Which of the following statements are correct about the following reaction :
`RCOOH + LiAlH_4 rarr`.

Options

A. First step in the above reaction is : `RCOOH + LiAlH_4 rarr RCOO^(Ө) Li^(oplus) + H_2 + AlH_3`.
B. Second step is the tranfer of `H^(Theta)` ion from `AIH_(3)` to `(C=O)` of `RCOO^(Theta)`.
C. The intermediate product `(RCH =O)` is formed
D. `(RCH=O)` is further reduced to `RCH_(2)O^(Theta)` which on acidification gives `RCH_(2)OH`.

Explanation

To analyze the reaction between a carboxylic acid (RCOOH) and lithium aluminum hydride (LiAlH4), we can break down the process step by step:

### Step 1: Reaction Overview
When a carboxylic acid reacts with lithium aluminum hydride, it undergoes reduction. Lithium aluminum hydride is a strong reducing agent that can convert carboxylic acids into alcohols.

### Step 2: Formation of Lithium Salt
Initially, the carboxylic acid (RCOOH) reacts with LiAlH4. During this reaction, a lithium salt of the carboxylic acid is formed, and hydrogen gas is evolved. The reaction can be represented as:
\[ RCOOH + LiAlH_4 \rightarrow RCOOLi + AlH_3 + H_2 \]

### Step 3: Proton Transfer
In the next step, the aluminum hydride (AlH4) donates a hydride ion (H-) to the carbonyl carbon of the carboxylic acid. This results in the formation of an intermediate, which is an aldehyde (RCHO):
\[ RCOOLi + H^- \rightarrow RCHO + LiOH \]

### Step 4: Formation of Aldehyde
The intermediate formed is an aldehyde. The carbonyl group (C=O) of the carboxylic acid is reduced to an aldehyde:
\[ RCHO \]

### Step 5: Further Reduction to Alcohol
The aldehyde can then be further reduced by another hydride transfer from lithium aluminum hydride:
\[ RCHO + H^- \rightarrow RCH2OH \]
This step converts the aldehyde into a primary alcohol (RCH2OH).

### Step 6: Acidification
Finally, upon acidification (addition of H+), the product is stabilized as an alcohol:
\[ RCH2OH \]

### Conclusion
The overall reaction can be summarized as:
\[ RCOOH + 2LiAlH_4 \rightarrow RCH2OH + LiOH + Al \]

### Correct Statements
Based on the above steps, we can confirm that:
1. The reaction produces a lithium salt of the carboxylic acid initially.
2. The aluminum hydride transfers a hydride ion to the carbonyl carbon of the carboxylic acid.
3. An aldehyde intermediate is formed during the reaction.
4. The aldehyde is further reduced to an alcohol upon the addition of hydride.

Thus, all statements regarding the reaction are correct.