Answer:
Explanation:
In the following reaction we have shown an example of aromatic substitution reaction .
C₆H₆ + RCl = C₆H₅R + HCl
This reaction takes place in the presence of catalyst like AlCl₃ which is a lewis acid .
First of all formation of carbocation is made as follows .
RCl + AlCl₃ = R⁺ + AlCl₄⁻
This R⁺ is carbocation which is also called electrophile . It attacks the ring to get attached with it .
C₆H₆ + R⁺ = C₆H₅R⁺H.
The complex formed is unstable , though it is stabilized by resonance effect . In the last step H⁺ is kicked out of the ring . The driving force that does it is the steric hindrance due to presence of two adjacent group of H and R⁺ at the same place . Second driving force is attack by the base AlCl₄⁻ that had been formed earlier . It acts as base and it extracts proton ( H⁺ ) from the ring .
C₆H₅R⁺H + AlCl₄⁻ = C₆H₆ + AlCl₃ + HCl .
The formation of a stable product C₆H₆ also drives the reaction to form this product .
Answer:
True
Explanation:
Friction is defined as the force that opposes motion when an object is sliding over a surface.
As a result of friction, all objects moving over a surface eventually come to rest over time.If we were to successfully create a friction-less surface, an object will remain in motion forever because it will encounter no opposition to its motion.
Hence, the resistance to the motion of objects over a surface which causes the objects to come to a halt after moving over the surface for some time is called friction.
<u>Answer:</u> Step 2 in the given mechanism is the rate determining step
<u>Explanation:</u>
In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.
The intermediate reaction of the mechanism follows:
<u>Step 1:</u> 
<u>Step 2:</u> 
As, step 2 is the slow step. It is the rate determining step
Rate law for the reaction follows:
![\text{Rate}=k[O_3][O]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BO_3%5D%5BO%5D)
Hence, step 2 in the given mechanism is the rate determining step
<span> this represents the relative overall energy of each orbital, and the energy of each orbital increases as the distance from the nucleus increases</span>
