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2 years ago

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Draw the major organic product for the reaction of 1-phenylpropan-1-one with the provided phosphonium ylide in part 1. In part 2

answer the question about the transformation.

2 answers:

ipn [44]2 years ago

5 0

Answer:

The major product formed is a benzyl allyl compound, namely 2-methylene propylbenzene

Explanation:

<u>Part 1.</u>

The product formed from this Wittig reaction with phosphonium ylide is a phenylallyl compound, 2-methylene propylbenzene. Structure and reaction synthesis is attached in file.

<u>Part 2</u>

The types of transformation involved are:

alkylation (as there is a transfer of alkyl group)
addition (reaction of ylide)

Ilia_Sergeevich [38]2 years ago

4 0

Answer:

2-methylene propylbenzene (major product)

Explanation:

This reaction yields olefins and organophosphorus compound and is also known as wittig olefination. The reactants fits the category of a ketone (1-phenylpropan-1-one) and a Wittig reagent (phosphonium ylide).

Part 1: Answer in the fig. Below.

1-phenylpropan-1-one + phosphonium ylide ➡ phenylallyl compound + 2-methylene propylbenzene (major product)

Part 2.

Classical mechanisms that happened during its transformation are thus;

• nucleophilic addition

• Elimination

• cycloaddition

• [2+2]/retro-[2+2] mechanism.

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2 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

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Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

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2 years ago

Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ivann1987 [24]

Answer:The formulas of ionic compounds are:

a) $CuBr$

b) $Mn_2O_3$

c) $Hg_2I_2$

d) $Mg_3(PO_4)_2$

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains $Cu^+$ ion.

$Cu^++Br^-\rightarrow CuBr$

(b) Manganese oxide : Given that it contains $Mn^{3+}$ ion.

$Mn^{3+}+O^{2-}\rightarrow Mn_2O_3$

(c)Mercury iodide :Given that it contains $Hg_2^{2+}$

$Hg_2^{2+}+I^-\rightarrow Hg_2I_2$

(d) Magnesium phosphate :Given that it contains $PO_4^{3-}$

$Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2$

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2 years ago