Isopropyl iodide can be prepared by the action of Hl on

Options

A. propane
B. propene
C. isopropyl alcohol
D. glycerol

Answer

To prepare isopropyl iodide (CH3CH(I)CH3) using hydroiodic acid (HI), we need to start with an appropriate precursor that can react with HI. The most suitable precursor is propene (CH3-CH=CH2), which is an unsaturated hydrocarbon. Here’s a step-by-step breakdown of the preparation:

### Step 1: Identify the Reactant
We begin with propene (CH3-CH=CH2). This is an alkene that can react with HI.

### Step 2: Protonation of the Alkene
When propene reacts with HI, the double bond in propene will be protonated. The hydrogen ion (H+) from HI adds to one of the carbon atoms of the double bond, resulting in a carbocation. The more stable carbocation formed here is a secondary carbocation because it is attached to two other carbon atoms.

### Step 3: Formation of the Carbocation
The protonation leads to the formation of a secondary carbocation:
- Propene (CH3-CH=CH2) → CH3-CH(+)-CH2

### Step 4: Nucleophilic Attack by Iodide Ion
Next, the iodide ion (I-) from HI will attack the positively charged carbon atom of the carbocation. This results in the formation of isopropyl iodide (CH3-CH(I)-CH3).

### Step 5: Final Product
The final product of this reaction is isopropyl iodide (CH3-CH(I)-CH3), which is what we aimed to prepare.

### Summary of the Reaction
The overall reaction can be summarized as:
\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{HI} \rightarrow \text{CH}_3\text{CH(I)}\text{CH}_3 \]