Compound A has the molecular formula C14H25Br and was obtained by reaction of sodium acetylide with 1,12-dibromododecane. On tre
atment 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.
1 answer:
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.
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Answer:
The balanced equation for this reaction will be
→
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.