Consider the following sequence of reaction and identify the final product (Z). `CH_3-CH=CH_2overset(HBr)rarr(X)overset(Aq.NaOH)rarr(Y)overset(NaOH+l_2)rarr(Z)`

Options

A. `CH_3CHICH_3`
B. `CH_3CH_2CH_2I`
C. `CHI_3`
D. `CH_3COCH_3`

Answer

To solve the problem step by step, we will analyze the sequence of reactions provided:

1. **Starting Material**: The starting compound is propene, which is represented as \( CH_3-CH=CH_2 \).

2. **First Reaction (with HBr)**:
- Propene undergoes an electrophilic addition reaction with HBr.
- Since propene is an unsymmetrical alkene, the addition follows Markovnikov's rule, where the bromine atom will attach to the more substituted carbon atom.
- The reaction can be represented as follows:
\[ CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3 \]
- The product (X) is 2-bromopropane, \( CH_3-CH(Br)-CH_3 \).

3. **Second Reaction (with Aq. NaOH)**:
- The product (X) is then treated with aqueous NaOH, which leads to a nucleophilic substitution reaction.
- The hydroxide ion (\( OH^- \)) replaces the bromine atom:
\[ CH_3-CH(Br)-CH_3 + NaOH \rightarrow CH_3-CH(OH)-CH_3 + NaBr \]
- The product (Y) is 2-propanol (isopropanol), \( CH_3-CH(OH)-CH_3 \).

4. **Third Reaction (with NaOH + I2)**:
- The final product (Y) is treated with NaOH and iodine (\( I_2 \)), which indicates that it will undergo the iodoform reaction.
- The iodoform reaction occurs with compounds that have a methyl ketone group or secondary alcohols that can be oxidized to form methyl ketones.
- In this case, 2-propanol can be oxidized to acetone, which is a methyl ketone:
\[ CH_3-CH(OH)-CH_3 \xrightarrow{NaOH + I_2} CH_3-C(=O)-CH_3 \]
- The acetone then reacts with iodine to form iodoform (\( CHI_3 \)):
\[ CH_3-C(=O)-CH_3 \xrightarrow{I_2} CHI_3 + byproducts \]
- The final product (Z) is iodoform, \( CHI_3 \).

Thus, the final product (Z) is \( CHI_3 \).

### Summary of Steps:
1. Start with propene \( CH_3-CH=CH_2 \).
2. Add HBr to form 2-bromopropane \( CH_3-CH(Br)-CH_3 \).
3. Treat with aqueous NaOH to form 2-propanol \( CH_3-CH(OH)-CH_3 \).
4. React with NaOH and iodine to yield iodoform \( CHI_3 \).