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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Answer:

Explanation:
Nitrogen has the atomic number = 7
So, No. of electrons = 7
<u><em>Electronic Configuration:</em></u>

<u>Remember that:</u>
s sub shell holds upto 2 electrons while p sub shell upto 6
A solution is the answer to a problem
i think the greater the electric charge the atom decreases in size
Answer:
- Volume = <u>2.0 liter</u> of 1.5 M solution of KOH
Explanation:
<u>1) Data:</u>
a) Solution: KOH
b) M = 1.5 M
c) n = 3.0 mol
d) V = ?
<u>2) Formula:</u>
Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:
<u>3) Calculations:</u>
- Solve for n: M = n / V ⇒ V = n / M
- Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter
You must use 2 significant figures in your answer: <u>2.0 liter.</u>