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A sp³ hybridized carbon atom is the one which forms 4 single bonds with other atoms.
In the structure of Bilobalide, we can see 12 sp³ hybridized carbon atoms which are attached to other carbon, hydrogen and oxygen atoms.
In order to classify carbon atoms as 1°, 2°, 3°, or 4° we need to take a look at the atoms to which these carbon atoms are attached.
A 1° carbon atom is the one which is attached to only 1 carbon atom.
In the diagram, we can see 3 methyl groups ( -CH3) being attached to only 1 carbon atom. There is one more primary carbon atom.
There are 4 primary carbon atoms in Bilobalide.
A 2° carbon atom is the one which is attached to 2 other carbon atoms.
In Bilobalide we can see 4 secondary carbon atoms.
A 3° carbon atom is the one which is attached to 3 other carbon atoms.
In Bilobalide we can see only 1 tertiary carbon atom.
A 4° carbon atom is the one which is attached to 4 other carbon atoms.
In Bilobalide we can see 3 quaternary carbon atoms.
Please see the attached image for classification.
Answer:
r= 0.9949 (For 15,000)
r=0.995 (For 19,000)
Explanation:
We know that
Molecular weight of hexamethylene diamine = 116.21 g/mol
Molecular weight of adipic acid = 146.14 g/mol
Molecular weight of water = 18.016 g/mol
As we know that when adipic acid and hexamethylene diamine react then nylon 6, 6 comes out as the final product and release 2 molecule of water.
So
So
Mo= 226.32/2 =113.16 g/mol
Given that
Mn= 15,000 g/mol
So
15,000 = Xn x 113.16
Xn = 132.55
Now by using Carothers equation we know that
By calculating we get
r= 0.9949
For 19,000
19,000 = Xn x 113.16
Xn = 167.99
By calculating in same process given above we get
r=0.995