Identify the structures of P and Q in the fol | vedclass

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Question

(C) The reaction of an alkyne with $H_2O/H^+/Hg^{2+}$ follows Markovnikov's addition to form an enol,which undergoes keto-enol tautomerization to form

Vedclass · Accepted Answer

$P$ = $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_4-CH_3$,$Q$ = $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_3(Br)-CH_3$

Vedclass · Answer

(C) The reaction of an alkyne with $H_2O/H^+/Hg^{2+}$ follows Markovnikov's addition to form an enol,which undergoes keto-enol tautomerization to form a ketone $(P)$.$P$ is $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_4-CH_3$.Next,the reaction with $Br_2/FeBr_3$ is an electrophilic aromatic substitution $(EAS)$. The $-C(=O)CH_2Ar$ group is deactivating and meta-directing,while the $-OCH_3$ group is strongly activating and ortho/para-directing. The $-CH_3$ group is weakly activating and ortho/para-directing. The electrophilic substitution will occur on the ring containing the $-CH_3$ group,specifically at the position ortho to the $-CH_3$ group,as it is more activated than the ring with the deactivating carbonyl group. Thus,$Q$ is $CH_3O-C_6H_4-C(=O)-CH_2-C_6H_3(Br)-CH_3$.

Identify the structures of $P$ and $Q$ in the following reaction sequence:

Similar Questions

The product $(B)$ of the reaction is:
$CH(CO_2H)=CH(CO_2H)$ $\xrightarrow[2 \ mole]{NaOH} (A)$ $\xrightarrow{\text{electrolysis}} (B)$

In the reaction sequence:
$CH_3-C\equiv CH$ $\xrightarrow{CH_3MgBr}$ $CH_4 + (A)$ $\xrightarrow[ (ii) H_2O/H^{\oplus} ]{(i) CO_2}$ $(B)$
$(B)$ will be:

$CH \equiv CH + HBr \to X$,the product $X$ is:

$(A)$ $\xrightarrow[dil. H_2SO_4]{HgSO_4} (B)$ $\xrightarrow{LiAlH_4} \underset{\text{racemic mixture}}{(C)}$
$\therefore$ reactant $(A)$ is

The shortest $C-C$ bond length is present in:

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