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.
Answer:
The correct answer would be A.
Explanation:
The observable visible spectrum of Cr(acac)3 complex is different from that of
[Cr(en)3]Br3 due to strength of bonded ligand in the coordination sphere.
en is a strong field ligand compared to acac thus pairing occurs. In both the complexes
Cr is in +3 state, configuration [Ar] 3d^3. There are 3 unpaired e in Cr(acac)3 whereas 1 unpaired e in [Cr(en)3]Br3 due to pairing.
B. The Secondary side of the step down transformer.
Pollution and people littering. Can change everything.