Analytical techniques (A2): NMR
A-Level Chemistry Topic 37 8:06 English narration · English + 中文 subtitles burned in
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Here are seven different molecules.
这里有七种不同的分子。
Every one has five carbons, ten hydrogens and one oxygen, and every one contains a carbonyl group.
每一种都含有五个碳原子、十个氢原子和一个氧原子， 而且每一种都含有羰基。
Weigh them in a mass spectrometer and you get the same mass for all seven. Their infrared spectra look alike too.
把它们放进质谱仪去称重，七种得到的质量完全一样， 红外光谱看起来也差不多。
So how do you tell them apart?
那怎么把它们区分开呢？
You need something that looks at the atoms one environment at a time.
你需要一种能一个环境一个环境地去看原子的仪器。
That machine is NMR.
这台仪器就是核磁共振。
Welcome to the A2 analytical techniques.
欢迎来到 A2 的分析技术。
First we separate a mixture, with chromatography. Then we identify it, with nuclear magnetic resonance: carbon-13 first, then proton NMR, where most of the marks live.
我们先用色谱把混合物分离开， 再用核磁共振把物质鉴定出来：先讲碳-13 核磁共振，再讲质子核磁共振—— 大部分分数都在质子核磁共振这里。
First you have to get your compound on its own. Thin-layer chromatography does that with two phases.
在鉴定一种化合物之前，你得先把它单独分离出来，薄层色谱靠的是两个相。
The stationary phase stays still: the aluminium oxide layer on the plate.
固定相不动——这里是涂在板上的一薄层氧化铝。
The mobile phase moves: the solvent.
流动相会移动——这里就是溶剂。
Spot the mixture on the baseline, then let the solvent rise up the plate.
把混合物点在基线上，让溶剂沿着板往上爬。
The R f value is the distance moved by the spot, divided by the distance moved by the solvent front.
比移值等于斑点移动的距离，除以溶剂前沿移动的距离。
A substance that sticks more has a smaller R f; a bigger R f means more soluble in the mobile phase.
粘得更牢的物质比移值更小； 比移值越大，说明这种物质在流动相中越易溶。
Gas liquid chromatography does the same for anything you can turn into a gas, and it starts with the phases: a high boiling point liquid on a solid, and an unreactive carrier gas.
气液色谱做的是同一件事，只是用在能变成气体的物质上，而且同样从两个相说起。 固定相是附着在固体上的高沸点液体，流动相是一种不活泼的载气。
The retention time is the time between injection and detection, and a component held more strongly moves slower, so it takes longer.
保留时间是从进样到检测所经过的时间；被固定相吸引得越强的组分走得越慢， 保留时间就越长。
The area under each peak gives the amount: for peak A, one hundred times twenty-eight over a total of one hundred and forty-one, about twenty per cent by mass.
而每个峰下面的面积就代表含量。 对于峰 A：一百乘以二十八，除以总面积一百四十一，约等于百分之二十。
Now the main event.
现在进入正题。
Certain nuclei behave like tiny magnets, so in a strong magnetic field they line up, and radio waves of the right energy make them resonate.
某些原子核就像一个个小磁铁，放进很强的磁场里它们就会排列起来。 再射入能量合适的无线电波，它们就会吸收能量，发生共振。
Here is the part the exam cares about. A nucleus in a different chemical environment feels different electrons around it, so it absorbs at a slightly different point.
下面是考试真正关心的部分：原子核还会感受到周围的电子， 所以处在不同化学环境中的原子核，吸收的位置会略有不同。
One environment gives one peak.
一个环境给出一个峰。 在碳-13 核磁共振里，这一句话就能回答大部分题目。
So what is a chemical environment?
那么什么叫化学环境？
Two atoms share an environment when the rest of the molecule looks the same from each of them.
如果从两个原子各自看出去，分子其余部分完全一样， 它们就处在同一个环境。
Take ethanol: its two carbons are clearly different, so carbon-13 gives two peaks. Now propanone.
看乙醇：两个碳明显不同，一个连着氧，另一个没有， 所以碳-13 给出两个峰。
It has three carbons, but the two methyls sit one on each side of the carbonyl, and swapping them changes nothing.
再看丙酮：它也有三个碳，但两个甲基碳分别位于羰基两侧， 把它们互换毫无变化。
They are equivalent, so propanone gives two peaks, not three.
它们是等价的，共用一个峰， 所以丙酮给出的是两个峰，不是三个。
Always hunt for symmetry before you count.
数之前，一定先找对称性。
Let's use that.
我们来用一用。
These two are isomers: five carbons, ten hydrogens, one oxygen.
这两种是同分异构体，都是五个碳、十个氢、一个氧。
Pentan-2-one first.
先看戊-2-酮。
Walk along the chain and label each carbon: a methyl, a carbon with two hydrogens, another one, the carbonyl carbon, then the far methyl.
沿着碳链走，把每个碳都标出来：一端是一个甲基，然后是一个带两个氢的碳， 再一个，接着是羰基碳，最后是另一端的甲基。
No two are alike, so five environments, and five peaks.
没有任何两个是相同的， 所以是五个环境、五个峰。
Now pentan-3-one. Draw a mirror line down through the carbonyl carbon.
再看戊-3-酮：在羰基碳处画一条对称线。
The two end methyls match, and the two middle carbons match, so we are left with three environments, and three peaks.
两端的甲基彼此相同，中间的两个碳也彼此相同，于是只剩下三个环境、三个峰。
Same formula, different spectra.
分子式相同，谱图不同。
Now proton NMR, which looks at the hydrogen atoms.
接下来是质子核磁共振，它看的是氢原子。
One spectrum, three clues.
一张谱图，三条独立的线索。
The chemical shift tells you what each proton is joined to. The peak area — integration — gives the relative numbers of each type of proton.
化学位移告诉你每个质子连在什么上面；峰面积——积分——给出各类质子的相对数目； 裂分则告诉你它们旁边有几个氢。
And the splitting pattern tells you how many neighbours they have.
三条要一起读，因为大多数题目至少要用到两条。
Start with the chemical shift.
先讲化学位移。
Good news: you never memorise these numbers, because the paper prints the table for you. Match a peak to a row.
先说个好消息：这些数字你从来不用背，因为试卷会把表印给你， 你要做的只是把一个峰对应到表里的一行。
A proton on a plain alkane chain sits low, zero point nine to one point seven.
普通烷基链上的质子位置很低， 在零点九到一点七之间。
Next to a carbon oxygen double bond it moves up, two point two to three.
接在碳氧双键旁边就往上移，到二点二至三之间。
Joined to an oxygen or a nitrogen, further still: three point two to four.
如果连在氧或氮上，还要更高：三点二到四。
And watch the axis: like infrared, it runs backwards, with zero on the right.
还要注意横轴—— 和红外一样，它是倒着排的，零在右边。
The second clue is the area under each group of peaks: it is proportional to the number of protons there.
第二条线索是每一组峰下面的面积：它正比于该环境中质子的数目。
Take ethanol again.
再看乙醇。
Three proton environments: the methyl, the middle carbon with two hydrogens, and the oxygen hydrogen.
它有三个质子环境：甲基、带两个氢的中间碳，还有羟基上的氢。
The areas come out three to two to one, exactly how many hydrogens sit in each.
三者的面积是三比二比一，正好等于每个环境中氢原子的个数。
But be careful: that is a ratio, not a count.
但要小心： 面积给出的是比例，不是个数。
It could equally mean six, four and two, so check the molecular formula.
三比二比一同样可能是六、四、二， 所以一定要对照分子式检查。
The third clue carries the most marks: splitting.
第三条线索分值最高：裂分。
A peak is split by the hydrogen atoms on the neighbouring carbon, following the n plus one rule — n neighbours give n plus one lines.
一个峰会被相邻碳上的氢原子裂开，遵循 n 加一规则—— n 个相邻氢给出 n 加一条谱线。
No neighbours at all gives a single line, a singlet. One neighbour gives a doublet. Two neighbours give a triplet.
一个相邻氢都没有，就是一条线，叫单峰； 有一个相邻氢，得到双峰；有两个，得到三峰；有三个，得到四峰；超过三个，叫多重峰。
Three neighbours give a quartet. More is simply a multiplet. And notice what you count: the hydrogens on the carbon next door, never the same carbon.
还要看清楚数的是什么：数的是隔壁那个碳上的氢，绝不是同一个碳上的氢。
Now go the other way and predict a spectrum.
现在反过来，去预测一张谱图。
Pentan-3-one again.
还是戊-3-酮。
First find the proton environments: the two end methyls are equivalent, and the two middle groups are equivalent, so there are only two peaks.
先找出质子环境： 两端的甲基彼此等价，中间的两个基团也彼此等价，所以只有两个环境，只有两个峰。
The six methyl hydrogens sit next to a carbon carrying two hydrogens, so they give a triplet, near one point one.
六个甲基上的氢，旁边那个碳带着两个氢，所以给出三峰，位置很低，在一点一附近。
The four middle hydrogens sit next to a methyl, so they give a quartet, near two point four.
中间的四个氢旁边是一个甲基，也就是三个相邻氢，所以给出四峰，在二点四附近。
And the areas are six to four, which is three to two.
面积则是六比四，也就是三比二。
Now the full deduction.
现在做一次完整的推断。
A compound with four carbons, eight hydrogens and two oxygens gives three peaks: a triplet of area three, a singlet of area three, and a quartet of area two.
某化合物含四个碳、八个氢、两个氧，给出三个峰： 面积为三的三峰、面积为三的单峰，以及面积为二的四峰。
Start with the areas: three plus three plus two makes eight hydrogens, matching the formula.
先从面积入手： 三加三加二等于八个氢，与分子式相符。
Now apply the n plus one rule in reverse. The triplet has two neighbours and the quartet has three, so each sits next to the other, and that pair can only be an ethyl group. Now the shift.
再反向使用 n 加一规则：三峰有两个相邻氢， 四峰有三个相邻氢，说明它们彼此相邻，这一对只能是乙基。
The quartet is at four point one, well to the left, so that carbon is joined to an oxygen. The singlet has no neighbours and sits at two point zero, beside the carbonyl.
再看位移：四峰在四点一，靠左很多，说明那个碳连着氧； 单峰没有相邻氢，位于二点零，就挨着羰基。
Put it together and you get ethyl ethanoate.
把这些拼起来，答案就是乙酸乙酯。
Two practical points the exam asks about directly.
有两个实际操作要点，考试会直接考。
First, every sample has a little TMS added, tetramethylsilane.
第一，每个样品里都加了少量 TMS， 也就是四甲基硅烷。
All twelve of its hydrogens are equivalent, so it gives one sharp peak, and we define that peak as zero.
它的十二个氢完全等价，所以只给出一个尖锐的峰， 我们把这个峰定为零。
A small peak at zero is always TMS, never your compound.
其他所有位移都以它为基准来测量， 所以零处的小峰永远是 TMS，绝不是你的化合物。
Second, the solvent must not add proton peaks of its own, so we use a deuterated solvent.
第二，溶剂本身不能再添出质子峰，所以要用氘代溶剂——里面的氢已经换成了氘。
One more trick, and it is a favourite question.
还有一个小技巧，也是常考题。
Shake the sample with heavy water.
把样品和重水一起振荡。
Any hydrogen attached to an oxygen or a nitrogen will swap places with a deuterium atom, and deuterium gives no signal in a proton spectrum.
任何连在氧上或氮上的氢，都会和一个氘原子交换位置， 而氘在质子谱里不给出信号。
So the oxygen hydrogen peak simply vanishes.
所以把振荡前后的谱图各测一次： 羟基上那个氢的峰就直接消失了。
Whatever disappeared was the oxygen hydrogen or the nitrogen hydrogen, and that is how you prove a molecule contains an alcohol or an amine group.
消失的那个峰就是羟基氢或者氨基氢， 这正是证明分子中含有醇羟基或者胺基的方法。
Four marks students throw away.
四个学生常丢的分。
First, in carbon-13, count environments, not carbon atoms, and look for symmetry first.
第一，碳-13 数的是环境，不是碳原子的个数， 写答案之前先找对称性。
Second, when you name a splitting pattern, always give the reason with it: a triplet, because there are two hydrogen atoms on the neighbouring carbon atom.
第二，写裂分峰型时，一定要连原因一起写： 三峰，因为相邻碳上有两个氢原子。 只写一半是不得分的。
Third, peak areas are a ratio, not a count.
第三，峰面积是比例，不是个数，要拿分子式去核对。
Fourth, the small peak at zero is TMS.
第四，零处那个小峰是 TMS。
Get those four right, and this topic is yours.
这四点做对，这个专题就是你的了。