MP Board · Class 12 · Chemistry · Haloalkanes and HaloarenesAn organic compound 'A' with molecular formula $C4H9Cl$ on treatment with aqueous KOH gives compound 'B' with molecular formula $C4H{10}O$. Compound 'A' on treatment with alcoholic KOH gives an alkene 'C' ($C4H8$). Alkene 'C' on ozonolysis gives two different aldehydes. Identify compounds A, B, and C with complete chemical reactions and explain the steps clearly.

1. Analysis of Molecular Formula and Reactions

• The molecular formula of compound 'A' is $C_4H_9Cl$, which corresponds to an alkyl chloride (general formula $C_nH_{2n+1}Cl$).
• Reaction 1 (Aqueous KOH): Compound 'A' reacts with aqueous KOH to give compound 'B' ($C_4H_{10}O$), which indicates a nucleophilic substitution reaction where the chlorine atom is replaced by an $-OH$ group, forming an alcohol. Thus, 'A' is an alkyl chloride and 'B' is a butyl alcohol.
• Reaction 2 (Alcoholic KOH): Compound 'A' reacts with alcoholic KOH to undergo a dehydrohalogenation reaction (elimination reaction) to yield an alkene 'C' ($C_4H_8$).
• Reaction 3 (Ozonolysis of C): Alkene 'C' undergoes ozonolysis to form two different aldehydes. This is a crucial clue. Ozonolysis cleaves a carbon-carbon double bond to form carbonyl compounds. If 'C' ($C_4H_8$) on ozonolysis gives two different aldehydes, the structure of 'C' must be an unsymmetrical alkene, specifically but-2-ene ($CH_3-CH=CH-CH_3$) or but-1-ene ($CH_3-CH_2-CH=CH_2$).

2. Identifying the Isomer of Alkyl Chloride 'A'\nIf alkene 'C' is but-2-ene, its ozonolysis yields two molecules of ethanal ($CH_3CHO$), which are identical. However, the problem states that it gives two different aldehydes. Therefore, alkene 'C' must be but-1-ene ($CH_3-CH_2-CH=CH_2$).

• Since alkene 'C' is but-1-ene, the parent alkyl chloride 'A' must be 2-chlorobutane or 1-chlorobutane.
• Let us check with 2-chlorobutane ($CH_3-CH_2-CH(Cl)-CH_3$): Elimination via Saytzeff rule would majorly yield but-2-ene. But-1-ene would be the minor product.
• Let us check with 1-chlorobutane ($CH_3-CH_2-CH_2-CH_2Cl$): Elimination strictly gives but-1-ene as the sole alkene product. \nThus, compound 'A' is 1-chlorobutane ($CH_3-CH_2-CH_2-CH_2Cl$).

3. Step-by-Step Chemical Reactions

Step 1: Conversion of A to B $$CH_3-CH_2-CH_2-CH_2Cl + KOH (aq) \xrightarrow{\Delta} CH_3-CH_2-CH_2-CH_2OH + KCl$$ (Compound A: 1-Chlorobutane, Compound B: Butan-1-ol)

Step 2: Conversion of A to C $$CH_3-CH_2-CH_2-CH_2Cl + KOH (alc) \xrightarrow{\Delta} CH_3-CH_2-CH=CH_2 + KCl + H_2O$$ (Compound C: But-1-ene)

Step 3: Ozonolysis of C $$CH_3-CH_2-CH=CH_2 + O_3 \xrightarrow{CCl_4} [Molozonide] \xrightarrow{Zn/H_2O} CH_3-CH_2-CHO + HCHO$$ (Products: Propanal and Methanal — two different aldehydes)

4. Final Conclusion

• Compound A: 1-Chlorobutane ($C_4H_9Cl$)
• Compound B: Butan-1-ol ($C_4H_{10}O$)
• Compound C: But-1-ene ($C_4H_8$)