Which among the following compounds give yellow solid when reacted with NaOI/NaOH?
Choose the correct answer from the options given below:
When alcohols are reacted with NaOI/NaOH, a yellow solid is produced due to the formation of iodine complexes.
Upon examining the compounds:
- (A) \( \text{CH}_3 \text{CH}_2 \text{C}_2 \text{H}_5 \) (an alcohol) reacts with NaOI/NaOH, producing a yellow solid.
- (B) \( \text{CH}_3 \text{CH}_2 \text{C}_2 \text{H}_2 \text{OH} \) (an alcohol) reacts with NaOI/NaOH, producing a yellow solid.
- (C) \( \text{CH}_3 \text{C}_2 \text{H}_5 \) (an aldehyde) does not react with NaOI/NaOH to produce a yellow solid.
- (D) \( \text{CH}_3 \text{OH} \) does not react with NaOI/NaOH to form a yellow solid.
- (E) \( \text{CH}_3 \text{CH}_2 \text{H} \) does not react with NaOI/NaOH to form a yellow solid.
Thus, the correct answer is (2).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: