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pshichka [43]
3 years ago
14

Draw the structure(s) of the major organic product(s) of the following reaction. 1. lithium diisopropylamide / hexane 2. 1 eq. C

H3I You do not have to consider stereochemistry. Omit products derived from the acidic or basic reagent itself, e.g. HN(i-Pr)2 derived from N(i-Pr)2-. If no reaction occurs, draw the organic starting material. If substantial starting material is present at the end of the reaction, include it in the products. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with signs from the drop-down menu.
Chemistry
1 answer:
Jobisdone [24]3 years ago
6 0

Answer:

See explanation below

Explanation:

You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.

Now let's analyze what is happening in the reaction so we can predict the final product.

We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.

Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.

Hope this helps

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Convection: the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat.

hope that helps :)

3 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
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nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

  • Concentration of the base (Cb): 0.300 M
  • Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8  \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M

Step 4: Calculate the pOH

We will use the following expression.

pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6

Step 5: Calculate the pH

We will use the following expression.

pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4

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