(1) Isobutyraldehye + formaldehyde + NaOH/H2O → II (C5H10O2)

what is the structure?

(CH3)2CHCHO will react with formadehyde under basic conditions to give

(CH3)2C(CH2OH)(CHO)

Normally during this reaction the alcohol would be expected to dehydrate but there is no second proton adjacent to the CHO group

For mechanism see
http://en.wikipedia.org/wiki/Aldol_reaction

its difficult recognizing what rxn to use

To determine the structure of the compound II (C5H10O2), we can start by examining the reactants involved in the reaction.

The reaction involves isobutyraldehyde (C4H8O) and formaldehyde (CH2O) in the presence of sodium hydroxide (NaOH) and water (H2O).

Isobutyraldehyde has the formula C4H8O. It consists of a four-carbon chain with a carbonyl group (C=O) at the second carbon and a methyl group (-CH3) at the third carbon.

Formaldehyde has the formula CH2O. It consists of a single carbon atom with a carbonyl group (C=O) and two hydrogen atoms attached.

The reaction of isobutyraldehyde and formaldehyde suggests the formation of a larger compound, II (C5H10O2), which contains five carbon atoms, ten hydrogen atoms, and two oxygen atoms.

To determine the structure of II (C5H10O2), we can consider various possibilities. One possible structure is a five-carbon chain with a carbonyl group (C=O) at one end and an alcohol functional group (-OH) at the other end. This structure can be represented as:

O
|
C - C - C - C - C
| |
O H

Another possibility is the formation of a cyclic compound with five carbon atoms and two oxygen atoms. The specific structure of the cyclic compound depends on the exact conditions of the reaction.

It is important to note that without further information, it is challenging to determine the precise structure of compound II (C5H10O2). Additional experimental data, such as spectroscopic analysis or other chemical tests, would be necessary for a more accurate identification of the compound's structure.