Let’s say you have 3 UV active spots in the crude material and co-spot TLC plate. One has the same Rf value as the starting mate
rial and the other 2 are very different. What could these 2 nonstarting material spots be? Use structures and words. Hint: think about the nucleophilic addition of the hydride.
2 answers:
Answer:
Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)
Explanation:
Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.
In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.
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Answer:
The factor of increasing reaction rate is 1,85x10¹².
Explanation:
Using arrhenius formula:
Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K
Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:
k = 1,25x10⁻²²
When activation energy is 55kJ/mol:
k = 2,31x10⁻¹⁰
Thus, the factor of increasing reaction rate is:
2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>
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I hope it helps!
Answer:
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Explanation:
Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g
1 kg = 1000 g
Molar mass of fat = M
M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]
Moles of fat =
According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;
of water
Mass of 148.31 moles of water ;
148.31 mol × 18 g/mol = 2,669.58 g
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.