How can pyrene be aromatic when it has 4n pi electrons?
441962
At first glance you might make this assumption - but pyrene is an exception to Huckel's rule.  Huckel's rule of 4N + 2 is valid for mono cyclic ring structures (not in the case of the 4-fused benzene structure of pyrene).
477755
Aromatic groups are also those which obey Huckels rule, and contain 4n+2 (and n>1) electrons in the ring.\n\nBenzene is aromatic since it has 6 electrons, which solves the equation for when n = 1.\n\nAlkyl/aliphatic groups are fully saturated carbon chains, which can be either cyclic, non-cyclic, branched or unbranched.  Alkyl and aliphatic mean the same thing.\n\nAryl is generally used to describe benzene derivatives.
3700
Carbon is "special" for organic chemistry because of its ability to form molecules of arbitrary numbers of carbon atoms and other elements, primarily hydrogen, nitrogen and oxygen. Carbon can have up to four single bonds. Besides, it can have double or triple bonds, allowing for even more variations than its heavier homologue Silicon.\n\nOne of my old chemistry professors gave three "fundamental theorems" for organic chemistry:\n\n- Carbon always has four bonds\n- Nucleophile will react with electrophile (slightly  negative ends with slightly positive ones)\n- Whenever the opportunity to form a ring of five or six atoms offers itself during a reaction, it will be taken.\n\nBetter than 95% of organic chemistry can be deduced using these rules.
789677
3 is incorrect.  Three of the carbon electrons are located in the three sp2 hybrid orbitals and the remaining electron sits in the lone p orbital (orthogonal/perpendicular) to the molecular plane.  It is this electron in the p orbital which "interacts" with the oxygen lone electron in it's unhybidized p orbital to form the pi bond.\n\nThere are no electrons left in the original carbon 2s orbital because these orbitals have been hybridized to the sp2 orbitals in order to allow for the three sigma bonds (one to each hydrogen and one to the oxygen atom)
136157
Mostly because of its sp3 hybridization - that is, the fact that 4 atoms or molecules can bond with great stability to the carbon atom. For example, methane (CH4) is the simplest of the carbon compounds. But now, replace any of those H's with almost any other kind of molecule and you get the meat & potatoes of what organic chemistry is all about.
273309
wat r u ppl talking about....u ppl r absolutely wrong.....its alkynes n i am 100% sure about\n\nthey r more reactive becoz of there triple bonds...\n\n\n2 ppl in a row a row giving a wrong answer can confuse a person completely
92358
Just count the (sigma) bonds and the lone electron pairs and memorize the cooresponding geometric shapes.  Sorry, but it's the only way.
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chemical properties of benzene\nElectrophilic aromatic substitution is a general method of derivatizing benzene. Benzene is sufficiently nucleophilic that it undergoes substitution by acylium ions or alkyl carbocations to afford give substituted derivatives. \n\nThe Friedel-Crafts acylation is a specific example of electrophilic aromatic substitution. The reaction involves the acylation of benzene (or many other aromatic rings) with an acyl chloride using a strong Lewis acid catalyst such as aluminium chloride.. \n\nLike the Friedel-Crafts acylation, the Friedel-Crafts alkylation involves the alkylation of benzene (and many other aromatic rings) usng an alkyl halide in the presence of a strong Lewis acid catalyst. \n\nsulfonation. \nNitration: Benzene undergoes nitration with nitronioum ions (NO2+) as the electrophile. Thus, warming benzene with a combination of concentrated sulphuric and nitric acid gives nitrobenzene. \nHydrogenation: Benzene and derivatives convert to cyclohexane and derivatives when treated with hydrogen at high hydrogen pressures. \nBenzene is an excellent ligand in the organometallic chemistry of low-valent metals. Important examples include the sandwich and half-sandwich complexes respectively Cr(C6H6)2 and [RuCl2(C6H6)]2
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Most plastics are made of organic compounds.  The inter-atomic bonds of organic chemicals have quantum energies that correspond precisely with the energy levels in the infrared region of the electromagnetic spectrum.  \n\nIf you irradiate an organic chemical with IR, it will absorb quanta of IR energy that correspond to specific vibrational energies in the molecule -- this excites the bond and makes it vibrate more rapidly and with greater amplitude.  By scanning across the entire range of IR frequencies, you get a "fingerprint" of absorptions that can help to identify the nature of the organic chemical being scanned.\n\nThat's it in a nutshell -- I hope it helps.  There are lots of instrumental analysis books available that will explain it in more detail.
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There are several reactions that have ortho selectivity. \n\nElectrophilic aromatic substitution:\nHere the ortho effect could be explained three different ways:\n1. Electron donating substituents increase electron density on the ring, specifically at the ortho and para positions, which makes the reaction faster.\n2. Electron donating substituents have lone pairs which stabilize the carbocation intermediate.\n3. Electron donating substituents stabilize the positive charge on the ring through an inductive dipole effect.\n\nThe Kolbe-Schmidt, aromatic Claisen, and Fries rearrangement reactions essentially operate the same way, but usually are not covered (if at all) until the second semester of organic chemistry.\n\nThe newest type of reaction is called "directed ortho metalation". Unless you are in an advanced organic synthesis, probably at the graduate level, you won't encounter this. The concept is easy to understand, but a little difficult to get to work in an undergraduate lab.\n\nIs this what you were looking for?
414105
I presume you asking for the N15 shift since there is no proton or carbon.  There is variability of the shift depending on it's environment (is it aliphatic or aromatic?)
95279
Answer is?  none... PCl5 P only has five valence electrons by which to bond with.  A pi bond is defined as a bond with no electron cloud density along the internuclear separation.  However, all five of the chlorines are bonded directly to the Phosphorus.  It has five sigma bonds.
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Flourines electron configuration is [He]2s2 2p5.  In the p-orbitals there are three subshells, each subshell is capable of taking only two electrons a piece.  This leaves you with one electron that is unpaired.  Because in the p-orbitals you can only have a total of 6 electrons.  So, 6-5=1.
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I am not exactly sure what answers you are looking for but here is what I can tell you.\n\nBenzene rings are very very stable because they are aromatic structures.  The Pi system of the benzene ring can however react with an electrophile.\n\nBenzene rings can also undergo substitution reactions with a number of constraints.  Because rings can not undergo SN1 and SN2 reactions like alkanes because their sructure does not allow for a backside attack with the nucleophile, nor do they have the ability to form a carbocation upon the dissociation of a leaving group.\n\nBenzene rings undergo two specific types of substitution.  \n1) Addition-Elimination\n-The nucleophile attacks the benzene ring at the carbon containing the leaving group.  A negative charge results.  The negative charge form resonance back in the ring by kicking out the leaving group.\n\n2) Elimination-Addition\n-A base grabs a proton in the ortho position to the leaving group and the electrons from the C-H bond are thrown into the ring forming a triple bond (benzyne).  A nucleophile attacks one carbon of the triple bond causing it to form a double bond and a negative charge in the ortho position to the nucleophile.  Upon acidic workup, the negative charge is protonated.\n\n---\nWith all of this said hopefully, you can deduce the correct answers to your question under the context of what you are currently doing in class.  Many different answers could fit the blanks.\n\nIf I were to answer your question as given, I would say: Benzene rings are very unreactive, but undergo substitution relatively high speed when an electron withdrawing group is present, on the carbon containing the best leaving group, but unlike alkanes do not participate in SN1 or SN2 reactions.\n\nI hope this helps!
806015
three principles that govern this process: \n1) the Aufbau (build-up) principle, \n2) the Pauli exclusion principle, \n3) Hund's rule\n\nExclusion principle \nNo more than one electron can have all four quantum numbers the same. What this translates to in terms of our pictures of orbitals is that each orbital can only hold two electrons, one "spin up" and one "spin down". \n\nBuild-up principle \nYou may consider an atom as being "built up" from a naked nucleus by gradually adding to it one electron after another, until all the electrons it will hold have been added. Much as one fills up a container with liquid from the bottom up, so also are the orbitals of an atom filled from the lowest energy orbitals to the highest energy orbitals. \n\nHowever, the three p orbitals of a given shell all occur at the same energy level. So, how are they filled up? Is one of them filled full with the two electrons it can hold first? Or do each of the three orbitals receive one electron apiece before the any single orbital is double occupied. As it turns out, the latter situation occurs. \n\nHund's rule \nThis rule is applicable only for those elements that have d electrons, and so is less important in organic chemistry (though it is important in organometallic chemistry). It says that filled and half-filled shells tend to have additional stability. In some instances, then, for example, the 4s orbitals will be filled before the 3d orbitals.
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unpaired electrons are 2 according to mollecular orbital (2) theory. that's why oxygen is paramagnetic. but from the VSEPR theory, this charactoristics of O2 can not be recognised. \n\nI can not draw the diagram of MO theory otherwise I would have shown you how there is 2 unpaired electrons. In short, in the pie 2P orbital, there are 2 unpaired electrons. So, the answer is 2.
842873
the simple answer to your question is just putting unidirectional electrons in the p-orbital. but a more complicated way of how the aufbau principle is disregarded is when there is hybridization of the orbitals together. you get s orbitals and p orbitals fuzing to create sp, sp2 and sp3 orbitals. but dealing with those are only needed when knowing the actual 3-D shape of the molecule.