- A$MeCOCH _{2} Cl$
- B

- C$C _{6} H _{5} CH _{2} CH _{2} Cl$
- ✓$MeOCH _{2} Cl$

$MeOCH _{2} Cl$ $\xrightarrow{Slow}$$\underset{(i)}{\mathop{Me\underset{\,\,\centerdot \,\,\centerdot }{\overset{\centerdot \,\,\centerdot }{\mathop{O}}}\,\overset{+}{\mathop{C{{H}_{2}}}}\,+C{{l}^{-}}}}\,$ $\leftrightarrow $ $\underset{(ii)}{\mathop{Me-\underset{\,\,\centerdot \,\,\centerdot }{\overset{+}{\mathop{O}}}\,=C{{H}_{2}}}}\,$
Though $(ii)$ contains positive charge on oxygen, octet around each atom in $(ii)$ is complete, structure $(ii)$ is more stable than $(i)$.
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$(A)$ All are isoelectronic
$(B)$ All have the same nuclear charge
$(C)$ $\mathrm{O}^{2-}$ has the largest ionic radii
$(D)$ $\mathrm{Mg}^{2+}$ has the smallest ionic radii
Choose the most appropriate answer from the options given below :
Product $(B)$ is
$CH_3-CH_2-C \equiv CH + HCl \to B \xrightarrow{{HCl}} C$