Consider the following reaction and identify X and Y.
`{:(CH_3CH_2CH_2Ioverset(alc.KOH)rarrXoverset(HBr)rarrY),(" "X" "Y):}`

Explanation

To solve the given reaction and identify compounds X and Y, we will follow these steps:

### Step 1: Identify the starting compound
The starting compound is 1-iodopropane (CH3CH2CH2I).

### Step 2: React with alcoholic KOH
When 1-iodopropane is treated with alcoholic KOH, a strong base, it undergoes an elimination reaction (beta-elimination). In this reaction, the base removes a beta hydrogen (the hydrogen on the carbon adjacent to the carbon bearing the leaving group, iodine in this case).

- The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{I} + \text{alc. KOH} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KI} \]
- The product formed is propene (CH3CH=CH2), which is compound X.

### Step 3: React X with HBr
Next, we react the alkene (propene) with HBr. This reaction is an electrophilic addition where HBr adds across the double bond.

- According to Markovnikov's rule, the hydrogen (H+) will add to the carbon with more hydrogen atoms already attached (the less substituted carbon), and the bromine (Br-) will add to the more substituted carbon.
- The reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CHBrCH}_3 \]
- The product formed is 2-bromopropane (CH3CHBrCH3), which is compound Y.

### Summary of Results
- Compound X is propene (CH3CH=CH2).
- Compound Y is 2-bromopropane (CH3CHBrCH3).

### Final Answer
- X = Propene (CH3CH=CH2)
- Y = 2-bromopropane (CH3CHBrCH3)

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