Problem 39 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Compound (A) on bromination gives (B), which gives (C) with alcoholic KOH. (C) decolouries \(1 \%\) alkaline \(\mathrm{KMnO}_{4}\) solution and on ozonolysis, it gives two molecules of smallest carbonyl compound. Compound (A) will be (a) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\)

Step-by-Step Solution

Answer

Compound (A) is C2H2 (option a).

Step 1: Identifying Key Reactions and Compounds

We start by examining the given reactions and what each compound transforms into. Compound (A) undergoes bromination to form (B), then (B) transforms into (C) with alcoholic KOH. (C) decolorizes alkaline KMnO4 and on ozonolysis gives two molecules of the smallest carbonyl compound. The smallest carbonyl compound is formaldehyde (CH2O).

Step 2: Understanding Bromination Process

Bromination typically results in the addition of bromine (Br2) to unsaturated bonds, such as double or triple bonds. This suggests that compound (A) should have multiple bonds. Therefore, we can eliminate saturated compounds like C2H6.

Step 3: Analyzing the Role of Alcoholic KOH

Alcoholic KOH generally causes dehydrohalogenation, where a halogen and a hydrogen atom are removed from adjacent carbon atoms, typically producing an alkene if starting from an alkyl halide. This means (B) must have halogens present and (C) must be an alkene.

Step 4: Examining the Decolorization of KMnO4

The ability of (C) to decolorize alkaline KMnO4 indicates the presence of a carbon-carbon double bond since this reagent acts on alkenes. Hence, (C) must have a C=C bond.

Step 5: Considering Ozonolysis Outcome

Ozonolysis splits alkenes (C=C bonds) to form carbonyl compounds. Given that the product is two molecules of formaldehyde, the alkene in (C) must be ethylene (C2H4) as it has the structure H2C=CH2.

Step 6: Determining Compound (A)

Since (B) forms from (A) through bromination and (B) must yield ethylene when treated with alcoholic KOH, (A) is most likely acetylene (C2H2), which forms dibromoethylene upon bromination, converting to ethylene by alcoholic KOH. Hence, the correct answer is option (a).

Key Concepts

Alkaline KMnO4 reaction

The alkaline potassium permanganate (KMnO4) reaction is a fundamental chemical test that helps identify unsaturation in organic compounds, such as alkenes and alkynes. When an alkene or alkyne is treated with a dilute solution of alkaline KMnO4, it undergoes oxidation.
This reaction is recognized by the characteristic decolorization of the purple KMnO4 solution.
Here's what happens during the reaction:

• Alkenes with a carbon-carbon double bond (\( C=C \)) and alkynes with a triple bond (\( C\equiv C \)) are oxidized to diols.
• The purple color of KMnO4 fades to brown MnO2, indicating the reaction.

This oxidation-reduction reaction makes alkaline KMnO4 a useful tool for distinguishing alkenes from saturated hydrocarbons like alkanes, which do not react and hence do not decolorize the solution.

Ozonolysis of alkenes

Ozonolysis is a crucial reaction in organic chemistry used to break carbon-carbon double bonds in alkenes.
It involves the treatment of alkenes with ozone (O3) followed by reductive work-up usually with zinc and acetic acid.
The steps involved in ozonolysis are:

• During the reaction, ozone adds to the carbon-carbon double bond forming an unstable ozonide intermediate.
• Upon reductive cleavage, the ozonide splits to yield carbonyl compounds such as aldehydes or ketones, depending on the substitution pattern of the original alkene.
• When applied to simple alkenes like ethylene, the reaction yields simple molecules such as formaldehyde.

This reaction provides a direct method to determine the position of double bonds within a molecule, making it an essential tool in structural elucidation.

Dehydrohalogenation reaction

The dehydrohalogenation reaction is a chemical process that involves the elimination of a hydrogen halide from an alkyl halide to form an alkene.
Typically, this reaction is carried out using a strong base like alcoholic potassium hydroxide (KOH).
Key steps in dehydrohalogenation include:

• An alkyl halide is treated with a strong base, causing the halogen atom and a hydrogen from an adjacent carbon to be removed.
• This elimination of \( HX \) leads to the formation of a double bond, converting the alkyl halide to an alkene.

The reaction is driven by the stability of the resultant alkene and often follows Zaitsev's rule, which predicts that the more substituted alkene will be the major product. It is widely used in synthetic organic chemistry to introduce unsaturation into a compound through elimination.