Question no 1 with reasoning
Is option a the correct answer ?
Because when you’ll draw enol form of A then it will have just resonance , but when you’ll draw enol form of B then B will be aromatic
Yes and what about question 11 i have read than more steric hindrance enol form is less then why c option is correct
Because of more resonance enol of C will be highest .
Steric hinderance at C between the two C=O will decrease the enol content because hinderance over there will lead to bad H bonding ,for eq in option D the middle most c is hindered so that will decrease the enol content but in option C the side C are hinderance so that will not effect much so just see resonance
In q 18 i think it should be B because all the alpha H will be replaced but 1 of the alpha H is bridge head , so it can’t be replaced
So only 2 of the alpha H will be replaced
19th -> during tautomerism the 2 forms are keto form and enolate form
In Enolate form OH attached to a C having double bond
So basically this question is , which of them can’t tautomerise
I am right now in coaching i will confirm the answers in evening
Answer given in answer key is A for 18th and D for 19th
Yes , for 19th D doesn’t have any alpha H so it can’t tautomerise
And for 18th as shown in post#6 there are 2 alpha H i.e option A
Do you have solution booklet ? If yes , then you can check which of the above structure which i drew is missing there
I have shown 4 possibilities, if any of those seems incorrect then do let me know .
I think solution booklet has mentioned wrong
Because in enol form of phenol there is benzene ring and in keto form it is non aromatic
Q2 ) it is exactly the same question you asked
So 1st compund should have more enol content due to more resonance
I asked because answer key didnt provided answer of this question but i knew first will be more