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.

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.

h let me think