Question

An optically active alkyl halide C\(_4\)H\(_9\)Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH\(_2\). During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :

• A. But-2-yne
• B. Butan-2-ol
• C. Butan-2-one
• D. Butan-1-al

Hint:

For alkene to alkyne conversions, always check for the necessary reagents such as NaNH\(_2\) and the correct conditions for hydration (HgSO\(_4\)).

Answer 1

The Correct Option is C

The reaction sequence is as follows:

(A) Reacts with KOH to form alkene [B] (Elimination of HBr).
(B) Reacts with Br\(_2\) to give dibromide [C].
(C) Dibromide reacts with NaNH\(_2\) to form [D] (Alkyne formation).
(D) The alkyne [D] undergoes hydration with mercuric sulfate to give [E].

Answer 2

The problem describes a series of reactions starting from an optically active alkyl halide, C\(_4\)H\(_9\)Br [A], and asks for the IUPAC name of the final product, [E]. We will identify each intermediate compound ([A], [B], [C], and [D]) to determine the structure of the final compound [E].

Concept Used:

This problem involves several key organic reactions:

1. Elimination Reaction (E2): Treatment of an alkyl halide with a strong base like alcoholic KOH leads to dehydrohalogenation. According to Zaitsev's rule, the major product is the more substituted (more stable) alkene.
2. Electrophilic Addition: Alkenes react with halogens like bromine (Br\(_2\)) to form vicinal dihalides through an electrophilic addition mechanism.
3. Double Dehydrohalogenation: Vicinal dihalides react with a very strong base like sodamide (NaNH\(_2\)) to undergo two successive elimination reactions, forming an alkyne.
4. Hydration of Alkynes (Kucherov Reaction): Alkynes undergo hydration (addition of water) in the presence of mercuric sulphate (HgSO\(_4\)) and dilute acid (H\(_2\)SO\(_4\)). This reaction initially forms an unstable enol, which tautomerizes to a more stable carbonyl compound (a ketone or an aldehyde).

Step-by-Step Solution:

Step 1: Identify the structure of the starting alkyl halide [A].

The molecular formula is C\(_4\)H\(_9\)Br. The compound is stated to be optically active, which means it must contain a chiral carbon atom (a carbon atom bonded to four different groups). Let's examine the isomers of C\(_4\)H\(_9\)Br:

• 1-Bromobutane: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \) (achiral)
• 2-Bromobutane: \( \text{CH}_3\text{CH}_2\overset{*}{C}\text{H(Br)CH}_3 \) (chiral, the C2 atom is a stereocenter)
• 1-Bromo-2-methylpropane: \( (\text{CH}_3)_2\text{CHCH}_2\text{Br} \) (achiral)
• 2-Bromo-2-methylpropane: \( (\text{CH}_3)_3\text{CBr} \) (achiral)

Since [A] is optically active, it must be 2-Bromobutane.

\[ \text{[A] = } \text{CH}_3\text{CH}_2\text{CH(Br)CH}_3 \]

Step 2: Identify the major product alkene [B].

Compound [A] (2-Bromobutane) reacts with hot alcoholic KOH, which is a strong base, causing an E2 elimination reaction. According to Zaitsev's rule, the major product is the more substituted alkene.

Elimination of HBr can form two possible alkenes:

• Elimination from C1: \( \text{CH}_2\text{=CHCH}_2\text{CH}_3 \) (But-1-ene, monosubstituted)
• Elimination from C3: \( \text{CH}_3\text{CH=CHCH}_3 \) (But-2-ene, disubstituted)

The major product [B] is the more stable, disubstituted alkene, But-2-ene.

\[ \text{CH}_3\text{CH}_2\text{CH(Br)CH}_3 \xrightarrow[\text{heat}]{\text{alc. KOH}} \text{CH}_3\text{CH=CHCH}_3 \quad \text{(Major Product [B])} \]

Step 3: Identify the dibromide [C].

Alkene [B] (But-2-ene) reacts with bromine (Br\(_2\)) in an addition reaction. The double bond breaks, and a bromine atom adds to each carbon of the former double bond.

\[ \text{CH}_3\text{CH=CHCH}_3 + \text{Br}_2 \rightarrow \text{CH}_3\text{CH(Br)CH(Br)CH}_3 \quad \text{(Compound [C])} \]

The product [C] is 2,3-Dibromobutane.

Step 4: Identify the gas [D].

The vicinal dibromide [C] (2,3-Dibromobutane) is treated with alcoholic NaNH\(_2\), a very strong base. This causes double dehydrohalogenation to form an alkyne.

\[ \text{CH}_3\text{CH(Br)CH(Br)CH}_3 + 2 \text{NaNH}_2 \xrightarrow{\text{alcohol}} \text{CH}_3\text{C}\equiv\text{CCH}_3 + 2 \text{NaBr} + 2 \text{NH}_3 \]

The gaseous product [D] is But-2-yne.

Step 5: Identify the final compound [E].

The alkyne [D] (But-2-yne) undergoes hydration on warming with mercuric sulphate and dilute acid. This reaction adds a molecule of water across the triple bond. (Note: 18 grams of water is 1 mole, consistent with the 1:1 stoichiometry of the reaction).

The initial addition of H\(_2\)O forms an enol intermediate:

\[ \text{CH}_3\text{C}\equiv\text{CCH}_3 + \text{H}_2\text{O} \xrightarrow[\text{dil. H}_2\text{SO}_4]{\text{HgSO}_4, 333\text{K}} \left[ \text{CH}_3\text{C(OH)=CHCH}_3 \right] \quad \text{(Unstable enol)} \]

This enol immediately tautomerizes to the more stable keto form:

\[ \text{CH}_3\text{C(OH)=CHCH}_3 \rightleftharpoons \text{CH}_3\text{C(=O)CH}_2\text{CH}_3 \quad \text{(Compound [E])} \]

The final compound [E] is a ketone.

Step 6: Determine the IUPAC name of compound [E].

The structure of [E] is \( \text{CH}_3\text{COCH}_2\text{CH}_3 \). The longest carbon chain contains four carbon atoms (butane). The functional group is a ketone (suffix "-one"). Numbering from the left, the carbonyl group is at position 2. Therefore, the IUPAC name is Butan-2-one.