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3 years ago

What product is obtained from the aldol condensation of cyclohexanone?

Physics

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-Second product is FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Explanation:

Please see the attached image for complete chemical reaction of aldol condensation of cyclohexanone

Aldol Condensation is a form of electrophilic substitution reaction in which the alpha carbon in enols or enolate anions is substituted by an electrophile to form carbon-carbon bond. Cyclohexanone also known as the first ketone consists of two alpha-carbons and four potential substitutions i.e alpha-hydrogens but none of the hydrogen on the ring is substituted. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbon.

First product is FCH-OH chemically known as 2-[2-furyl(hydroxyl)methyl]-cyclohexanone that further undergoes dehydration resulting into FCH i.e (2E)-2-[2-furyl-methylene]-cyclohexanone

Based on the explanations above, the compound formed is shown in the image.

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a

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√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

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Solving for t':

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Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

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3 years ago

Ohm’s law describes the relationship between which quantities?

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3 years ago

Read 2 more answers

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Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (zero movement) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

graph B, and

graph D.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

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2 years ago