推荐回答(4个)
1、原理:甲苯,乙苯都属于含有α氢原子的苯同系物。苯同系物中若含有α氢原子,则α碳原子可以被高锰酸钾酸性溶液氧化生成苯甲酸,即α碳原子被氧化。
2、甲苯和乙苯都是苯的同系物,甲苯和乙苯中的α碳原子都可以被氧化,因此都生成苯甲酸。
3、甲苯被高锰酸钾氧化时断裂α碳原子与苯相连的碳碳单键,然后被氧化生成羧基。

扩展资料
一、苯甲酸用氢化铝锂还原成苯甲醇,苯甲醇在铜作催化剂的条件下加热脱水生成苯甲醛,苯甲醛用克莱门森或者黄鸣龙法还原可以得到甲苯。
二、能使酸性高锰酸钾褪色的有:
1、碳碳双键:碳碳双键是指由碳的一个2s亚层和两个2P亚层杂化为三个sp2杂化轨道。这三个sp2杂化轨道分布在同一平面上。
2、碳碳三键:三键,在化合物分子中两个原子间以三对共用电子构成的重键。
3、羟基:羟基化学式为-OH。羟基主要分为醇羟基,酚羟基等。
三、苯的同系物的性质:
(1)使得苯环邻位和对位上的氢原子活泼,易被其它的原子或原子团取代。
(2)使得侧链跟苯环直接相连的碳原子上的氢原子活泼,易被氧化,且无论侧链长短,其氧化产物都为苯甲酸。如甲苯、乙苯均能使酸性高锰酸钾溶液褪色。
参考资料来源:百度百科——高锰酸钾
参考资料来源:百度百科——苯的同系物
原理:甲苯的甲基被酸性高锰酸钾溶液氧化成羧酸,生成苯甲酸,钾盐(K₂SO₄)、锰盐(MnSO₄)和水。
甲苯上的甲基,被苯环激发,所以它比较容易反应。反应的时候断裂的是侧链上甲基的两个C-H键,生成两个C-OH键;继续断裂生成的C-OH键和O-H键,生成一个C=O键,即为-CHO;生成的-CHO上的C-H键断裂,生成一个C-OH键,即为最终的-COOH。
化学方程式为:5C₆H₅-CH₃+6KMnO₄+9H₂SO₄→5C₆H₅-COOH+3K₂SO₄+6MnSO₄+14H₂O。

扩展资料
用途
1、甲苯大量用作溶剂和高辛烷值汽油添加剂,也是有机化工的重要原料,但与同时从煤和石油得到的苯和二甲苯相比,目前的产量相对过剩,因此相当数量的甲苯用于脱烷基制苯或岐化制二甲苯。
2、甲苯衍生的一系列中间体,广泛用于染料、医药、农药、火炸药、助剂、香料等精细化学品的生产,也用于合成材料工业。
3、甲苯进行侧链氯化得到的一氯苄、二氯苄和三氯苄,包括它们的衍生物苯甲醇、苯甲醛和苯甲酰氯(一般也从苯甲酸光气化得到)。
4、甲苯氧化得到苯甲酸,是重要的食品防腐剂(主要使用其钠盐),也用作有机合成的中间体。
参考资料来源:百度百科--甲苯
参考资料来源:百度百科--苯甲酸
断裂的是侧链上甲基的两个C-H键,生成两个C-OH键;继续断裂生成的C-OH键和O-H键,生成一个C=O键,即为-CHO;生成的-CHO上的C-H键断裂,生成一个C-OH键,即为最终的-COOH
反应式为C6H5-CH3+O2=KMnO4=C6H5-COOH+H2
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