Analytical techniques (AS)
A-Level Chemistry Topic 22 8:45 English narration · English + 中文 subtitles burned in
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Transcript
You have just made a new compound.
你刚刚做出了一种新化合物。
It is a clear liquid in a flask, with no label, and you cannot see a single molecule.
它是烧瓶里的一种澄清液体，没有标签，而你又看不见任何一个分子。
How do you prove what it is?
怎样证明它到底是什么？
You ask the sample two questions.
我们向样品提出两个问题。
Infrared spectroscopy asks which bonds are inside.
红外光谱问的是：里面有哪些化学键。
Mass spectrometry asks how heavy the molecule is, and what pieces it breaks into.
质谱问的是：这个分子有多重，它会碎成哪些碎片。
Together, they leave it nowhere to hide.
两个答案放在一起，分子就无处可藏了。
Welcome to analytical techniques.
欢迎来到分析技术。
We will read an infrared spectrum and name the bond behind every dip.
我们会先读懂红外光谱，说出每一个吸收凹陷背后的化学键。
Then a mass spectrum: the molecular ion, the fragments, and the isotope peaks that count carbons and give away a halogen.
然后再读质谱：分子离子、碎片，以及那些不起眼的同位素小峰——它们能数出碳原子的个数， 还能暴露分子里的卤素。
Let's begin.
我们开始吧。
Start with infrared.
先看红外。
Infrared spectroscopy helps you find the functional group in a molecule.
红外光谱帮助你找出分子里的官能团。
Every bond is vibrating, stretching and bending like a tiny spring, and each kind absorbs infrared over its own range — that absorption is the dip you see on the spectrum.
每一根化学键都在振动，像一根小弹簧一样伸缩和弯曲， 而每一种键都在自己特定的范围内吸收红外辐射——那个吸收就是谱图上的凹陷。
Shine infrared through the sample and measure how much gets out: where a bond absorbs, the line drops.
把红外光照射到样品上，再测量透过来多少：凡是有键吸收的地方，曲线就往下掉。
So a spectrum is a line near the top with dips hanging down.
所以一张红外光谱就是一条靠近顶部的线，下面挂着一个个凹陷。
And note the bottom axis — the wavenumber runs backwards, from 4000 on the left to 500 on the right.
注意横轴—— 波数是倒着排的，左边是 4000，右边是 500。
Good news: you never memorise these numbers.
好消息是：这些数字你从来不用背。
Every paper prints the table; your job is to match a dip to a row.
每一份试卷都会把这张表印给你， 你要做的只是把一个凹陷对应到表里的一行。
Four rows do most of the work.
四行就能解决大部分题目。
An oxygen to hydrogen bond in an alcohol, 3200 to 3650.
醇里的氧氢键，3200 到 3650。
In a carboxylic acid the same bond sits lower and much broader, 2500 to 3000.
在羧酸里，同样是氧氢键，位置更低、峰形宽得多， 2500 到 3000。
A carbon to oxygen double bond, a sharp deep dip near 1700.
碳氧双键，一个又尖又深的凹陷，在 1700 附近。
A carbon to carbon double bond, 1500 to 1680.
碳碳双键，1500 到 1680。
Shape counts too: broad or sharp is part of the evidence.
峰形也算证据：是宽还是尖，本身就是线索。
A real one, from a recent paper.
来做一道真题。
Compound S is an ester made from propanoic acid and an alcohol that also has a double bond.
化合物 S 是一种酯，由丙酸和一种同时含有双键的醇生成。
Three dips prove both of those things.
光谱里有三个凹陷，正好证明了这两件事。
The first is at about 1750.
第一个大约在 1750。
Down the table, that is a carbon to oxygen double bond, in the ester row.
查表可知，那是碳氧双键，属于酯那一行。
The second is at about 1650, inside the carbon to carbon double bond range, so the molecule really is unsaturated.
第二个大约在 1650，落在碳碳双键的范围内， 所以这个分子确实是不饱和的。
The third is at about 1170: a single carbon to oxygen bond.
第三个大约在 1170：那是碳氧单键。
Three dips, three bonds, three marks.
三个凹陷，三根键，三分。
Now turn it around.
现在反过来问。
Propene is converted into propan-2-ol by adding water.
丙烯加水变成丙-2-醇。
How does the spectrum show that no propene is left?
光谱怎样说明产物里已经没有丙烯了？
Students hunt for what appeared; the mark is in what disappeared.
学生总去找"多出来了什么"，可分数在"少了什么"里。
Any propene left would absorb between 1500 and 1680, from its carbon to carbon double bond.
只要还剩下丙烯，它的碳碳双键就会在 1500 到 1680 之间给出吸收。
The evidence is a blank: no dip in that range.
所以证据是一段空白：那个范围里根本没有凹陷。
The new broad oxygen to hydrogen dip proves the alcohol formed — but only the missing dip proves the alkene has gone.
新出现的宽的氧氢凹陷能证明醇生成了——但只有"消失的那个凹陷"才能证明烯烃已经反应完。
Now the second machine.
再看第二台仪器。
The sample is knocked into positive ions, they all get the same push, and a magnetic field bends their paths — light ions bend most.
样品先被打成正离子，所有离子受到同样的推力， 然后磁场让它们的路径发生偏转——轻的离子偏得最多。
Each mass lands in a different place, and that is the spectrum.
不同质量的离子落在不同的位置，这就是质谱图。
A gift from the syllabus: you do not need to know how the machine works — only how to read what it prints.
大纲送了你一份礼物：你不需要知道这台机器是怎么工作的，只需要会读它打印出来的图。
Mass spectrometry works on elements too.
质谱也能用在元素上。
Chlorine gives two peaks because chlorine has two isotopes: 75 per cent of its atoms are chlorine-35, 25 per cent chlorine-37.
氯会给出两个峰，因为氯有两种同位素： 75% 的原子是氯-35，25% 是氯-37。
From the isotopic abundance of each, the relative atomic mass is simply the weighted average.
相对原子质量就是它们的加权平均值。
Multiply each isotope mass by its abundance, add them, and divide by the total abundance: 35 times 75, plus 37 times 25, all over 100, which gives 35.5.
把每种同位素的质量乘以它的丰度，加起来，再除以总丰度： 也就是 35 乘 75，加上 37 乘 25，再除以 100，得到 35.5。
Note what you needed: masses alone are never enough.
注意你用到了什么：光有质量是远远不够的。
Now organic molecules.
再看有机分子。
Inside the machine most break apart, but some survive whole.
在仪器内部，大多数分子会碎裂，但有一部分能完整地保留下来。
That whole molecule, with one electron knocked off, is the molecular ion, and it gives the peak at the highest mass to charge ratio.
这个失去一个电子、依然完整的分子，就是分子离子，它给出质荷比最大处的那个峰。
That value is the relative molecular mass.
这个数值就是相对分子质量。
Here is ethanol.
这里是乙醇。
Careful: the tallest peak is not the answer — height only says how common a fragment is.
注意：最高的峰不是答案—— 峰高只说明某种碎片有多常见。
Look to the far right instead, at 46.
要看最右边那个峰，也就是 46。
The smaller peaks earn marks too.
那些较小的峰同样能得分。
The molecular ion shatters into pieces — that is fragmentation — and each charged piece makes its own peak.
分子离子会碎成许多片段——这就是碎裂—— 每一个带正电荷的碎片都会给出自己的峰。
Now subtract.
现在做减法。
In ethanol the molecular ion is at 46 and there is a tall peak at 31.
乙醇的分子离子在 46， 而 31 处有一个很高的峰。
The gap is 15, and 15 is the mass of a methyl group, so a methyl group was lost.
相差 15，而 15 正是一个甲基的质量，所以分子失去了一个甲基。
Learn the two gaps the exam repeats: a loss of 15 is a methyl group, a loss of 29 is an ethyl or an aldehyde group.
记住考试反复考的两个差值：少 15 是甲基，少 29 是乙基或者醛基。
Look closely at the molecular ion and you find a small extra peak one unit to its right: the M plus one peak.
仔细看分子离子峰，你会发现它右边一个质量单位处还有一个小峰：这就是 M 加一峰。
It is there because about 1.1 per cent of all carbon atoms are carbon-13, one unit heavier than normal.
它之所以存在，是因为大约 1.1% 的碳原子是碳-13，比普通的碳重一个单位。
The more carbon atoms a molecule has, the more likely one is carbon-13, so that little peak grows taller.
分子里碳原子越多，其中有一个是碳-13 的可能性就越大，所以那个小峰就越高。
The number of carbon atoms equals 100 times the M plus one abundance, divided by 1.1 times the molecular ion abundance.
碳原子数等于 100 乘以 M 加一峰的丰度，再除以 1.1 乘以分子离子峰的丰度。
Use it twice. Same ester S.
我们用两次。
Its molecular ion peak has relative abundance 4.7, and its M plus one peak has relative abundance 0.31.
还是那个酯 S。 它的分子离子峰相对丰度是 4.7，M 加一峰相对丰度是 0.31。
How many carbon atoms?
它有几个碳原子？
Put the numbers straight in: 100 times 0.31, over 1.1 times 4.7.
把数字直接代进去：100 乘 0.31，除以 1.1 乘 4.7。
That is 31 divided by 5.17, which comes to 5.996 — so S has 6 carbon atoms. Always round to a whole number.
也就是 31 除以 5.17，等于 5.996——所以 S 含有 6 个碳原子。
Now the same idea backwards, on another paper: a compound has 6 carbons and a molecular ion of abundance 34.7.
永远要取整数。 现在把同一个想法反过来用，这是另一份试卷的题：某化合物有 6 个碳， 分子离子峰的丰度是 34.7，求它的 M 加一峰。
Find its M plus one peak. Rearrange: 1.1 times 6 times 34.7, over 100, gives 2.29.
把公式变形： 1.1 乘 6 乘 34.7，再除以 100，得到 2.29。
One more isotope clue — learn the common isotope patterns.
还有一个同位素线索——也是这个专题里最快拿到的分。
Look two units above the molecular ion.
看分子离子峰上方两个单位的位置。
A real peak there means the molecule contains chlorine or bromine.
如果那里确实有一个峰，就说明分子里含有氯或者溴。
Which one?
是哪一个？
Read the ratio, never just the peak.
看比例，别只看有没有峰。
From those same chlorine abundances, one chlorine gives two peaks in a ratio of about 3 to 1.
还是刚才那组氯的丰度：一个氯原子给出两个峰，比例大约是 3 比 1。
Bromine is nearly half bromine-79 and half bromine-81, so one bromine gives two peaks two units apart with roughly equal heights.
溴则接近一半溴-79、一半溴-81，所以一个溴原子给出两个相隔两个单位、高度大致相等的峰。 要学会常见的同位素峰型。
Last example, and it needs both machines.
最后一道题，它需要两台仪器一起用。
An unknown compound gives a molecular ion at 108 and a peak of almost equal height at 110.
某未知化合物的分子离子在 108， 在 110 处还有一个高度几乎相同的峰。
Its infrared spectrum has no broad dip near 3300.
它的红外光谱在 3300 附近没有宽的凹陷。
What is it?
它是什么？
Start with the pair: two units apart, almost equal — that is bromine; one chlorine would have given 3 to 1.
先看这一对峰：相差两个单位、高度几乎相同——那就是溴；如果是氯，比例应该是 3 比 1。
Now take the bromine away: 108 minus 79 leaves 29, and 29 is an ethyl group.
再把溴去掉：108 减 79 等于 29，而 29 是一个乙基。
Finally the infrared: no broad dip near 3300 means no oxygen to hydrogen bond, so no alcohol.
最后看红外：3300 附近没有宽的凹陷，说明没有氧氢键，也就没有醇羟基。
The compound is bromoethane.
这个化合物就是溴乙烷。
Four marks students throw away.
四个学生常丢的分。
First, on an infrared question, give the wavenumber and the bond together.
第一，红外题一定要把波数和化学键一起写出来。
Half an answer scores nothing.
半个答案是不得分的。
Second, the molecular ion is the peak at the highest mass to charge ratio, not the tallest peak on the page.
第二，分子离子是质荷比最大处的那个峰，不是页面上最高的那个峰。
Third, for fragments, quote the gap between two peaks and name the piece that left.
第三，碎片题要写出两个峰之间的差值，并说出失去的是哪一个基团。
Fourth, for the peak two units above, quote the ratio: 3 to 1 chlorine, 1 to 1 bromine.
第四，遇到上方两个单位处的峰，要写出比例：3 比 1 是氯，1 比 1 是溴。
Get those right, and this topic is yours.
把这四点做对，这个专题就是你的了。