STATEMENT-1 : The order of acidic character of the given compounds
STATEMENT-2 : -I and -R effect of the group stabilizes the conjugate base decreasing the -ve charge, whereas +I effect of the group destabilises the conjugate base by intensifying the charge

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

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STATEMENT-1 is True, STATEMENT-2 is False

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STATEMENT-1 is False, STATEMENT-2 is True

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Solution

The correct option is B STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
The order of acidity is correct and hence, statement-1 is correct.
This is because the -I and -R effect of $- N O_{2}$ and $- C l$ group stabilises the conjugate base whereas +I effect of $- M e$ destabilises the conjugate base.
Hence, statement -2 is correct and is the correct reason.

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Similar questions

- 1 :

0 < x < y \Rightarrow {log}_{a} x > {log}_{a} y

a > 1

- 2 :

a > 1, {log}_{a} x

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} = (n + 2) 2^{n - 1} .

\sum_{r = 0}^{n} (r + 1)^{n} C_{r} x^{r} = (1 + x)^{n} + n x (1 + x)^{n - 1} .

n \geq 2,

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n}

\sqrt{n (n + 1)} < n + 1

(p \lor q) \land \sim p

\sim p \land q

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