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zysi [14]
3 years ago
13

Which solvent, water or hexane (C6H14), would you choose to dissolve each of the following?1. NaCl2. HF3. Octane4. (NH4)2SO4

Chemistry
1 answer:
hjlf3 years ago
3 0

Answer:

            Water for;

                            (a) NaCl

                            (b) HF

                            (c) (NH₄)₂SO₄

            Hexane for;

                             (a) Octane

Explanation:

                     In chemistry there is a Rue of Thumb, "<em><u>Like Dissolves Like</u></em>" according to which polar compounds will dissolve in polar solvents and non-polar compounds will dissolve in  non-polar solvents.

In given statement two solvents are given.

             (1) Water: Water is a polar compound. It has a polar bond and a bent shape, hence has a net dipole moment.

             (2) Hexane: Hexane is nonpolar because the electronegativity difference between the carbon atom and hydrogen atom is very less i.e. 0.4.

Therefore, polar solutes like NaCl (ionic), HF (highly polar) and (NH₄)₂SO₄ (ionic compound) will dissolve in water and Octane (C₈H₁₈ nonpolar) will dissolve in hexane respectively.

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Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene
kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}, 2-bromo-<em>2</em>-methylbutane;
  • \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}, 2-bromo-<em>1</em>-methylbutane.

It is expected that \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3} 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 \text{H}^{+} and negatively-charged bromide ions \text{Br}^{-}.

\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}

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:

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3};
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}.

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 \text{H}^{+} 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.  

  • \text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}
  • \text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}

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.

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