How to Calculate DBE-1 - UCLA

HowtoCalculateDBE Hellofriends!HaveproblemswithcalculatingDBE?Noworries!Hereisthetutorialwhichwill helpyoustepbystep.Hopefullyafterreadingthistutorial,youcancalculateDBEfasterand moreaccurately. ? WhatisDBE? DBE=doublebondequivalent.Itisalsocalleddegreeofunsaturation.Fromthestructureofthe chemicals,eachpibondorringwillgenerateoneDBE.

? Forexample: 1. Ethylene,C2H6,isasaturatedacyclicalkaneanditdoesnothaveanypibondor ring,soDBE=0. 2.Propylene,C3H6,containsapibond,soDBE=1. 3.Cyclohexane,C6H12,containsaring,soDBE=1.

Ethylene.Propylene.Cyclohexane

DBE=0DBE=1DBE=1

Thechemicalformulaandthechemicalstructurearehighlyrelated:eachpibondorringwill

causeapairofHlessthanthemaximumnumber.Therefore,forhydrocarbon,eachpairofH

lessthanmaxHcount,whichisequaltothesaturatedacyclicalkane,iscountforoneDBE.

Thinkquestion:HowmanyDBEisinC2H5Cl?

? WhyweneedtoknowDBE?

Infraredspectroscopyisveryusefultooltoanalyzethestructureofanunknownchemical. CalculatingDBEisanimportantstepbeforeanalyzingthefunctionalgroupsanditwillofferus morecluestonarrowdowntheoptionsoffunctiongroupsandmaketheworkeasier.

? HowtocalculatetheDBEifwedonotknowthestructureofthechemicals?

ForChem14C,alltheproblemswehaveevermettalkabouttheorganicchemicalswhichonly containcarbon,oxygen,hydrogen,nitrogen,andhalogens.Therefore,peoplesummarizeda DBEformulaforourconvenience.

Inthisformula,Cmeansthenumberofcarbon.Hmeansthenumberofhydrogenandhalogen. Nmeansthenumberofthenitrogen.

Let'sapplytheformulatothechemicalsthatwementionedbefore.

Ethylene(C2H6):DBE=C-(H/2)+(N/2)+1=2-(6/2)+(0/2)+1=0 Propylene(C3H6):DBE=C-(H/2)+(N/2)+1=3-(6/2)+(0/2)+1=1 Cyclohexane(C6H12):DBE=C-(H/2)+(N/2)+1=6-(12/2)+(0/2)+1=1 Thinkquestion:Howdidpeoplegetthisformula? ? OtherimportantthingsyouneedtoknowaboutDBE: 1.DBEisanonnegativeinteger.IfyougotanegativeorafractionDBE,youwillhavetogo backandcheckyourmathmistake! 2.MemorizethefollowingfactsanditwillbemoreconvenientwhenyouapplyDBEto figureoutthechemicalstructure: DoublebondcontainsoneDBE. TriplebondcontainstwoDBE. AbenzeneringcontainsfourDBE. 3.DoNOTdoublecounttheringsinthechemicalstructure. E.g.

Onlycounttworingsinthisstructure

5pibonds+2rings=>DBE=5+2=7

Beforewestartthepracticeproblem,letmeexplainthethinkquestionsfirst:

Question1:HowmanyDBEisinC2H5Cl? Inthiscase,C2H5Clisnothydrocarbon,sowecannotsimplytelltheDBEfromhowmuchH

doesitlack.However,halogensusuallyformonesinglebondtowithotheratoms,whichisthe

sameashydrogen.WecanimaginethatthestructureofC2H5ClissimilartoC2H6,inwhichcase

DBE=0. Question2:Howdowegetthisformula?

AccordingtotheHydrogen/HalogenRule,maxnumberofhydrogens+halogens=2C+N+2.We

discussedthateachpairofhydrogenandhalogenslessthanmaxhydrogensandhalogenscount

equalsDBE.

? Practiceproblem 1. LookatthechemicalstructurebelowandcalculatetheDBE.

1)

2)

3)

4)

5)

6)

2.UsetheformulatocalculatetheirDBE.

1)C8H102)C6H12O63)C3H4Cl2

4)C8H7N5)C13H106)C4H9NO

? Answerkeyforthepracticeproblems: 1. 1)Onepibond.DBE=1

2)Twopibonds.DBE=2

3)Onepibond.DBE=1

4)Tworings.DBE=2

5)Onepibondsandthreerings.DBE=4

6)Threepibondsandoneringinthemiddleandthreepibondsonsubstituents.DBE=7

2.1)DBE=C-(H/2)+(N/2)+1=8-(10/2)+(0/2)+1=4

2)DBE=C-(H/2)+(N/2)+1=6-(12/2)+(0/2)+1=1

3)DBE=C-(H/2)+(N/2)+1=3-(6/2)+(0/2)+1=1

4)DBE=C-(H/2)+(N/2)+1=8-(7/2)+(1/2)+1=6

5)DBE=C-(H/2)+(N/2)+1=13-(10/2)+(0/2)+1=9

6)DBE=C-(H/2)+(N/2)+1=4-(9/2)+(1/2)+1=1

Reference:

Hardinger, Steven. Chemistry 14C: Structure of Organic Molecules : Course Thinkbook : Concept Focus Questions, OWLS Problems, Practice Problems. Plymouth, MI: Hayden-McNeil, 2011. Print.

Hardinger, Steven. Chemistry 14C: Organic Molecular Structures and Interactions : Lecture Supplements. Plymouth, MI: Hayden-McNeil Pub., 2008. Print.

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