Which of the reagents listed below would efficiently accomplish the transformation of ethyl-3-pentenoate into 3-penten-1-ol?
1 answer:
Reactions of Ethyl-3-pentenoate with all given reagents are given below.
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄:
Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄:
Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result:
The correct answer is Option-A (Highlighted RED below).
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄:
Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄:
Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result:
The correct answer is Option-A (Highlighted RED below).
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Answer:
19,26 kJ
Explanation:
The work done when a gas expand with a constant atmospheric pressure is:
W = PΔV
Where P is pressure and ΔV is the change in volume of gas.
Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:
500,0g Zn(s)××
= 7,648 moles of H₂
At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:
V = nRT/P
V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm
V = 190,1L
That means that ΔV is:
190,1L - 0L = <em>190,1L</em>
And the work done is:
W = 1atm×190,1L = 190,1atmL.
In joules:
190,1 atmL× = <em>19,26 kJ</em>
I hope it helps!