39Is that 1-bromo-2-fluorobenzene? or 4?
1)which one reacts fastest with aq. methanol ?
a. Me3C-Br b. Me3C-Cl
c.Me2CH-Br c. Me-CH2-CH2-Br
2)find the product of the following reaction -
1-bromo-4fluoro-benzene + Mg (+ether) --->
-
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17 Answers
71Don't confuse with EA and Electron Gain Enthalpy.
And I think 1-bromo-4-fluorobenzene should form Grignard Reagent. While 1-2 Substituted benzene does not forms..
262@ pritish - it is 4.
@subhomoy -
how is x >y ?
as EA of Cl is More than EA of Br , x should be less than y
1i hv a confusion regarding the 2nd qstn....
will a Grignard reagent be formed as a product?
1but when v go across a period then size does not change much....so then v hav to look at electronegativity
49I gave thermodynamic picture..
Ashish gave the bond length picture...
Rishab said same thing as me! :)
49First of all.. we need to know, lower the energy content of a species, (lower is the ability to do work), higher is the stability.. (u can easily ignore the part in brackets.. and stop any confusion, if any)..
And also we need to know, that there is no absolute value of energy content in a compound.. but in a reaction, we measure the energy content of the products w.r.t the reactants...(thus, it is ΔHreaction that accounts for stability, and not the meaning less term H) (If you have not done thermodynamics yet, ignore the sentence in brackets)
Also, we need to realise, everything in chemistry drools for stability, i.e. wants to goto a state of higher stability..(lets forget the problem with activation energy for now -- again just a term, do not confuse over this)
And due to this, we can say, stabler the products, more is the tendency of the reactant to convert to the products, and thus the reactants are unstable...
And also, stabler the reactants, less is the stability of products.. (same logic, just remember every reaction is reversible)
Okay so.. back to the original controversy..
R-Cl or R-Br, which of these two break??
R-Cl → R+ + Cl-
R-Br → R+ + Br-
Suppose, R is the same for both..
So which one will be favoured, is determined by the stability of Cl- and Br-..
Now, consider these reactions..
Br + e → Br- ΔH=-x kJ/mol
Cl + e → Cl- ΔH=-y kJ/mol
(question: why is ΔH negative?)
Now, realise that these x and y are the electron-gain enthalpy of Br and Cl..
And thus, as u have said, x>y..
Thus, if we consider w.r.t the reactants.. Br- is stabler than Cl-..
Thus in,
R-Cl → R+ + Cl-
R-Br → R+ + Br-
The second reaction is more spontaneous, owing to the more stability of Br- [1]
And hence, Br breaks up with R faster..(dukhi parivaar :P) [1]
1Methanol means Polar Protic solvent.......means will favor SN1 reaction......thus answer of first should be (a) .. as Carbocation will be most stable ...and Br is a good Leaving group....! ....
1for (2) just add a 'mg' between c-br bond..as c-br bond is weaker than c-f bond
1if -x leaves then it will acquire a negative charge since c-x bond is polar nd e- are more on the -x side.
now Br- is more stable than cl- as bromine is larger than cl.. so the negative charge is distributed and thus stabilised.
where as cl is smaller than br so the negative chege will be concentrated in that small area. so unstable.
=> br is more polarisable due to its large size therefore better l.g.
30Also because of the ineffective sigma overlap right? The orbital sizes of C and Br differ greatly and hence overlap is not effective. Cl having a smaller orbital has a stronger and more effective overlap making C-Cl bond stronger than C-Br!
49Here, it is not about electron-affinity(not directly) ..
It is all about stability of C-Br and C-Cl bonds..
Since, C-Br bond is weaker, we have Br- go out...
Why is C-Cl bond stronger than C-Br bond, answer comes from electron gain enthalpy..
I will be posting the reason shortly!
262that is where my confusion lies.
isnt the electron affinity of Cl more than Br ?
then Cl will be easily able to attract the electrons towards itself comparing with Br
1since it is aq methanol hence the solvent(water) is polar protic.so SN1 will be favoured.
now if we set our eyes on the compounds....den we can see that the intermediates in case of (a) and (b) will be tertiary carbocations....hence hyperconjugation more in case of them.
Now between Br and Cl,Br is a better leaving grp bcoz the size Br is more than that of Cl.hence in Br- anion electronic repulsion will be less hence more stable.so the answr will be (a)