(
23Q1]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
