The following compounds can be used as monomers for chain polymerization in polymers:

ethylene, propylene, isobutylene, styrene, butadiene, methyl acrylate, vinyl chloride, vinyl acetate, methyl vinyl ether, acrylic acid, acrylonitrile  

For the above substances, the following propositions are given:  
1. All the above monomers can undergo free radical chain polymerization to form polymers.  
2. Among the above monomers, at least four can undergo cationic polymerization to form polymers, with one of them being difficult to form high-molecular-weight polymers.  
3. Among the above monomers, there are four species that can undergo anionic polymerization.  
4. The number of monomers among the above that contain nitrogen atom substituents is equal to the number of monomers that can undergo all four major types of polymerization (free radical, cationic, anionic, coordination).  
5. Among the above monomers, the several olefin monomers that do not contain aromatic rings can all undergo the same type of polymerization.  

Analyze and determine the number of correct propositions among the above.
现有以下化合物，可被用作高分子链式聚合的单体：

乙烯，丙烯，异丁烯，苯乙烯，丁二烯，丙烯酸甲酯，氯乙烯，醋酸乙烯酯，甲基乙烯基醚，丙烯酸，丙烯腈

对于以上物质，有以下命题：
1. 以上单体均可进行自由基链式聚合形成高分子
2. 以上单体中，至少有四种可以进行正离子聚合形成高分子，其中有一种难以形成高分子量聚合物
3. 以上单体中，可以进行负离子聚合的物种有四种。
4. 以上单体中，含氮原子取代基的单体数量，与可以进行所有四种主要类型聚合（自由基、阳离子、负离子、配位）的单体数量相等
5. 以上单体中，不含芳香环的几种烯烃单体，它们都能进行同一种聚合反应

经过分析判断以上命题中正确的命题数量。
0
1
2
3
4
5
0
1
2
3
4
5
Analyze each proposition one by one:

Proposition 1: All the above monomers can undergo free radical chain polymerization to form polymers.  

Counterexample: Isobutylene, whose double-bond carbons are connected to two methyl groups, exhibits significant steric hindrance and is highly prone to chain transfer reactions, making it difficult to form polymers. It primarily undergoes cationic polymerization. Therefore, this proposition is incorrect.  

<Checkpoint>[Isobutylene has high steric hindrance and mainly undergoes cationic polymerization](1)</Checkpoint>  

<Checkpoint>[Proposition 1 is incorrect, as exemplified by the counterexample of isobutylene](0.5)</Checkpoint>  

### Proposition 2: Among the above monomers, at least four can undergo cationic polymerization, with one of them being difficult to form high-molecular-weight polymers.  

Monomers capable of cationic polymerization require substituents that can stabilize carbocations, such as isobutylene (forming stable tertiary carbocations), styrene (forming conjugation-stabilized benzyl carbocations), butadiene (forming conjugation-stabilized allyl carbocations), and methyl vinyl ether (where the oxygen atom's p-π conjugation stabilizes the carbocation).  

<Checkpoint>[Monomers capable of cationic polymerization require substituents that can stabilize carbocations](1)</Checkpoint>  

<Checkpoint>[Isobutylene, styrene, butadiene, and methyl vinyl ether all have substituents that stabilize carbocations and can undergo cationic polymerization](1)</Checkpoint>  

Although propylene has an electron-donating methyl group, the resulting secondary carbocation is insufficiently stable and highly prone to chain transfer and rearrangement, leading to chain termination. Thus, it is difficult to form high-molecular-weight polymers.  

<Checkpoint>[The secondary carbocation formed by propylene is not stable enough to form high-molecular-weight polymers, making Proposition 2 correct.](1)</Checkpoint>  

Proposition 3: Among the above monomers, there are four species that can undergo anionic polymerization.  

Monomers capable of anionic polymerization require strong electron-withdrawing groups or conjugated systems. Examples include styrene, butadiene, methyl acrylate, and acrylonitrile.  

<Checkpoint>[Monomers capable of anionic polymerization require strong electron-withdrawing groups or conjugated systems](1)</Checkpoint>  

<Checkpoint>[Styrene, butadiene, methyl acrylate, and acrylonitrile can undergo anionic polymerization](1)</Checkpoint>  

The count is four, so Proposition 3 is correct.  

<Checkpoint>[There are four species capable of anionic polymerization, so Proposition 3 is correct.](0.5)</Checkpoint>  

Proposition 4: Among the above monomers, the number of monomers with nitrogen-containing substituents equals the number of monomers capable of undergoing all four major types of polymerization.  

The only nitrogen-containing monomer is acrylonitrile. The count is one.  

<Checkpoint>[The only nitrogen-containing monomer is acrylonitrile](1)</Checkpoint>  

Styrene and butadiene, due to their conjugated systems, can stabilize radicals, carbocations, and carbanions, and can also participate in coordination polymerization. The count is two.  

<Checkpoint>[Styrene and butadiene can undergo all four types of polymerization, making Proposition 4 incorrect](1)</Checkpoint>  

Proposition 5: Among the above monomers, the non-aromatic olefin monomers can all undergo the same type of polymerization reaction.  

The non-aromatic olefin monomers are ethylene, propylene, isobutylene, and butadiene.  

<Checkpoint>[Non-aromatic olefin monomers include ethylene, propylene, isobutylene, and butadiene](0.5)</Checkpoint>  

Isobutylene cannot undergo free radical polymerization, ethylene cannot undergo cationic polymerization, and ethylene, propylene, and isobutylene cannot undergo anionic polymerization. However, for coordination polymerization, suitable catalytic systems exist for ethylene (Ziegler-Natta catalyst), propylene (Ziegler-Natta catalyst), isobutylene, and butadiene.  

<Checkpoint>[The common polymerization mechanism for ethylene, propylene, isobutylene, and butadiene is coordination polymerization, making Proposition 5 correct](1)</Checkpoint>  

After analysis, the correct propositions are Proposition 2, Proposition 3, and Proposition 5. Therefore, there are three correct propositions in total.  

<Checkpoint>[There are three correct propositions in total](0.5)</Checkpoint>
逐一分析每个命题：

命题 1：以上单体均可进行自由基链式聚合形成高分子

反例：异丁烯，其双键碳上连接两个甲基，空间位阻很大，且极易发生链转移反应，难以形成高分子。它主要通过阳离子聚合。故该命题错误。

<Checkpoint>[异丁烯空间位阻很大，主要通过阳离子聚合](1)</Checkpoint>

<Checkpoint>[命题1错误，可举出反例异丁烯](0.5)</Checkpoint>

### 命题 2：以上单体中，至少有四种可以进行正离子聚合，其中有一种难以形成高分子量聚合物

能够进行阳离子聚合的单体需要其取代基能稳定碳正离子，如异丁烯（生成稳定的叔碳正离子）、苯乙烯（生成共轭稳定的苄基正离子）、丁二烯（生成共轭稳定的烯丙基正离子）、甲基乙烯基醚（氧原子p-π共轭效应稳定碳正离子）。

<Checkpoint>[能够进行阳离子聚合的单体需要其取代基能稳定碳正离子](1)</Checkpoint>

<Checkpoint>[异丁烯、苯乙烯、丁二烯、甲基乙烯基醚均有取代基稳定正离子的能力，可进行正离子聚合](1)</Checkpoint>

丙烯虽然其甲基有给电子效应，但生成的仲碳正离子不够稳定，极易发生链转移重排，导致链终止，因此难以形成高分子量聚合物。

<Checkpoint>[丙烯生成的仲碳正离子不够稳定，难以形成高分子量聚合物，命题2正确。](1)</Checkpoint>

命题 3：以上单体中，有四种可以进行负离子聚合的物种。

可进行负离子聚合的单体需要强吸电子基或共轭体系。例如苯乙烯、丁二烯、丙烯酸甲酯、丙烯腈。

<Checkpoint>[可进行负离子聚合的单体需要强吸电子基或共轭体系](1)</Checkpoint>

<Checkpoint>[苯乙烯、丁二烯、丙烯酸甲酯、丙烯腈可进行负离子聚合](1)</Checkpoint>

数量为4，因此命题3正确。

<Checkpoint>[可以进行负离子聚合的物种为4，因此命题3正确。](0.5)</Checkpoint>

命题 4：以上单体中，含氮原子取代基的单体数量，与可以进行所有四种主要类型聚合的单体数量相等

含氮原子的单体只有丙烯腈一种。数量为 1。

<Checkpoint>[含氮原子的单体只有丙烯腈一种](1)</Checkpoint>

苯乙烯和丁二烯 因其共轭体系既能稳定自由基，也能稳定正负离子，同时也能参与配位聚合。数量为 2。

<Checkpoint>[苯乙烯与丁二烯可进行所有四种聚合，因此命题4错误](1)</Checkpoint>

命题 5：以上单体中，不含芳香环的几种烯烃单体，它们都能进行的同一种聚合反应

不含芳香环的烯烃单体有乙烯、丙烯、异丁烯、丁二烯。

<Checkpoint>[不含芳香环的烯烃单体有乙烯、丙烯、异丁烯、丁二烯](0.5)</Checkpoint>

异丁烯不能自由基聚合，乙烯不能阳离子聚合， 乙烯、丙烯、异丁烯均不能阴离子聚合，但对于配位聚合，乙烯（Ziegler-Natta催化剂）、丙烯（Ziegler-Natta催化剂）、异丁烯和丁二烯均可找到合适的配位聚合催化体系。

<Checkpoint>[乙烯、丙烯、异丁烯、丁二烯共同的聚合机理为配位聚合，命题5正确](1)</Checkpoint>

经过分析，正确的命题为 命题2、命题3 和 命题5。因此，正确的命题共有3个。

<Checkpoint>[正确的命题共有 3个](0.5)</Checkpoint>