综合nmr练习题

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核磁共振习题答案

核磁共振习题答案【篇一：核磁共振氢谱专项练习及答案】1 ．核磁共振波谱法与红外吸收光谱法一样，都是基于吸收电磁辐射的分析法。

( )2 ．质量数为奇数，核电荷数为偶数的原子核，其自旋量子数为零。

( )3 ．自旋量子数i=1 的原子核在静磁场中，相对于外磁场，可能有两种取向。

( )4 ．氢质子在二甲基亚砜中的化学位移比在氯仿中要小。

( )5 ．核磁共振波谱仪的磁场越强，其分辨率越高。

( )6．核磁共振波谱中对于och3 、cch3 和nch3 ，nch3 的质子的化学位移最大。

( )7 ．在核磁共振波谱中，耦合质子的谱线裂分数目取决于邻近氢核的个数。

( )8 ．化合物ch3ch2och(ch3)2 的1h nmr 中，各质子信号的面积比为9:2:1 。

( )9 ．核磁共振波谱中出现的多重峰是由于邻近核的核自旋相互作用。

( )10 ．化合物cl2ch —ch2cl 的核磁共振波谱中，h 的精细结构为三重峰。

( )12 ．氢键对质子的化学位移影响较大，所以活泼氢的化学位移在一定范围内变化。

( )13 ．不同的原子核产生共振条件不同，发生共振所必需的磁场强度(bO)和射频频率(v)不同。

()14 ．(ch3)4si 分子中1h 核共振频率处于高场，比所有有机化合物中的1h 核都高。

( )(一)判断题(二)选择题(单项选择)1 ．氢谱主要通过信号的特征提供分子结构的信息，以下选项中不是信号特征的是( ) 。

a .峰的位置；b .峰的裂分；c .峰高；d .积分线高度。

2．以下关于“核自旋弛豫”的表述中，错误的是( )。

a .没有弛豫，就不会产生核磁共振；b.谱线宽度与弛豫时间成反比；c ．通过弛豫，维持高能态核的微弱多数；d ．弛豫分为纵向弛豫和横向弛豫两种。

3 ．具有以下自旋量子数的原子核中，目前研究最多用途最广的是( ) 。

a. i=1 / 2 ； b . i=0 ; c . i=1 ; d . i > 1。

仪器分析之核磁共振波谱法试题及答案

核磁共振波谱法一、填空题1. NMR法中影响质子化学位移值的因素有：__________，___________，__________、，，。

2. 1H 的核磁矩是2.7927核磁子, 11B的核磁矩是2.6880核磁子, 核自旋量子数为3/2，在1.000T 磁场中, 1H 的NMR吸收频率是________MHz, 11B的自旋能级分裂为_______个, 吸收频率是________MHz(1核磁子=5.051×10-27J/T, h=6.626×10-34J·s)3. 化合物C6H12O,其红外光谱在1720cm-1附近有1个强吸收峰，1HNMR谱图上,有两组单峰d a=0.9, d b=2.1,峰面积之比a:b =3:1, a为_______基团, b为_________基团,其结构式是__________________。

4. 苯、乙烯、乙炔、甲醛,其1H化学位移值d最大的是_______最小的是_________,13C的d值最大的是_________最小的是____________。

二、选择题1. 自旋核7Li、11B、75As, 它们有相同的自旋量子数Ι=3/2, 磁矩μ单位为核磁子，m Li =3.2560, m B=2.6880, m As =1.4349 相同频率射频照射, 所需的磁场强度H大小顺序为 ( )(1) B Li>B B>B As (2) B As>B B>B Li (3) B B>B Li>B As (4) B Li>B As>B Li2.在O - H 体系中，质子受氧核自旋－自旋偶合产生多少个峰? ( )(1) 2 (2) 1 (3) 4 (4) 33. 下列化合物的1HNMR谱, 各组峰全是单峰的是 ( )(1) CH3-OOC-CH2CH3(2) (CH3)2CH-O-CH(CH3)2(3) CH3-OOC-CH2-COO-CH3(4) CH3CH2-OOC-CH2CH2-COO-CH2CH34.一种纯净的硝基甲苯的NMR图谱中出现了3组峰, 其中一个是单峰, 一组是二重峰，一组是三重峰. 该化合物是下列结构中的 ( )5. 自旋核7Li、11B、75As, 它们有相同的自旋量子数Ι=3/2, 磁矩μ单位为核磁子，m Li =3.2560, m B=2.6880, m As =1.4349 相同频率射频照射, 所需的磁场强度H大小顺序为( )(1) B Li>B B>B As (2) B As>B B>B Li (3) B B>B Li>B As (4) B Li>B As>B Li6. 化合物CH3COCH2COOCH2CH3的1HNMR谱的特点是( )(1) 4个单峰 (2) 3个单峰, 1个三重峰(3) 2个单峰 (4) 2个单峰, 1个三重峰和1 个四重峰7.核磁共振波谱法中乙烯, 乙炔, 苯分子中质子化学位移值序是( )(1) 苯 > 乙烯 > 乙炔 (2) 乙炔 > 乙烯 > 苯(3) 乙烯 > 苯 > 乙炔 (4) 三者相等8. 在下列因素中, 不会使NMR谱线变宽的因素是 ( ) (1) 磁场不均匀 (2) 增大射频辐射的功率(3) 试样的粘度增大 (4) 种种原因使自旋-自旋弛豫(横向弛豫)的速率显著增大9. 将(其自旋量子数I=3/2) 放在外磁场中，它有几个能态( 4 )10. 在下面四个结构式中哪个画有圈的质子有最大的屏蔽常数？11. 下图四种分子中，带圈质子受的屏蔽作用最大的是 ( )12.核磁共振的弛豫过程是 ( )(1) 自旋核加热过程 (2) 自旋核由低能态向高能态的跃迁过程(3) 自旋核由高能态返回低能态, 多余能量以电磁辐射形式发射出去(4) 高能态自旋核将多余能量以无辐射途径释放而返回低能态三、问答题1. 只有一组1H-NMR谱图信号和特定分子式的下列化合物, 可能的结构是什么?(1) C2H6O (2) C3H6Cl2(3) C3H6O2. 试推测分子式为C3H6Cl2, 且具有下列NMR谱数据的化合物结构.d 质子数信号类型2.2 2 五重峰3.8 4 三重峰3. 一个沸点为72O C的碘代烷, 它的1H-NMR谱: d=1.8(三重峰, 3H)和3.2(四重峰, 2H)试推测其结构并加以说明。

仪器分析之核磁共振波谱法试题及答案

核磁共振波谱法一、填空题1. NMR法中影响质子化学位移值的因素有：__________，___________，__________、，，。

4. 苯、乙烯、乙炔、甲醛,其1H化学位移值d最大的是_______最小的是_________,13C的d值最大的是_________最小的是____________。

HNMR习题及答案

1H-NMR作业题及答案1.图1-a、1-b、1-c是分子式C4H8O的三个化合物的1H-NMR谱，在它们的IR谱图3000cm-1附近都给出一个红外吸收峰，试推测这三个化合物的结构。

解：分子式C4H8O，峰面积比为1:2:2:3 即为1H、2H、2H、3H U=1δ9.74 1H -CHOδ2.45 2H -CH2-δ1.70 2H -CH2-δ1.0 3H -CH3至此，分子式中的原子及不饱和度全部消耗故图1-a的结构为：CH3CH2CH2CHO解：分子式C4H8O，峰面积比为2:3:3 即为2H、3H、3HU=1δ2.45 2H -CH2-δ2.05 3H -CH3δ1.0 3H -CH3至此，扣掉以上结构片段，分子式中还剩1个C、1个O及1个不饱和度则应有1个C=O，故图1-b的结构为：CH3CCH2CH3O解：分子式C4H8O，峰面积比为1:1 即为4H、4HU=1δ1.85 4H -CH2-（2个）δ3.80 4H -CH2-（2个）至此，扣掉以上结构片段，分子式中还剩1个O及1个不饱和度则这1个不饱和度应该只能让“环”来消耗故图1-c的结构为：O2. 某未知物经元素分析证明含有C 、H 、O ，其中C 62.2%，H 10.3%。

1H-NMR谱（CDCl 3作溶剂）：δ1.2（单峰，6H ），δ2.2（单峰，3H ），δ2.6（单峰，2H ），δ4.0（单峰，1H ）；IR 谱在1700cm -1及3400cm -1处有强吸收，试推测该化合物的结构。

解：根据分子中C%、H%及O%，可计算得该分子的分子式为C 6H 12O 2则U=1根据IR 吸收：1700cm -1 表示有C=O 、3400cm -1表示有-OH 根据1H-NMR ：δ1.2 6H 单峰 → -CH 3（2个） δ2.2 3H 单峰 → -CH 3 δ2.6 2H 单峰 → -CH 2- δ4.0 1H 单峰 → -OH至此，扣掉以上结构片段，分子式中还剩2个C 、1个O 及1个不饱和度则应有1个C=O 和 1个季 C 故该化合物的结构为：CH 3CCH 2C(CH 3)2OOH3.普鲁卡因（Procaine）是一种局部麻醉药，其分子式为C13H20N2O2，而医药上用的是其盐酸盐C13H21N2O2Cl[商业名称奴弗卡因（Novocain）]。

核磁共振氢谱专项练习及答案

核磁共振氢谱专项练习及答案(一)判断题(正确的在括号内填“√”号；错误的在括号内填“×”号。

)1．核磁共振波谱法与红外吸收光谱法一样，都是基于吸收电磁辐射的分析法。

( )2．质量数为奇数，核电荷数为偶数的原子核，其自旋量子数为零。

( )3．自旋量子数I=1的原子核在静磁场中，相对于外磁场，可能有两种取向。

( )4．氢质子在二甲基亚砜中的化学位移比在氯仿中要小。

( )5．核磁共振波谱仪的磁场越强，其分辨率越高。

( )6．核磁共振波谱中对于OCH3、CCH3和NCH3，NCH3的质子的化学位移最大。

( )7．在核磁共振波谱中，耦合质子的谱线裂分数目取决于邻近氢核的个数。

( )8．化合物CH3CH2OCH(CH3)2的1H NMR中，各质子信号的面积比为9:2:1。

( )9．核磁共振波谱中出现的多重峰是由于邻近核的核自旋相互作用。

( )10．化合物Cl2CH—CH2Cl的核磁共振波谱中，H的精细结构为三重峰。

( )11．苯环和双键氢质子的共振频率出现在低场是由于π电子的磁各向异性效应。

( )12．氢键对质子的化学位移影响较大，所以活泼氢的化学位移在一定范围内变化。

( )13．不同的原子核产生共振条件不同，发生共振所必需的磁场强度(B0)和射频频率(v)不同。

( ) 14．(CH3)4Si分子中1H核共振频率处于高场，比所有有机化合物中的1H核都高。

( )15．羟基的化学位移随氢键的强度变化而移动，氢键越强，δ值就越小。

( )答案(一)判断题1．√ 2．×3．×4．×5．√ 6．×7．√ 8．×9．√l0．√11．√ l2．√ l3．√ l4．×l5．×(二)选择题(单项选择)1．氢谱主要通过信号的特征提供分子结构的信息，以下选项中不是信号特征的是( )。

A．峰的位置；B．峰的裂分；C．峰高；D．积分线高度。

2．以下关于“核自旋弛豫”的表述中，错误的是( )。

核磁共振部分习题及答案_2

NMR problems 2nd part1.Below are the 1H and 13C NMR spectra of 2-hexanone (CH3COCH2CH2CH2CH3). Explain carefully how, using homonuclear and heteronuclear decoupling experiments, you could assign the each of the resonances in the 1H and 13C spectra to which nuclei give rise to them.1H NMR spectra of 2-hexanone13C NMR spectra of 2-hexanoneAnswer This is a commonly encountered problem. You obtain a 1H and a 13C NMR spectrum and need to assign each of the resonances in both 1H and 13C spectra.It is generally straight-forward to assign the 1H spectrum using a combination ofdirect inspection (characteristic chemical shifts and multiplicities) in combination with homonuclear decoupling experiments. For 1H spectra, homonuclear decoupling gives you connectivity (which protons are on adjacent carbon atoms) because 1H is 100% abundant. Note that homonuclear decoupling cannot be used for 13C spectra because 13C is isotopically dilute there essentially no possibility that there will be 2 13C nuclei adjacent to each other in an organic molecule.For 2-hexanone, the 1H spectrum contains 5 resonances. You would expect to see one singlet (integral 3H) for the CH3 group attached to the ketone (ie at C1). The CH2 group adjacent to the ketone (ie at C3) would appear as a triplet (with coupling to the adjacent CH2. The CH2 at C4 would appear as a multiplet (a triplet of triplets with coupling to both the CH2 at C3 and the CH2 at C5). Likewise the CH2 at C5 would appear as a multiplet (a triplet of quartets with coupling to both the CH2 at C4 and the CH3 at C6). The resonance of the CH3 at C6 would be a triplet (integral 3H). Intuitively, you would also expect the CH3 and CH2 groups adjacent to the carbonyl to occur at low field (between 1.5 and 2.5 ppm) and the CH3 at C6 to be at high field (between 0.5 and 1 ppm). By inspection, you can assign the proton signals for C1 (δ 1.7), C3(δ 1.9) and C6(δ 0.8). The multiplet signals atδ1.1 andδ 1.3 must belong protons at C4 and C5 and to distinguish these you would use homonuclear decoupling experiments. If the protons at C3 (δ 1.9) are irradiated, the multiplet due to protons at C4 would collapse from a triplet of triplets to a triplet and hence its shift would be known. Similarly, if the protons at C6 (δ 0.8) were irradiated, the multiplet due to protons at C5 would collapse from a triplet of quartets to a triplet and hence its shift would be known.Having assigned the 1H spectrum, the 13C spectrum can be assigned using selective heteronuclear decoupling experiments.The 13C spectrum contains 6 resonances, the resonance due to the carbonyl carbon is obvious from its shift. For the remaining 5 carbons you would expect to have signals from 2 x CH3 and 3 x CH2 groups and in the absence of any 1H decoupling these would appear as 2 quartets and 3 triplets. In the heteronuclear decoupling experiment you would irradiate each of the resonances in the 1H spectrum and observe the 13C spectrum. As each of the 1H signals is irradiated, the resonance of the 13C coupled to it would collapse to a singlet - the multiplicity of the other signals would remain essentially unchanged.The correlation of 1H and 13C NMR spectra can also be achieved using two-dimensional NMR using a heteronuclear shift correlation (HSC) experiment.2.The simple molecules below contain the NMR-active nuclei 14N (I=1), 31P (I=1/2), 1H (I=1/2) and 13C (I=1/2). Considering only the coupling constants through one bond (1J ax) as being significant, construct diagrams which schematically represent the splitting pattern you would see in :1The 1H NMR spectrum of the ammonium ion [N H4+].2The 14N NMR spectrum of the [N H4+].3The 13C NMR spectrum of trimethylamine [(C H3)3N]4The 31P NMR spectrum of phosphine [P H3].5The 1H NMR spectrum of phosphine [P H3].6The 13C NMR spectrum of H2P-C H2-PH2.7The 13C NMR spectrum of [(PH2)3C H].Answer This is simply an exercise in predicting the multiplicity observed using the formula: multiplicity = 2nI + 1.1The 1H NMR spectrum of CH4 will be a singlet (no multiplicity) since the molecule is tetrahedral in shape and all of the protons are equivalent. There is no coupling from carbon since almost all C is 12C and this is NMR silent.2The 13C NMR spectrum of CH4 is a quintet. The C is coupled to 4 equivalent protons and the spin of 1H is ½.(2nI + 1) = (2 x 4 x ½) + 1 = 53The 1H NMR spectrum of NH4+ will have 3 lines. The molecule is tetrahedral in shape and all of the protons are equivalent and these will be coupled to 14N (which has a spin I = 1).(2nI + 1) = (2 x 1 x 1) + 1 = 34The 14N NMR spectrum of NH4+ is a quintet. The 14N is coupled to 4 equivalent protons and the spin of 1H is ½.(2nI + 1) = (2 x 4 x ½) + 1 = 55The 1H NMR spectrum of PH3 is a doublet. The molecule (like ammonia) is pyramidal and all 3 protons are equivalent. The protons (with spin I = ½) are coupled to one 31P nucleus.(2nI + 1) = (2 x 1 x ½) + 1 = 26The 31P NMR spectrum of PH3 will have 4 lines (quartet). The molecule is pyramidal in shape and all of the protons are equivalent (with a spin I = ½). and these will be coupled to 31P.(2nI + 1) = (2 x 3 x ½) + 1 = 47The 13C NMR spectrum of H2PCH2PH2 will be a triplet of triplets. The 13C will be coupled to 2 x 31P (giving a triplet splitting) and the 13C will be coupled to 2 x 1H (giving a triplet splitting).8The 13C NMR spectrum of (H2P)3CH will be a doublet of quartets. The 13C will be coupled to 3 x 31P (giving a quartet splitting) and the 13C will be coupled to 1 x 1H (giving a doublet splitting).3 The 1H NMR spectrum of (E)-2-pentenal (given below) has 5 distinct resonances: δ9.5 (1H, doublet), δ 6.9 (1h, doublet of triplets), δ 6.0 (1H, doublet of doublets), δ 2.2 (2H, multiplet) and δ 1.0 (3H, triplet). Sketch and clearly describe the 1H spectra you would obtain while applying strong Rf. irradiation at :1δ9.52δ 6.93δ 6.04δ 2.25δ 2.26δ 1.0Answer The signals in the spectrum can be readily assigned by inspection: •δ 9.5 is due to the aldehydic proton - it appears as a doublet due to coupling with the vinylic proton on the adjacent carbon.•δ 6.0 is due to the vinyl proton at C2. It appears as a doublet of doublets with one doublet splitting due to the aldehyde proton and the other doublet splitting due to the vinylic proton at C3.•δ 6.9 is due to the vinyl proton at C3. It appears as a doublet of triplets with the doublet splitting due to the vinylic proton at C2 and the triplet splitting due to coupling to the CH2 group at C4.•δ 2.2 is due to the CH2 group at C4. It appears as a multiplet but must be adoublet of quartets with the doublet splitting due to coupling to the vinylic proton at C3 and the quartet splitting due to the CH3 group at C5.•δ 1.0 is due to the CH3 group at C5. It appears as a triplet due to coupling to the CH2 group at C4 (δ 2.2).The effect of irradiation at each of these frequencies is to decouple the nuclei which have signals at these frequencies. Decoupling effectively removes the nucleus from the spin system.1 With irradiation at δ 9.5, the multiplicity of the vinylic CH resonance at δ 6.0is simplified from a doublet of doublets to a doublet. The remaining resonances in the spectrum are unchanged.2With irradiation at δ 6.9, the multiplicity of the vinylic CH resonance at δ 6.0 is simplified from a doublet of doublets to a doublet and the CH2 resonance at C4 (δ 2.2) is simplified from a doublet of quartet to a simple quartet. The remaining resonances are unchanged.3With irradiation at δ 6.0, the multiplicity of the vinylic CH resonance at δ 6.9 is simplified from a doublet of triplets to a simple triplet and the resonance of the aldehyde proton is simplified from a doublet to a singlet.4With irradiation at δ 2.2, the multiplicity of the vinylic CH resonance δ 6.9 is simplified from a doublet of triplets to a simple doublet and the multiplicity of the CH3 resonance (δ 1.0) is simplified from a triplet to a singlet.5With irradiation at δ 1.0, the multiplicity of the CH2 group at C4 (δ 2.2) is simplified from a doublet of quartets to a simple doublet. The remaining resonances in the spectrum are unchanged.4.The 1H NMR spectrum of 1,4-dioxan-d7 at room temperature consists of a singlet (with broadband decoupling of 2H). At -150o C the spectrum appears as two singlets of equal intensity. Using your knowledge about the conformation (and conformational mobility) of 6-membered rings, give a rationalisation as to why the room temperature and low temperature spectra appear as they do.Answer 1,4-dioxan is a 6-membered ring and adopts a chair conformation (like other 6-membered rings such as cyclohexane). In the chair conformation, there are two proton environments and protons can be either axial or equatorial. At room temperature, rapid ring flipping causes protons to exchange between axial and equatorial sites so the proton spectrum is a singlet.In 1,4-dioxan-d7, seven out of the 8 protons have been substituted deuterium, leaving only 1 proton. Again this proton would be expected to be in either the axial or equatorial position and at low temperature (slow exchange) the protons in axial positions will have a different chemical shift to protons in the equatorial sites.At low temperature where the ring flip is slow, there will be molecules with the single proton axial some with the proton equatorial. There would be two signals one corresponding to axial protons and one corresponding to equatorial protons. The fact that these signals are of approximately equal in intensity indicates that there is no significant thermodynamic preference to have the proton to be equatorial or axial.At high temperature, the ring flip is rapid and the proton is rapidly exchanged between axial and equatorial sites and the shift is averaged to a single resonance at a shift half way between the signals between the signals in the slow exchange spectrum.5.The 1H free induction decay (FID) was acquired following a single 90o pulse to a sample of chloroform (CHCl3). The spectrum following Fourier transformation (FT) is given on the right.Describe what would happen to the FID and to the spectrum obtained following FT when :1The pulse angle was increased to 150o.2 A small amount of an Fe(III) salt was dissolved in the solution before acquisitionof the FID.316 FIDs were accumulated and added together before FT.4 A second FID was acquired using a 270o pulse instead of a 90o pulse and added tothe first before FT.5The sample was dissolved in a viscous solvent before acquisition of the FID.6The FID was accumulated for twice the acquisition time before FT.7The proton spectrum was saturated for several seconds immediately before (but not during) the acquisition of the FID.What would happen to the spectrum if 16 FIDs were accumulated, each wastransformed and the resulting spectra were added together to give the final spectrum ? Answer The 1H FID and spectrum are derived from a sample of CHCl3.1The response of the sample is a sinusoidal function of the pulse angle. Sin(90) = 1 and Sin(150) = Sin(30) = ½ so the intensity of the signal in the FID and in the spectrum will be decreased by a factor of 2.2Addition of a small amount of a paramagnetic salt will increase the efficiency of relaxation. Nuclei would be expected to relax more quickly and this would be reflected in a change in the appearance of the FID and the spectrum. The signal in the FID would decay to zero more rapidly (i.e. in less time); signals in the spectrum would be broader.3The signal-to-noise ratio (S/N) of the spectrum increases as the square root of the number of acquisitions that are added together. If 16 acquisitions are added the S/N increases by a factor of √ 16 = 4.4If an FID was accumulated with a 270o pulse, it would be a mirror image (about the zero level in the FID) of the FID accumulated with a 90o pulse. This means that at every point where the 90o FID was above zero, the 270o FID would be below zero and addition of the 90o and 270o FID's would mean that the signals would exactly cancel - only noise would remain. The spectrum would contain only noise.5 A viscous solvent will slow the rate of molecular tumbling and this in turnincreases the efficiency of relaxation. Nuclei would be expected to relax more quickly and this would be reflected in a change in the appearance of the FID and the spectrum. The signal in the FID would decay to zero more rapidly (ie in less time); signals in the spectrum would be broader.6The FID as presented has clearly decayed to zero well before the accumulation of the FID has stopped. By the end of the acquisition time, there is no signal only noise being accumulated. Increasing the acquisition time by a factor of two will double the time for which the FID signal is accumulated however the real signal in the FID will still decay at the same rate so the real signal will be proportionally less of the FID and there would be proportionally more noise. In the spectrum, the S/N would be decreased.7Saturation of the 1H spectrum would mean that the 1H spectrum would not be observable, providing there was no significant delay between the saturation and recording the FID. Nuclei take several T1's to recover following saturation.8If 16 FIDs were accumulated then each transformed and the spectra added, there would be no difference to the situation where 16 FID's were added and the summed FID was transformed. In practice since the FT actually takes some time it is more time-economical to minimise the number of tiems that the FT has to be performed.6.The 13C NMR spectrum of 2-methyl-4,5-dihydrofuran (A) is given below. The spectrum was the result of 32 scans with 1 sec. delay between acquisitions and no proton decoupling.1What would happen to the spectrum if 128 scans were accumulated and added together ?2What would happen to the spectrum if the spectrum was acquired with a long delay (say 300 seconds) between acquisitions?3What would happen to the spectrum if a small amount of soluble paramagnetic salt was added to the sample prior to acquisition ?4What would happen to the spectrum if the 1H spectrum was irradiated with broad-band Rf radiation before (but not during) each acquisition ?5Could the 1H NMR spectrum of this molecule be analysed by first order splitting rules? Why?Answer The 13C spectrum of the compound (A) contains 5 signals and from the multiplicity these can be assigned to the two CH2 carbons, one CH carbon, one CH3 signal and one quaternary carbon.1The signal-to-noise ratio (S/N) of the spectrum increases as the square root of the number of acquisitions that are added together. If 128 acquisitions are added, this4 times the number of scans as the 32 used to accumulate the basic spectrum. TheS/N increases by a factor of √ 4 = 2.2With only a 1 sec delay between acquisitions, it very likely that all of the carbons in the sample are not fully relaxed between scans and hence the relative intensities of the signals are not correct. A long delay between acquisitions would ensure that all the nuclei are fully relaxed and therefore all 5 signals should all be of the same intensity.3 A paramagnetic species added to the sample will cause more efficient relaxationof all nuclei. Any differences in intensity resulting from differences in relaxation times would be reduced. All of the lines in the spectrum would be broadened.4Irradiation of the proton spectrum before (but not during) the acquisition would provide an NOE enhancement to the carbon spectrum (without decoupling carbon from protons). So the intensity of the signals from protonated carbons in the sample would be increased. The spectrum would not be proton decoupled so the multiplicities of signals would remain the same.5There are 8 protons in this molecule and the spin system can be assigned as an AA'GG'RX3 spin system. Because the spin system contains nuclei which arechemically equivalent but magnetically non equivalent it cannot be analysed by first order rules.7.The 1H free induction decay (FID) below was acquired following a single 45o pulse to a sample of containing two species. The spectrum obtained following Fourier transformation (FT) is given on the right.Describe what would happen to the FID and to the spectrum obtained following FT when:1The pulse angle was doubled.2The pulse angle was increased by a factor of 4.3The pulse angle was increased by a factor of 3.464 FIDs were accumulated and added together before FT.Answer The FID and the spectrum are derived from a sample containing 2 species. The spectrum clearly contains one sharp resonance and one broad resonance and you would predict that the broad signal arises from a species with a short relaxation time and the narrow resonance arises from a species with a long relaxation time.1The pulse angle is 45o so increasing this by a factor of 2 will mean that a 90o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(45) = 1/√ 2 and Sin(90) = 1 so the intensity of the signal in the FID and in the spectrum will be increased by a factor of √ 2 (1.414).2The pulse angle is 45o so increasing this by a factor of 4 will mean that a 180o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(180) = 0 so the intensity of the signal in the FID and in the spectrum will be zero. Both the FID and the spectrum would contain only noise.3The pulse angle is 45o so increasing this by a factor of 3 will mean that a 135o pulse is delivered to the sample. The response of the sample is a sinusoidal function of the pulse angle. Sin(135) = Sin(45) = 1/ 2 so the intensity of the signal in the FID and in the spectrum will be unchanged.4The signal-to-noise ratio (S/N) of the spectrum increases as the square root of the number of acquisitions that are added together. If 64 acquisitions are added the S/N increases by a factor of √ 64 = 8.。

核磁共振谱习题问题详解

核磁共振谱习题一．选择题1．以下五种核，能用以做核磁共振实验的有（ACE ）A：19F9B：12C6C：13C6D：16O8E：1H12．在100MHz仪器中，某质子的化学位移δ=1ppm，其共振频率与TMS相差（A ）A ：100Hz B：100MHz C：1HzD:50Hz E：200Hz3．在60MHz仪器中，某质子与TMS的共振频率相差120Hz则质子的化学位移为（E ）A：1.2ppm B：12ppm C：6ppmD：10ppm E：2ppm4．测试NMR时，常用的参数比物质是TMS，它具有哪些特点（ABCDE ）A：结构对称出现单峰B：硅的电负性比碳小C：TMS质子信号比一般有机物质子高场D：沸点低，且容易溶于有机溶剂中E：为惰性物质5．在磁场中质子周围电子云起屏蔽作用，以下说确的是（ACDE ）A：质子周围电子云密度越大，则局部屏蔽作用越强B：质子邻近原子电负性越大，则局部屏蔽作用越强C：屏蔽越大，共振磁场越高D：屏蔽越大，共振频率越高E：屏蔽越大，化学位移δ越小6．对CH3CH2OCH2CH3分子的核磁共振谱，以下几种预测正确的是（ACD ）A：CH2质子周围电子云密度低于CH3质子B：谱线将出现四个信号C：谱上将出现两个信号D：<E：>7．CH3CH2Cl的NMR谱，以下几种预测正确的是（D）A：CH2中质子比CH3中质子共振磁场高B：CH2中质子比CH3中质子共振频率高C：CH2中质子比CH3中质子屏蔽常数大D：CH2中质子比CH3中质子外围电子云密度小E：CH2中质子比CH3中质子化学位移δ值小8．下面五个化合物中，标有横线的质子的δ最小的是（A）A：CH4B：CH3F C：CH3ClD：CH3Br E：CH 3l9．下面五个化合物中，标有横线的质子的共振磁场H0最小者是（A）A：RCH2OH B：RCH2CH2OH C：RCH2ClD：CHBr E：ArCH2CH310．下面五个结构单元中，标有横线质子的δ值最大的是（E）A：CH3-C B：CH3-N C：CH3-OD：CH3F E：CH2F211．预测化合物的质子化学位移，以下说确的是（C）A：苯环上邻近质子离C=O近，共振在高磁场B：苯环上邻近质子离C=O近，屏蔽常数大C：苯环上邻近质子离C=O近，化学位移δ大D：苯环上邻近质子外围电子云密度大12．氢键对化学位移的影响，以下几种说确的是（BCE）A 氢键起屏蔽作用B：氢键起去屏蔽作用C：氢键使外围电子云密度下降D：氢键使质子的屏蔽增加E：氢键使质子的δ值增加13．对于羟基的化学位移，以下几种说确的是（ABE）A：酚羟基的δ随溶液浓度而改变B：浓度越大δ值越大C：浓度越大，δ值越小D：邻羟基苯乙酮的羟基δ值也随溶液的浓度改变而明显改变E：邻羟基苯乙酮的δoH与浓度无明显关系二．填充题1．在磁场H0的作用下，核能级分裂，(u为核磁矩)，已知，在同一频率条件下，使氟，磷，氢发生共振，所需磁场强度最大的是：＿磷＿。

核磁共振试题.

选择题：1．下列哪一组原子核的核磁矩为零；不产生核磁共振信号的是（ D ）A 2H、14N B 19F、12C C 1H、13 C D 16O、12C2．在外磁场中，其核磁矩只有两个取向的核是( D )A 2H 19F 13CB 1H、2H、13C C 13C、19F、31 PD 19F 31P 12C3、下列有机物分子在核磁共振氢谱中只给出一种信号的是（ D ）A HCHOB CH3OH C HCOOH D CH3COOCH34. 不影响化学位移的因素是（ A ）A 核磁共振仪的磁场强度B 核外电子云密度C 磁的各向异性效应Ｄ内标试剂5．自旋量子数I=1/2的原子核在磁场中，相对于外磁场，有多少种不同的能量状态？（ B ）A 1B ２C ４D 06．下面四个化合物中在核磁共振谱中出现单峰的是（ C ）A CH3CH2Cl B CH3CH2OH C CH3CH3D CH3CH(CH3)27．下面四个化合物质子的化学位移最小的者是（ A ）A CH3F B CH4C CH3Cl D CH3Br8. 使用60MHz 核磁共振仪，化合物中某质子和四甲基硅烷之间的频率差为120 Hz，其化学位移值δ为（ D ）A 120B 1.20C 0.20D 2.09. 某化合物中两种相互偶合质子，在100兆周的仪器上测出其化学位移δ差为1.1，偶合常数(J)为5.2Hz，在200兆周仪器测出的结果为( C )。

A δ差为2.2，J为10.4HzB 共振频率差为220Hz，J为5.2HzC δ差为 1.1，J为0.4HzD 共振频率差为110Hz，J为5. 2Hz10. HF的质子共振谱中可看到（ A ）A 质子的双峰B 质子的单峰C 质子的三峰D 质子和19F 的两个双峰11. 一化合物分子式为 C5H8，在它的氢谱中仅有一个单峰，它的结构可能是（ D ）A B CD12. 有一样品，从它的氢谱中得知，有两种不同的羟基氢：分别为δ5.2和10. 5，它的结构为（ C ）A BＣＤ13. 某化合物中三种质子相互耦合成AM2X2系统JAM=10Hz，JXM=4Hz，它们的峰形为（）A A为单质子三重峰，M为双质子4重峰，X为双质子三重峰B A为单质子三重峰，M为双质子6重峰，X为双质子三重峰C A为单质子单峰，M为双质子6重峰，X为双质子三重峰D A为单质子二重峰，M为双质子6重峰，X为双质子三重峰14. 下列化合物结构中标记的质子化学位移归属正确的为（ B ）A δ 1 在1.5 - 2.0，δ 2 和δ 3 在1.8 - 3，δ 4 在9 - 10B δ 1 在1.5 - 2.0，δ 2 和δ 3 在5.5 – 7.5，δ 4 在10 - 15C δ 1 在0 - 1，δ 2 和δ 3 在5.5 – 7.5，δ 4 在9 - 10D δ 1 在1.5 - 2.0，δ 2 和δ 3 在5.5 – 7.5，δ 4 在9 – 1015. 在下列三个结构式中，标记的质子的屏蔽常数大小顺序是（）A1>2>3B2>1>3C3>2> 1 D 3>1>216. 在100MHz仪器中，某质子的化学位移δ＝l，其共振频率与TMS相差（）A100H z B60H zC 1HzD 200H z17. 若外加磁场的磁场强度H0逐渐增大时，则使质子从低能级E1跃迁至高能级E2所需的能量（）A 不发生变化B 逐渐变小C 逐渐变大D 不变或逐渐变小18. 下述哪一种核磁共振技术不能简化图谱（）A 加大磁场强度B 化学位移试剂C 去偶法D 改变内标试剂19.在化合物中偶合常数最小的是（）A Ja b B Jbc C JadD Jbc20. 影响偶合常数的主要因素是( )A 浓度B 键角C 核磁共振仪的磁场强度D 温度21. 下列系统中，哪种质子和其它原子之间能观察到自旋裂分现象( ) abA 19F-HB 35Cl-HC 75Br-H D 127I-H22. 不影响邻位偶合常数的因素是( )A 两面角B 核磁共振仪的磁场强度C 取代基的电负性D 取代基的空间取向23. 化合物Cl—CH2—O—CH3中，Ha和 Hb质子峰的裂分数及强度分别为（）A Ha ：5和l:4:6:4:1， Hb：3和1:2:1 B Ha：2和1:1，Hb：2和1:1C Ha ：4和1:3:3:1，Hb：3和1:2:1 D Ha：1和2, Hb:1和324. 某化合物的分子式为C4H8Br2，核磁共振谱图给出以下信息：δ4.2多重峰，δ3.6三重峰，δ2.3四重峰，δ1.8双峰；丛低场到高场积分线高度比为1：2：2：3。

核磁共振谱习题

1、三个不同的质子Ha 、Hb、Hc，其屏蔽常数的大小次序为σb＞σa＞σc，这三种质子在共振时外加磁场强度的次序如何？这三种质子的化学位移次序如何？σ增大化学位移如何变化？2、下列每个化合物中质子Ha 和Hb哪个化学位移大？为什么？3、指出下列化合物属于何种自旋体系：a. CH2Br-CH2Cl b. CH3CH2Fc. d.4、异香草醛（I）与一分子溴在HOAc中溴化得（Ⅱ），（Ⅱ）的羟基被甲基化，主要产物为（Ⅲ），（Ⅲ）的NMR图谱如下，溴的位置在何处？5、丙酰胺的图谱如下，说明图谱中各组峰对应分子中哪类质子。

6、下列一组NMR图谱内标物皆为TMS，试推测结构。

ab.7.下图给出的是某一化合物的门控去偶（非NOE方式）测定的 13CNMR图谱。

已知分子式为C10H12O，试推测其结构。

8.化合物分子式为C4H7NO，其碳谱和氢谱如下，试推测其结构。

（溶剂为CDCl3）9.苯甲醛中，环的两个质子共振在δ7.72处，而其他三个质子在δ7.40处，说明为什么？10. 释2-碘丙烷中异丙基的分裂型式和强度。

11.某化合物在300MHz谱仪上的1H NMR谱线由下列三条谱线组成，它们的化学位移值分别是0.3，1.5和7.3，在500MHz谱仪上它们的化学位移是多少？用频率（单位用Hz）来表示其值分别是多少？12.判断下列化合物1H化学位移的大小顺序，并说明理由：CH3Cl，CH3I，CH3Br，CH3F。

13.在常规13C谱中，能见到13C-13C的偶合吗？为什么？14.试说出下面化合物的常规13C NMR谱中有几条谱线？并指出它们的大概化学位移值。

15.从DEPT谱如何区分和确定CH3、CH2、CH和季碳？16.下图为L-薄荷醇（L-menthol）的2D-INADEQUATE谱及解析结果，试在L-薄荷醇的结构上标出相应字母。

答案：1、外加磁场强度H b>H a>H c；化学位移δHc>δHa>δHb；σ增大化学位移减小。

光谱部分综合练习题

光谱部分综合练习题一、推测结构1．化合物A，C9H18O2，对碱稳定，经酸性水解得到B，C7H14O2和C，C2H6O，B与Ag(NH3)2+反应后酸化得到D，D经碘仿反应后再酸化得到E，E加热得到F，C6H8O3，F的IR主要特征是：1755cm-1和1820 cm-1；F的1HNMR数据为：δH1.0（3H，d）；2.1（1H，m）；2.8（2H，q）。

推出A到F各化合物的结构。

（中科院2006）2．化合物A，C10H12O2，其IR在1735 cm-1有强吸收，3010 cm-1有中等强度的吸收；其1HNMR数据如下：δH：1.3（3H，t）；2.4（2H，q）；5.1（2H，s）；7.3（5H，m）。

写出A的结构式，并指出IR、HNMR各峰的归属。

（中科院2006）3．某化合物的元素分析表明只含有C、H、O，最简式为C5H10O，IR在1720 cm-1有强吸收，2720 cm-1无吸收，该化合物的质谱图如下：试推测该化合物的结构，并说明m/e=43、58和71等主要碎片的生成途径。

（中科院2006）4．有一个化合物分子式为C5H10O，红外和核磁共振数据如下：1HNMRδ：1.10（6H，d）、2.10（3H，s）、2.50（1H，m）；IR：1720cm-1。

请推导其结构，并表明各峰的归属。

（东华大学2004）第3题图（中科院2006）5．化合物A的分子式为C6H12O3，其IR在1710cm-1有一强吸收峰；当A在NaOH中用碘处理时，则有黄色沉淀生成。

当A用Tollen试剂处理时无反应发生。

假如A先用稀硫酸处理后再与Tollen试剂反应时，则在试管中有银镜形成。

化合物A的1HNMR谱图数据如下：2.1（s）、2.6（d）、3.2（6H，s）、4.7（t）。

请给出化合物A的结构式及各峰的归属（东华大学2005）。

6．有一化合物X（C10H14O）溶于NaOH水溶液中而不溶于碳酸氢钠水溶液中。

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练习题
1．什么是核磁共振?要满足什么条件才能观察到核磁共振现象？
2．影响化学位移的主要因素？
3．比较下面化学基团质子的δ的大小？
（1） CH3CH2F,CH3CH2OH,CH3CH2COOH
（2）
H

F
H

H
H

OH
H

H
H

CO2H
H

H
(3)

H
OH
H

H
H

CO2H
H

H
(4)

CHO

H
4.什么是自旋偶合？
5．影响偶合常数的因素？
6.比较下面化合物的J值的大小
（1）（3JH-F）

F
F
FHH

F
(2) CH3CH2F CH3CH2Cl CH3CH2Br CH3CH2I (3JH-H)
(3) CH3F CH2F2 CHF3 (1JH-C)
(4) (3JH-H)
H
H
H

F
H

H
H

CO2H
H

H
H

Li
（5）（2JH-H）
H

H
H

F
H

H
H

CO2H
H

H
H

Li
（6）（3JH-H）

HHFH
H
OHHHNH
2

（7）（3J2-3）
3

2
S

3
2
O

3
2
N
H

3
2

(8) （3JH-H）
H

H
H

HHHg-Mg-Li

7.核自旋体系与NMR类型之间有何内在的联系？
8．下面化合物属于哪些NMR的类型？
（1）
H
Cl
H
H

H
F

H
H

（2）
H

CH
3

H
H
CH2CH
3

H
H
H

（3）
ClCH2CH2Br FCH2CH2I
(4)
Cl

Cl
Br

8．化合物CH3CH2NHCO-CH=CH-CO2H的NMR如图所示，求出烯质子
的NMR参数，并判断分子构型。（60MHz）

9．一级谱图与高级谱图有何区别？
10．某化合物的分子式为C14H14S,请根据下面的HNMR鉴定化合物的结构。
11．某化合物分子式是C8H9OCl，该有机物的核磁共振氢谱
图如图：请确定化合物的结构。

12．比较标量偶合和偶极偶合。
13．名词简述
双照射
NOE效应
14．简化核磁共振谱图有哪些方法？
15．碳谱去偶有哪些方法？
16。

17
18
19
20.
20.
21.
22
23
24.
25
26
27
27
28
28.
30.
31..根据下面化合物的谱图推导化合物的结构，并归属核磁的信号。（300MHZ）

氘代HNMR
1H-1HCOSY

EI质谱图

13
CNMR
33.根据下面化合物的谱图，确定化合物的结构，并对信号归属。
(500MHZ)

化合物的HNMR
化合物的13CNMR
DEPT

1H-1HCOSY
HMQC

化合物HMBC
化合物的FABMS
0
0.2
0.4
0.6
0.8
1

350300250200
Wavelength(nm)

A
b
s

化合物的紫外光谱

化合物的IR
34化合物YGM-3-9，（40mg），高分辨HRFABMASS确定化
合物的分子式为C11H11N5OBr2. HCl，淡黄色针状固体，mp. 222
0
C（分解）。不溶于氯仿，易溶于甲醇，根据下面的谱图确定化

合物的结构。
0
0.1
0.2
0.3
0.4
0.5
400350300250200
Wavelength (nm)

A
b
s

化合物的HNMR
13CNMR

化合物的H,HCOSY
化合物的HMQC

化合物的HMBC
37.化合物的分子式为C19H30O4,根据下面的谱图推导化合物的结
构。
13
CNMR
H-HCOSY
HMQC
HMBC
NOESY