Animated Solution for Chemistry - Organic Chemistry: $A(C_4{H}_{8}C{l}_2)\underset{\text{373 K}}{\overset{\text{Hydrolysis}}{\to }}B(C_4{H}_{8}O)$ $B$ reacts with hydroxyl amine but does not give Tollen's test. Identify $A$ and $B$.

The problem presents us with a fascinating organic chemistry puzzle. We start with an unknown dihalide, compound A ($C_4{H}_{8}C{l}_2$), which undergoes hydrolysis to form compound B ($C_4{H}_{8}O$).

To identify these compounds, we must rely on the chemical tests provided for compound B.

First, we are told that compound B reacts with hydroxylamine ($N{H}_{2}OH$). This is a classic test for the presence of a carbonyl group ($>C=O$). Therefore, compound B must be either an aldehyde or a ketone.

Next, the problem states that compound B does not give a positive Tollen's test.

Tollen's reagent (ammoniacal silver nitrate) is a mild oxidizing agent. It is strong enough to oxidize aldehydes into carboxylic acids (forming a beautiful silver mirror in the process), but it is not strong enough to oxidize ketones.

Since compound B fails this test, we can definitively conclude that B is a ketone.

Now that we know B is a ketone, let's look at its molecular formula: $C_4{H}_{8}O$.

A ketone must have its carbonyl carbon attached to two other carbon atoms. With only four carbon atoms available, there is only one possible structure for a ketone: 2-butanone.

So, compound B is 2-butanone.

With the structure of B in hand, we can deduce the structure of A.

Compound A is a dihalide ($C_4{H}_{8}C{l}_2$) that yields a ketone upon hydrolysis. When a dihalide is hydrolyzed to form a carbonyl compound, it must be a geminal dihalide (both halogen atoms attached to the same carbon).

Specifically, to form a ketone, the halogens must be on a non-terminal carbon. Since the carbonyl group in 2-butanone is at the second carbon (C-2), the two chlorine atoms in compound A must also be attached to C-2.

Therefore, compound A is 2,2-dichlorobutane.

Let's trace the chemical journey from A to B.

When 2,2-dichlorobutane is heated with water at 373 K, it undergoes nucleophilic substitution. The two chlorine atoms are replaced by two hydroxyl ($-OH$) groups, forming an intermediate called butane-2,2-diol.

This intermediate is a gem-diol. Having two highly electronegative oxygen atoms attached to the same carbon creates intense electronic repulsion and steric hindrance, making the molecule highly unstable.

To achieve stability, the gem-diol spontaneously loses a molecule of water (dehydration) to form the stable carbon-oxygen double bond of the ketone.

This elegant sequence perfectly explains why A is 2,2-dichlorobutane and B is 2-butanone.