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$C_{4} H_{10} O (A) \overset{K M n O_{4} / H^{+}}{\to} C_{4} H_{8} O_{2} .$ $‘ A ’ w h e n t r e a t e d w i t h P C l_{3} f o l l o w e d b y r e d u c t i o n w i t h$ $Z n / H C l g i v e s n - b u \tan e . ‘ A ’ i s$

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n-butyl alcohol

20 butyl alcohol

Isobutyl alcohol

30 butyl alcohol

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Correct option is A

On Strong OxidationAll compounds contain same number of carbon atoms∴ ‘A’ can be (1) (or) (3) If A is option (3)So ‘A’ cannot be (3)CH3CH2CH2CH2OH→PCl3 CH3CH2CH2CH2Cl→ Zn/HClCH3CH2CH2CH3n−butane

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C4H10OA →KMnO4 / H+ C4H8O2.‘A’ when treated with PCl3 followed by reduction with Zn / HCl gives n – butane. ‘A’ is

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$C_{4} H_{10} O (A) \overset{K M n O_{4} / H^{+}}{\to} C_{4} H_{8} O_{2} .$ $‘ A ’ w h e n t r e a t e d w i t h P C l_{3} f o l l o w e d b y r e d u c t i o n w i t h$ $Z n / H C l g i v e s n - b u \tan e . ‘ A ’ i s$