Draw An Example Of A Saturated Four Carbon Compound

Alkanes and Cycloalkanes

Robert J. Ouellette , J. David Rawn , in Principles of Organic Chemistry, 2015

3.2 Alkanes

Hydrocarbons that have no carbon-carbon double or triple bonds are called saturated hydrocarbons . They have a continuous chain of carbon atoms, and do not have any "branches." They are called normal alkanes. Their structures are often drawn with the carbon chain in a horizontal line.

The names and condensed structural formulas of 20 normal alkanes are given in Table 3.1. The first four compounds have common names. The names of the higher molecular weight compounds are derived from Greek numbers that indicate the number of carbon atoms in the chain. Each name has the suffix -ane, which identifies the compound as an alkane.

Table 3.1. Names of Normal Alkanes

Number of Carbon Atoms	Name	Molecular Formula
1	Methane	CH₄
2	Ethane	C₂H₆
3	Propane	C₃H₈
4	Butane	C₄H₁₀
5	Pentane	C₅H₁₂
6	Hexane	C₆H₁₄
7	Heptane	C₇H₁₆
8	Octane	C₈H₁₈
9	Nonane	C₉H₂₀
10	Decane	C₁₀H₂₂
11	Undecane	C₁₁H₂₄
12	Dodecane	C₁₂H₂₆
13	Tridecane	C₁₃H₂₈
14	Tetradecane	C₁₄H₃₀
15	Pentadecane	C₁₅H₃₂
16	Hexadecane	C₁₆H₃₄
17	Heptadecane	C₁₇H₃₆
18	Octadecane	C₁₈H₃₈
19	Nonadecane	C₁₉H₄₀
20	Eicosane	C₂₀H₄₂

Saturated hydrocarbons in which one or more groups are bonded to a secondary carbon are called branched alkanes. The carbon atom bonded to three or four other carbon atoms is the branching point. The carbon atom attached to the chain of carbon atoms at the branching point is part of an alkyl group. For example, isobutane is the simplest example of a branched alkane. It has three carbon atoms in the main chain and one branch, a –CH₃ group.

Isopentane and neopentane are isomers of pentane. Isopentane is a branched alkane with four carbon atoms in the main chain and one branching methyl group. Neopentane has three carbon atoms in the main chain and two methyl groups bonded to the central carbon. If an alkane does not have branches, it is said to be a normal alkane.

Both normal and branched alkanes have the general molecular formula C_nH_2n+2. For example, the molecular formula of hexane is C₆H₁₄.

Each carbon atom in this normal alkane, where n = 6, has at least two hydrogen atoms bonded to it, which accounts for the 2n in the general formula. Each of the two terminal carbon atoms has another hydrogen atom bonded to it, which accounts for the +2 in the subscript on hydrogen in the general formula.

Classification of Carbon Atoms

Hydrocarbon structures are classified according to the number of carbon atoms directly bonded to a specific carbon atom. We will use this classification in later chapters to describe the reactivity of functional groups attached at the various carbon atoms in a structure.

A carbon atom bonded to only one other carbon atom is a primary carbon atom, which is designated by the symbol 1°. The carbon atom at each end of a carbon chain is primary. For example, butane has two primary carbon atoms. A carbon atom that is bonded to two other carbon atoms is a secondary carbon atom, designated by the symbol 2°. For example, the middle carbon atoms of butane are secondary (Figure 3.1a).

A carbon atom bonded to three other carbon atoms is tertiary, and is designated by 3°. For example, when we examine the structure of isobutane, we see that one of the four carbon atoms is tertiary; the other three are primary (Figure 3.1b). A quaternary carbon atom (4°) is bonded to four other carbon atoms.

Problem 3.1

The following compound is a sex attractant released by the female tiger moth. Classify the carbon atoms in this compound as primary, secondary, or tertiary.

Solution

Each of the two terminal carbon atoms and the branching –CH ₃ group are primary carbon atoms, because each is bonded to only one other carbon atom. The second carbon atom from the left is bonded to two atoms in the chain as well as to the branching –CH₃ group, so it is tertiary. All 14 remaining carbon atoms are bonded to two carbon atoms, so they are secondary.

Problem 3.2

Pentaerythritol tetranitrate is a drug used to reduce the frequency and severity of angina attacks. Classify the carbon atoms in this compound.

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Alkanes and Cycloalkanes

Robert J. Ouellette , J. David Rawn , in Organic Chemistry, 2014

4.2 Alkanes

Saturated hydrocarbons that have a continuous chain of carbon atoms do not have any "branches." They are called normal alkanes. Their structures are often drawn with the carbon chain in a horizontal line.

The names and condensed structural formulas of 20 normal alkanes are given in Table 4.1. The first four compounds have common names. The names of the higher-molecular-weight compounds are derived from Greek numbers that indicate the number of carbon atoms in the chain. Each name has the suffix -ane, which identifies the compound as an alkane.

Table 4.1. Names of Normal Alkanes

Number of Carbon Atoms	Name	Molecular Formula
1	Methane	CH₄
2	Ethane	C₂H₆
3	Propane	C₃H₈
4	Butane	C₄H₁₀
5	Pentane	C₅H₁₂
6	Hexane	C₆H₁₄
7	Heptane	C₇H₁₆
8	Octane	C₈H₁₈
9	Nonane	C₉H₂₀
10	Decane	C₁₀H₂₂
11	Undecane	C₁₁H₂₄
12	Dodecane	C₁₂H₂₆
13	Tridecane	C₁₃H₂₈
14	Tetradecane	C₁₄H₃₀
15	Pentadecane	C₁₅H₃₂
16	Hexadecane	C₁₆H₃₄
17	Heptadecane	C₁₇H₃₆
18	Octadecane	C₁₈H₃₈
19	Nonadecane	C₁₉H₄₀
20	Eicosane	C₂₀H₄₂

Ball-and-stick structures of butane and isobutane.

Both normal and branched alkanes have the general molecular formula C _n H_{2n+ 2}. For example, the molecular formula of hexane is C₆H₁₄.

Each carbon atom in this normal alkane, where n = 6, has at least two hydrogen atoms bonded to it, which accounts for the 2n in the general formula. Each of the two terminal carbon atoms has another hydrogen atom bonded to it, which accounts for the + 2 in the subscript on hydrogen in the general formula.

Classification of Carbon Atoms

A carbon atom bonded to three other carbon atoms is tertiary and is designated by 3°. For example, when we examine the structure of isobutane, we see that one of the four carbon atoms is tertiary; the other three are primary (Figure 4.1b). A quaternary carbon atom (4°) is bonded to four other carbon atoms.

Problem 4.1

One of the components of the wax of a cabbage leaf is a normal alkane containing 29 carbon atoms. What is the molecular formula of the compound?

Sample Solution

For n = 29, there must be (2 × 29) + 2 hydrogen atoms. The molecular formula is C₂₉H₆₀.

Problem 4.2

The following compound is a sex attractant released by the female tiger moth. Classify the carbon atoms in this compound as primary, secondary, or tertiary.

Sample Solution

Each of the two terminal carbon atoms and the branching ─CH ₃ group are primary carbon atoms, because each is bonded to only one other carbon atom. The second carbon atom from the left is bonded to two atoms in the chain as well as to the branching ─CH₃ group, so it is tertiary. All 14 remaining carbon atoms are bonded to two carbon atoms, so they are secondary.

Problem 4.3

The octane number is a scale used to rate gasoline. The octane number of the following compound, called iso -octane, is 100. Classify the carbon atoms of this compound.

Sample Solution

The carbon atoms in all five ─CH ₃ groups are primary. The carbon atom bonded to three ─CH₃ groups near the left side of the structure is also bonded to a ─CH₂ unit. This carbon atom is quaternary. The ─CH₂ and CH carbon atoms are secondary and tertiary, respectively.

Problem 4.4

Pentaerythritol tetranitrate is a drug used to reduce the frequency and severity of angina attacks. Classify the carbon atoms in this compound.

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Alkanes and Cycloalkanes: Structures and Reactions

Robert J. Ouellette , J. David Rawn , in Organic Chemistry (Second Edition), 2018

Classification of Carbon Atoms

A carbon atom bonded to only one other carbon atom is a primary carbon atom, designated by the symbol 1°. The carbon atom at each end of a carbon chain is primary. For example, butane has two primary carbon atoms. A carbon atom that is bonded to two other carbon atoms is a secondary carbon atom, designated by the symbol 2°. For example, the middle carbon atoms of butane are secondary (Figure 4.1a).

A carbon atom bonded to three other carbon atoms is tertiary, and is designated by 3°. For example, when we examine the structure of isobutane, we see that one of the four carbon atoms is tertiary; the other three are primary (Figure 4.1b). A quaternary carbon atom, 4°, is bonded to four other carbon atoms (Figure 4.1c).

Problem 4.1

One of the components of the wax of a cabbage leaf is a normal alkane containing 29 carbon atoms. What is the molecular formula of the compound?

Sample Solution

For n = 29, there must be (2 × 29) + 2 hydrogen atoms. The molecular formula is C₂₉H₆₀.

Problem 4.2

The following compound is a sex attractant released by the female tiger moth. Classify the carbon atoms in this compound as primary, secondary, or tertiary.

Sample Solution

Each of the two terminal carbon atoms and the branching —CH ₃ group are primary carbon atoms because each is bonded to only one other carbon atom. The second carbon atom from the left is bonded to two atoms in the chain as well as to the branching —CH₃ group, so it is tertiary. All 14 remaining carbon atoms are bonded to two carbon atoms, so they are secondary.

Problem 4.3

The octane number is a scale used to rate gasoline. The octane number of the following compound, called isooctane, is 100. Classify the carbon atoms of this compound.

Sample Solution

The carbon atoms in all five —CH ₃ groups are primary. The carbon atom bonded to three —CH₃ groups near the left side of the structure is also bonded to a —CH₂ unit. This carbon atom is quaternary. The —CH₂ group is secondary, the carbon atom bound to three alkyl groups, on the right, is tertiary.

Problem 4.4

Pentaerythritol tetranitrate is a drug used to reduce the frequency and severity of angina attacks. Classify the carbon atoms in this compound.

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Structure and Bonding in Organic Compounds

Robert J. Ouellette , J. David Rawn , in Organic Chemistry, 2014

Multiple Covalent Bonds

A carbon atom forms four bonds in stable organic compounds such as ethane, ethene (ethylene), and ethyne (acetylene).

Each carbon atom in ethane forms four single bonds, one to each of three hydrogen atoms and one to the neighboring carbon atom. However, in some organic molecules, a carbon atom shares two or three pairs of electrons with another bonded atom. If two electron pairs are shared, a double bond exists. For example, ethene has a carbon–carbon double bond. Each carbon atom in ethene forms two single bonds to hydrogen atoms and one double bond to the neighboring carbon atom. If two bonded atoms share three electron pairs, a triple bond exists. For example, each carbon atom in ethyne forms a single bond to a hydrogen atom, and the two carbon atoms share a triple bond. This triple bond contains six electrons. In ethane, ethene, and ethyne, each carbon atom makes a total of four bonds.

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Aldehydes and Ketones

Robert J. Ouellette , J. David Rawn , in Organic Chemistry, 2014

Carbonyl Compounds

When a carbonyl carbon atom is bonded to at least one hydrogen atom, the resulting compound is an aldehyde. The aldehyde with the simplest structure is formaldehyde, in which the carbonyl carbon atom is bonded to two hydrogen atoms. The carbonyl group is bonded to one hydrogen atom and either an alkyl group (R) or an aromatic group (Ar) in other aldehydes. Although the bond angles around the carbonyl carbon atom are approximately 120°, structures are often written with a linear arrangement of carbon atoms.

When a carbonyl carbon atom is bonded to two other carbon atoms, the compound is a ketone. The bonded groups may be any combination of alkyl or aromatic groups. A ketone has 120° bond angles at the carbonyl carbon atom, but structures are often written with a linear arrangement of carbon atoms.

An aldehyde can be written with the condensed formula RCHO or ArCHO, where the symbol CHO indicates that both hydrogen and oxygen atoms are bonded to the carbonyl carbon atom. A ketone has the condensed formula RCOR. In this condensed formula, the symbol CO represents the carbonyl group, and the two R groups flanking the CO group are bonded to the carbonyl carbon atom.

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Condensation Reactions of Carbonyl Compounds

Robert J. Ouellette , J. David Rawn , in Organic Chemistry Study Guide, 2015

22.1 The α Carbon Atom of Carbonyl Compounds

The carbon atom bonded to the carbonyl carbon atom is known as the α carbon atom. It is a reactive site because its hydrogen atom is acidic. Extraction of the α hydrogen atom results in formation of a carbanion that serves as a nucleophile. The p K _a of the α hydrogen atom is approximately 18, which means that the K _a is approximately 30 powers of 10 larger than the K for hydrocarbons. The increased acidity is the result of resonance stabilization of the enolate ion. One of the two resonance forms of the enolate ion has the negative charge on the α carbon atom; the other resonance form has the negative charge on the oxygen atom.

Enolates are formed by reaction of a carbonyl compound with a base. The concentration of enolate formed depends on the K _b of the base, which in turn is related to the K _a of the conjugate acid of the base. Sodium hydroxide is a weaker base than the enolate ion, and it is not sufficiently basic to give a high concentration of enolate. Sodium amide is a much stronger base, and it quantitatively converts carbonyl compounds to their enolates.

Enolates can react as nucleophiles that attack an electrophilic center at the oxygen atom or the carbon atom of a second enolate. Although the electron density of the enolate is highest at the oxygen atom, the most common reaction site of enolates is at the carbon atom. This selectivity is related to the bonds formed in the transition state. Reaction at the oxygen atom forms an enol product that contains a carbon–carbon double bond. Reaction at the carbon atom forms a keto product that contains a carbon–oxygen double bond. The greater stability of the carbonyl group favors formation of the keto product.

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Carboxylic Acid Derivatives

Robert J. Ouellette , J. David Rawn , in Organic Chemistry, 2014

Esters

The reaction of an ester with a Grignard reagent to give a tertiary alcohol that contains two equivalents of the alkyl group of the organometallic reagent is an important synthetic procedure.

The first step of the reaction is addition of the Grignard reagent to the carbonyl group to give a tetrahedral intermediate. It releases an alkoxide ion complexed with magnesium bromide. Thus, the reaction is a typical nucleophilic acyl substitution reaction.

The carbon atom of the carbonyl group of ketones is more electrophilic than the acyl carbon atom of esters. Therefore, a second mole of Grignard reagent adds to the ketone to give a tertiary alcohol.

Problem 21.21

Propose a synthesis of 3-phenyl-3-pentanol using the addition reaction of a Grignard reagent to an ester.

Sample Solution

The carbon atom that bears the hydroxyl group has two ethyl groups and a phenyl group bonded to it.

The two ethyl groups can come from a Grignard reagent that can add to the carbonyl carbon of an ester. The ester must have a phenyl group bonded to the carbonyl carbon atom; it is a benzoate ester. The benzoate ester can contain any alkyl group such as in methyl benzoate or ethyl benzoate. Adding two equivalents of ethylmagnesium bromide followed by addition of aqueous acid gives the product, 3-phenyl-3-pentanol.

Problem 21.22

Explain how symmetrical secondary alcohols of the type R₂CHOH can be prepared by adding a Grignard reagent to an ester.

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Structure of Six-membered Rings

Alan R. Katritzky , ... Viktor V. Zhdankin , in Handbook of Heterocyclic Chemistry (Third Edition), 2010

2.2.3.4.1 Aromatic systems: Chemical shifts

Chemical shift data for a number of monocyclic, unsubstituted six-membered heteroaromatic compounds are given in Table 12 .

Table 12. ¹³C NMR chemical shifts of the simple monocyclic azines (cf. benzene, δ = 128.5)

	δ(¹³C) (ppm, TMS) (position of N atoms indicated)
Position	Pyridine	Pyridazine	Pyrimidine	Pyrazine	1,2,3-Triazine	1,2,4-Triazine	1,3,5-Triazine	1,2,4,5-Tetrazine
1	N	N	N	N	N	N	N	N
2	149.5	N	158.4	145.9	N	N	166.1	N
3	125.6	153.0	N	145.9	N	158.1	N	161.9
4	138.7	130.3	156.9	N	149.7	N	166.1	N
5	125.6	130.3	121.9	145.9	117.9	149.6	N	N
6	149.5	153.0	156.9	145.9	149.7	150.8	166.1	161.9

Ring carbon atoms α to a heteroatom are most heavily deshielded, those γ to a heteroatom are also deshielded relative to benzene, while those in a β-position are more benzene-like. Introduction of a second nitrogen atom α or γ to a ring carbon atom results in further deshielding by ∼10 and 3 ppm, respectively, whereas the effect on a β-carbon atom is a shielding of ∼3 ppm. Substituent effects follow the same general trend as in substituted benzenes. The chemical shifts of ring carbon atoms which either carry the substituent or are para to it differ in a predictable way relative to the unsubstituted heterocycle, whereas the shifts of ring carbon atoms meta to the substituent are little affected. These effects are conveniently exemplified in the pyridine series; typical data for a variety of monosubstituted pyridines are listed in Table 13 . Fusion of an aromatic or heteroaromatic ring to an azine changes the electronic distribution and hence the chemical shifts of remaining ring carbon atoms in the azine portion of the molecule, although the difference from those in the parent azine is usually less than 10 ppm. Shift data for a number of common bicyclic azine systems are given in Figure 15 .

Table 13. ¹³C NMR chemical shifts of monosubstituted pyridines

		δ (¹³C) (ppm)
Substituent position	Substituent	C(2)	C(3)	C(4)	C(5)	C(6)
	H	150.6	124.5	136.4	124.5	150.6
2	Br	142.9	129.0	139.5	123.7	151.0
2	CHO	153.1	121.6	137.5	128.3	150.3
2	CN	133.8	129.2	137.9	127.8	151.5
2	COMe	153.9	121.4	136.9	127.5	149.3
2	Me	158.7	123.5	136.1	120.8	149.5
2	NH₂ ^a	160.9	109.5	138.5	113.6	148.7
2	OH ^a,b	162.3	119.8	140.8	104.8	135.2
2	OMe ^a	163.1	110.5	138.7	116.7	146.6
3	Br	151.7	121.6	139.1	125.4	148.7
3	CHO	152.0	132.1	136.2	124.8	155.0
3	CN	153.2	110.5	140.6	124.8	153.8
3	COMe	150.1	123.9	132.5	121.5	153.8
3	Me	150.9	133.1	136.4	123.4	147.3
3	NH₂ ^a	137.7	145.7	122.0	125.1	138.8
3	OH ^a	137.8	153.5	121.4	123.8	140.0
3	OMe ^a	137.3	155.2	120.0	123.8	141.4
4	Br	152.6	127.6	133.2	127.6	152.6
4	CHO	151.3	123.6	141.7	123.6	151.3
4	CN	151.7	126.4	120.5	126.4	151.7
4	COMe	151.2	121.6	143.0	121.6	151.2
4	Me	150.1	125.0	147.0	125.0	150.1
4	NH₂ ^a	148.5	110.4	155.8	110.4	148.5
4	OH^a,b	139.8	115.9	175.7	115.9	139.8
4	OMe ^a	150.7	109.8	164.9	109.8	150.7

Neat liquids, unless otherwise specified; values for ipso position (carrying substituent) italicized; data from <1972CPB429, 1973OMR(5)551>.

a: In DMSO-d ₆.
b: Compound in NH form.

Protonation of azines results in shielding of the α carbon atoms and deshielding of the β- and γ-carbon atoms ( Table 14 ), particularly the latter, and these effects have been accounted for in terms of additivity parameters. The upfield protonation parameter for the α-carbon atom has been assigned to changes in the C–N bond order, while the β- and γ-parameters have been assigned to charge-polarization effects. The parameters are highly reproducible for monoprotonation but deviate significantly from additivity for diprotonated heterocycles. A related effect is observed on quaternization, but in this case the operation of a β-substituent effect results in the overall change at the α-carbon atom normally being small ( Table 14 ).

Table 14. ¹³C NMR chemical shifts (δ, ppm from TMS) and one-bond ¹³C–¹H coupling constants (Hz) of some simple heterocyclic cations (cf. pyridine, column 2)

Position	Chemical shift (coupling constant)
1	N	O⁺	NH⁺	NMe⁺	NPh⁺	S⁺
2	150.3 (178)	169.3 (218)	148.3 (192)	145.8	145.3 (191)	158.8
3	124.3 (162)	127.7 (180)	128.6 (173)	128.5	129.5 (178)	138.3
4	136.4 (162)	161.2 (180)	142.2 (173)	145.8	147.8 (174)	150.8
Anion	–	C1O₄ ⁻	CF₃CO₂ ⁻	I⁻	Cl⁻	BF₄ ⁻
Solvent	DMSO-d ₆	CD₃CN	D₂O	DMSO-d ₆	D₂O	CD₃CN
Ref.	1970MP573	1973OMR(5)251	1980CCC2766	1976OMR(8)21	1980CCC2766	2003JFC(120)49
	1976OMR(8)21	1977OMR(9)16	1970MI20100			2001EJO2477

A further important general trend in the azines arises on N-oxidation, which results in shielding of the α- and γ-carbon atoms, especially the former, and clearly indicates the higher electron density at these ring positions ( Figure 16 ). The corresponding conjugate acids of the N-oxides have chemical shifts very similar to those of the protonated parent heterocycles.

The effect of benzo substitution on thiinium tetrafluoroborate is shown in Figure 17 . In a study of thiinium salts with different counterions (BF₄ ⁻, BPh₄ ⁻, I⁻, TfO⁻) and solvents (CD₃CN, DMSO-d ₆) it was evident that these changes have the least effect on the C(2) chemical shift <CHEC-III(7.10.2.2.3)775>.

The pyrones and thiinones show general ¹³C NMR spectral characteristics similar to the pyridones which reflect charge distributions in the heterocyclic rings. Thus, carbon atoms α or γ to the heteroatom are deshielded relative to benzene, while those β are shielded. Substituent effects are in general as expected, although fewer detailed studies have been carried out in this area with the oxygen and sulfur heterocycles than with the azines. Chemical shift data for representative oxygen and sulfur compounds are given in Figure 18 .

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Carbon in β-rhombohedral boron

Helmut Werheit , ... Torsten Lundström , in C,H,N and O in Si and Characterization and Simulation of Materials and Processes, 1996

Abstract

Carbon atoms in the β-rhombohedral boron structure substitute for boron atoms at the polar sites of the B ₁₂ icosahedra. The donor level of the excess electron of carbon coincides with the upper valence band, thus raising the hopping probability within this band. This assumption is consistent with various experimental results. The ionisation energy of the intrinsic electron trapping levels is slightly different for B₁₂ and B₁₁C icosahedra. The photoconductivity and quotient spectra of the reflectivities of differently doped samples indicate carbon-induced changes in the density of states within the energy bands.

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Physics and Fundamental Theory

Y. Iye , in Comprehensive Semiconductor Science and Technology, 2011

1.10.1.2 Bond and Local Coordination

Carbon atom has four valence electrons in the 2s and 2p atomic orbitals. The carbon–carbon (C–C) bond can be either sp ³- or sp²-hybridized bond. The richness of carbon allotropes stems from various combinations of the sp³- and sp²-hybridized bonds.

In diamond crystals, each carbon atom is tetrahedrally coordinated. Strong C–C bond is formed with each of the neighboring four atoms by sp³-hybridization. The C–C bond length in diamond is 0.154 nm. The crystal structure is cubic with lattice constant 0.357 nm. The diamond lattice can be viewed as consisting of two face-centered cubic (f.c.c.) sublattices of which one is shifted by $(\frac{1}{4} \frac{1}{4} \frac{1}{4})$ relative to the other.

Carbon atoms in a graphitic sheet form a honeycomb network so that each atom is trigonally coordinated. Three of the four valence electrons are used for the sp²-hybridized σ-bonds with the three neighbors. The remaining one is in the π-orbital and can be easily delocalized. The C–C bond length in the graphitic sheet is 0.142 nm. The three-dimensional (3D) graphite is made of a stack of graphitic sheets. The ordinary graphite (Bernal graphite) exhibits the stacking order ABAB…. A thermodynamically unstable variant called rhombohedral graphite has a stacking sequence ABCABC….

In the case of amorphous carbon, the C–C bonds consist of mixture of sp³- and sp²-hybridized bonds. Actually, one important parameter that characterizes the amorphous carbon is the ratio of sp³- and sp²-hybridized bonds. Materials that are high in sp³-hybridized bonds are referred to as tetrahedral amorphous carbon or as diamond-like carbon owing to the similarity of many physical properties to those of diamond. Raman spectroscopy provides a useful means of characterization. The Raman spectrum of diamond shows a sharp peak at ω = 1332 cm⁻¹, while the frequency of the Raman-active in-plane mode of graphite is ω =1582 cm⁻¹. The Raman spectra of amorphous carbon generally exhibit broadened mixture of these two peaks.

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