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mr Goodwill [35]
3 years ago
6

10.Give the possible values for the magnetic quantum number for each of the following orbitals (a) 4p: (b) 3d: (c) 2s (d) 5P

Chemistry
1 answer:
andre [41]3 years ago
5 0

Answer:

(a) 4p: ml = -3, -2, -1, 0, +1, +2, +3

(b) 3d: ml = -2, -1, 0, +1, +2

(c) 2s: ml = -1, 0, +1

(d) 5p: ml = -4, -3, -2, -1, 0, +1, +2, +3, +4

Explanation:

Manetic quantum numbers determine the spatial orientation and number of orbitals in the subshell. ml is from -l to +l.

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Electrophilic addition of hbr to 3-methyl-2-hexene creates an asymmetric center at c-2. what is the product of this reaction?
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This reaction would give rise to two products.

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  • 3-bromo-3-methylhexane.

However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.

The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton \text{H}^{+} and a bromide ion \text{Br}^{-}.

The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.

  • \phantom{\text{H}_3\text{C}-\text{CH}-\;}+\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;}|\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \text{H}\phantom{\text{CH}}\;\;}\text{CH}_3
  • \phantom{\text{H}_3\text{C}-\text{CH}-}\;\;\text{H}\\\phantom{\text{H}_3\text{C}-\;}+\phantom{\text{H}-\;}|\\\text{H}_3\text{C}-\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\\phantom{\text{H}_3\text{C}-\;\;\;}\phantom{\text{CH}-}\;}|\\\phantom{\text{H}_3\text{C}-\;} \phantom{\text{H}}\phantom{\text{CH}}\;\;}\text{CH}_3

Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon.   The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.

Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.

The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.

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