Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

We first write the name of the required ozonolysis product exactly as given in the question:

$$\text{5-keto-2-methyl hexanal}$$

This name tells us four vital pieces of information.

1. “hex-” shows that the longest carbon skeleton in the required fragment contains $$6$$ carbon atoms.

2. “-anal’’ tells us that carbon-1 of that skeleton is an aldehydic carbon, that is, $$\,-CHO$$.

3. “5-keto’’ tells us that carbon-5 of the same chain is a ketonic carbonyl $$\bigl(>C\! =\! O\bigr)$$.

4. “2-methyl’’ tells us that a $$CH_3$$ group is attached to carbon-2 of the chain.

So in the product we must have

$$\begin{aligned}

\text{C-1}&: && \; -CHO\\

\text{C-2}&: && \; -CH(CH_3)-\\

\text{C-3}&: && \; -CH_2-\\

\text{C-4}&: && \; -CH_2-\\

\text{C-5}&: && \; -CO-\\

\text{C-6}&: && \; -CH_3

\end{aligned}$$

The structure obtained by placing these pieces in order is therefore

$$\boxed{\,OHC-CH(CH_3)-CH_2-CH_2-CO-CH_3\,}$$

Ozonolysis is nothing but cleavage of a $$C=C$$ double bond, each vinylic carbon being converted into a carbonyl centre:

• if the olefinic carbon possessed at least one hydrogen, it ends up as an aldehyde $$\bigl(-CHO\bigr);$$
• if the olefinic carbon possessed no hydrogen, it ends up as a ketone $$\bigl(>C\!=\!O\bigr).$$

Thus our required product contains both kinds of carbonyls. Evidently, during the cleavage

• one alkene carbon must have borne at least one H atom (that carbon is now C-1 of the product and appears as $$-CHO$$);
• the other alkene carbon must have carried no hydrogen (that carbon is now C-5 of the product and appears as $$>C\!=\!O$$).

We therefore back-translate the product into the only double bond that can furnish it:

remove the oxygen of the aldehyde group and remove the oxygen of the ketone group, then join the two carbon centres thus freed by a double bond. This “reverse ozonolysis” step gives

$$

\begin{aligned}

&O\;{\Large\strikethrough{\text{HC}}}-CH(CH_3)-CH_2-CH_2-{\Large\strikethrough{\text{CO}}}-CH_3\\[4pt]

&\;\;\;\;\;\;\;\downarrow\;\text{join the two centres}\\[4pt]

&CH_2\;=\;C(CH_3)-CH_2-CH_2-CO-CH_3

\end{aligned}

$$

The fragment we have reconstructed is

$$\boxed{\,CH_2= C(CH_3)-CH_2-CH_2-CO-CH_3\,}$$

Notice carefully:

• the left-hand alkene carbon ($$CH_2$$) carries two hydrogens and will therefore give an aldehyde (specifically an $$-CHO$$) on ozonolysis;
• the right-hand alkene carbon is tertiary, $$C(CH_3)-$$, and therefore possesses no hydrogen; on ozonolysis it must therefore give a ketone.

Exactly that pattern - a terminal $$CH_2= C<$$ group flanked two carbons away from a pre-existing $$>C\!=\!O$$ - occurs in compound (3) of the options. The other three alternatives all fail on at least one count:

• either both alkene carbons possess hydrogen (so two aldehydes would be produced), or
• neither alkene carbon possesses hydrogen (so two ketones would result), or
• cleavage of their double bond does not leave a six-carbon chain that can still contain the embedded ketone at the right position.

Because only option (3) contains the exact $$CH_2= C(CH_3)-$$ sub-unit and the correctly placed pre-existing keto group, it alone can furnish the target fragment together with formaldehyde as the co-product, exactly as demanded by the rules stated above.

Hence, the correct answer is Option C (3).