STATEMENT-1 : $α$ -D-glucose and $β$ -D-glucose are anomers.
STATEMENT-2 : These are mirror images and superimposable.

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 C STATEMENT-1 is True, STATEMENT-2 is False
-D-glucose and -D-glucose are anomers.
They differ only at C-1 carbon atom. Hence, they are not mirror images of each other.

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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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