Chemistry Reference
In-Depth Information
d.
2.51
The four possible structures are
CH
3
CH
2
CHCl
2
(1,1-dichloropropane)
CH
3
CH
2
ClCH
2
Cl
(1,2-dichloropropane)
CH
2
ClCH
2
CH
2
Cl
(1,3-dichloropropane)
CH
3
CCl
2
CH
3
(2,2-dichloropropane)
Only the last structure has all hydrogens equivalent and can give only
one
trichloro
compound. This structure must, therefore, be C:
Cl
2
CH
3
CCl
2
CH
3
CH
3
CCl
2
CH
2
Cl
1,3-Dichloropropane has only two different “kinds” of hydrogens. It must be D:
Cl
2
CH
2
ClCH
2
CH
2
Cl
CH
2
ClCHClCH
2
Cl and CH
2
ClCH
2
CH
Cl
2
Next, A must be capable of giving 1,2,2-trichloropropane (the product from C). This
is not possible for the 1,1-isomer since it already has two chlorines on carbon-1.
Therefore, A must be 1,2-dichloropropane; it can give the 1,2,2-trichloro product (as
well as 1,1,2- and 1,2,3-). By elimination, B is CH
3
CH
2
CHCl
2
.
2.52
The equations follow the same pattern as in eqs. 2.16 through 2.21.
heat
initiation
Cl
Cl
2Cl
or light
propagation
CH
3
CH
3
+Cl
CH
3
CH
2
+
HCl
CH
3
CH
2
+
Cl
Cl
CH
3
CH
2
Cl
+
Cl
termination
2
Cl
Cl
Cl
2CH
3
CH
2
CH
3
CH
2
CH
2
CH
3
CH
3
CH
2
+Cl
CH
3
CH
2
Cl
By-products resulting from the chain-termination steps would be butane and any
chlorination products derived from it.
