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 identify the final product (Z) from the given sequence of reactions, let's break down the steps involved in the reaction sequence.

### Step 1: Addition of HBr to CH₃-CH=CH₂
- The starting compound is propene (CH₃-CH=CH₂).
- When HBr is added to propene, the double bond reacts with HBr. The H⁺ from HBr adds to the less substituted carbon (the terminal carbon), leading to the formation of a more stable carbocation on the second carbon.
- The bromide ion (Br⁻) then attacks the carbocation, leading to the formation of the product X: **CH₃-CH(Br)-CH₃** (2-bromopropane).

### Step 2: Reaction of X with Aqueous NaOH
- The next step involves treating the product X (2-bromopropane) with aqueous sodium hydroxide (NaOH).
- NaOH acts as a nucleophile and performs an SN2 reaction, where the bromine is replaced by a hydroxyl group (OH).
- The product Y formed is: **CH₃-CH(OH)-CH₃** (2-propanol).

### Step 3: Reaction of Y with NaOH and I₂
- The next step involves treating Y (2-propanol) with NaOH and iodine (I₂).
- The combination of NaOH and I₂ acts as a mild oxidizing agent, converting the secondary alcohol (2-propanol) into a ketone.
- The product formed is: **CH₃-CO-CH₃** (acetone).

### Step 4: Reaction of Acetone with NaOH and I₂
- The final step involves treating the ketone (acetone) with NaOH and I₂ again.
- In this step, the NaOH will deprotonate the alpha hydrogen of acetone, leading to the formation of an enolate ion.
- The enolate ion will then react with I₂, leading to the formation of a triiodo compound.
- The final product Z will be: **CH₃-CO-CH(I₃)** (1,1,1-triiodoacetone).

### Final Product
Thus, the final product Z is **CH₃-CO-CH(I₃)**.

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