Match List-I with List-II and then select the correct answer from the codes given below the lists-
List -I
`(A) CH_3MgI+CH_3CHO to "product " overset(H_3O^(Theta))to`
(B) `(CH_3)_2C=CH_2overset(Dil.H_2SO_4)to`
(C ) `CH_3COOC_2H_5 overset(Na^+)underset("alcohol")to`
(D) `CH_3CHOHC_2H_5`
List -II
(A) Shows optical isomerism
(B) A secondary alcohol giving iodoform test
(C) Product is a tertiary alcohol
(D) Product is primary alcohol

Options

A. Ab, Bd, Cc, Da
B. Ab, Bc, Cd, Da
C. Ab, Bc, Ca, Dd
D. Ab, Ba, Cd, Dc

Answer

To match List-I with List-II, we will analyze each item in List-I and determine the corresponding item in List-II based on the reactions and products formed.

### Step-by-Step Solution:

**Step 1: Analyze (A) CH₃MgI + CH₃CHO → Product (H₃O⁺)**
- Here, CH₃MgI (a Grignard reagent) reacts with CH₃CHO (acetaldehyde).
- The Grignard reagent acts as a nucleophile and attacks the carbonyl carbon of acetaldehyde, forming a tertiary alcohol after hydrolysis.
- The product formed is a secondary alcohol (CH₃-CHOH-CH₃).
- **Match:** (A) → (B) (A secondary alcohol giving iodoform test)

**Step 2: Analyze (B) (CH₃)₂C=CH₂ (Dil. H₂SO₄) → Product**
- The compound is an alkene that undergoes hydration in the presence of dilute sulfuric acid.
- The reaction leads to the formation of a tertiary alcohol (due to the more stable carbocation formed during the reaction).
- **Match:** (B) → (C) (Product is a tertiary alcohol)

**Step 3: Analyze (C) CH₃COOC₂H₅ (Na⁺ in alcohol) → Product**
- This is an ester undergoing hydrolysis in the presence of sodium in alcohol.
- The hydrolysis leads to the formation of two primary alcohols (ethanol and methanol).
- **Match:** (C) → (D) (Product is a primary alcohol)

**Step 4: Analyze (D) CH₃CHOHC₂H₅**
- This compound has a chiral center (due to the presence of four different groups around a carbon).
- It can show optical isomerism since it has an asymmetric carbon atom.
- **Match:** (D) → (A) (Shows optical isomerism)

### Final Matches:
- (A) → (B)
- (B) → (C)
- (C) → (D)
- (D) → (A)

### Codes:
1. (A) → (B)
2. (B) → (C)
3. (C) → (D)
4. (D) → (A)