- A(empty) and electron delocalisations
- Band electron delocalisations
- C(filled) and electron delocalisations
- Dp (filled) and electron delocalisations
View written solutionFree
Correct answer: A
Introduction to Hyperconjugation
Hyperconjugation is a stabilizing interaction that involves the delocalization of electrons of a C-H or C-C bond into an adjacent empty or partially filled p-orbital or a -orbital. It is also known as "no-bond resonance". The stability it provides is related to the number of contributing hyperconjugative structures, which in turn depends on the number of -hydrogens.
Step-by-Step Solution
1. Stability of tert-butyl cation
- Structure: The tert-butyl cation consists of a central carbon atom with a positive charge, which is hybridized. This hybridized carbon atom has an empty p-orbital perpendicular to the plane of the three C-C bonds.
- Condition for Hyperconjugation: This empty p-orbital is adjacent to three methyl groups (). The C-H bonds in these methyl groups are bonds. The carbons of the methyl groups are called -carbons.
- Mechanism: The electrons from the C-H bonds of the methyl groups can delocalize into the adjacent empty p-orbital of the positively charged carbon atom. This delocalization stabilizes the carbocation.
- Orbital Interaction: The interaction involves the overlap of the filled orbital of the C-H bond with the empty
porbital of the carbocation. - Notation: This type of electron delocalization is represented as $$\sigma \to p.
For the tert-butyl cation, there are nine such -hydrogens, leading to nine hyperconjugative structures and significant stability.
2. Stability of 2-butene
- Structure: 2-butene is an alkene with a carbon-carbon double bond (
C=C). The carbons of the double bond are hybridized. TheC=Cbond consists of a bond and a bond. The formation of the bond also results in a corresponding empty antibonding molecular orbital. - Condition for Hyperconjugation: The
C=Cdouble bond is flanked by two methyl groups (). The C-H bonds in these methyl groups are bonds, and they are adjacent to the system. - Mechanism: The electrons from the C-H bonds of the methyl groups can delocalize into the adjacent empty antibonding orbital of the double bond. This delocalization strengthens the C-C single bond and slightly weakens the C=C double bond, leading to overall stabilization of the molecule.
- Orbital Interaction: The interaction involves the overlap of the filled orbital of the C-H bond with the empty orbital of the
C=Cbond. - Notation: This type of electron delocalization is represented as .
For 2-butene, there are six -hydrogens (three on each methyl group), leading to six hyperconjugative structures.
3. Evaluation of Options
- For tert-butyl cation: The stabilization is due to $$\sigma \to p delocalization.
- For 2-butene: The stabilization is due to delocalization.
Let's analyze the given options based on our findings:
- A: $$\sigma \to p and electron delocalisations: This matches our analysis for both tert-butyl cation and 2-butene, respectively. This is the correct option.
- B: and electron delocalisations: Incorrect. delocalization is not possible because the bonding orbital is already filled.
- C: $$\sigma \to p and electron delocalisations: Incorrect. The carbocation has an empty p-orbital, not a filled one. Delocalization into a filled orbital would be destabilizing.
- D: p (filled) \to \sigma^*$$ and \sigma \to \pi^*$ electron delocalisations: Incorrect. The tert-butyl cation does not have a filled p-orbital to donate electrons from. This describes negative hyperconjugation, which is not relevant here.
Conclusion
The correct description for the hyperconjugative stabilities of tert-butyl cation and 2-butene is $$\sigma \to p and electron delocalisations, respectively.
More from Basics of Organic Chemistry
- Which of the following molecules, in pure form, is(are) unstable at room temperature?2012 · Multiple correct
- When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is . Includes diagram2012 · Numerical
- Which of the given statement(s) about N, O, P and Q with respect to M is(are) correct? Includes diagram2012 · Multiple correct
- Among the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any) is(are)2011 · Multiple correct
- Among the following compounds, the most acidic is2011 · MCQ
- The total number of contributing structure showing hyper-conjugation (involving C-H bonds) for the following carbocation is . Includes diagram2011 · Numerical
- In the Newman projection for 2,2-dimethylbutane, X and Y can, respectively, be Includes diagram2010 · Multiple correct
- The total number of cyclic isomers possible for a hydrocarbon with the molecular formula is2010 · Numerical