The compound 'X' reacts with ammonical $A g N O_{3}$ to give a white precipitate and on oxidation with hot alkaline $K M n O_{4}$ gives the acid $(C H_{3})_{2} C H C O O H$ . Therefore, 'X' is:

C H_{2} = C H - C H = C H - C H_{2}

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C H_{3} (C H_{2})_{2} C \equiv C H

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(C H_{3})_{2} C H - C \equiv C H

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(C H_{3})_{2} C = C = C H_{2}

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Solution

The correct option is C $(C H_{3})_{2} C H - C \equiv C H$
Ammonical $A g N O_{3}$ reacts with acidic hydrogen and white ppt. forms.
So, here reactant is end alkyne (as they have acidic hydrogen).Also on oxidation with alkaline $K M n O_{4}$ it forms acid.
$(C H_{3})_{2} C H \equiv C H h o t a l k a l i n e K M n O_{4} - ----------- \to (C H_{3})_{2} C H C O O H$
$(C H_{3})_{2} C H \equiv C H T o l l e n^{'} s R e a g e n t - --------- \to W h i t e p p t$

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C H_{2} = C H - C H_{2} - C H_{2} - C \equiv C H

(C H_{3})_{2} C = C H - C H_{2} - C H_{3}

C H_{2} = C H - C \equiv C H

C H \equiv C - C \equiv C H

C H_{3} - C H = C H_{2}

C H_{2} = C H - C H = C H_{2}

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