Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:

Ozonolysis is the reaction in which an alkene (or alkyne) is cleaved by ozone $$\left(O_3\right)$$ to give carbonyl compounds after reductive work-up.

For an alkene segment $$C_1=C_2$$, the products are:

$$C_1=C_2 \xrightarrow[Zn/H_2O]{O_3} \; C_1=O + C_2=O$$

Therefore, by inspecting the carbonyl products we can identify the carbon skeleton that was present before cleavage.

Formaldehyde $$(H-CHO)$$ contains only one carbon. It can arise in ozonolysis only when one of the double-bonded carbons was a terminal $$=CH2$$ group because:

$$CH_{2} = CR_{2} + {O_3} \rightarrow  {HCHO} + {O=CR_{2}}$$

The fragment $$CH2 =$$ directly attached to the rest of the molecule is called a vinyl group (also written as $$-CH=CH_2$$).

Checking the options:

Option A — a vinyl group: Contains $$\ce{-CH=CH2}$$; its ozonolysis indeed yields formaldehyde, so this option is correct.

Option B — an isopropyl group: $$\ce{-(CH3)2CH-}$$ has no double bond; if part of a double bond $$\ce{(CH3)2C=CH2}$$, the products would be acetone $$\left(\ce{(CH3)2C=O}\right)$$ and formaldehyde. Presence of acetone would also be detected, so isopropyl alone is not the confirming feature.

Option C — an acetylenic triple bond: Alkyne ozonolysis gives carboxylic acids (or ketones/CO2) but not formaldehyde from a $$\ce{-C#C-}$$ fragment, so this is wrong.

Option D — two ethylenic double bonds: Merely having two double bonds does not guarantee formaldehyde; only a terminal $$\ce{=CH2}$$ double bond does.

Hence, formaldehyde formation specifically confirms the presence of a vinyl group.

Final answer: Option A which is: a vinyl group