Find the final product 'E'
`CH_(2)OH-CH=CH_(2)overset((i)OsO_(4))underset((ii)NaSO_(3)H)rarr(A)overset(KHSO_(4))rarr (B) overset(Zn-Hg//"conc. HCl")rarr (C ) overset(NBS//hv)underset(C Cl_(4))rarr (D ) overset((CH_(3))_(2)"CuLi")rarr (E )`

Options

A. But - 2 - ene
B. But -1 - ene
C. Isopentene
D. Pen - 2 - ene

Answer

To find the final product 'E' from the given reaction sequence, we will go through each step systematically.

### Step 1: Reaction with OsO4 and NaSO3H
The starting compound is CH₂OH-CH=CH₂. When this compound reacts with osmium tetroxide (OsO₄) in the presence of sodium bisulfite (NaSO₃H), it undergoes syn-dihydroxylation. This means that the double bond (C=C) will be converted into a diol (two hydroxyl groups will be added across the double bond).

**Product A:**
CH₂OH-CH(OH)-CH(OH)

### Step 2: Reaction with KHSO4
Next, the product A reacts with KHSO₄. This step involves dehydration (removal of water) to form an alkene. The hydroxyl groups will be converted back to a double bond.

**Product B:**
CH₂=CH-CHOH

### Step 3: Reaction with Zn-Hg and concentrated HCl
The product B is then treated with zinc amalgam in the presence of concentrated HCl. This reaction reduces the carbonyl group (if present) and adds hydrogen across the double bond.

**Product C:**
CH₂=CH-CH₂

### Step 4: Reaction with NBS in the presence of hv and CCl₄
Next, product C undergoes allylic bromination when treated with N-bromosuccinimide (NBS) in the presence of light (hv) and carbon tetrachloride (CCl₄). This results in the addition of a bromine atom to the allylic position.

**Product D:**
CH₂=CH-CH₂Br

### Step 5: Reaction with (CH₃)₂CuLi
Finally, product D reacts with lithium diorganocopper reagent, (CH₃)₂CuLi. This reagent will replace the bromine atom with a methyl group through a nucleophilic substitution reaction.

**Final Product E:**
CH₃-CH=CH-CH₃ (Butyne)

### Summary of the Reaction Sequence:
1. CH₂OH-CH=CH₂ + OsO₄/NaSO₃H → CH₂OH-CH(OH)-CH(OH) (Product A)
2. Product A + KHSO₄ → CH₂=CH-CHOH (Product B)
3. Product B + Zn-Hg/HCl → CH₂=CH-CH₂ (Product C)
4. Product C + NBS/hv/CCl₄ → CH₂=CH-CH₂Br (Product D)
5. Product D + (CH₃)₂CuLi → CH₃-CH=CH-CH₃ (Final Product E)

### Final Answer:
The final product 'E' is **Butyne** (CH₃-CH=CH-CH₃).