Alkenes combine with carbon monoxide and hydrogen in presence of octacarbonyldicobalt as catalyst under high temperature and pressure to form

Options

A. aldehydes which can be further reduced to alcohols by hydrogen
B. alkanes which are formed by addition of hydrogen
C. alcohols formed byb reaction CO and hydrogen
D. Ketones which can be further reduced to aldehydes by hydrogen.

Explanation

To solve the question regarding the reaction of alkenes with carbon monoxide and hydrogen in the presence of octacarbonyldicobalt as a catalyst, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have alkenes (general formula: R-CH=CH2), carbon monoxide (CO), and hydrogen (H2).
- The catalyst used is octacarbonyldicobalt (Co2(CO)8).

2. **Understand the Reaction Type**:
- The reaction described is known as hydroformylation. This process involves the addition of a formyl group (CHO) and a hydrogen atom to the double bond of the alkene.

3. **Write the Reaction**:
- The general reaction can be represented as:
\[ R-CH=CH2 + H2 + CO \xrightarrow{\text{Co}_2(\text{CO})_8, \text{high temp, high pressure}} R-CH2-CH2-CHO \]
- Here, R-CH2-CH2-CHO is the product formed, which is an aldehyde.

4. **Product Formation**:
- The product formed from this reaction is an aldehyde (R-CH2-CH2-CHO).
- The aldehyde can undergo further reduction if additional hydrogen (H2) is added, converting it into an alcohol (R-CH2-CH2-CH2OH).

5. **Conclusion**:
- The main product of the reaction is an aldehyde, which can further be reduced to an alcohol.

### Final Answer:
The product formed from the reaction of alkenes with carbon monoxide and hydrogen in the presence of octacarbonyldicobalt as a catalyst under high temperature and pressure is **aldehydes** (which can further be reduced to alcohols).