Ochem question

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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





D) (CH 3 ) 3C-Br





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?
 
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.
 
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.

Thanks!
so steric hindrance depends on size also? i thought it was only on the geometry
 
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.
 
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
 
Last edited:
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

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
 
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

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
 
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

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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