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Question Number 57785 by Tinkutara last updated on 11/Apr/19

Commented by rahul 19 last updated on 14/Apr/19

$$1$$

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

$$b^{\prime }cozofthepresenceofE.W.Gwhich$$$$resiststhefurther{rex}^n.\left(incompound1\right).$$

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!☺��

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