Draw the structures of the major and minor organic products formed when hbr reacts with 2,4,4-trimethyl-2-pentene in the presenc
e of peroxides. when drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be marked incorrect.
1 answer:
Answer : The major and minor products of 2,4,4-trimethy-2-pentene will be a tertiary radical of Br which will be in major quantity whereas secondary radical of Br will be in minor quantity.
As the reaction follows anti-markonikov's rule the radical formed in tertiary state will be more stable and the secondary state will be much unstable as compared to tertiary state.
Note - The structures are arranged in the attachment
As the reaction follows anti-markonikov's rule the radical formed in tertiary state will be more stable and the secondary state will be much unstable as compared to tertiary state.
Note - The structures are arranged in the attachment
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Answer:
a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
MM: 17.03 32.00 30.01
4NH₃ + 5O₂ ⟶ 4NO + 6H₂O
Mass/g: 1.5 1.85
2. Calculate the moles of each reactant
3. Calculate the moles of NO we can obtain from each reactant
From NH₃:
The molar ratio is 4 mol NO:4 mol NH₃
From O₂:
The molar ratio is 4 mol NO:5 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO.
The excess reactant is NH₃.
5. Calculate the mass of NO formed
6. Calculate the moles of NH₃ reacted
The molar ratio is 4 mol NH₃:5 mol O₂
7. Calculate the mass of NH₃ reacted
8. Calculate the mass of NH₃ remaining
Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃