Soonnnn pleaseeee 5 POINTSS EXTRAAAA!!! Please answer all 3 questionsss
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Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.
Another way to look at this problem is to use Newton's second law of motion. The first law states that , where
is the acceleration
is the net force and
is the mass of the object.
When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.
The proton which is easily abstracted in 1-Benzyl-3-propylbenzene is the proton which is present on carbon atom in between two phenyl rings, or the central carbon which is shared by two benzene rings.
This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the acetate ion.
In our case the stability of conjugate base arises due to stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.
I have shown the resonance of negative ion on both Phenyl rings with Blue and Pink colors.<span />
This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the acetate ion.
In our case the stability of conjugate base arises due to stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.
I have shown the resonance of negative ion on both Phenyl rings with Blue and Pink colors.<span />