Which of the following will give an alkane having same number of carbon atoms on reacting with an alkyl halide ?
(A ) RMgX
(b) Na + Ether
(c ) `"LiAIH"_(4)`
(d) Sn + HCl

Options

A. [A], [C]
B. [B], [D]
C. [C], [D]
D. [A], [D]

Answer

To solve the question of which option will give an alkane with the same number of carbon atoms when reacting with an alkyl halide, we will analyze each option step by step.

### Step 1: Understand the Reaction
We need to determine which of the given reagents will react with an alkyl halide (RX) to produce an alkane (C_nH_(2n+2)) that has the same number of carbon atoms as the alkyl halide.

### Step 2: Analyze Each Option
- **(A) RMgX**: This is a Grignard reagent. When a Grignard reagent reacts with an alkyl halide, it typically forms a new carbon-carbon bond, resulting in a product with more carbon atoms than the original alkyl halide. Therefore, this option does not yield an alkane with the same number of carbon atoms.

- **(B) Na + Ether**: This combination is known for the Wurtz reaction. In this reaction, two molecules of an alkyl halide react with sodium in ether to form an alkane. The general reaction is:
\[ 2 RX + 2 Na \rightarrow R-R + 2 NaX \]
Here, the number of carbon atoms in the product alkane (R-R) is the same as that in the two alkyl halides (2R), thus this option is likely to yield an alkane with the same number of carbon atoms.

- **(C) LiAlH4**: Lithium aluminum hydride is a strong reducing agent. It reduces carbonyl compounds and other functional groups but does not typically react with alkyl halides to form alkanes. Thus, this option does not yield an alkane with the same number of carbon atoms.

- **(D) Sn + HCl**: This reaction typically involves the reduction of alkyl halides to alkanes, but it does not involve the formation of a new carbon-carbon bond. Instead, it replaces the halide with hydrogen, resulting in a decrease in the number of carbon atoms if starting from a di- or tri-substituted halide. Therefore, this option does not yield an alkane with the same number of carbon atoms.

### Conclusion
After analyzing all the options, the correct answer is **(B) Na + Ether**, as it leads to the formation of an alkane with the same number of carbon atoms as the alkyl halide.

### Final Answer:
**(B) Na + Ether**