Electrophilic Aromatic Substitution (Haloalkanes and Haloarenes) MCQ Questions & Answers in Organic Chemistry | Chemistry

This is Sandmeyer's reaction and the product $$Y$$ is chlorobenzene.

\[\begin{align} & C{{H}_{3}}-C{{H}_{2}}-CHC{{l}_{2}}\xrightarrow[\Delta ]{NaN{{H}_{2}}} \\ & C{{H}_{3}}-CH=CHCl\xrightarrow[\Delta ]{NaN{{H}_{2}}}\underset{\text{Final Product}}{\mathop{C{{H}_{3}}-C\equiv CH}}\, \\ \end{align}\]

Chloroform gets oxidised by air in sunlight to poisonous phosgene gas.
\[2CHC{{l}_{3}}+{{O}_{2}}\xrightarrow{h\nu }\underset{\text{Phosgene}}{\mathop{2COC{{l}_{2}}+2HCl}}\,\]

Since $${S_N}1$$  reactions involve the formation of carbocation as intermediate in the rate determining step, more is the stability of carbocation higher will be the reactivity of alkyl halides towards $${S_N}1$$  route. Since stability of carbocation follows order.

Hence correct order is (II) < (I) < (III)

\[\underset{\left( {{1}^{\circ }}\,\text{halide} \right)}{\mathop{{{H}_{3}}C\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{2}}-C{{H}_{2}}Cl}}\,\]

$${S_N}1$$  reaction proceeds with racemisation.

The correct order of increasing bond length is $$C{H_3}F < C{H_3}Cl < C{H_3}Br < C{H_3}I$$

No explanation is given for this question. Let's discuss the answer together.

\[C{{H}_{3}}C{{H}_{2}}Br+P{{h}_{3}}P\to \]      \[C{{H}_{3}}C{{H}_{2}}{{P}^{+}}P{{h}_{3}}B{{r}^{-}}\xrightarrow[{{C}_{2}}{{H}_{5}}ONa]{{{C}_{6}}{{H}_{5}}Li\,\,\text{or}}\]       \[C{{H}_{3}}CH=PP{{h}_{3}}\underset{{}}{\overset{{}}{\longleftrightarrow}}C{{H}_{3}}\overset{-}{\mathop{C}}\,H\overset{+}{\mathop{P}}\,P{{h}_{3}}\]

The compound $$A$$  is
\[C{{l}_{3}}C.C{{H}_{2}}C{{H}_{2}}Cl\xrightarrow{aq.\,KOH}{{\left( OH \right)}_{3}}C.C{{H}_{2}}C{{H}_{2}}OH\xrightarrow{-2{{H}_{2}}O}HO\,C{{H}_{2}}.C{{H}_{2}}COOK.\]
The compounds $$B$$  is has chiral centre and can be resolved into $$d-$$  and $$l-$$  form.