How many monochloro and monobromo isomers are there with 3-ethyl-4-methylhexane?
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Answer:
To determine the % of each possible monochloro or monobromo isomer formed, the number of hydrogens must first be found. For 3-ethyl-4-methylhexane, the number of hydrogens is 20. Therefore, monosubstitution could occur at any of these 20 positions.
The best way to approach this answer is to use the experimental evidence that the positions on the alkane have relative reactivities towards free radical chlorination are primary < secondary < tertiary at relative ratios of 1.0 : 3.8 : 5.0 respectively.
The 7 possible chlorination products are
A) 1-chloro-3-ethyl-4-methylhexan... (6 primary hydrogen positions)
B) 2-chloro-3-ethyl-4-methylhexan... (4 hydrogen secondary positions)
C) 3-chloro-3-ethyl-4-methylhexan... (1 tertiary hydrogen position)
D) 4-chloro-3-ethyl-4-methylhexan... (1 tertiary hydrogen position)
E) 5-chloro-3-ethyl-4-methylhexan... (2 secondary hydrogen positions)
F) 6-chloro-3-ethyl-4-methylhexan... (3 primary hydrogen positions)
G) 3-ethyl-4-chloromethylhexane (3 primary hydrogen positions)
To calculate the % of each isomer, the number of hydrogen positions and the relative reactivity at each position must be taken into account.
A) 6 x 1.0 = 6.0
B) 4 x 3.8 = 15.2
C) 1 x 5.0 = 5.0
D) 1 x 5.0 = 5.0
E) 2 x 3.8 = 7.6
F) 3 x 1.0 = 3.0
G) 3 x 1.0 = 3.0
TOTAL = 6.0 + 15.2 + 5.0 + 5.0 + 7.6 + 3.0 + 3.0 = 44.8
The predicted relative % for each product would be as follows.
A) (6.0 / 44.8) x 100% = 13.4%
B) (15.2 / 44.8) x 100% = 33.9%
C) (5.0 / 44.8) x 100% = 11.2%
D) (5.0 / 44.8) x 100% = 11.2%
E) (7.6 / 44.8) x 100% = 17.0%
F) (3.0 / 44.8) x 100% = 6.7%
G) (3.0 / 44.8) x 100% = 6.7%
For free radical bromination, substitution occurs at the tertiary positions >99%. The 2 products predicted would be
A) 3-bromo-3-ethyl-4-methylhexane (1 tertiary hydrogen position)
B) 4-bromo-3-ethyl-4-methylhexane (1 tertiary hydrogen position)
The predicted percentages of these products would be
A) >49.5%
B) >49.5%
all other brominated products <1%
Remember that any substitutions on the 3-ethyl will be the same as substitutions on the 'main' branch. So that is 2. Then substitute the single hydrogen on C3 makes 3. Substitute one of the hydrogens on C4 - 4. One of the carbons on the methyl group - 5. One of the hydrogens on C5 - 6, and one of the hydrogens on C6 - 7. That is monosubstitutions - 7 for Cl and 7 for Br.
For bromo-chloro disubstitutions (C7 and C8 are the ethyl-)
1. C1(Cl,Br) 1-Bromo-1-Chloro-3 ethyl-4 methylhexane
2. C1(Cl) C2(Br) 1-Chloro-2-Bromo-3 ethyl-4 methylhexane
3. C1(Cl) C3(Br) etc
4. C1(Cl) C4(Br)
5. C1(Cl) C5(Br)
6. C1(Cl) C6(Br)
7. C1(Cl) C7(Br)
8. C1(Cl) C8(Br)
and keep permutating the combinations.
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