Hydrogen chloride gas is dissolved in water to produce a solution of hydrochloric acid.
1 answer:
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Answer:- 21.4 grams of are formed.
Solution:- The balanced equation is:
From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.
The molar mass of Li is 6.94 gram per mol and for It is 28.02 gram per mol. The molar mass of
is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:
=
=
From above calculations, Li gives least amount of the product. So, 21.4 g of are formed.
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
