The correct order of basic strengths for the given compounds is:

R < P < Q

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Q < P < R

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P < R < Q

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Q < R < P

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Solution

The correct option is D $Q < R < P$
Due to ortho effect (steric inhibition to protonation), compound Q will be the weakest base. In both P and R, $N O_{2}$ is at the para position which exhibits both $- R$ and $- I$ effects. Since the presence of two methyl groups at meta positions in P diminishes the $- R$ effect of $N O_{2}$ makes it the strongest base.
Hence the order will be: $Q < R < P$

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Similar questions

(p - q)^{3} + (q - r)^{3} + (r - p)^{3}

p (p - q), q (q - r)

r (r - p)

p (p - q), q (q - r)

r (r - p)

The factors of $(p - q)^{3} + (q - r)^{3} + (r - p)^{3}$ are:

\frac{p - q}{q - r}

\frac{q - r}{p - q}

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