Please be as detailed as possible, I\'m on the brink of failing and I need these
ID: 952786 • Letter: P
Question
Please be as detailed as possible, I'm on the brink of failing and I need these problems which I will use as my study guide for our next exam. Thank you so much greatly appreciate it.
TABLE 4-2 Bond-Dissociation Enthalpies for Homolytic Cleavages Bond-Dissociation Enthalpy Bond-Dissociation Enthalpy Bond kJ/mol keal/mol Bond kJ/mol keal/mal H-X bonds and X- X bonds Bonds to secondary carbons CH)2CH-H CH)2CH F CH CH C (CHs) CHBr (CH) CH I 435 104 106 10s 159 242 192 151 Cl-Cl 58 285 Br Br 53 381 136 103 431 Bonus to tertiary carbons H CI 381 S8 (CHs)CH 71 (CH)C F CH3)C CI 368 498 213 331 272 HO OH Methyl bonds CH,H CH3-F CH, C (CHs) CI 381 104 (CH3))C-OH 109 Other- bonds 84 293 234 381 PhCH,-H(benzylic) CH2 CHCH H fallylic) CH: CHH (vinyl) Ph-H (aromatic) HC=C-H (acetylenic) 87 CH3 I CH, QH Bonds to primary carbons 464 473 523 91 113 98 107 CC bonds 81CH CH 339 CHCH Ci CHCH,CH CHCH CH,CH 35fi 343 351 85 82 84 CH3CH2-CH2CH3 CH: CH CH (CHC CH 53 CH,CH OH CH,CH CHH CHCH,CH2 CH:CH CH-C 410 448 339 107 53 CH CH-CH2I CH,CH,CH OHExplanation / Answer
Since all the given species here are free radicals,so it should be desirable to review about " the generation of free radicals".
Free radicals are generated through a homolytic cleavage of a covalent bond in a manner so that one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus , such a cleavage results in the formation of neutral species ( atom or group ) which contain an unpaired electron. These species are called free radical. Free radicals so generated are also very reactive like carbocations or carbanions. Again the formation of free radicals depend on bond dissociation energies ( consideringa homolytic fission ,of course ). Hence , a free radical resulting with greater bond dissociationenthalpy (considering consistent unit of bond enthalpy) should be more active , or in other words less stable.
With these considerations a review of the given table of bond dissociation enthalpies of related bonds follow the order:
PhC. H2 (356)* < CH= CH- C. H2 (364)*< (CH3)3 C.(381)*< (CH3)2C. H (397)* < CH3C.H2 (410 )* < .CH3 (435)*..............[* the values indicate bond dissociation enthalpies in kJ/ mole]
therefore ,C. H3 has the maximum bond dissociation energy / enthalpy and should, therefore, be least stable.
Thus the stability of the given species (free radicals ) should follow the order :
.CH3 < . CH2CH3 < .CH (CH3)2 < .C(CH3)3 < CH2 = CH - C.H2 < Ph.C. H2