7
$\begingroup$

Have the various $\ce{C-C}$ bonds in straight chain hydrocarbons, have the same dissociation energies? Or does it depend on the location of the carbon in the chain? In other words, are the terminal bonds weaker or stronger than the internal bonds?

$\endgroup$
8
$\begingroup$

The terminal bonds are stronger than the internal bonds.

In simple hydrocarbons where strain or electronegative substituents are absent, the bond strengths will correlate with the stability of the resultant radicals that would be formed when the bond breaks.

Because radicals are electron deficient species (they are one electron short of an octet) they are stabilized by electron donating substituents (alkyl groups for example). Hence, a tertiary radical is more stable than a secondary radical which is more stable than a primary radical which is more stable than a methyl radical.

Using this information we can see that $$\ce{R-CH2. + .CH2-R}$$ is more stable than $$\ce{CH3. + .CH2-R}$$ Hence the internal bond will have a lower bond dissociation energy and break more readily.

$\endgroup$
  • $\begingroup$ then another question arises: Which of the R-CH2 radicals is more stable? (R=C2 or R=C3)? or (R=C3 or R=C4)? and so on $\endgroup$ – Mohammad Askarian Sep 4 '16 at 16:16
  • $\begingroup$ Does R=C2 mean that R is an ethyl group and R=C3 is an isopropyl group and R=C4 is a t-butyl group? If so, then this earlier answer may provide some guidance. Inductive effects are likely similar for R=C2, C3 and C4. Resonance effects would be different due to hyperconjugation and they would favor R=C2 as explained in the earlier answer. However, the effects are very small in the earlier answer where a positive charge is being developed. They would be even smaller here since we $\endgroup$ – ron Sep 4 '16 at 17:13
  • $\begingroup$ are dealing with a radical rather that a carbocation. $\endgroup$ – ron Sep 4 '16 at 17:13
  • $\begingroup$ R=C2, R=C3 and R=C4 means ethyl, n-propyl and n-butyl group, respectively. $\endgroup$ – Mohammad Askarian Sep 6 '16 at 5:44
  • $\begingroup$ C2 might be a bit more stable than C3, no significant difference between C3 and C4 $\endgroup$ – ron Sep 6 '16 at 14:41
3
$\begingroup$

According to here:

There are three places where a butane molecule ($\ce{CH3-CH2-CH2-CH3}$) might be split. Each has a distinct likelihood:

48%: break at the $\ce{CH3-CH2}$ bond.

$\ce{CH3* / *CH2-CH2-CH3}$

38%: break at a $\ce{CH2-CH2}$ bond.

$\ce{CH3-CH2* / *CH2-CH3}$

This seems to suggest that internal bonds are weaker than terminal bonds (there are two terminal bonds, so we need to divide 48% by 2).


From data that I gather:

  • The enthalpy of formation of methyl radical is 146 kJ/mol (source)
  • The enthalpy of formation of ethyl radical is 119 kJ/mol (source)
  • The enthalpy of formation of propyl radical is 100 kJ/mol (source)

Therefore:

$$\ce{CH3*+*CH2CH2CH3 -> CH3CH2*+*CH2CH3 (\Delta H= -8kJ/mol)}$$

Again showing that the internal bond is weaker.


Surely this only applies to butane.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.