Concept explainers
Videos
The relative rate of radical bromination is 1;82;1640 for 1°;2°;3° hydrogens, respectively. Draw all of the monobrominated products that you might obtain from the radical bromination of the compounds below. Calculate the relative percentage of each.
(a) methylcyclobutane
(b) 3,3-dimethylpentane
(c) 3-methylpentane
a) Methylcyclobutane
Interpretation:
The relative rate of radical bromination is 1:82:1640 for 1°, 2° and 3° hydrogens respectively. The structures of all the monobrominated products obtainable from the radical bromination of methylcyclobutane are to be drawn. The relative percentage of their formation is also to be calculated.
Concept introduction:
During radical bromination all types of hydrogens present in a compound are replaced by bromine to yield different products. The number of each type of hydrogen present in the compound and their relative rates of bromination are calculated seperately. The relative percentage of formation of a particular type of hydrogen can be calculated from the total rate of bromination of all types of hydrogens and that of the particular hydrogen.
To draw:
The structures of all the monobrominated products obtainable from the radical bromination of methylcyclobutane.
To calculate:
The relative percentage of formation of each monobromination product.
Answer to Problem 48AP
Four different monochlorination products are possible by the radical bromination of methylcyclohexane. They are bromomethylcyclobutane(I), 1-bromo-1-methylcyclobutane(II), 1-bromo-2-methylcyclobutane(III) and 1-bromo-3-methylcyclobutane(IV).
The relative percentage of 1°, 2° and 3° hydrogens in methylcyclobutane is 0.15: 23.1: 76.8.
Explanation of Solution
Methylcyclobutane has four types of hydrogens, one in methyl, a second on C1 to which methyl group is attached, a third one on C2 and C4 and a fourth one on C3. It has three 1° hydrogens, six 2° hydrogens and one 3° hydrogen. Hence four monochlorination products are possible.
The relative rate of radical bromonation of 1° hydrogens = 3 x 1 = 3.
The relative rate of radical bromonation of 2° hydrogens = 6 x 82 = 492.
The relative rate of radical bromonation of 3° hydrogens = 1 x 1640 = 1640.
Total rate of radical bromonation of all hydrogens = 3+492+1640 = 2135.
Therefore the relative percentage of
Radical bromonation of 1° hydrogens = 3/2135 x 100= 0.15%.
Radical bromonation of 2° hydrogens = 492/2135 x 100= 23.1%.
Radical bromonation of 3° hydrogens = 1640/2135 x 100= 76.8%.
Four different monochlorination products are possible by the radical bromination of methylcyclohexane. They are bromomethylcyclobutane(I), 1-bromo-1-methylcyclobutane(II), 1-bromo-2-methylcyclobutane(III) and 1-bromo-3-methylcyclobutane(IV).
The relative percentage of 1°, 2° and 3° hydrogens in methylcyclobutane is 0.15: 23.1: 76.8.
b) 3,3-dimethylpentane
Interpretation:
The relative rate of radical bromination is 1:82:1640 for 1°, 2° and 3° hydrogens respectively. The structures of all the monobrominated products obtainable from the radical bromination of 3,3-dimethylpentane are to be drawn. The relative percentage of their formation is also to be calculated.
Concept introduction:
During radical bromination all types of hydrogens present in a compound are replaced by bromine to yield different products. The number of each type of hydrogen present in the compound and their relative rates of bromination are calculated seperately. The relative percentage of formation of a particular type of hydrogen can be calculated from the total rate of bromination of all types of hydrogens and that of the particular hydrogen.
To draw:
The structures of all the monobrominated products obtainable from the radical bromination of 3,3-dimethylpentane.
To calculate:
The relative percentage of formation of each monobromination product.
Answer to Problem 48AP
Three different monochlorination products are possible by the radical bromination of 3-bromomethyl-3-methylpentane. They are bromomethylcyclobutane(I), 2-bromo-3,3-dimethylpentane(II) and 1-bromo-3,3-dimethylpentane (III).
The relative percentage of 10and 20 hydrogens in 3,3-dimethylpentane is 3.6:96.4.
Explanation of Solution
3,3-Dimethylpentane has two types of hydrogens, one in methyl groups and other in CH2 attached to methyl group. Hence two monochlorination products are possible. It has twelve 1° hydrogens and four 2° hydrogen atoms.
The relative rate of radical bromonation of 1° hydrogens = 12 x 1 = 12.
The relative rate of radical bromonation of 2° hydrogens = 4 x 82 = 328.
Total rate of radical bromonation of all hydrogens = 12+328 = 340.
Therefore the relative percentage of
Rdical bromonation of 1° hydrogens = 12/340 x 100= 3.6%.
Rdical bromonation of 2° hydrogens = 328/340 x 100=96.4%.
Three different monochlorination products are possible by the radical bromination of 3-bromomethyl-3-methylpentane. They are bromomethylcyclobutane(I), 2-bromo-3,3-dimethylpentane(II) and 1-bromo-3,3-dimethylpentane (III).
The relative percentage of 1° and 2° hydrogens in 3,3-dimethylpentane is 3.6:96.4.
c) 3-methylpentane
Interpretation:
The relative rate of radical bromination is 1:82:1640 for 1°, 2° and 3° hydrogens respectively. The structures of all the monobrominated products obtainable from the radical bromination of 3-methylpentane are to be drawn. The relative percentage of their formation is also to be calculated.
Concept introduction:
During radical bromination all types of hydrogens present in a compound are replaced by bromine to yield different products. The number of each type of hydrogen present in the compound and their relative rates of bromination are calculated seperately. The relative percentage of formation of a particular type of hydrogen can be calculated from the total rate of bromination of all types of hydrogens and that of the particular hydrogen.
To draw:
The structures of all the monobrominated products obtainable from the radical bromination of 3-methylpentane.
To calculate:
The relative percentage of formation of each monobromination product.
Answer to Problem 48AP
Four different monochlorination products are possible by the radical bromination of 3-methylpentane. They are bromomethylpentane(I), 3-bromo-3-methylpentane(II), 2-bromo-3-methylpentane(III) and 1-bromo-3-methylpentane(IV).
The relative percentage of 1°, 2° and 3° hydrogens in 3,3-dimethylpentane is 0.46:16.6:82.9.
Explanation of Solution
3,3-Dimethylpentane has four types of hydrogens, one in methyl at C1, another methyl groups attached to CH2, a third in C2 and a fourth in CH2. It has nine 1° hydrogens, four 2° hydrogens and one 3° hydrogen. Hence four monochlorination products are possible.
The relative rate of radical bromonation of 1° hydrogens = 9 x 1 = 9.
The relative rate of radical bromonation of 2° hydrogens = 4 x 82 = 328.
The relative rate of radical bromonation of 3° hydrogens = 1 x 1640 = 1640.
Total rate of radical bromonation of all hydrogens = 9+328+1640 = 1977.
Therefore the relative percentage of
Radical bromonation of 1° hydrogens = 9/1977x100= 0.46%.
Radical bromonation of 2° hydrogens = 328/1977x100= 16.6%.
Radical bromonation of 3° hydrogens = 1640/1937x100= 82.9%.
Four different monochlorination products are possible by the radical bromination of 3-methylpentane. They are bromomethylpentane(I), 3-bromo-3-methylpentane(II), 2-bromo-3-methylpentane(III) and 1-bromo-3-methylpentane(IV).
The relative percentage of 1°, 2° and 3° hydrogens in 3,3-dimethylpentane is 0.46:16.6:82.9.
Want to see more full solutions like this?
Chapter 10 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- we were assigned to dilute 900ppm in to 18ppm by using only 250ml vol flask. firstly we did calc and convert 900ppm to 0.9 ppm to dilute in 1 liter. to begin the experiment we took 0,225g of kmno4 and dissolved in to 250 vol flask. then further we took 10 ml sample sol and dissolved in to 100 ml vol flask and put it in to a spectrometer and got value of 0.145A . upon further calc we got v2 as 50ml . need to find DF, % error (expval and accptVal), molarity, molality. please write the whole report. thank you The format, tables, introduction, procedure and observation, result, calculations, discussion and conclusionarrow_forwardQ5. Predict the organic product(s) for the following transformations. If no reaction will take place (or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state is present for each reaction (think Hammond Postulate). I Br₂ CH3 F2, light CH3 Heat CH3 F₂ Heat Br2, light 12, light CH3 Cl2, light Noarrow_forwardNonearrow_forward
- In the phase diagram of steel (two components Fe and C), region A is the gamma austenite solid and region B contains the gamma solid and liquid. Indicate the degrees of freedom that the fields A and B have,arrow_forwardFor a condensed binary system in equilibrium at constant pressure, indicate the maximum number of phases that can exist.arrow_forwardPart V. Label ad match the carbons in compounds Jane and Diane w/ the corresponding peak no. in the Spectra (Note: use the given peak no. To label the carbons, other peak no are intentionally omitted) 7 4 2 -0.13 -0.12 -0.11 -0.10 -0.08 8 CI Jane 1 -0.09 5 210 200 190 180 170 160 150 140 130 120 110 100 -8 90 f1 (ppm) 11 8 172.4 172.0 f1 (ppr HO CI NH Diane 7 3 11 80 80 -80 -R 70 60 60 2 5 -8 50 40 8. 170 160 150 140 130 120 110 100 90 -0 80 70 20 f1 (ppm) 15 30 -20 20 -60 60 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -0.01 10 -0.17 16 15 56 16 -0.16 -0.15 -0.14 -0.13 -0.12 -0.11 -0.10 -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 17.8 17.6 17.4 17.2 17.0 f1 (ppm) -0.03 -0.02 550 106 40 30 20 20 -0.01 -0.00 F-0.01 10 0arrow_forward
- n Feb 3 A T + 4. (2 pts) Draw the structure of the major component of the Limonene isolated. Explain how you confirmed the structure. 5. (2 pts) Draw the fragment corresponding to the base peak in the Mass spectrum of Limonene. 6. (1 pts) Predict the 1H NMR spectral data of R-Limonene. Proton NMR: 5.3 pon multiplet (H Ringarrow_forwardPart VI. Ca H 10 O is the molecular formula of compound Tom and gives the in the table below. Give a possible structure for compound Tom. 13C Signals summarized C1 C2 C3 C4 C5 C6 C7 13C shift (ppm) 23.5 27.0 33.0 35.8 127 162 205 DEPT-90 + DEPT-135 + +arrow_forward2. Using the following data to calculate the value of AvapH o of water at 298K. AvapH o of water at 373K is 40.7 kJ/mol; molar heat capacity of liquid water at constant pressure is 75.2J mol-1 K-1 and molar heat capacity of water vapor at constant pressure is 33.6 J mol-1 K-1.arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning