Question Number 57785 by Tinkutara last updated on 11/Apr/19 | ||
Commented by rahul 19 last updated on 14/Apr/19 | ||
$$\mathrm{1} \\ $$ | ||
Commented by Tinkutara last updated on 13/Apr/19 | ||
How? | ||
Answered by rahul 19 last updated on 13/Apr/19 | ||
Commented by rahul 19 last updated on 16/Apr/19 | ||
$${b}'{coz}\:{of}\:{the}\:{presence}\:{of}\:{E}.{W}.{G}\:{which} \\ $$$${resists}\:{the}\:{further}\:{rex}^{{n}} .\:\left({in}\:{compound}\:\mathrm{1}\right). \\ $$ | ||
Commented by Tinkutara last updated on 14/Apr/19 | ||
Why the second product is chosen not first? | ||
Commented by Tinkutara last updated on 16/Apr/19 | ||
Ok but why O of benzene ring not get bonded to AlCl3 initially? | ||
Commented by rahul 19 last updated on 16/Apr/19 | ||
1)it will lead to the same compound you started with. at the end Oh will be form . 2) show me your final product what exactly you want to do. | ||
Commented by Tinkutara last updated on 18/Apr/19 | ||
How do you removed OH on the carbonyl carbon? Please check whether I am correct: | ||
Commented by Tinkutara last updated on 18/Apr/19 | ||
Commented by rahul 19 last updated on 18/Apr/19 | ||
read my 1st comment.... I'm saying that C=O grp is electron withdrawing grp. → deactivates the ring → E.A.S not possible or we can say it will never be the major product as asked in the Question. | ||
Commented by Tinkutara last updated on 18/Apr/19 | ||
Ohh yes, and how did you remove the OH group on the carbonyl carbon on the product with ✔ ? | ||
Commented by rahul 19 last updated on 18/Apr/19 | ||
AlCl3 which is regenerated will remove OH . | ||
Commented by Tinkutara last updated on 19/Apr/19 | ||
Thanks a lot!☺ | ||