The end product Y for the following reaction is
`(C_(6)H_(5))_(2)CHCH_(2)OHoverset(Na)rarrXoverset(CH_(3)l)rarrY`

Options

A. `(C_(6)H_(5))_(2)C=CH_(2)`
B. `(C_(6)H_(5))_(2)CHCH_(2)ONa^(+)`
C. `(C_(6)H_(5))_(2)CHCH_(2)O-overset(O)overset(||)(C)-S^(-)Na^(+)`
D. `(C_(6)H_(5))_(2)CHCH_(2)OCH_(3)`

Answer

To determine the end product Y from the given reaction sequence, we can break down the steps as follows:

### Step 1: Identify the starting material
The starting material is `(C6H5)2CHCH2OH`, which is a secondary alcohol with two phenyl groups attached to a carbon chain.

### Step 2: Reaction with Sodium (Na)
When the secondary alcohol reacts with sodium (Na), it forms a sodium alkoxide. The reaction can be represented as:
\[ (C6H5)2CHCH2OH + Na \rightarrow (C6H5)2CHCH2O^{-}Na^{+} \]
In this step, the hydroxyl group (-OH) of the alcohol is converted to an alkoxide ion (-O^{-}) and sodium ion (Na^{+}) is formed.

### Step 3: Reaction with Methyl Iodide (CH3I)
Next, the alkoxide ion reacts with methyl iodide (CH3I). The alkoxide ion, which has a negative charge on the oxygen, acts as a nucleophile and attacks the methyl group of methyl iodide, resulting in the substitution of the iodine atom. The reaction can be represented as:
\[ (C6H5)2CHCH2O^{-}Na^{+} + CH3I \rightarrow (C6H5)2CHCH2OCH3 + NaI \]
In this step, the alkoxide ion forms an ether by attaching a methyl group (from CH3I) to the oxygen, and sodium iodide (NaI) is released as a byproduct.

### Step 4: Identify the end product Y
The final product Y is `(C6H5)2CHCH2OCH3`, which is a methyl ether of the original compound.

### Conclusion
Thus, the end product Y for the reaction is:
\[ Y = (C6H5)2CHCH2OCH3 \]