Question

Compound A has the molecular formula C14H25Br and was obtained by reaction of sodium acetylide with 1,12-dibromododecane. On treatment of compound A with sodium amide, it was converted to compound B (C14H24). Ozonolysis of compound B gave the compound with the following chemical formula: HO2C(CH2)12COOH (a diacid because it has two acid functional groups on each end). Catalytic hydrogenation of compound B over Lindar palladium gave compound C (C14H26) and hydrogenation of compound B over platinum gave compound D (C14H28). Sodium-ammonia reaction of compound B gave compound E (C14H26). Both C and E yielded O=CH(CH2)12CH=O on ozonolysis. Assign structures to compounds A through E so as to be consistent with the observed transformations.

Answer 1

Compound B contains an internal alkyne, which can be either partially hydrogenated to form alkenes (compounds C and E) or fully hydrogenated to yield alkane (compound D). Both C and E have the same double bond position, as evidenced by their identical ozonolysis products.

Step-by-step explanation:

The challenge revolves around elucidating the structures of the series of compounds A to E based on their chemical reactions and resulting transformation products. Sodium acetylide's reaction with 1,12-dibromododecane yields compound A (C14H25Br), which upon treatment with sodium amide transforms into compound B (C14H24). Ozonolysis of B produces a diacid (HO2C(CH2)12COOH) while catalytic hydrogenation over Lindar palladium and over platinum produces compounds C (C14H26) and D (C14H28), respectively. Compound B's reaction with sodium in ammonia leads to compound E (C14H26). Both C and E when subjected to ozonolysis yield the same compound O=CH(CH2)12CH=O, indicating the presence of a double bond at identical positions in C and E. Considering that ozonolysis of B yields a diacid with a 12 carbon chain between the acid groups, B must contain an internal alkyne, which upon catalytic hydrogenation can be fully or partially hydrogenated to yield the corresponding alkenes in C and E. The full hydrogenation to an alkane in D indicates the complete saturation of the double bonds.

Answer 2

Look at the pictures. On the 1 are compounds A and B. Compound c from b is on the 2nd image. Compound D is on 3rd image. Compound E is the same for compound C.

Step-by-step explanation:

So for compound A sodium acetylide substitutes nucleophilicaly one Br on 1,12-dibromododecane. Then to obtain compound B sodium amide eliminates another Br. So for acetylene and alkene groups ozonolysis works the same way and we obtain diacid. Lyndlar catalyst works only on alkynes and make cis-alkenes from them. but we have a terminal alkyne for wich no isomers may occur. Pt reduction provides alkanes from both alenes and akynes. And sodium ammonia reduction works only on alkynes to provide trans-alkenes but, as I've said, isomers are not our case. So compounds E and C are the same and undergo same reaction with ozone.