Draw the products of Claisen condensation of (a) ethyl propanoate; (b) ethyl-3-methylbutanoate; (c) ethyl pentanoate. For each,
the base used is sodium ethoxide and the solvent is ethanol.
1 answer:
Answer:
Structures of products has been shown below
Explanation:
- In claisen condensation, two ester molecules combines with each other through nucleophilic acyl substitution resulting formation of a
-ketoester.
- At first, one ester molecule gets deprotonated in presence of a strong base e.g. NaOEt yielding a carbanion.
- The carbanion then gives nucleophilic acyl substitution reaction with another ester molecule to replace alkoxy group present in ester group
- General reaction mechanism of claisen condensation along with products of given esters are shown below.
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It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.
By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then
=patm+
= 101325+ρ
It is given that height is 125ft. Put the value of h in above formula:
h1 =125ft=38.1m
ρ=1.04g/mL=1040kg/
g=9.81
=101325Pa+388711.44
=490036.44Pa
=p atm =101325Pa
It is known that volume and pressure can be expressed as:
V*P=const.
where, V is volume and P is pressure.
Now,
=
=
=490036.44/101325
=4.84
Assume constant temperature
d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.
now =p atm+
=490036.44Pa
V*p=const
where, V is volume and P is pressure.
Now,
=
=
=490036.44/X
=490036.44pa/(V2/V1) =326690.96Pa
=patm +p
=101325Pa+ρ
326690.96Pa=101325Pa+ρ
ρgh1 =151987.5-101325=225365.96Pa
ρ=1,04g/mL=1040kg/m3
g=9.81
=225365.96/ρ∗g
=225365.96 / 1040∗9.81
=22.09m= 72.47ft
ΔH=
=125-72.47
=52.53ft
So she can safely ascend up to 52.53 ft without Breathing out
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