Consider the following reaction $C H_{3} - (C H_{2})_{14} - C O O H L i A l H_{4} ⟶ X H C l ⟶$
Palmitic acid
$Y 1. M g / E t_{2} O ⟶ 2. O x i r a n e Z K M n O_{4} ⟶ W$
The correct identity of the compounds is/are :

X i s C H_{3} - (C H_{2})_{14} - C H_{2} O H

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Y i s C H_{3} - (C H_{2})_{14} - C H_{2} C l

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Z i s C H_{3} - (C H_{2})_{16} - C H_{2} O H

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W i s C H_{3} - (C H_{2})_{16} - C H O

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Solution

The correct options are
A $X i s C H_{3} - (C H_{2})_{14} - C H_{2} O H$
B $Y i s C H_{3} - (C H_{2})_{14} - C H_{2} C l$
C $Z i s C H_{3} - (C H_{2})_{16} - C H_{2} O H$
Reaction:
$C H_{3} - (C H_{2})_{14} - C O O H L i A l H_{4} ⟶ C H_{3} - (C H_{2})_{14} - C H_{2} O H H C l ⟶ C H_{3} - (C H_{2})_{14} - C H_{2} C l$
$C H_{3} - (C H_{2})_{14} - C H_{2} C l 1. M g / E t_{2} O ⟶ 2. O x i r a n e C H_{3} - (C H_{2})_{16} - C H_{2} O H K M n O_{4} ⟶ C H_{3} - (C H_{2})_{16} - C O O H$

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Which of the following compounds gives fastest reaction with HBr and why

(i) (CH3)3 COH

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(iii) CH3 CHOH CH3

(iv) CH3 (CH3) CH CH2OH

C H_{2} C H_{3} | C | O H - C H_{3} | C H C H_{3} I I \leftarrow - C H_{3} C H_{3} | C | C H_{3} C H = C H_{2} I \to C H_{3} - C H_{3} | C | C H_{3} - C | O H H C H_{3}

- C H O, - C H_{2} O H, - C H_{3}, - O H

{C H}_{3} - C | {C H}_{3} = C | B r - C H_{2} O H

{C H}_{3} - {C H}_{2} - C H | {C H}_{2} C I - {C H}_{2} O H | C H - C H | {C H}_{3} - {C H}_{3}

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