[savy2020]

Hi,
Please help me with this question

Question: Which of the following statements is true with respect to the following pair of compounds?
(i) D is more basic than E
(ii) E is more basic than D
(iii) D and E are of comparable basicity.

My explanation: Since both are tertiary amines I've no reason to prefer one over the other in deciding basic strength.
So I go with (iii)

But that's not the answer given.
Any opinions please.

[sjb]

Both are tertiary amines, true, but how available is the lone pair to accept protons, or other (Lewis) acids?

[savy2020]

Both are tertiary amines, true, but how available is the lone pair to accept protons, or other (Lewis) acids?

Well, {may be} will there be any steric hindrance in D due to which lone pair is not so easily available for other lewis acids to bond with?

[handsome_21]

you can dissolve them with water and check the PH value of each, I think at least E is miscible.
From the angle of hindrance, I think D is more basic than E

Hi,
Please help me with this question

Question: Which of the following statements is true with respect to the following pair of compounds?
(i) D is more basic than E
(ii) E is more basic than D
(iii) D and E are of comparable basicity.

My explanation: Since both are tertiary amines I've no reason to prefer one over the other in deciding basic strength.
So I go with (iii)

But that's not the answer given.
Any opinions please.

[savy2020]

you can dissolve them with water and check the PH value of each, I think at least E is miscible.

Oh I didn't know that 
If asked in an examination, I'll get out of the examination hall go to lab, synthesise those chemicals, disolve in water and find their pH. RIght??

From the angle of hindrance, I think D is more basic than E

Are you kidding?

[stewie griffin]

Don't ask for help if you're not willing to listen and give others a chance to help  
(BTW, both of those amines are commercially available and quite common in almost any and every organic synthesis lab. So if you are able to step out of the exam room, at least you'll save some time in not having to synthesize them )
Anyway, sorry for the rant.  
Although you may think that D's lone pairs are more sterically hindered since D simply looks bigger, the lone pairs are actually more available.
Notice that in compound D, that nitrogen can't do an amine inversion. So it's lone pairs are in a sp³ orbital that point straight down from the nitrogen (as you've drawn it in the original post). In compound E, we do have amine inversion. Furthermore, in compound E, those ethyl groups aren't tied back like the alkyl groups in D are. That means those CH₃ groups at then end and those CH₂ are allowed to spin/rotate all over the place. This is a problem b/c it effectively blocks access to E's nitrogen's lone pairs.
Therefore, it doesn't matter that D looks big and mean when it comes to the nitrogen lone pairs.
Although the difference in basicity is probably small, I would answer D is more basic.
Good luck on the final.  

[savy2020]

Don't ask for help if you're not willing to listen and give others a chance to help  .

Therefore, it doesn't matter that D looks big and mean when it comes to the nitrogen lone pairs.
Although the difference in basicity is probably small, I would answer D is more basic.

 
@handsome_21
I am extremely sorry for that.
@stewie griffin
Thanks for your explanation.
I feel guilty about my self.

[handsome_21]

I agree with these words. 

Don't ask for help if you're not willing to listen and give others a chance to help  


(BTW, both of those amines are commercially available and quite common in almost any and every organic synthesis lab. So if you are able to step out of the exam room, at least you'll save some time in not having to synthesize them )
Anyway, sorry for the rant.  
Although you may think that D's lone pairs are more sterically hindered since D simply looks bigger, the lone pairs are actually more available.
Notice that in compound D, that nitrogen can't do an amine inversion. So it's lone pairs are in a sp³ orbital that point straight down from the nitrogen (as you've drawn it in the original post). In compound E, we do have amine inversion. Furthermore, in compound E, those ethyl groups aren't tied back like the alkyl groups in D are. That means those CH₃ groups at then end and those CH₂ are allowed to spin/rotate all over the place. This is a problem b/c it effectively blocks access to E's nitrogen's lone pairs.
Therefore, it doesn't matter that D looks big and mean when it comes to the nitrogen lone pairs.
Although the difference in basicity is probably small, I would answer D is more basic.
Good luck on the final.