Step 1: Deduce Compound C.

Here are the structural formulas for compounds A, B, C, and D based on the provided information:

• Step 1: Deduce Compound C. Compound C is an alkyne that, upon hydrogenation with a nickel catalyst, yields 2-methylpentane. The structure of 2-methylpentane is CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-CH$_3$. Compound C does not react with sodamide (NaNH$_2$) or ammoniacal silver nitrate, indicating it is an internal alkyne (the triple bond is not at the end of the carbon chain). Considering the 2-methylpentane skeleton, the only possible internal alkyne is 4-methylpent-2-yne.

  $$\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}(\text{CH}_3)-\text{CH}_3$$

  Hydrogenation of 4-methylpent-2-yne gives 2-methylpentane.

• Step 2: Deduce Compound B. Compound B (C$_6$H$_{12}$Cl$_2$) is treated with alcoholic KOH, followed by NaNH$_2$, to form compound C (4-methylpent-2-yne). This is a double dehydrohalogenation reaction. For an alkyne to be formed, B must be a dihaloalkane where the halogens are either on adjacent carbons (vicinal dihalide) or on the same carbon (geminal dihalide) that correspond to the triple bond in C. Since C is CH$_3$-C≡C-CH(CH$_3$)-CH$_3$, the halogens in B must be on the carbons that form the triple bond.

  $$\text{CH}_3-\text{CHCl}-\text{CHCl}-\text{CH}(\text{CH}_3)-\text{CH}_3$$

  This is 2,3-dichloro-4-methylpentane.

• Step 3: Deduce Compound A. Compound A (C$_6$H$_{12}$) is treated with chlorine in the presence of carbon tetrachloride (CCl$_4$) to give compound B (2,3-dichloro-4-methylpentane). This is an addition reaction of Cl$_2$ across a double bond. Therefore, A must be an alkene, and the double bond must be located where the chlorine atoms were added in B.

  $$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}(\text{CH}_3)-\text{CH}_3$$

  This is 4-methylpent-2-ene. Its formula is C$_6$H$_{12}$, which matches the given information.

• Step 4: Deduce Compounds D and E. Ozonolysis of compound A (4-methylpent-2-ene) gives two aldehydes, D and E. Compound E is identified as acetaldehyde (CH$_3$CHO). Ozonolysis cleaves the double bond and forms carbonyl compounds.

  $$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}(\text{CH}_3)-\text{CH}_3 \xrightarrow{\text{O}_3, \text{ then H}_2\text{O/Zn}} \text{CH}_3\text{CHO} + \text{CHO}-\text{CH}(\text{CH}_3)-\text{CH}_3$$

  So, E is acetaldehyde (CH$_3$CHO). And D is 2-methylpropanal (CH$_3$CH(CH$_3$)CHO).

Here are the structural formulas:

Structure of A:

$$\boxed{\text{CH}_3-\text{CH}=\text{CH}-\text{CH}(\text{CH}_3)-\text{CH}_3}$$

Structure of B:

$$\boxed{\text{CH}_3-\text{CHCl}-\text{CHCl}-\text{CH}(\text{CH}_3)-\text{CH}_3}$$

Structure of C:

$$\boxed{\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}(\text{CH}_3)-\text{CH}_3}$$

Structure of D:

$$\boxed{\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CHO}}$$