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The following question was given in chemistry today , August 2020 edition , page 47 , problem 9. The answer to this problem is (a). enter image description here

My Attempt

enter image description here

In the third step , the carbon electrophile can attack at ortho position to carbon substituent to give a 6 membered ring.

But , the O-Methyl group is ortho para directing. The resonance structures are as follows.

enter image description here

The given answer is , a , is possible only if we consider O-Methyl group on Benzene as meta directing group.

My Question:

Is methoxy group on ANISOLE Meta directing ?

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  • $\begingroup$ The presence of the OMe activates the whole ring to electrophilic attack. The o & p positions are most activated but m is also activated. Your attempted mechanism is correct $\endgroup$
    – Waylander
    Nov 19, 2020 at 8:05
  • $\begingroup$ Anisole is a whole. The question in the last line is wrong in terms, Even if we know what you mean. $\endgroup$
    – Alchimista
    Nov 19, 2020 at 8:16
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    $\begingroup$ My question is how did the CH2 group para to the methoxy shift to a meta position? Unless this question represents a surprising mechanism, answer d seems to make the most sense. $\endgroup$
    – James Gaidis
    Nov 19, 2020 at 14:26
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    $\begingroup$ Are you not concerned how an E-double bond becomes Z in the cyclized product? $\endgroup$
    – user55119
    Nov 19, 2020 at 15:58
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    $\begingroup$ The double bond geometry is not specified in the original question. $\endgroup$
    – Waylander
    Nov 20, 2020 at 11:52

1 Answer 1

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Now that the general consensus of the Commenters is that meta cyclization is viable and that the answer is (d) and not (a), the OP and other readers should be concerned as to how an (E)-double bond can cyclize in a Friedel-Crafts (F-C) reaction. It doesn't! Prior isomerization to the (Z)-configuration is required.
(E)-Acyl chloride 1 must rotate about the red β-γ single bond and $\ce{AlCl3}$ must complex with the carbonyl oxygen forming species 2. [One may also invoke $\ce{AlCl3}$ complexation with the chlorine and rationalize the subsequent steps via an acylium ion and conjugated ketenes. The principle is the same.] Loss of a proton at the γ-position of conformation 2 provides the β,γ-(Z)-double bond of 3. Now deconjugated acyl chloride 4 is ready for F-C cyclization to intermediate 7 and onto 7-methoxynaphthalen-1-ol (9), aka (d). Alternatively, rotation about the red bond in 3 leads to conformation 5, which, upon γ-protonation leads to α,β-(Z)-conjugated acyl chloride 6. Subsequent F-C cyclization via intermediate 8 affords 9. Incidently, base is not required to produce 9 from 7 or 8. F-C conditions are sufficient.



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