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SOVA2 [1]
3 years ago
6

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

Chemistry
1 answer:
prohojiy [21]3 years ago
4 0

Answer:

2-methylene propylbenzene

Explanation:

The Wittig Reaction is a reaction that converts aldehydes and ketones into alkenes through reaction with a phosphorus ylide.

The ketone in this case is 1-phenylpropan-1-one. The provided phosphonium ylide is shown in the image attached. The reaction involves;

i) alkylation  

ii) addition

The product of the major organic product of the reaction is 2-methylene propylbenzene.

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Provide a stepwise synthesis of 1-cyclopentylethanamine using the Gabriel synthesis. Collapse question part Testbank, Question 0
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Answer:

Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.

an azide ion

An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,

imide linkage

You should note the commonly used trivial names of the following compounds.

phthalic acid, phthalic anhydride, and phthalimide

The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.

If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).

Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).

The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.

Explanation:

8 0
3 years ago
What is the atomic mass of neon to the nearest tenth<br><br><br>help me pls
motikmotik

Answer:

20.2

Explanation:

6 0
3 years ago
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n =
lys-0071 [83]

Answer:

\large \boxed{\text{4650 nm}}

Explanation:

The second line in the Pfund series corresponds to a transition from n = 7 to n = 5.

To calculate the wavelength of the transition, we can use the Rydberg equation:

\dfrac{1}{\lambda} = R_{H}\left ( \dfrac{1 }{n_{1}^{2}} - \dfrac{1 }{n_{2}^{2}} \right )

where

R_{H} = 1.097 \times 10^{7} \text{ m}^{-1}

If n₁ = 5 and n₂ = 7

\begin{array}{rcl}\dfrac{1}{\lambda} & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{1 }{5^{2}} - \dfrac{1 }{7^{2}} \right )\\\\ & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{1 }{25} - \dfrac{1 }{49} \right )\\\\ & = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{49 - 25 }{49 \times25}\right )\\\\& = & 1.097 \times 10^{7} \text{ m}^{-1}\left ( \dfrac{24 }{1225}\right )\\\\ & = & 2.149 \times 10^{7}\text{ m}^{-1}\\\end{array}\\

\begin{array}{rcl}\lambda & = & \dfrac{1}{2.149 \times 10^{7}\text{ m}^{-1}}\\\\ & = & 4.65 \times 10^{-6} \text{ m}\\ & = & \mathbf{4650} \textbf{ nm}\\\end{array}\\\text{The wavelength of the line is $\large \boxed{\textbf{4650 nm}}$, which is in the $\textbf{infrared region}$}.

 

8 0
3 years ago
To calculate percent by mass, use the equation
DiKsa [7]

Answer:

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Explanation:that’s the answer on edge2020

7 0
3 years ago
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Which statement is true according to the kinetic theory?
uranmaximum [27]
The correct answer is option D. i.e.<span>All states of matter show random motion of particles.ding to the kinetic theory

According to the kinectic theory of matter, all the particles show random motion at the room temperature. Decreasing the temperature to absolute zero temperature i.e. 0 K will cease the motion of particles.</span>
8 0
3 years ago
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