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:
(102 900 ÷ 12) + (170 × 1.27) = 8800
Step 1. Evaluate the expressions inside the parentheses (PEMDAS)
102 900 ÷ 12 = 8575
170 × 1.27 = 215.9
In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.
Thus, the underlined digits are not significant, but we keep them in our calculator to avoid roundoff error.
Step 2. Do the addition (PEMDAS).
8575
+ 215.9
= 8790.9
Everything that you add to an insignificant digit gives an insignificant digit as an answer.
Thus, the underlined digits are not significant.
We must drop them and round up the answer to 8800.
Explanation:
2HF + Na2SiO3 ----> 2NaF + H2SiO3 is the balanced equation.
2 moles of HF form 2 moles of NaF
therefore 0.508 Mol HF forms 0.508 mol of NaF
In grams this is 0.508( rel atomic mass of Na + RAM of F)
= 0.508 (22.99+19)
= 21.33 grams
Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more. These molecules then bump into nearby particles and transfer some of their energy to them. This then continues and passes the energy from the hot end down to the colder end of the substance.
Explanation:
Answer:
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Explanation:
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