Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
Answer:
1281.25 Moles of glucose
Explanation:
Multiply 1.25x1025= 1281.25
Answer:
see below
Explanation:
1. Predicting products (double replacement): ab + cd ---> ad + cb
KNO₃(aq) + Fe(OH)₃(s)
2. balance the equation
3KOH (aq) + Fe(NO3)₃ (aq) ---> 3KNO₃(aq) + Fe(OH)₃(s)
3. I don't know if you need this but ionic equation: only aqueous things get split into ions; gas, liquid, and solids stay together
3K⁺(aq) + 3(OH)⁻(aq) + Fe³⁺(aq) + 3NO₃⁻(aq) ---> 3K ⁺(aq) + 3NO₃⁻(aq) + Fe(OH)₃(s)
removing things on both product and reactant side
3(OH)⁻(aq) + Fe³⁺(aq) --->Fe(OH)₃(s)
I don’t have a picture but I can describe it to you.
The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.
The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.
Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.
Hope this helps!
