Question

I have the answer already from textbook soultions so please please please answer me from your words , your experience don't copy the answer of the textbook soultion I don't need that.

And this photo for the refernces for problems it may help!

Answer 1

When 1 equivalent of KNH2 is employed, the more acidic proton at C2 carbon gets deprotonated. It attacks the benzylchloride in an SN2 fashion producing 2,3,3-triphenylpropanonitrile.

However, when the base amount is increased to 2 equivalent, after deprotonation at C2 position, sucessive deprotonation at C3 position is observed. As the proton at C3 is less acidic than the proton at C2, its reactivity is higher. In otherwords, we can say, it is comparably unstable. As can be seen, the carbanion at C2 can be stabilized by the electron withdrawing nitrile group at C1 position. Hence, the carbanion at C3 reacts with benzylchloride producing 2,3,3,4-tetraphenylbutanonitrile in higher yield. After quenching with acid/water, the unreacted carbanion at C2 gets protonated.