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Oxidative Hydroboration of Alkynes

What is the e...

Question

What is the end product of the following sequences of operations

Methyl alcohol

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Acetaldehyde

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C₂ H₅ OH

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C₂ H₄

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Solution

The correct option is C
C₂ H₅ OH

Calcium carbide on Reaction with $H_{2} O$ gives acetylene:

$C a C_{2} + H_{2} \to C_{2} H_{2} + C a (O H)_{2}$

Addition of $H_{2} O$ or Hydration of Alkyne: Acetylene on reaction with dilute $H_{2} S O_{4}$ (or $20 % H_{2} S O_{4}$ ) in presence of $1 % H g S O_{4}$ (or mercuric salt or $H g_{2 +}$ ion at $80^{\circ}$ C or no temperature mentioned) gives acetaldehyde.

The reaction is believed to take place via the formation of a three membered ring involving $H_{2} O$ ion.

Therefore,

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Similar questions

10

C a C_{2}

C a C_{2} + H_{2} O \to C a O + C_{2} H_{2}

C_{2} H_{2} \to C_{2} H_{4}

n C_{2} H_{4} \to {(C_{2} H_{4})}_{n}

C_{2} H_{4} + H_{2} O = C_{2} H_{5} O H

{Δ H}_{f o r m}^{o} (C_{2} H_{4}) = 55, 28 K J / m o l

{Δ H}_{f o r m}^{o} (H_{2} O) = - 241.8 K J / m o l

{Δ H}_{f o r m}^{o} (C_{2} H_{5} O H) = - 235.3 K J / m o l

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- {C H}_{2} - {C H}_{2} -)}_{n}

64.1 k g

C a C_{2}

K_{p}

C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}

5 \times 10^{18} {a t m}^{- 1}

K_{p}

C_{2} H_{2} + H_{2} ⇌ C_{2} H_{4}

5 \times 10^{26} {a t m}^{- 1}

50

K_{p}

2 H_{2} + C_{2} H_{2} ⇌ C_{2} H_{6}

x \times 10^{45}

10 x

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- C H_{2} - C H_{2} -)}_{n}

C a C_{2}

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