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
Organic Chemistry
- For each alkane, which mono brominated derivatives could you form in good yield by free-radical bromination?(a) cyclopentane (b) methylcyclopentane(c) 2-methylpentane (d) 2,2,3,3-tetramethylbutanearrow_forwardDraw and name all of the monochorination products that you might obtain from the radical chlorination of the compounds below. Which of the products are chiral? Are any of the products optically active? (a) 2-methylbutane (b) methylcyclopropane (c) 2,2-dimethylpentanearrow_forwardReaction of this bicycloalkene with bromine in carbon tetrachloride gives a trans dibro- mide. In both (a) and (b), the bromine atoms are trans to each other. However, only one of these products is formed. CH3 CH3 CH3 Br Br CH,Cl, + Br2 or Br Br (a) (b) Which trans dibromide is formed? How do you account for the fact that it is formed to the exclusion of the other trans dibromide?arrow_forward
- Propene reacts with hydrogen bromide to form two isomers, the major product being, 2- bromopropane. 1(c) Draw the mechanism of the reaction of hydrogen bromide with propene to form 1- bromopropane and 2-bromopropane, showing the structure of both intermediates. (i) (ii) Explain why 2-bromopropane is the major product.arrow_forward1. Predict the elimination products of the following reactions. When two alkenes are possible, predict which one will be the major product. Explain your answers, showing the degree of substitution of each double bond in the products. 2. Which of these reactions are likely to produce both elimination and substitution products? (a) 2-bromopentane +NaOCH3 (b) 3-bromo-3-methylpentane +NaOMe(Me= methyl, CH3) (c) 2-bromo-3-ethylpentane +NaOH (d) cis-1-bromo-2-methylcyclohexane +NaOEt (Et= ethyl, CH2CH3)arrow_forwardCompounds X and Y are both C7H15Cl products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C7H₁4 alkene. Both X and Y undergo an SN2 reaction with sodium iodide in acetone solution to give C7H15l products; in this reaction Y reacts faster than X. What is the structure of X? • Do not use stereobonds in your answer. • In cases where there is more than one possible structure for each molecule, just give one for each. . Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate structures with + signs from the drop-down menu. наarrow_forward
- Draw the organic product obtained on treatment of each of the following two alkenes with bromine: (a) trans-2-pentene and (b) 1- methylcyclohexene. Having done this, draw the product of the reaction of these same alkenes with bromine in aqueous solution.arrow_forwardPlease give the main substitution product for each of the following reactions, and indicate the dominant mechanism: (a) 1-bromopropane + NaOCH3 → (b) 3-bromo-3-methylpentane + NaOC2H5 →arrow_forwardWrite the appropriate reagents, conditions and products for the following transformations, in a single step. KMNO, OH', heatarrow_forward
- Treatment of propadiene (an allene) with hydrogen bromide produces 2-bromopropene as the major product. This suggests that the more stable carbocation intermediate is produced by the addition of a proton to Br HBr. H2C=C=CH, H3C CH2 a terminal carbon rather than to the central carbon. Propadiene 2-Bromopropene (a) Draw both carbocation intermediates that can be produced by the addition of a proton to the allene. (b) Explain the relative stabilities of those intermediates. Hint: Draw the orbital picture of the intermediates and consider whether the CH, groups in propadiene are in the same plane.arrow_forwardPara-substituted product was produced when phenol reacts with cyclohexanecarbonyl bromide in the presence of AIB13. -Br Cyclohexanecarbonyl bromide (i) Outline the mechanism for this reaction. (ii) Draw the alternative substituted product formed.arrow_forwardIndicate the letter of the correct answer and kindly briefly justify the letter of answer. Cyclohexene undergoes hydrobromination. Which of these is a possible product? (A) Bromocyclohexane (B) All of these (C) Trans 1,2-dibromocyclohexane (D) Cis 1,2-dibromocyclohexanearrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY