An introduction to A Level organic chemistry
A-Level Chemistry Topic 29 8:28 English narration · English + 中文 subtitles burned in
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Two molecules. Same atoms, same bonds. The only difference: one is the mirror image of the other, like your left hand and your right hand.
两个分子：原子相同，化学键也相同，唯一的区别是一个是另一个的镜像， 就像你的左手和右手。
And yet one of them smells of oranges, and the other smells of lemons.
可是其中一个闻起来是橙子味，另一个却是柠檬味。
Your nose can tell them apart. So can your body — which is why a medicine has to be the right hand.
你的鼻子能分辨它们，你的身体也能——所以一种药必须是那只"正确的手"。
This topic is the toolkit for all of A2 organic chemistry. New families of compounds, and how to name a ring. Two new mechanism words. The flat, very stable ring called benzene. And handedness. Let's begin.
这个专题是整个 A2 有机化学的工具箱：新的化合物家族，以及如何给环命名； 两个新的反应机理名称；那个平面而且非常稳定的环——苯；还有手性。 让我们开始吧。
First, the new families.
先看新的家族。
A2 adds four to the ones you met at AS. The arene family, which means the molecule contains a benzene ring.
在 AS 学过的基础上，A2 又加了四种。
Phenol, the ring with an oxygen hydrogen group joined straight to it. An acyl chloride, a very reactive relative of a carboxylic acid. And the amide, a carbon oxygen double bond joined to nitrogen.
芳香烃家族，意思是分子中含有一个苯环；苯酚，就是这个环上直接连着一个羟基； 酰氯，它是羧酸非常活泼的"亲戚"；还有酰胺，把碳氧双键和氮连在一起。
No rule changes: the functional group still decides the properties.
规则本身没有变：官能团仍然决定性质。
And read all four ways to draw a molecule: general, structural, displayed and skeletal.
四种分子式写法你都要会读——通式、结构式、结构简式和键线式。
Naming a ring follows one pattern.
给环命名只有一套套路。
If the molecule contains a benzene ring, the ring is the parent.
如果分子中含有苯环，那么这个环就是母体。
Number the six carbons, starting at the carbon with the main group, and go round the way that gives the lowest numbers.
给六个碳编号，从连着主要官能团的那个碳开始，并且沿着能让取代基编号最小的方向数。
An acid group with a nitro group two carbons further round is 3-nitrobenzoic acid. An oxygen hydrogen group with three bromines is 2,4,6-tribromophenol.
环上带一个羧基、再隔两个碳带一个硝基，就叫 3-硝基苯甲酸； 带一个羟基和三个溴原子，就叫 2，4，6-三溴苯酚。
But careful: a ring of six carbons is not always aromatic — cyclohexanol is not.
但要小心：六个碳的环不一定是芳香的——环己醇就不是。
Now the ring itself.
现在看这个环本身。
Benzene is six carbons and six hydrogens, and it is completely flat — a regular hexagon.
苯由六个碳和六个氢组成，而且完全是平面的——一个正六边形。
Every bond angle is one hundred and twenty degrees.
每一个键角都是一百二十度。
That angle is the clue: each carbon is s p two hybridised, just like the carbons in an alkene.
这个角度就是提示：每个碳都是 sp2 杂化的， 和烯烃里的碳一样。
That hybridisation uses three hybrid orbitals to make three sigma bonds — two to its neighbouring carbons, one to its hydrogen.
那种杂化用三个杂化轨道形成三条 σ 键： 两条连到相邻的碳，一条连到自己的氢。
But each carbon still has one electron left, in a p orbital that does not lie in the ring.
但每个碳还剩下一个电子，在一个不在环平面内的 p 轨道里。
Those p orbitals stand up at right angles to the flat hexagon.
这些 p 轨道垂直地"立"在平面六边形上。
Here is the ring seen edge on.
这是从侧面看这个环。
They overlap sideways with their neighbours, all the way round, and merge into one cloud above the ring and one below — a single delocalised pi system.
p 轨道向两侧的邻居侧向重叠，一圈都是这样，融合成环上方的一片电子云和环下方的一片—— 一个完整的离域 π 体系。
That is why we draw a circle inside the hexagon.
这就是我们在六边形里画一个圆的原因。
So all six carbon carbon bonds are exactly the same length, and the ring is unusually stable.
于是六条碳碳键长度完全相同，而且这个环异常稳定。
How do we know?
我们怎么知道？
The enthalpy change of hydrogenation.
靠氢化焓变。
Add hydrogen to the double bond in cyclohexene, and about one hundred and twenty kilojoules per mole are given out.
给一个碳碳双键加氢，比如环己烯，大约放出一百二十千焦每摩尔。
For three separate double bonds the model predicts three times that: three hundred and sixty.
如果是三个独立的双键，模型预测应该放出三倍，也就是三百六十。
But the measured value for benzene is only two hundred and eight.
但苯实测只放出二百零八。
It gives out far less than it should, so it started far lower down — about one hundred and fifty two kilojoules per mole more stable than the model predicts.
它放出的能量比"应该"的少得多，说明它一开始就低得多—— 比模型预测的大约稳定一百五十二千焦每摩尔。
Now a real one. Describe and explain the shape of benzene, four marks.
我们来做一道真题：描述并解释苯的形状，四分。
Write separate lines, one idea each.
把答案写成一行一个要点。
Line one: planar — a regular hexagon.
第一行：平面——正六边形。
Line two: every bond angle is one hundred and twenty degrees.
第二行：每个键角都是一百二十度。
Line three: each carbon is s p two hybridised.
第三行：每个碳都是 sp2 杂化。
Now the part most students miss: the orbitals.
接下来是大多数学生漏掉的地方：轨道。
Hybrid orbitals overlap end-on, making the sigma bonds. The leftover p orbitals overlap sideways, above and below the ring, making the pi system.
杂化轨道彼此"头碰头"重叠，形成 σ 键； 剩下的 p 轨道在环的上方和下方侧向重叠，形成 π 体系。
Two marks for the first three lines, one for sigma, one for pi.
前三行合起来两分，σ 一分，π 一分。
The syllabus adds two mechanism words here.
本专题只增加两个机理名称。
The first word is electrophilic substitution: an electrophile is attracted to the ring and replaces one of the ring hydrogens.
第一个是亲电取代：亲电试剂被环吸引，取代掉环上的一个氢。
The second word is addition elimination: something adds on first, then a small molecule is thrown out.
第二个是加成—消去：先有东西加上去，然后又脱去一个小分子。
Learn both — the exam often just asks you to name the mechanism, for one mark.
两个都要记住——考试常常只是让你写出机理名称，就有一分。
Now the arrows.
现在看箭头。
Stage one: the delocalised cloud is rich in electrons, so the ring attacks the electrophile. The arrow starts inside the ring and points at the positive group.
第一步：离域电子云电子密度很高，所以是环去进攻亲电试剂—— 箭头从环内部出发，指向带正电的基团。
Stage two: that gives an unstable intermediate, with the delocalised system over only five carbons and one carbon holding both a hydrogen and the new group.
第二步：得到一个不稳定的中间体， 此时离域体系只覆盖五个碳，分子带正电荷，有一个碳同时连着一个氢和新基团。
Stage three: an arrow from that carbon hydrogen bond goes back into the ring, the hydrogen leaves as a positive ion, and the full ring is restored.
第三步：一支箭头从这个碳氢键出发，回到环里，氢以正离子的形式离去， 完整的环重新形成。
The ring is kept, so it is substitution.
环保留下来了，所以这是取代。
The second mechanism is two steps in one.
第二个机理是两步合成一个过程。
Take a carbonyl compound and a reagent with a nitrogen hydrogen group.
取一个羰基化合物和一个含氮氢基团的试剂。
First the nitrogen adds on to the carbon of the carbon oxygen double bond. Then water is thrown out, leaving a carbon nitrogen double bond.
首先氮加到碳氧双键的碳上；然后脱去一分子水，留下一个碳氮双键。
That is the test for aldehydes and ketones, which gives an orange precipitate.
这正是检验醛和酮时发生的事，会生成橙色沉淀。
Acyl chlorides behave the same way: a nucleophile adds on, then hydrogen chloride is lost.
酰氯也是一样：亲核试剂先加上去，然后脱去氯化氢。
Back to the hook.
回到开头的问题。
A carbon carrying four different groups is a chiral centre.
一个碳上连着四个不同的基团，就是一个手性中心。
Draw it, draw its reflection, and slide one onto the other. They never match, just like your two hands.
画出它，再画出它的镜像，然后把一个滑到另一个上面——它们永远重合不了， 任何旋转都做不到，就像你的两只手。
The two forms are enantiomers.
这两种形式就叫对映体。
Now the exam's favourite fact.
这是考试最爱考的事实。
Two enantiomers are completely identical: same melting point, same boiling point, same chemical reactions.
两个对映体几乎完全相同：熔点相同、沸点相同、化学反应也相同。 只有两点不同。
Only two things differ. First, shine plane polarised light through each. Both rotate the plane by the same angle — but in opposite directions.
第一，让平面偏振光分别通过它们：两者都会使偏振面旋转， 而且旋转的角度相同——但方向相反。
A substance that does this is optically active.
能做到这一点的物质就叫具有旋光性。
Second, they can behave differently inside a living thing.
第二，它们在生物体内的作用可能不同。
Mix the two fifty fifty and you get a racemic mixture: the rotations cancel, so it turns the light not at all.
把两者按五十比五十混合， 就得到外消旋混合物：两个旋转互相抵消，所以完全不使光旋转。
That second difference is why chirality matters in medicines.
第二个差别，正是制药时手性如此重要的原因。
Your body's receptors are chiral too, so of the two mirror forms, one may cure the illness while the other does nothing, or even does harm.
你体内的受体本身也是手性的， 所以两种镜像形式中，一种可能治好疾病，另一种却毫无用处，甚至造成伤害。
So either make both and separate the mixture into two pure enantiomers — slow, costly, half the batch wasted — or use a chiral catalyst, which builds only the enantiomer you want.
于是要么把两种都做出来，再把混合物拆分成两个纯对映体——又慢又贵，还要扔掉一半； 要么使用手性催化剂，只生成你想要的那一个对映体。
That is what the mark scheme wants you to praise: no need to separate optical isomers.
第二条路正是评分标准希望你写的：不需要分离旋光异构体。
One more worked example.
再做一道例题。 化合物 Q。
Compound Q. Part a: how many carbons are s p, s p two and s p three?
第 a 问：有多少个碳是 sp、sp2 和 sp3 杂化？
You only need one rule: look at the bonds on each carbon.
你只需要一条规则：看每个碳上的键。
A triple bond means s p, a double bond means s p two, single bonds only means s p three. So count them.
在三键上的碳是 sp，在双键上的碳是 sp2， 只有单键的碳是 sp3。
The nitrile carbon is in a triple bond, so s p is one. The amide carbon has a carbon oxygen double bond, so s p two is one. The other three are s p three.
那就数一数：腈基上的碳在三键上，所以 sp 是一个； 酰胺上的碳有碳氧双键，所以 sp2 是一个；其余三个碳都是 sp3。
Part b: a chiral centre is a carbon with four different groups.
第 b 问：手性中心是连着四个不同基团的碳。
Two of them qualify.
逐个检查，有两个符合。
So the number of optical isomers is two to the power two: four.
所以旋光异构体的数目就是二的二次方——四个。
Three marks students throw away.
三个学生常丢的分。
First, on any benzene bonding question, say which orbitals overlap and how: hybrid orbitals end-on for sigma, p orbitals sideways for pi.
第一，凡是苯的成键题，要写清楚哪些轨道怎样重叠： 杂化轨道头碰头形成 σ，p 轨道侧向重叠形成 π。
Second, never say the two enantiomers have different properties. Only the direction of rotation and the behaviour in the body differ, and a racemic mixture rotates nothing.
第二，绝不要说两个对映体性质不同——只有旋光方向和在体内的作用不同， 而外消旋混合物完全不使光旋转。
Third, a chiral centre needs four different groups — two identical methyl groups on one carbon destroy the chirality there. Check all four, and each extra centre doubles the number of isomers.
第三，手性中心必须有四个不同的基团——同一个碳上 两个相同的甲基就会毁掉那里的手性；四个都要检查； 每多一个手性中心，异构体数目就翻一倍。
Get those right, and this topic is yours.
做对这三点，这个专题就是你的了。