112.
In the following reaction, product $$(P)$$ is \[R\overset{\begin{smallmatrix}
O \\
\parallel
\end{smallmatrix}}{\mathop{-C-}}\,Cl\xrightarrow[\frac{Pd}{BaS{{O}_{4}}}]{{{H}_{2}}}P\]
$$C{H_3}CHO$$ and $${C_6}{H_5}C{H_2}CHO$$ both being aliphatic aldehydes react with Tollen’s reagent, Fehling solution and Benedict solution. So, these reagents cannot be used to distinguish them.
$$C{H_3}CHO$$ due to the presence of group
reacts with $$NaOH$$ and $${I_2}$$ to give yellow crystals of iodoform while $${C_6}{H_5}C{H_2}CHO$$ does not react with it.
$$C{H_3}CHO + 3{I_2} + 4NaOH \to $$ $$CH{I_3} + HCOONa + 3NaI + 3{H_2}O$$
$${C_6}{H_5}C{H_2}CHO + {I_2} + NaOH \to $$ $${\text{No reaction}}$$
Thus, $$C{H_3}CHO$$ and $${C_6}{H_5}C{H_2}CHO$$ can be distinguished by iodoform test.
117.
The organic product formed in the reaction
\[{{C}_{6}}{{H}_{5}}COOH\xrightarrow[II\,{{H}_{3}}{{O}^{+}}]{I\,LiAl{{H}_{4}}}\]
A
\[{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\]
B
\[{{C}_{6}}{{H}_{5}}COOH\,\And C{{H}_{4}}\]
C
\[{{C}_{6}}{{H}_{5}}C{{H}_{3}}\,\And \,C{{H}_{3}}OH\]
D
\[{{C}_{6}}{{H}_{5}}C{{H}_{3}}\,\And \,C{{H}_{4}}\]
TIPS/FORMULAE :
$$LiAl{H_4}$$ is a reducing agent, it reduces $$ - COOH$$ group to $$ - C{H_2}OH$$ group.
\[{{C}_{6}}{{H}_{5}}COOH\xrightarrow{LiAl{{H}_{4}}}{{C}_{6}}{{H}_{5}}C{{H}_{2}}OH\]
118.
The product formed by the reaction of an aldehyde with a primary amine is
Aldehydes and ketones with $$\alpha $$ - hydrogen atom, when reacted with a base yields aldol which on heating loses water molecule to give $$\alpha ,\beta $$ -unsaturated aldehydes or ketones. This reaction is called aldol condensation reaction.
120.
A compound $$'A'$$ having the molecular formula $${C_5}{H_{12}}O,$$ on oxidation gives a compound $$'B'$$ with molecular formula $${C_5}{H_{10}}O.$$ Compound $$'B'$$ gave a 2, 4-dinitrophenylhydrazine derivative but did not answer haloform test or silver mirror test. The structure of compound $$'A'$$ is
Since $$(B)$$ on reaction with $$2,4{\text{ - }}DNP$$ forms a derivative, it implies that $$(B)$$ has group.
$$(B)$$ gives $$-ve$$ Tollens' test, hence it is not an aldehyde, but it is a ketone.
$$(B)$$ gives $$-ve$$ haloform test, thus it is not a methyl ketone.
$$(B)$$ is formed from the oxidation of $$(A),$$ thus $$(A)$$ is a $${2^ \circ }$$ alcohol, and among the given options,
$$(A)$$ is \[C{{H}_{3}}-C{{H}_{2}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
OH\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}-C{{H}_{3}}\]
and $$(B)$$ is \[C{{H}_{3}}-C{{H}_{2}}\underset{\begin{smallmatrix}
\parallel \\
O
\end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{2}}-C{{H}_{3}}\]