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

Answer:

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.

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.