Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
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In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.
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Explanation:
30 lb is 480 ounces
34 mi/second is 54.718 kilometre/ second
455 lb/ gal is 54521.024 grams / litre
50 cl is 500 millilitres
55nm is 5.5 × 10^-6 centimetre
Yes you can. in order to do this however, you would need to observe the hummingbird and study it closely