A chemistry student mixes 25 milliliters of 3.0 M hydrochloric acid (HCl) with excess sodium hydroxide. The chemicals react acco
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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.
Answer:
Explanation:
Zn + 2HCl = ZnCl₂ + H₂
A ) mole of Zn = 2.65 / 65
= .04 mol
mole of HCl required = .04 x 2 mol
.08 mol
If v be the volume required
v x 6 = .08
v = .0133 liter
= 13.3 cc
B )
volume of gas at NTP :
moles of gas obtained = .04 moles
= 22.4 x .04 liter
= .896 liter
we have to find this volume at given temperature and pressure
= 1.175 liter.
C )
.04 mole of zinc chloride will be produced
mol weight of zinc chloride
= 65 + 35.5 x 2
= 136 gm
.04 mole = 136 x .04
= 5.44 gm