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Mice21 [21]
4 years ago
8

The nucleophilic addition reaction depicted below involves a prochiral ketone carbon atom reacting with a nucleophilic hydride i

on source (LiAlH4 or NaBH4) and, subsequently, a proton source (e.g., H2O or dilute aq. HCl). Consequently, the reaction produces a racemic mixture of an alcohol. Finish drawing the structures of the products resulting from nucleophilic attack upon the front and back faces of the carbonyl group, being careful to specify the stereochemistry via wedge-and-dash bonds.

Chemistry
1 answer:
Morgarella [4.7K]4 years ago
3 0

Answer:

we are given the 3-methyl2 butanone and upon the reduction with LiAIH4 there is formed alcohol, there are two possible side attack,

  1. from the back side
  2. from the front side.

therefore, whenever the front side attacks then the -CH3 moves back  the plain and the H will be above the plane. More so, when  attack from the back side the H moves below the plane and -CH3 moves above the plane. -OH is evident in the plane. see the attachment below to view the structure.

Explanation:

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Explanation:

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4 years ago
Only 235u can be used as fuel in a nuclear reactor, so uranium for use in the nuclear industry must be enriched in this isotope.
Yuki888 [10]

There are two isotopes of uranium abundant in nature

U235 and U238

As given that the sample has average molar mass of 237.482 amu

Let the amount of U235 in 100g sample = x

the amount of U238 in 100 g sample = 100-x

the average molar mass = [235(x) + 238 (100-x)] / 100

237.482  =  [235(x) + 238 (100-x)] / 100

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23748.2 = 235 x + 23800 - 238x

51.8 = 3x

x = 17.27 g

So percentage of U-235 = 17.27 %


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3 years ago
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Answer:

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Which of these solutes raises the boiling point of water the most?
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7 0
3 years ago
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Suppose a grill lighter contains 50.0 g of butane. How many grams of butane in the lighter would have to be burned to produce 17
Hitman42 [59]

<span>Answer is: mass of burned butane is 11.6 g.</span>

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<span> V(CO</span>₂) = 17,9 L.<span>
n(CO</span>₂) = V(CO₂) ÷ Vm.<span>
n(CO</span>₂) = 17,9 L ÷ 22,4 L/mol.<span>
n(CO</span>₂) = 0,8 mol.<span>
From chemical reaction n(CO</span>₂) : n(C₄H₁₀) = 8 : 2.<span>
n(C</span>₄H₁₀) = 0,8 mol ÷ 4.<span>
n(C</span>₄H₁₀) = 0,2 mol.<span>
m(C</span>₄H₁₀) = n(C₄H₁₀) · M(C₄H₁₀).<span>
m(C</span>₄H₁₀) = 0,2 mol · 58 g/mol.<span>
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5 0
4 years ago
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