This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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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. CH_3-C-CH(CH_3)-CH_3 Hydrogenation of 4-methylpent-2-yne gives 2-methylpentane. Step 2: Deduce Compound B. Compound B (C_6H_12Cl_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. CH_3-CHCl-CHCl-CH(CH_3)-CH_3 This is 2,3-dichloro-4-methylpentane. Step 3: Deduce Compound A. Compound A (C_6H_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. CH_3-CH=CH-CH(CH_3)-CH_3 This is 4-methylpent-2-ene. Its formula is C_6H_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_3CHO). Ozonolysis cleaves the double bond and forms carbonyl compounds. CH_3-CH=CH-CH(CH_3)-CH_3 O_3, then H_2O/Zn CH_3CHO + CHO-CH(CH_3)-CH_3 So, E is acetaldehyde (CH_3CHO). And D is 2-methylpropanal (CH_3CH(CH_3)CHO). Here are the structural formulas: Structure of A: CH_3-CH=CH-CH(CH_3)-CH_3 Structure of B: CH_3-CHCl-CHCl-CH(CH_3)-CH_3 Structure of C: CH_3-C-CH(CH_3)-CH_3 Structure of D: CH_3-CH(CH_3)-CHO That's 2 down. 3 left today — send the next one.