Question

(a) Write the structures of A and B in the following reactions:
$\text{C}{{\text{H}}_{\text{3}}}\text{COCl}\xrightarrow{{{\text{H}}_{\text{2}}}\text{Pd-BaS}{{\text{O}}_{\text{4}}}}\text{A}\xrightarrow{\text{N}{{\text{H}}_{\text{2}}}\text{OH}}\text{B}$
\[\text{C}{{\text{H}}_{\text{3}}}\text{MgBr}\xrightarrow[\text{2}\text{.}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}]{\text{1}\text{.C}{{\text{O}}_{\text{2}}}}\text{A}\xrightarrow{\text{PC}{{\text{l}}_{\text{5}}}}\text{B}\]
B. Distinguish between:
(i) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\text{COC}{{\text{H}}_{\text{3}}}$and ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\text{CHO}$
(ii) $\text{C}{{\text{H}}_{3}}\text{COOH}$and $\text{HCOOH}$
(c) Arrange the following in the increasing order of boiling points.
 $\text{C}{{\text{H}}_{3}}\text{CHO}$, $\text{C}{{\text{H}}_{3}}\text{COOH}$, and $\text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}$

Answer 1

Hint:The Grignard’s reagent is used to prepare different compounds as it reacts with alkanes, alcohols, carboxylic acids, amine, and aldehydes and ketones to form a number of compounds.

Complete answer:
$\text{C}{{\text{H}}_{\text{3}}}\text{COCl}\xrightarrow{{{\text{H}}_{\text{2}}}\text{Pd-BaS}{{\text{O}}_{\text{4}}}}\text{A}\xrightarrow{\text{N}{{\text{H}}_{\text{2}}}\text{OH}}\text{B}$
In the above reaction, the combination of palladium metal along with barium sulphate is called “Rosenmund Reduction'' and it is used along with hydrogen gas for the hydrogenation of acyl chloride to acetaldehyde. Acetaldehyde in reaction with hydroxylamine forms Acetaldeoxime.
$\text{C}{{\text{H}}_{\text{3}}}\text{COCl}\xrightarrow{{{\text{H}}_{\text{2}}}\text{Pd-BaS}{{\text{O}}_{\text{4}}}}\text{C}{{\text{H}}_{\text{3}}}\text{CHO}\xrightarrow{\text{N}{{\text{H}}_{\text{2}}}\text{OH}}\text{C}{{\text{H}}_{\text{3}}}\text{- CH = N - OH}$
\[\text{C}{{\text{H}}_{\text{3}}}\text{MgBr}\xrightarrow[\text{2}\text{.}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}]{\text{1}\text{.C}{{\text{O}}_{\text{2}}}}\text{A}\xrightarrow{\text{PC}{{\text{l}}_{\text{5}}}}\text{B}\]
In the above reaction, the “Methyl Magnesium Bromide” is called the Grignard reagent that is an organometallic compound that adds onto a carbonyl group. The carbon atom attached to the magnesium ion acts as a nucleophile, attacking the electrophilic carbon that is present within the polar bond of the carbonyl group. Here methyl magnesium bromide reacts with carbon dioxide and then the product is hydrolysed to form acetic acid. Then acetic acid gets chlorinated by phosphorous pentachloride to form acetyl chloride.
\[\text{C}{{\text{H}}_{\text{3}}}\text{MgBr}\xrightarrow[\text{2}\text{.}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}]{\text{1}\text{.C}{{\text{O}}_{\text{2}}}}\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\xrightarrow{\text{PC}{{\text{l}}_{\text{5}}}}\text{C}{{\text{H}}_{\text{3}}}\text{COCl}\]
B. Distinguish between:
(i) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\text{COC}{{\text{H}}_{\text{3}}}$And${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\text{CHO}$- Benzaldehyde reacts with Tollen’s Reagent to form the silver mirror, because it is an aldehyde while acetophenone being a ketone does not form the silver mirror with the Tollen’s Reagent.
(ii) $\text{C}{{\text{H}}_{3}}\text{COOH}$ and $\text{HCOOH}$- Due to the presence of the “keto-methyl” group in acetic acid, it responds to the iodoform test while acetic acid does not respond to the same.
(c) Arrange the following in the increasing order of boiling points.
$\text{C}{{\text{H}}_{3}}\text{CHO}$, $\text{C}{{\text{H}}_{3}}\text{COOH}$, and$\text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{\text{2}}}\text{OH}$: Among the given compounds, the highest boiling point will be that of acetic acid as there can be both intermolecular and intramolecular hydrogen bonding and it has the highest molecular weight among the three compounds. The second highest boiling point will be that for ethyl alcohol and finally the boiling point will be the lowest for that of acetaldehyde due to the absence of hydrogen bonding.

Note:
Hydrogen bonding is a special type of dipole-dipole interaction between molecules having a hydrogen atom bonded to strongly electronegative atoms and the electronegative atom of the adjacent molecule.