Question

`0.40 g` of an organic compound (A), `(M.F.-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)`

Answer 1

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