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【交流】MS图谱解析导论(来源:美国ARIZONA大学化学系)

气质联用(GCMS)

  • Introduction to Mass Spectrometry -- Interpretation
    Here are a list of steps to follow when interpreting a mass spectrum. This simplified list will help you to interpret many spectra, however there are other mechanisms of fragmentation which cannot be covered in this brief tutorial.

    Steps to interpret a mass spectrum:

    1. Look for the molecular ion peak.

    This peak (if it appears) will be the highest mass peak in the spectrum, except for isotope peaks.
    Nominal MW (meaning=rounded off) will be an even number for compounds containing only C, H, O, S, Si.
    Nominal MW will be an odd number if the compound also contains an odd number of N (1,3,...).
    2. Try to calculate the molecular formula:
    The isotope peaks can be very useful, and are best explained with an example.
    Carbon 12 has an isotope, carbon 13. Their abundances are 12C=100%, 13C=1.1%. This means that for every 100 (12)C atoms there are 1.1 (13)C atoms.
    If a compound contains 6 carbons, then each atom has a 1.1% abundance of (13)C.
    Therefore, if the molecular ion peak is 100%, then the isotope peak (1 mass unit higher) would be 6x1.1%=6.6%.
    If the molecular ion peak is not 100% then you can calculate the relative abundance of the isotope peak to the ion peak. For example, if the molecular ion peak were 34% and the isotope peak 2.3%: (2.3/34)x100 = 6.8%. 6.8% is the relative abundance of the isotope peak to the ion peak. Next, divide the relative abundance by the isotope abundance: 6.8/1.1=6 carbons.
    Follow this order when looking for information provided by isotopes: (A simplified table of isotopes is provided in the introduction, more detailed tables can be found in chemistry texts.)
    Look for A+2 elements: O, Si, S, Cl, Br
    Look for A+1 elements: C, N
    "A" elements: H, F, P, I
    3. Calculate the total number of rings plus double bonds:
    For the molecular formula: CxHyNzOn
    rings + double bonds = x - (1/2)y + (1/2)z + 1
    4. Postulate the molecular structure consistent with abundance and m/z of fragments.
    More information on specific fragmentation can be found in the quiz for each functional group.
    这是个比较系统的MS课件,有GC-MS简介,按基础化合物类型(10个)分别用图谱例子展示,是一个MS入门讲座.请连接:http://www.chem.arizona.edu/massspec/contents.html
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  • 第1楼2005/06/12

    好东西啊,就是看不懂啊,可以汉化吗?

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  • 第2楼2005/06/13

    这一篇还算是简单的了,你可以用金山词霸翻译着阅读,
    看多了英语阅读能力说不定能逐渐提高,你以后读研也要看很多外文资料的。

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  • 第3楼2005/06/13

    说的很对,专业英语是少不了的.坚持读,必有收获.

    icefirel 发表:这一篇还算是简单的了,你可以用金山词霸翻译着阅读,
    看多了英语阅读能力说不定能逐渐提高,你以后读研也要看很多外文资料的。

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  • 第4楼2005/06/13

    谢谢你们了,会努力的,呵呵

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  • 第5楼2005/06/13

    我试着翻译了一部分,剩下的不会翻译了,请感兴趣的朋友补全吧!

    1.寻找分子离子峰

    分子离子峰(如果能打出)一般是质量数最大的峰,同位素峰除外。Nominal MW在化合物包含C,H,O,S,Si原子时为偶数。Nominal MW在化合物包含奇数个N原子时为偶数。

    2.尝试计算分子式

    同位素峰在计算分子式时会非常有用,下面有一个很好的例子:
    碳12有一个同位素碳13,它们的丰度比是100:1.1,这意味着如果有100个C12就有1.1个C13。如果一个化合物含六个C,由于每个原子含有1.1%的C13,因此,如果分子离子峰是100%,那么同位素峰(质量数大1的)就是6×1.1%=6.6%。如果分子离子峰不是100%,你可以计算出同位素峰相对于分子离子峰的丰度。举个例子,如果分子离子峰是34%,同位素峰是2.3%:(2.3/34)x100= 6.8%,6.8%就是同位素峰相对于分子离子峰的丰度。然后用相对丰度除以同位素丰度:6.8/1.1=6个C

    3.计算环数和双键的总数

    对于分子式:CxHyNzOn
    环数+双键数= x - (1/2)y + (1/2)z + 1

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  • 第6楼2005/06/13

    版主,你也太认真了吧,强,不过还是要赞扬你哦

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  • 第7楼2005/06/13

    不要光赞扬!帮我把剩下的翻译了呀!

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  • 第8楼2005/06/13

    Postulate the molecular structure consistent with abundance and m/z of fragments.
    More information on specific fragmentation can be found in the quiz for each functional group.
    4.根据碎片离子得丰度和质量数推导分子结构
    从每个官能团中可以获得更多得特征碎片信息,我翻译得对吗?

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  • 第9楼2005/06/13

    好像不太对 ,汪老师能不能给我们两个学生指正一下呀?

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  • 第10楼2005/06/13

    那个链接怎么找不到服务器?试了几次都不行

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