最新典型题高考化学二轮复习+知识点总结+有机化学基础优秀名师资料.docx
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最新典型题高考化学二轮复习+知识点总结+有机化学基础优秀名师资料
(典型题)2014高考化学二轮复习知识点总结有机化学基础
【典型题】2014高考化学二轮复习名师知识点总结:
有机化学
基础
一、选择题
1(药物贝诺酯可由乙酰水杨酸和对乙酰氨基酚在一定条件下反应制得:
下列有关叙述正确的是()
A(贝诺酯分子中有三种含氧官能团
B(可用FeCl溶液区别乙酰水杨酸和对乙酰氨基酚3
C(乙酰水杨酸和对乙酰氨基酚均能与NaHCO溶液反应3
D(贝诺酯与足量NaOH溶液共热,最终生成乙酰水杨酸钠和对乙酰氨基酚钠
解析:
结合三种有机物的结构确定含有的官能团及具有的性质。
贝诺酯分子中含有两种含氧官能团,A错。
对乙酰氨基酚分子中含有酚羟基,能与FeCl溶液发生显色反应,乙酰水杨酸则不能,可利用FeCl33溶液区别这两种有机物,B对。
乙酰水杨酸分子中含有—COOH,能与NaHCO溶液发生反应,3而对乙酰氨基酚则不能,C错。
贝诺酯与足量NaOH溶液共热发生水解反应,生成
,D错。
答案:
B
点拨:
知识:
有机物的结构及性质。
能力:
考查考生的综合分析能力和迁移应用能力。
试题难度:
中等。
2((2013?
浙江卷?
10)下列说法正确的是()
Weherebyapplyfor!
Applicant:
datedatecase,agreetowriteoffloans.Loanreviewsignature:
daterecordedform17:
creditcancellationapplicationformnameofborrowerIDnumberandcreditlimitexpirationdateyearmonthdaytoyearmonthdayloanmaximumamount...ComputertransferqilubankfileqiluSilver(2011)39th,onissuedqiluBankpersonalguaranteeloanmanagementapproachofnotificationthebranch,andjurisdictionlineandtheby'sbusinessline,andJinan'swithinthebranch,andheadofficethesector:
forpromotepersonalguaranteeloanbusinessofdevelopment,specificationbusinessoperation,effectivecontrolrisk,HeadOfficeOrganizationamendmenthasqiluBankpersonalguaranteeloanmanagementapproach(followingreferredto"thisapproach"),nowwillaboutrequirementsandtheamendmentcontentnotificationfollowing,pleaseseriouslyimplementation:
a,andSeriouslystudythedocuments,andstrictimplementationofthisapproachinallaspectsofthebusinessoperationsrequirementstoeffectivelymanagerisks.II,andthisapproachmainamendmentcontent:
(a)accordingtoChinabankingsupervisionManagementCommitteepersonalloanmanagementprovisionalapproach,willloanprocesssubdivisionforloanapplication,andacceptedandsurvey,andriskevaluationandreview,andloanapproval,andcontractsigned,andloanissued,andloanpaid,andloanHoumanagement,andrecyclinganddisposal,ninealink,andonthelinkofoperationproposedhasspecificrequirements.(B)thesimplification,theannexoftherelevantamendments.(C)requestingtheExecutivedoubleinterview,doublechecking,doublesystem.(D)loanfactors:
1theborrower
(1)borrowe
A(按系统命名法,化合物的名称是2,3,5,5,四甲基,4,4,二乙基己烷
B(等物质的量的苯与苯甲酸完全燃烧消耗氧气的量不相等
C(苯与甲苯互为同系物,均能使KMnO酸性溶液褪色4
D(结构片段为的高聚物,其单体是甲醛和苯酚
解析:
从有机物的组成和结构入手,分析其化学性质,得出合理答案。
该有机物的名称中碳原子编号错误,应从右端碳原子开始编号,正确名称应为2,2,4,5,四甲基,3,3,二乙基己烷,A项错。
苯、苯甲酸的分子式分别为CH、CHO,而苯甲酸66762可写成CH?
CO,显然等物质的量的两种有机物完全燃烧时,消耗O的物质的量相等,B6622
项错。
苯和甲苯互为同系物,甲苯能使KMnO酸性溶液褪色,而苯不能,C项错。
由高聚物4
的结构片段可知,该高聚物应由单体苯酚和甲醛发生缩聚反应生成,D项正确。
答案:
D
点拨:
知识:
有机物的命名,同系物及其性质,高聚物及其单体的推断。
能力:
考查考生对有机化学基础知识的理解及迁移应用能力。
试题难度:
中等。
3((2013?
东北三省四市第二次联考?
8)乙苯的一氯代物的结构共有()
A(3种B(4种
C(5种D(6种
解析:
乙苯乙基上氢原子环境有2种,苯环上氢原子环境有3种,所以乙苯的一氯代物共5种,C项正确。
答案:
C
点拨:
本题考查常见有机物命名及同分异构体知识。
难度较小。
4((2013?
高考命题研究专家原创卷)14.5g某烷烃完全燃烧生成1.25molHO,则该2烃的一氯代物共有(不考虑立体异构)()
A(2种B(3种
C(4种D(5种
解析:
设该烷烃的分子式为CH,则:
x2x,2
CH,(x,1)HOx2x,22
(14x,2)g(x,1)mol
14.5g1.25mol
hecking,doublesystem.(D)loanfactors:
1theborrower
(1)borroweements.(B)thesimplification,theannexoftherelevantamendments.(C)requestingtheExecutivedoubleinterview,doublecrequirnpaid,andloanHoumanagement,andrecyclinganddisposal,ninealink,andonthelinkofoperationproposedhasspecificon,andacceptedandsurvey,andriskevaluationandreview,andloanapproval,andcontractsigned,andloanissued,andloalicatingsupervisionManagementCommitteepersonalloanmanagementprovisionalapproach,willloanprocesssubdivisionforloanapperationsrequirementstoeffectivelymanagerisks.II,andthisapproachmainamendmentcontent:
(a)accordingtoChinabankissopimplementation:
a,andSeriouslystudythedocuments,andstrictimplementationofthisapproachinallaspectsofthebusinereferredto"thisapproach"),nowwillaboutrequirementsandtheamendmentcontentnotificationfollowing,pleaseseriouslylowingon,effectivecontrolrisk,HeadOfficeOrganizationamendmenthasqiluBankpersonalguaranteeloanmanagementapproach(folanch,andheadofficethesector:
forpromotepersonalguaranteeloanbusinessofdevelopment,specificationbusinessoperatithebreeloanmanagementapproachofnotificationthebranch,andjurisdictionlineandtheby'sbusinessline,andJinan'swithinthdayloanmaximumamount...ComputertransferqilubankfileqiluSilver(2011)39th,onissuedqiluBankpersonalguaranttcancellationapplicationformnameofborrowerIDnumberandcreditlimitexpirationdateyearmonthdaytoyearmonWeherebyapplyfor!
Applicant:
datedatecase,agreetowriteoffloans.Loanreviewsignature:
daterecordedform17:
credi2
列式求得x,4,则该烃的分子式为CH,可能是正丁烷,也有可能是异丁烷,各有2410
种一氯代物,故共有4种一氯代物。
答案:
C
点拨:
本题考查简单有机物同分异构体的数目判断,意在考查考生对简单有机物结构的掌握情况和推理能力。
5((2013?
高考命题研究专家原创卷)某有机物的化学式为CHO,现有0.1mol该有机483
物分别与足量的钠、足量的碳酸氢钠溶液反应,生成标准状况下的气体分别为2.24LH、22.24LCO。
则该有机物的同分异构体有(不考虑立体异构)()2
A(3种B(4种
C(5种D(6种
解析:
由—COOH,NaHCO―?
CO知,1mol羧基生成1mol二氧化碳;由—OH(或—COOH)32
12.24L,Na―?
H知,2mol羟基(或羧基)生成1molH。
依题意,n(H),n(CO),2222,1222.4L?
mol,0.1mol。
所以CHO分子中含有1个羧基和1个羟基。
CHO拆去—COOH、—OH剩余“CH”,48348336可以看成是丙烷(CH)上两个氢原子被羧基、羟基取代。
第一步,羧基取代丙烷上一个氢原38
—CH—COOH、(?
)子得两种同分异构体:
(?
)CH—CH223
答案:
C
点拨:
本题考查有机物同分异构体的数目判断,意在考查考生对中学化学基础知识的掌握情况及复述、再现、辨认的能力。
6((2013?
高考命题研究专家原创卷)某品牌白酒中含有的塑化剂的主要成分为邻苯二甲酸二丁酯,其结构简式为
ents.(C)requestingtheExecutivedoubleinterview,doublechecking,doublesystem.(D)loanfactors:
1theborrower
(1)borendmalink,andonthelinkofoperationproposedhasspecificrequirements.(B)thesimplification,theannexoftherelevantamloanapproval,andcontractsigned,andloanissued,andloanpaid,andloanHoumanagement,andrecyclinganddisposal,ninew,andvisionalapproach,willloanprocesssubdivisionforloanapplication,andacceptedandsurvey,andriskevaluationandrevieapproachmainamendmentcontent:
(a)accordingtoChinabankingsupervisionManagementCommitteepersonalloanmanagementprothistimplementationofthisapproachinallaspectsofthebusinessoperationsrequirementstoeffectivelymanagerisks.II,andheamendmentcontentnotificationfollowing,pleaseseriouslyimplementation:
a,andSeriouslystudythedocuments,andstricandtsqiluBankpersonalguaranteeloanmanagementapproach(followingreferredto"thisapproach"),nowwillaboutrequirementsloanbusinessofdevelopment,specificationbusinessoperation,effectivecontrolrisk,HeadOfficeOrganizationamendmentharanteesdictionlineandtheby'sbusinessline,andJinan'swithinthebranch,andheadofficethesector:
forpromotepersonalguaqiluSilver(2011)39th,onissuedqiluBankpersonalguaranteeloanmanagementapproachofnotificationthebranch,andjurifilenumberandcreditlimitexpirationdateyearmonthdaytoyearmonthdayloanmaximumamount...Computertransferqilubankeetowriteoffloans.Loanreviewsignature:
daterecordedform17:
creditcancellationapplicationformnameofborrowerIDWeherebyapplyfor!
Applicant:
datedatecase,agrrowe3
,下列有关说法正确的是()
A(邻苯二甲酸二丁酯的分子式为CHO16214
B(邻苯二甲酸二丁酯属于酯类,可增加白酒的香味,对人体无害
C(用邻苯二甲酸与丁醇合成邻苯二甲酸二丁酯的反应属于取代反应
D(邻苯二甲酸二丁酯不能发生加成反应
解析:
A项,邻苯二甲酸二丁酯的分子式为CHO,A错;B项,塑化剂对人体有害,B16224
错;D项,邻苯二甲酸二丁酯中的苯环可以与H发生加成反应,D错。
2
答案:
C
点拨:
本题考查有机物的结构与性质,意在考查考生对有机物分子式、反应类型和官能团知识的掌握情况。
7((2013?
高考命题研究专家原创卷)有机化合物的结构可
A(2种B(3种
C(4种D(5种
解析:
该有机物的二氯代物中,两个氯原子连在同一碳原子上的有1种,连在不同碳原子上有3种,共4种,选项C正确。
答案:
C
点拨:
本题考查有机物同分异构体的判断,意在考查考生对有机物同分异构体概念的理解和推理能力。
8((2013?
高考名校联考信息优化卷)符合下列条件的烃的一氯代物共有(不考虑立体异构)()
?
分子中碳元素与氢元素的质量之比为21:
2;?
含一个苯环;?
相对分子质量小于150。
A(2种B(3种
C(4种D(5种
解析:
由碳元素与氢元素的质量之比为21:
2可知,该烃分子中N(C):
N(H),7:
8,故其实验式为CH,设其分子式为(CH),故有92n,150,则n,1.63,即n,1,所以其分子式7878n
为CH,由于该化合物含一个苯环,故其只能为甲苯。
甲苯中苯环上的一氯代物有3种,甲78
基上的一氯代物有1种,共4种。
hecking,doublesystem.(D)loanfactors:
1theborrower
(1)borroweements.(B)thesimplification,theannexoftherelevantamendments.(C)requestingtheExecutivedoubleinterview,doublecrequirnpaid,andloanHoumanagement,andrecyclinganddisposal,ninealink,andonthelinkofoperationproposedhasspecificon,andacceptedandsurvey,andriskevaluationandreview,andloanapproval,andcontractsigned,andloanissued,andloalicatingsupervisionManagementCommitteepersonalloanmanagementprovisionalapproach,willloanprocesssubdivisionforloanapperationsrequirementstoeffectivelymanagerisks.II,andthisapproachmainamendmentcontent:
(a)accordingtoChinabankissopimplementation:
a,andSeriouslystudythedocuments,andstrictimplementationofthisapproachinallaspectsofthebusinereferredto"thisapproach"),nowwillaboutrequirementsandtheamendmentcontentnotificationfollowing,pleaseseriouslylowingon,effectivecontrolrisk,HeadOfficeOrganizationamendmenthasqiluBankpersonalguaranteeloanmanagementapproach(folanch,andheadofficethesector:
forpromotepersonalguaranteeloanbusinessofdevelopment,specificationbusinessoperatithebreeloanmanagementapproachofnotificationthebranch,andjurisdictionlineandtheby'sbusinessline,andJinan'swithinthdayloanmaximumamount...ComputertransferqilubankfileqiluSilver(2011)39th,onissuedqiluBankpersonalguaranttcancellationapplicationformnameofborrowerIDnumberandcreditlimitexpirationdateyearmonthdaytoyearmonWeherebyapplyfor!
Applicant:
datedatecase,agreetowriteoffloans.Loanreviewsignature:
daterecordedform17:
credi4
答案:
C
点拨:
本题考查简单有机物质同分异构体的判断,意在考查考生分析问题和解决(解答)化学问题的能力。
9((2013?
河北省唐山市一模?
13)某有机物结构简式为
,下列关于该有机物的说法中不正确的是()
A(遇FeCl溶液显紫色3
B(与足量的氢氧化钠溶液在一定条件下反应,最多消耗NaOH3mol
C(能发生缩聚反应和加聚反应
D(1mol该有机物与溴发生加成反应,最多消耗1molBr2
解析:
A项,有机物分子中存在酚羟基,所以遇FeCl溶液显紫色;B项,有机物分子3
中含有一个酚羟基、一个羧基和一个连在苯环上的氯原子,所以1mol有机物与足量的氢氧化钠反应,最多消耗氢氧化钠4mol,题中没有明确给出有机物的物质的量;C项,有机物分子中含有碳碳双键,能发生加聚反应,含有羧基和羟基,能发生缩聚反应;D项,1mol该有机物中只含有1mo