Question

How will you convert: (i) 3 — Nitrotoluene into 3 — Methylanilhie (ii) 3 — Methylaniline into 3 — Nitrotoluene.

Answer 1

When 3-methylaniline treated with $NaNO_2$  and hydrogen chloride , HCl it gets converted into chlorine complex. When that complex reacted with $HBF_4$ It gets converted into Barium Fluoride complex. This complex reacts with $NaNO_2$ in presence of copper to give 3-Nitrotoluene.

reaction is shown in figure.

Answer 2

Answer:

ANSWER

We know, p(x)=[g(x)×q(x)]+r(x)

∴p(x)−r(x)=g(x)×q(x)

∴p(x)+{−r(x)}=g(x)×q(x)

It is clear that RHS is divisible by g(x). ∴LHS is also divisible by g(x)

Thus, if we add −r(x) to p(x), then the resulting polynomial is divisible by g(x).

Let us divide p(x)=x

4

+2x

3

−2x

2

+x−1 by g(x)=x

2

+2x−3 to find the remainder r(x).

∴r(x)=−x+2⇒{−r(x)}=x−2

Hence, we should add (x−2) to p(x)=x

4

+2x

3

−2x

2

+x−1 so that the resulting polynomial is exactly divisible by x

2

+2x−3.