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dk00

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20. Which of the following molecules will undergo an SN2 reaction the fastest?

A) CH 3 CH 2 --Br​


C) CH3 ( CH2)4CH2-Br





B) CH 3-CHBr-CH3​
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D) (CH 3 ) 3C-Br​
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i know why its not B or D due to the tertiary and secondary steric hindrance but what about A and C dont they both have the same geometry?​
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Molecular weight. A is smaller thus less hindrance and faster to react. Same as the difference in tertiary and secondary, albeit to a smaller degree.
 
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Yes, just to a lesser degree. Imagine its like the molecule is searching a train to find the correct compartment to react with. 4 trains takes longer than 3 trains takes longer than 2 trains takes longer than 1 train. But within that one train, the longer it is the longer it takes still.
 
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one more question​

How many unpaired electrons are there for an element whose
atomic number is 23?​
A) 1 B) 3 C) 5 D) 7


without looking at the periodic table wouldnt this element have 5 valence electrons? 23-2-8-8=5 with 4 bieng paired and 1 unpaired?
- -

correct answer is B
 
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dk00 said:
one more question​

How many unpaired electrons are there for an element whose
atomic number is 23?​
A) 1 B) 3 C) 5 D) 7


without looking at the periodic table wouldnt this element have 5 valence electrons? 23-2-8-8=5 with 4 bieng paired and 1 unpaired?
- -

correct answer is B
Click to expand...

It has to do with electron configuration.

Vanadium [V] (#23): 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Notice how adding all the bold/underlined numbers = 23 electrons. All the orbitals are full up until the 3d orbital, where there are 3 valence electrons. They are all unpaired because 1 electron must be in each orbital before they can begin to pair. 3d usually goes up to 10 (when it's full).

Check out this link for a detailed explanation:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/eleconfg.html
 
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Guitarmacy said:
It has to do with electron configuration.

Vanadium [V] (#23): 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Notice how adding all the bold/underlined numbers = 23 electrons. All the orbitals are full up until the 3d orbital, where there are 3 valence electrons. They are all unpaired because 1 electron must be in each orbital before they can begin to pair. 3d usually goes up to 10 (when it's full).

Check out this link for a detailed explanation:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/eleconfg.html
Click to expand...

Thanks!
so then is it safe to assume that the S orbital electrons pair before the P orbital electrons? and is this the case with all elements and sub shells(S pairs before P, P pairs before D etc.? i was under the assumption that the electrons of an entire shell are paired at the same time
 
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dk00 said:
Thanks!
so then is it safe to assume that the S orbital electrons pair before the P orbital electrons? and is this the case with all elements and sub shells(S pairs before P, P pairs before D etc.? i was under the assumption that the electrons of an entire shell are paired at the same time
Click to expand...

Yes, S orbitals pair before P orbitals, but it is better to memorize the Diagonal/Madelung's Rule instead:

http://www.wyzant.com/Help/Science/Chemistry/Electron_Configuration/

The electron pairing assumption is one that several people make, but it is incorrect. Also, as explained in that link, some heavy metals only have 1 electron in their S orbital and the other is bumped to the next orbital. Ex: 5s1 4d11
 
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