AL Organic Chem

Before studying organic reactions, we need to familiarize ourselves with the structral features of functional groups.

In alkanes, like methane, there is no unsaturated carbon-carbon bond so the addition reaction cannot take place. The possible reaction is substitution.
Also, the difference in electronegativity between carbon and hydrogen atom is small so the bond-breaking process is homolysis.

Concept of free radical substitution
In the presence of radiation of frequency similar to that of the covalent bond in Cl2, the resonance takes place and the oscillation is strong enough to make the Cl─Cl bond breaks homolyticaly to give a．Cl free radical.

圖片參考：http://us.f6.yahoofs.com/hkblog/d1WWTL6FGRtxq8qXRSSQITaWWQVB_2/blog/20090923115056272.jpg?ib_____Djd2xLfVo

The．Cl free radical initiates the next step of the reaction, producing methyl radical and HCl.

圖片參考：http://us.f6.yahoofs.com/hkblog/d1WWTL6FGRtxq8qXRSSQITaWWQVB_2/blog/20090923115103847.jpg?ib_____D6q.EdSpc

The methyl radical bonds to the Cl atom and gives a．Cl free radical.

圖片參考：http://us.f6.yahoofs.com/hkblog/d1WWTL6FGRtxq8qXRSSQITaWWQVB_2/blog/20090924123307657.jpg?ib_____Dgf3Ujuio

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In carbonyl compounds, there is a polar C=O bond as O atom is more electronegative than C atom. The C atom in C=O bond is electron poor and susceptible to the attack of nucleophiles(Nu).

圖片參考：http://us.f6.yahoofs.com/hkblog/d1WWTL6FGRtxq8qXRSSQITaWWQVB_2/blog/20090924120747587.jpg?ib_____Da3wigFyl

Since the C=O bond is unsaturated, the C atom can accept the incoming nucleophile to form a new cavalent bond. The reaction is nucleophilic addition.

To conclude, the free radical substitution always takes place in saturated alkanes while the nucleophilic addition always takes place in carbonyl compounds.
你只需要辨認出functional group係咩，就好容易知道它會行咩反應
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呢度，我唔明你第一句講咩
trimethylmethane is an alkane so the reaction is free radical substitution rather than nucleophilic addition.

2009-09-24 19:36:58 補充：
To michaelchan_wahyan

You name the molecule C(CH3)3H trimethylmethane, right?

The correct name of the molecule is 2-methylpropane.

1) Trimethylmethane is incorrect naming for CH3-CH(CH­3)-CH3. Its name should be methylpropane.

2) In the chlorination of methylpropane, it is hardly for the carbon atom attaching to three methyl groups as the reaction site because steric hindrance is high due to the three bulky methyl groups.

3) Inductive effect can only slightly increase/decrease the electron cloud density of the radical carbon atom, but can NEVER change a free radial into carbanion/carbocation. Therefore, the type of reaction is free radical substitution.