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Date: 5-9-2018
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Date: 5-9-2018
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Date: 20-7-2019
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Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples:
The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond. One of the important set of reactions of alkyl halides involves replacing the halogen by something else - substitution reactions. These reactions involve either:
You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one.
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تفوقت في الاختبار على الجميع.. فاكهة "خارقة" في عالم التغذية
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أمين عام أوبك: النفط الخام والغاز الطبيعي "هبة من الله"
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المجمع العلمي ينظّم ندوة حوارية حول مفهوم العولمة الرقمية في بابل
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