Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:
, 2-bromo-<em>2</em>-methylbutane;
, 2-bromo-<em>1</em>-methylbutane.
It is expected that
would end up being the dominant product.
Explanation
Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions
and negatively-charged bromide ions
.

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbo<em>cation</em> with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:
;
.
The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has <em>three</em> carbon-carbon single bonds after the addition of the proton
as opposed to <em>two</em> as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.
Both carbocations would then combine with bromide ions to produce a neutral halocarbon.
The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-<em>2</em>-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.
Umm. False i guess I don’t really know
Answer:
Lithium (Li)
Explanation:
Lithium has an atomic number of 3
Do you have a picture of the graph?
Covalent bonds do not involve electron transfer because in covalent bonds, the electrons are shared between the atoms to form the molecule rather than being transferred to form ions as it is in the case of ionic bonds.<span />