In the following reactions, {P}, {Q}, {R} and {S} are the major products. {CH}₃{CH}₂{CH}({CH}₃){CH}₂{CN}[(ii)\,{PhMgBr},\,{then}\,{H₂O}]{(i)\,{PhMgBr},\,{then}\,{H₃O}^{}}P {Ph-H}^{+}{{O}}{{}{{CH₃C}}}\,{Cl}[(ii)\,{PhMgBr},\,{then}\,{H₂O}]{(i)\,{anhyd.}\,{AlCl}₃Q {CH₃CH₂{{O}}{{}{{C}}}{Cl}[(ii)\,{PhMgBr},\,{then}\,{H₂O}]{1/2({PhCH₂)₂{Cd}}R {PhCH}₂{CHO}[(iii)\,{HCN}\;\;(iv)\,{H₂SO₄]{(i)\,{PhMgBr},\,{then}\,{H₂O}\;(ii)\,{CrO}₃,\,{dil.}\,{H₂SO₄}S The correct statement(s) about {P}, {Q}, {R} and {S} is(are) [JEE Adv 2023 P1]

Both $\mathbf{P}$ and $\mathbf{Q}$ have asymmetric carbon(s)
Both $\mathbf{Q}$ and $\mathbf{R}$ have asymmetric carbon(s)
Both $\mathbf{P}$ and $\mathbf{R}$ have asymmetric carbon(s)
$\mathbf{P}$ has asymmetric carbon(s), $\mathbf{S}$ does not have any asymmetric carbon

Solution:

\mathrm{CH_3CH_2CH(CH_3)CH_2CN}\xrightarrow[(ii)\,\mathrm{PhMgBr}_2,\mathrm{H_2O}]{(i)\,\mathrm{PhMgBr},\mathrm{H_3O}^{+}}\underset{(\mathrm{P})}{\mathrm{CH_3CH_2}-\underset{*}{\overset{\mathrm{CH}_3}{\overset{\vert}{\mathrm{C}}}}-\mathrm{CH_2}-\underset{\mathrm{Ph}}{\underset{\vert}{\overset{\mathrm{OH}}{\overset{\vert}{\mathrm{C}}}}}-\mathrm{Ph}}

\mathrm{CH_3-CH_2-}\overset{\mathrm{O}}{\overset{\parallel}{\mathrm{C}}}-\mathrm{Cl}\xrightarrow[(ii)\,\mathrm{PMgBr},\,\mathrm{H_2O}]{(i)\,\frac{1}{2}(\mathrm{PhCH_2})_2\mathrm{Cd}}\underset{(\mathrm{R})}{\mathrm{CH_3-CH_2}-\underset{\mathrm{Ph}}{\underset{\vert}{\overset{\mathrm{OH}}{\overset{\vert}{\mathrm{C}_{*}}}}}-\mathrm{CH_2-Ph}}

\mathrm{PhCH_2CHO}\xrightarrow[(ii)\,\mathrm{CrO}_3,\,\text{dil.}\,\mathrm{H_2SO}_4\;\;(iii)\,\mathrm{HCN}]{(i)\,\mathrm{PhMgBr},\,\mathrm{H_2O}}\mathrm{Ph-CH_2-}\underset{\mathrm{CN}}{\underset{\vert}{\overset{\mathrm{OH}}{\overset{\vert}{\mathrm{C}}}}}-\mathrm{Ph}\xrightarrow[\mathrm{H_2SO}_4,\;\Delta]{}\underset{(\mathrm{S})}{\mathrm{Ph-CH}=\underset{\mathrm{CN}}{\underset{\vert}{\mathrm{C}}}-\mathrm{Ph}}

Go to Question:

Prev Question

Next Question