$\ce{A ->[(i)EtMgCl][(ii)H2O] Et3C-OH}$

A can be:

a) $\ce{COCl2}$

b) $\ce{EtCOOEt}$

c) $\ce{Et2CO}$

d) $\ce{All}$

The answer should be (d) according to me, but is marked as (a) in the answer key.

Option (c) obviously gives the required product, as does (b). I derived this scheme for (a)-

$$\ce{2COCl2 + 2EtMgCl -> Et2CO + 2MgCl2}$$ $$\ce{Et2CO + EtMgCl -> Et3C-OMgCl ->[H2O] Et3C-OH + Mg(OH)Cl}$$

Could one please confirm if the above reaction scheme is correct? I personally could not find any flaw in it.

  • $\begingroup$ Seems like a goof up by the book: organic-chemistry.org/namedreactions/grignard-reaction.shtm $\endgroup$ – Yusuf Hasan Feb 21 at 6:48
  • 2
    $\begingroup$ I agree, certainly b and c are viable answers $\endgroup$ – Waylander Feb 21 at 8:06
  • $\begingroup$ I dont see how this is related to the 'synthesis' of Grignard reagent. Well if we were impractically too critical about the fact that the reagents aren't provided in excess, then also the answer should have been (c) and not (a). So I agree too, unless someone comes to enlighten us $\endgroup$ – Sir Arthur7 Feb 21 at 13:28

The answer is all of them, thus (d). In the case of $\ce{COCl2}$ (a), the Grignard reagent displaces both chloride ions in successive nucleophilic substitutions, forming diethyl ketone; then the reaction continues as it would for a typical ketone.

The displacement if a chloride ion in the case of an acyl chloride is described here. Reaction with $\ce{COCl2}$ would simply involve two stages of this process.

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