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2 years ago

7

Draw the major organic product in the reaction scheme. Be sure to clearly show stereochemistry (if applicable). The starting mat

erial is an alkyne where carbon 1 is bonded to cyclopentane and carbon 2 is bonded to H. Step 1 is N a N H 2. Step 2 is C H 3 I. Step 3 is sodium in liquid ammonia. Draw the major organic product.

1 answer:

never [62]2 years ago

5 0

Answer:

See answer below

Explanation:

The picture below will show you the final product and mechanism.

In the first step, the NaNH₂ is a strong base, so, this base will substract the hydrogen from carbon 2, to generate a negative charge there, and then, carbon 2 becomes a nucleophyle.

As a nucleophyle it will attack to the CH₃I in the next step, and it will attach to the CH₃.

The second step is just a regular step to reduce the triple bond of the alkyne to alkane or alkene, this will depend on the quantity of the reactant. In this case, an alkene.

Hope this helps,,,,,,k

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma

klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, $m_1$ = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, $m_2$ = $(M-m_1)=(39.664-21.577)g=18.087g$

To calculate density of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Mass of NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

$\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL$

Hence, the density of NaCl solution is 3.930 g/mL

7 0

2 years ago

In which state of matter would you expect to find water, at 20°C on Earth?

ad-work [718]

Liquid because 20 Celsius is 68 Fahrenheit<span />

5 0

2 years ago

9. At equilibrium a 2 L vessel contains 0.360

klio [65]

Answer:

Ke = 34570.707

Explanation:

H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

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3 years ago

What is the oh of a solution with a ph of 9.8

Veseljchak [2.6K]

It would be 4.2, hope this helps.

4 0

2 years ago