Introduction to organic chemistry

Why carbon shares instead of transferring electrons. Carbon's electronic configuration is $1s^2\,2s^2\,2p^2$ — four electrons short of neon's stable octet by loss and four short of it by gain. Losing four electrons to form $C^{4+}$ would need an enormous amount of energy (nothing removes four electrons from a small, tightly-held $2s^22p^2$ shell cheaply); gaining four to form $C^{4-}$ is equally unfavourable, since cramming four extra electrons onto so small a nucleus faces huge electron-electron repulsion. Carbon instead reaches a stable octet the third way: it shares all four valence electrons through covalent bonds. This is tetravalence, and it is true of essentially every stable carbon compound.

Why carbon bonds to itself so well. No other element matches carbon's catenation — long chains, branches and rings of atoms bonded only to each other. Two facts explain it: the C–C bond is unusually strong (bond enthalpy $\approx 348\ \text{kJ mol}^{-1}$, comparable to C–O or C–N), and carbon atoms are small, so neighbouring orbitals overlap well. Silicon sits directly below carbon and also self-bonds, but the longer, weaker Si–Si bond limits silicon chains to only a handful of atoms before they break down — catenation on carbon's scale is unique.

Why hybridisation happens at all. In its ground state, carbon has only two unpaired electrons ($2p^2$), which would predict just two bonds — but carbon almost never forms two bonds. Before bonding, one $2s$ electron is promoted to the empty second $2p$ orbital, giving four unpaired electrons; these four orbitals then mix mathematically into four new, energetically equivalent hybrid orbitals. How many $p$ orbitals mix in depends on how many $\pi$ bonds the carbon must also form:

where: the bond angle is the angle between two $\sigma$ bonds meeting at that carbon, and $\pi$ bonds form from unhybridised $p$ orbitals left over after hybridisation (one unhybridised $p$ orbital per $\pi$ bond needed). A useful trend follows directly: as more $s$-character goes into the hybrid orbital ($sp^3 \to sp^2 \to sp$, i.e. 25% $\to$ 33% $\to$ 50% $s$-character), the hybrid orbital is held closer to the nucleus, so bond length falls and bond strength rises in that same order.

Question: Buta-1,3-diene, $\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2$, is the next molecule after propene in this family. Identify the hybridisation of each carbon, predict the bond angles, and predict whether the central $C_2$–$C_3$ single bond should be longer, shorter, or the same length as the C–C single bond in ethane (154 pm).

Step 1. Map the bonding at each carbon. $C_1$: double bond to $C_2$, plus 2 H atoms (2 $\sigma$ + 1 $\pi$ to $C_2$). $C_2$: double bond to $C_1$, single bond to $C_3$, plus 1 H (3 $\sigma$ + 1 $\pi$ total). $C_3$: single bond to $C_2$, double bond to $C_4$, plus 1 H (3 $\sigma$ + 1 $\pi$ total). $C_4$: double bond to $C_3$, plus 2 H atoms.

Step 2. Assign hybridisation from the $\pi$-bond count. Every carbon here forms exactly one $\pi$ bond (one double bond each), so all four carbons are $sp^2$ — none is $sp^3$ or $sp$, unlike propene where $C_1$ was $sp^3$.

Step 3. Predict bond angles. Since all four carbons are trigonal planar $sp^2$, every bond angle at every carbon (H–C–H, H–C=C, C=C–C) is approximately $120^\circ$, and the whole carbon backbone is planar.

Step 4. Predict the central bond length using the s-character trend. The $C_2$–$C_3$ bond is a single ($\sigma$-only) bond, but it joins two $sp^2$ carbons rather than two $sp^3$ carbons. Since $sp^2$ orbitals carry more $s$-character (33%) than $sp^3$ orbitals (25%), an $sp^2$–$sp^2$ single bond should be shorter than an $sp^3$–$sp^3$ single bond.

Answer: All four carbons in buta-1,3-diene are $sp^2$, every bond angle is about $120^\circ$, and the central $C_2$–$C_3$ single bond (measured experimentally at about 148 pm) is indeed shorter than the 154 pm C–C bond of ethane, exactly as the s-character trend predicts.

• Believing carbon "wants" to lose or gain four electrons like an ionic-bond-forming metal or halogen — it always shares electrons through covalent bonds instead.
• Assuming hybridisation depends on how many atoms are attached rather than how many $\sigma$ bonds and $\pi$ bonds are formed; count bonds, not neighbouring atoms.
• Treating every C–C single bond as the same length regardless of the hybridisation of the carbons it joins, when in fact an $sp^2$–$sp^2$ single bond (as in buta-1,3-diene) is measurably shorter than an $sp^3$–$sp^3$ single bond.
• Rounding $109.5^\circ$, $120^\circ$ and $180^\circ$ as if they never shift — these are ideal values for identical substituents; real bond angles distort slightly when the groups attached differ in size or electronegativity.

Ask the tutor: Want to work through the hybridisation of every carbon in a bigger molecule, like isoprene or benzene, to build speed for the naming and isomerism topics coming next?