When heated with chlorine the hydrocarbon 2 2-dimethylbutane?
Answers
Answer:
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:(CH3)3−C−CH2−CH3
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:(CH3)3−C−CH2−CH3Initiation
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:(CH3)3−C−CH2−CH3InitiationCl−Cl → Cl· + Cl·
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:(CH3)3−C−CH2−CH3InitiationCl−Cl → Cl· + Cl·Propagation
In free radical halogenation, the propagation step involves the halogen radical pulling a hydrogen atom off your alkane to form an alkyl radical. Your alkane has many different hydrogens, so draw the alkane's structure to see how many different alkyl radicals you can get.Alkane's structure:(CH3)3−C−CH2−CH3InitiationCl−Cl → Cl· + Cl·PropagationBased on our alkane's structure, we get one distinct alkyl radical if one of the hydrogens on carbon 1 or either of the methyls is pulled off:
(CH3)3−C−CH2−CH3 + Cl· → ·CH2(CH3)2−C−CH2−CH3 + HCl
We get a second distinct alkyl radical if one of the hydrogens on carbon 3 is pulled off:
(CH3)3−C−CH2−CH3 + Cl· → (CH3)3−C−·CH−CH3 + HCl
And we get a third distinct alkyl radical if one of the hydrogens on carbon 4 is pulled off:
(CH3)3−C−CH2−CH3 + Cl· → (CH3)3−C−CH2−CH2· + HCl
The combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features. CH3-CH2-CH3 + 5 O2 ——> 3 CO2 + 4 H2O + heat
From the previous discussion, we might expect isomers to have identical heats of combustion. However, a few simple measurements will disabuse this belief. Thus, the heat of combustion of pentane is –782 kcal/mole, but that of its 2,2-dimethylpropane (neopentane) isomer is –777 kcal/mole. Differences such as this reflect subtle structural variations, including the greater bond energy of 1º-C–H versus 2º-C–H bonds and steric crowding of neighboring groups. In small-ring cyclic compounds ring strain can be a major contributor to thermodynamic stability and chemical reactivity. The following table lists heat of combustion data for some simple cycloalkanes and compares these with the increase per CH2 unit for long chain alkanes.
The chief source of ring strain in smaller rings is angle strain and eclipsing strain. As noted elsewhere, cyclopropane and cyclobutane have large contributions of both strains, with angle strain being especially severe. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane. Some examples are shown in the following diagram. The cyclopropane reactions are additions, many of which are initiated by electrophilic attack. The pyrolytic conversion of β-pinene to myrcene probably takes place by an initial rupture of the 1:6 bond, giving an allylic 3º-diradical, followed immediately by breaking of the 5:7 bond. Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride. The following facts must be accomodated by any reasonable mechanism for the halogenation reaction.