<h2>Heptene formed is -</h2><h2>
</h2>
Explanation:
The two possibilities when the peroxide is not present
+ HBr →
In presence peroxide,
≡
+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -
A
Is the correct answer
I’m 95% sure
Answer:
Explanation:
Hello!
In this case, since the enthalpy change for any process is computed by subtracting the enthalpy of the final state and the enthalpy of the initial state, for the given phase change, we subtract the enthalpy of the liquid (final state) and the enthalpy of the solid (initial state) considering this a melting process:
Which makes sense because this process absorbs energy.
Best regards!
Answer:
d
Explanation:
e works, a and b one of them is stronger but it works d is the only logical answer
Answer:
[Ar] 4s² 3d⁵ or 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
Explanation:
Manganese electronic configuration is [Ar] 3d⁵ 4s². Manganese has 5 electrons in its 3d shell and all five electrons and unpaired maintaining parallel spin since they must obey hund's rule.
Generally electrons first enter 4s² shell filling it first before entering the 3d shell due to the lower energy 4s orbital has compared to 3d orbital. Filling of the 3d orbital must be filled parallel first before pairing begins. However in the case of manganese, it only has five electrons in its 3d orbital which will only fill it with single spins.
Due to the fact that manganese has five electrons in its 3d orbital its highly magnetic and its classified as a "ferromagnetic" substance.
Note when manganese forms a bond, it loses electrons from its 4s orbital too first to form Mn(ii) or combined from both 4s and 3d to form Mn(iii), (iv) (vi) etc.
Attached is a diagram showing how manganese atoms are arranged in its shells
≡
→
