Yes. But how?
actually electronegatvity of sp carbon is same as that of nitrogen . so terminal hydrogen of alkyens can be assumed to be bonded to a hydrogen
Sorry, but I didn't get what ur trying to say
How can terminal hydrogen of alkynes can be assumed that they are bonded to hydrogen?
Sorry, actually I Just realised answer given for 12 is (c)
I mean the terminal hydrogen and hydrongen bonded to a nitrogen are equally acidic
sorry that is nitrogen not hydrogen @Lalit_Kumar4
Yes answer is c for Q14
Ohk so more is the electronegativity, the stronger hydrogen bond it will form . Right? @Mohammad_Monish
if still any doubt let me know.
Ok got it
For 16 wouldnt statement 3 be correct?
yes 1and2 are FI
7 .a) due to H bonding enol form of 1 will be more stable
b) enol of 3 will be phenol
But in 7.a) all 3 can for enol form, why would 1 have more percentage? Wouldn't the enol form of 2 be more stable as it would be surrounded to two C-O ( one is ketone and the other is ether)?
Also in b) would 2>3 as 2 wil have more resonating structure as compared to 3
firt of all draw the enol form
see if any H bonding
enol of 1 will be more stable due to H bonding and conjugation
while in second there will be cross conjugation
b) yes 2 will have more no of resonating structures and also H bonding
then why 3 will have more enol content ?
Got it bro.
And @pratyaksh_tyagi @Namish_Garg2 @Satyendra_Srivastva @Siddhant_Mudholkar @Anish_Saparia could u look at Q7 (b)
3 is aromatic , (phenol)
6)a x is shortest q is longest(totally single bond character)
7b) yes as @pratyaksh_tyagi bro told
keto form is non aromatic but its enol form is aromatic it have tendeny to exist in eonl form so more enol percentage
whats the answer for 6th one??
Got it bro
Answer for 6 is (c)