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0.40 g of an organic compound (A), (C(5)...

0.40 g of an organic compound (A), `(C_(5)H_(8)O)` reacts with x mole of `CH_(3)MgBr` to liberate 224 mL of a gas at STP with excess of `H_(2)`, (A) gives pentan-1-ol. The correct structure of (A) is :

A

`CH_(3)-C-=C-CH_(2)-CH_(2)-OH`

B

`CH_(3)-CH_(2)-C-=C-CH_(2)-OH`

C

`H-C-=C-CH_(2)-CH_(2)-CH_(2)-OH`

D

`H-C -= C-CH_(2)-underset(OH)underset(|)(CH)-CH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
mol. Mass of `A = 12 xx 5 + 8 + 16 = 84g`
moles of `A = (0.40 g)/(84 g) = 0.005` moles
moles of gas released at STP `= (224 mL)/(22.4 L) = 0.01` moles
`therefore` 0.005 moles of A release 0.01 moles of gas.
i.e., ratio of 1:2.
Hence the must be two acidic H in the compound A. One acidic H is from alcohol the other must be terminal alkyne as only H attached to terminal alkynes are acidic.
`A overset("Excess " H_(2))rarr` pentan-1-ol
Hence option (d) is incorrect.
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