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Q6.19(xiii):
How the following conversions can be carried out? (xiii) 2-Chloropropane to 1-propanol

Solution :

Step 1: Analyzing the Molecular Structure and Reactivity

The starting material is 2-chloropropane ($CH_3CHClCH_3$), a secondary alkyl halide. The desired product is 1-propanol ($CH_3CH_2CH_2OH$), a primary alcohol. The conversion requires two distinct chemical transformations:

  • Dehydrohalogenation: Removal of the halogen to generate an alkene.
  • Anti-Markovnikov Hydration: Regioselective addition of water to the alkene to position the hydroxyl group on the terminal carbon.

Step 2: Elimination Reaction (Dehydrohalogenation)

To convert 2-chloropropane into propene, we employ an elimination reaction using an alcoholic solution of potassium hydroxide ($KOH$). [Per the $E_2$ elimination mechanism]:

$CH_3-CHCl-CH_3 + KOH \text{ (alc.)} \xrightarrow{\Delta} CH_3-CH=CH_2 + KCl + H_2O$

The base abstracts a $\beta$-hydrogen from the 2-chloropropane, leading to the formation of the double bond between $C_1$ and $C_2$.

Step 3: Anti-Markovnikov Hydration (Hydroboration-Oxidation)

Direct hydration of propene with dilute acid would follow [Markovnikov's Rule], yielding 2-propanol. To achieve 1-propanol, we must utilize the Hydroboration-Oxidation sequence, which is regiospecific for anti-Markovnikov orientation.

  • Phase A (Hydroboration): Propene reacts with diborane ($BH_3)_2$ in tetrahydrofuran (THF) to form a trialkylborane.
  • Phase B (Oxidation): The trialkylborane is treated with hydrogen peroxide ($H_2O_2$) in an aqueous alkaline medium ($NaOH$).

Reaction mechanism: $3CH_3CH=CH_2 + \frac{1}{2}B_2H_6 \rightarrow (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} 3CH_3CH_2CH_2OH + B(OH)_3$

Step 4: Visualizing the Conversion Process

2-Chloropropane alc. KOH Propene 1. BH3, THF 2. H2O2, OH- 1-Propanol

Summary of Reagents

Transformation Reagent
2-Chloropropane $\rightarrow$ Propene $KOH$ (alc.), heat
Propene $\rightarrow$ 1-Propanol 1. $BH_3 \cdot THF$; 2. $H_2O_2/OH^-$

Final Solution: The conversion of 2-chloropropane to 1-propanol is achieved by first performing an $E_2$ elimination with alcoholic $KOH$ to form propene, followed by hydroboration-oxidation using $BH_3 \cdot THF$ and $H_2O_2/OH^-$.


More Questions from Class 12 Chemistry Haloalkanes and Haloarenes Exercises


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