IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

dev_22oct

ans

A>C>B>D

reason:for A aliphatic amines are stronger base than the aromatic amines(K_b = 10^-4)

For C due to -I effect of oxygen e^- w'll be pulled decreasing the basicity(K_b=2 x10^-6)

For B pyridine has a pair of electron that are available for sharing with acids. (K_b =2.3 x 10^-9)

For D in pyrrole e^-s are not available for acids , as they are delocalized.

(10^-14)

tarinbansal

I think it shud be A>B>C>D

In C there is O which is decreasing the basicity, whereas there is nothing present in B which can decrease its basicity. Even the l.p. of N in B is not involved in conjugation, so its easily available.
So, it means B>C.
And, in D, due to conjugation of lp of N, it is least basic and aliphatic amines are more basic than aromatic ones, so A is most basic.

So the overall order shud be A>B>C>D.

gaurav_2351

its a>c>b>d

dev_22oct

well tarin your ans is actually wrong, i was right. i gave a little thinking and i came to the conclusion that my ans is absolutely correct. check my previous post it has all the K_b values.

eistien

sir i have a doubt but i might have made a mistake but plz coorect me if A is greater than B obviously C must also be greater than B no why is B>C please explain sir and tarin bansal

i think the order is either

A>C>B>D

or

B>A>C>D

plz explain sir and why i am wrong!!!!????

tarinbansal

A>B
becoz, sp2 C has slight electron withdrawing effect which is not their in aliphatic(sp3) C.

B>C
Becoz, the electron pulling effect of O>>>> than that of sp2 C.
And there is no conjugation of lp of N in pyridine. some ppl think that the lp of N is in conjugation with double bond but there is no conjugation. So lp is available.

And hence the ans is A>B>C>D.

eistien

hey but conjugate acid of B is highly stable due to armaticity no???? what about that factor???

tarinbansal

No yaar. Not at all.
What R U saying?????????

How can +ve charge on sp2 hybridised N be stable? What is the use of aromaticity when there is no required conjugation.

If we consider ur point that, Ph+ must be very stable but it is HIGHLY unstable becoz the +ve charge is not in conjugation and secondly it is there on an sp2 hybridised C.

eistien

oh ok i accept!!!! you are rite tarin!!!!

Sushmi

Ok thank u all of u but answer given is a>c>b>d ...

However is Tarin correct? Wats d final conclusion?

eistien

even i thought A>C>B>D in my final post but looking at tarins explanation i am not able to clearly state the answer may be he might also be right!!!

anandghegde

C is a thousand times more basic than B.....

magia

the answers A>C>B>D

Comparing between C & A:
C has the -I of oxygen acting, therefore the basicity will be less.

B is less basic than A&C, since it has more s character.

D is very weakly basic since it loses its aromaticity on donating its lp.

Note: the answer is correct, we had this as a class illustration.

iberis22

Dev sir is correct.

The correct order is A>C>B>D

In A, B, C the lone pair of nitrogen is not involved in resonance and is available.

In D the lone pair is involved in resonance and is not available and is least available so D is least basic.

A will be most basic.

Comparing B and C; in B, nitrogen is sp²(33% s character) hybridised and in C it is sp³ (25% s character) hybridised. More the s character, more strongly it will hold the lone pair and hence less will be the availaibilty of lp and so less basic. Hence, C is more basic than B.

so, the order is A>C>B>D

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