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According to Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208[1], Kwang Sup Lee and Gerhard Wegner, attempting the synthesis and study of $n>100$, $\ce{C_nH_{2n}}$ cyclo-alkanes and linear alkanes, were successfully able to synthesize a linear alkane of $\ce{C384H770}$ and cycloalkanes with 288 carbons ($\ce{C288H576}$) Linear alkanes up to $\ce{... 10 You need to know the terms "staggered" and "eclipsed", because that's what we call these two conformations. Now to the point: the real ethane molecule rotates and changes shape all the time, so both geometries are possible, albeit the top one is more likely, because it is the energy minimum. To make the molecule stop in the staggered ... 10 If you use "cyclopenta-1,3-diene" in the end of the name (your first attempt), then this would be the stem system. In case like the one presented in the question -- two rings connected via a single bond -- it is recommended to recognize the cyclohexene as the stem (like parent), and the cyclopentadienyl moiety as a substituent (like child). ... 9 You have two choices for choosing the parent hydride here in this case. 1. Choosing the cyclohexene ring as the parent hydride If we consider the cyclohexene ring as the parent hydride, we get the cyclopentadiene ring to be a substituent leading the name to become 1-(cyclopenta-1,3-dien-1-yl)cyclohex-1-ene 2. Choosing the cyclopentadiene as the parent ... 8 LPG is propane, butane or a mixture of both. Paraffin candle wax is$(CH_2)_n$typically$C_{31}H_{64}$(other, nonparaffin, waxes are also used in candles, like stearin, beeswax, etc.) However, LPG is a gas from the start and can mix with air (i.e., oxygen) before it begins to burn, while candle wax has to be heated by the flame before it can begin to burn,... 8 Vegetable oils are mainly made of glycerol esters of oleic acid, linoleic acid and palmitic acid. For example, the fatty acids extracted from olive oil are a mixture of$74$% oleic acid,$11$% palmitic acid, and$10$% linoleum acid, plus 5% other acids. These proportions are average values that may change from one sample to the next one. The mass proportion ... 7 I have read some answers on web which says that it is due to the insufficient supply of oxygen, so I tried burning candle beside LPG ( so there is no difference in supply of oxygen for both candle and LPG ) , but still candle gives a yellow flame. Why? You cannot conduct this experiment by yourself without using a proper apparatus. Otherwise comparing the ... 7 Zaitsev's rules are nothing but a set of rules that were used to explain experimental results that were observed. It was not the other way around. The real rule you have to use here is the Hammond's Postulate. This states: The transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. So, in this ... 6 There is a comment elsewhere in this post saying: "(The mechanism for iodoform test given by 2-Aminoalkanes might be) similar to the iodoform reaction of secondary alcohols. It is oxidized to imine, maybe then similar procedure." I won't necessarily agree with that argument. To have a positive iodoform test, it is necessary to acitivate the ... 6 I don't think there is a "primary source" for types of isomerism. The types of isomerism are human-defined categories that different authors find conceptually or pedagogically useful in different ways. It isn't like isomerism classification emerges from some unsupervised analysis of all possible structures. If it were my opinion, I'd follow the ... 5 A compound or a molecule being polar or non-polar is rather a fuzzy term. Being non-polar does not mean no polarity of chemical bonds. It means this polarity is low enough for atoms of such a molecule not to interact with other molecules by relatively strong inter-molecular interactions. These include strong dipole-dipole interactions, hydrogen bonds or ... 5$\ce{HCOOCH3}$would likely be rendered as$\ce{H-C(O)-O-CH3}$, the lone hydrogen atom is atttached to the carbonyl group instead of the methyl group being there -- in other words, methyl formate instead of acetic acid. I suppose that$\ce{HOCOCH3}$could be interpreted as having the methyl group and hydroxyl group attached to the carbonyl, but you're better ... 5 The first name has the correct alphabetical order of groups: ethyl before dimethyl. The "d" of dimethyl is not involved in alphabetizing. The numbering is correct because the one-to-one comparisons of locants is {1,2,4 < 1,3,4}. The correct name is 4-ethyl-1,2-dimethylcyclohexane. For a brief set of IUPAC rules applicable to this question see, ... 5 The sp notation for hybridised orbitals applies only to the orbitals of single atoms. All orbitals that are hybridised must be centred on the same nucleus. Once you move to molecular orbitals, i.e. bonding between atoms, the sp notation no longer makes sense and is not used. Instead, localised bonds are typically referred to as σ, π, δ (depending on their ... 5 Oxidation of alkene to 1,2-syn-diol by cold basic$\ce{KMnO4}$is a common classification test for alkene in undergraduate organic chemistry laboratory courses. OP had correctly recognized that this reaction is a case of syn-dihydroxylation, and OP's conclusion of the product to be meso is also correct since the alkene in hand is a cis-but-2-ene. The ... 4 The boiling points of non-polar hydrocarbons are determined by the extent of van der Waals forces in between them. More specifically, molecules with a larger surface area have larger van der Waals forces of attraction between the molecules. From the structures, it is obvious that but-1-yne has a larger surface area than isobutane, and hence has the higher ... 4 A propane molecule$\ce{C3H8}$in the gas phase must find$7$molecules$\ce{O2}$to burn totally. A wax molecule$\ce{C_{31}H_{64}}$must find$63$molecules$\ce{O2}$to burn completely. During its trip through the flame (having maybe$10 - 20$cm length), the propane molecule will have enough time to meet$7$molecules$\ce{O2}$. The wax molecule may not ... 4 Conversion 1 Convert ethene to ethanol. Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300 °C and the pressure is about 60 to 70 atmospheres (chemguide — The mechanism for the acid catalysed hydration of ethene).$$\ce{CH2=CH2 + H2O <=>[H3PO4] CH3CH2OH}$...

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The rules in determining the aromaticity of a compound are as follows: The system must have 4n+2 π electrons The system must be planar The system must be completely conjugated The molecule must be cyclic. The cycloheptatriene anion does not follow two out of these four rules. Namely, The anion system has 4n electrons (whereas it should have had 4n+2 ...

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Here is the meaning of the words ""vulnerable attack" and "negative region of space". The nucleus of the six atoms (2 C and 4 H) are situated in the same plane, which is usually defined as horizontal. The electrons of the 4 H atoms and the three first electrons from each Carbon atom are all together included in sigma bonds, and their ...

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I wish I could convey the spirit of Professor Louis Fieser on the development of organic chemistry. I couldn't find it in his textbooks, but in class, it went something like this: Coal distillation led to a large number of different unsaturated compounds, most of which had strong odors, so they were classified as "aromatic". When more quantitative ...

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You can't compare them at all. If the ortho form dissociates at all then the carboxylate ion combines with the aldehyde function, forming a lactone ring. See Wikipedia. This question invokes a comparison of the similarly structured nitrobenzoic acids. There the ortho compound is found to be strongest, beating its isomers by about one $pK_a$ unit and ...

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The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate. The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity. The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]: It is not normally possible to add fluorine directly to alkenes as the ...

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MCPBA preferentially reacts at the more electron-rich alkene, in this case the more substituted alkene according to this entry on chemistryportal.net here and this review RSC Advances

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According to Markovnikov's rule we would get a vicinal diiodide, which is unstable. The iodine atoms would undergo elimination and once again form a double bonded structure that is 1-propene. As to why (excess) has been mentioned, 1-propene reacts with this $\ce{HI}$ undergoing an addition reaction. As per Markovnikov's rule, this would in fact form 2-...

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I'd say your mechanism is acceptable. However, the final product, pentafulvene (PIN: 5-Methylidenecyclopenta-1,3-diene) is said to be not stable at room temperature. Generally, fulvenes are thermally unstable, sensitive to oxygen and photosensitive (Ref.1). They are also prone to acid- and cation-catalysed polymerisations. For example, there have been ...

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You have misinterpreted what Wikipedia has said. I'd put whole quote here: Cyclobutadiene is an organic compound with the formula $\ce{C4H4}$. It is very reactive owing to its tendency to dimerize. Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable. Since the ...

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A basic principle in OChem says that stability of intermediate is the key factor in deciding the major product, in general. Another factor is stability of final product. Will a kinetically favoured (minor) product (c) form? It will form, as far as it's intermediate is known to exist (and it exists). However, that will be present in very minor amount, as ...

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According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes. Then, if there is a choice, ...

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