Step 1: Read the clue about the sodium derivative. Compound A reacts with $NaNH_2$ to give a sodium derivative. Only a terminal alkyne, with its acidic hydrogen on the carbon that carries the triple bond, can be deprotonated by sodamide. So A must have a $-C\equiv C-H$ group at the end of the chain. Step 2: Fix the structure of A. With molecular formula $C_4H_6$ and a terminal triple bond, the only fit is but-1-yne, $CH_3CH_2C\equiv CH$. This confirms A is a four carbon terminal alkyne. Step 3: Decide what B must be. The next step needs to turn the alkyne into an alkene without going all the way to the alkane. Full reduction with $H_2/Ni$ would give butane and destroy the double bond, so that is wrong. We need a poisoned (Lindlar type) catalyst. The matching choice is $H_2/Pd\text{-}C, quinoline$, which stops at the alkene stage. So B is $H_2/Pd\text{-}C, quinoline$. Step 4: Form the alkene. Partial hydrogenation of $CH_3CH_2C\equiv CH$ gives but-1-ene, $CH_3CH_2CH=CH_2$. Step 5: Carry out the ozonolysis to get D. Reductive ozonolysis cleaves the double bond of but-1-ene at the $CH=CH_2$ position. The terminal $=CH_2$ becomes formaldehyde $HCHO$, and the $CH_3CH_2CH=$ part becomes propanal $CH_3CH_2CHO$. So D is the pair $CH_3CH_2CHO + HCHO$. Step 6: Pick the matching option. We need $B = H_2/Pd\text{-}C, quinoline$ and $D = CH_3CH_2CHO + HCHO$, which is option (3). \[ \boxed{B=H_2/Pd\text{-}C,\ quinoline;\ \ D=CH_3CH_2CHO+HCHO\ \ \text{(Option 3)}} \]
