2Li+2H2O—>2LiOH+H2
1 answer:
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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.
