The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
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∆H = m x s x ∆T, where m is the mass of the reactants, s is the specific heat of the product, and ∆T is the change in temperature from the reaction.
Answer:
the answer is the second choice actual force
Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
