Answer:
Along the upper right side
Answer:
Methyl propyl ether protonates at oxygen, and this makes the oxygen a good leaving group for attack by iodide ion which is a good nucleophile.
We are aware that prefers 1 degree carbocation and here both methyl , propyl group is 1 degree.
So we have two possible carbon atoms to react with iodide which are methyl group and propyl group. They will react with iodide at similar rates, and so there is no great preference of one over the other.
As is present in excess the hydrogen of protonates the oxygen which leads to cleavage that can happen on both sides either methyl or propyl side leading to different products.
The reaction proceeds in the manner
+ → CH3I + CH3CH2CH2OH
OR
+ → CH3OH + CH3CH2CH2I
Answer:
-Being in the service of quality, safety, designing and problem solving.
-It plays an importnat part in our lives too, to measure any surface, object, etc.
Explanation:
Measurement is perhaps one of the most fundamental concepts in science. Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories.
Answer:
Compound A is 2-bromo-3-ethylpentane. Compound B is 3-ethylpent-2-ene. Compound C is 3-ethylpent-1-ene. Structures are given below.
Explanation:
Sodium ethoxide in ethanol gives E2 ellimination by removing HBr when it is treated with 2-bromo-3-ethylpentane.
2-bromo-3-ethylpentane ( has two possible positions to give ellimination products.
3-ethylpent-2-ene is the major product as it contains more substituted double bond and 3-ethylpent-1-ene is the minor product as it contains less substituted double bond.
Both 3-ethylpent-2-ene and 3-ethylpent-1-ene gives only one product due to catalytic hydrogenation through addition of in double bond.
Structures are given below.
CH₄ + 2O₂ ---> CO₂ + 2H₂O
.............64g.........44g...............
64g O₂ --- 44g CO₂
8,94g O₂ --- X
X = (44×8,94)/64
X = 6,15g CO₂
b)