halogenoalkanes and reactivity

right, in a bit of a muddle - my two chemistry books say conflicting things whoch is never useful :eek3:

basically, can someone tell me the pattern and why the reactivity/boiling point of halogenoalkanes increases as you go down group 7...
thanks

Reply 1

Dr_HI

www.4college.co.uk/as/index.php all there :smile:

even if you aren't OCR B it will answer your question on halagenoalkanes

Reply 2

MGIL

Van Der Waals forces increase as well as the size therefore the bp increases too.

Reply 3

Katermerang

Basically there are two things that determine the reactivity of Halgenoalkanes,

Polarity and Bond enthalpy.

Polarity- the Carbon fluorine bond is the most polar, so can attract a nuclephile most easily, so you it would lead you to believe that it is the most reactive.

Bond enthalpy- The carbon -Iodine bond is the easiest to brake, so thats the reason it should be more reactive.

In practise Bond enthalpy is more important than polarity so the reactivity of Halogenoalkanes increases down the group.

Also- the greater size and therefore greater electron and cloud and therfore greater Van der Waals forces is what causes the increase in boiling point.

Reply 4

mchammer

OK so a reactivity increases down the group according to salters.
is this because the halogens have lowr charge density and so form weaker bonds???

Reply 5

FallenPetal

Are you on edexcel?

Cause they copy stuff from the internet.

Reply 6

mchammer

no im doind salters

plllleaaasse someone put this friggin crappy stuff in a straightforward manner haha sorry

Reply 7

LearningMath

mchammer

no im doind salters

plllleaaasse someone put this friggin crappy stuff in a straightforward manner haha sorry

I am salters as well. You need to know this part to explain what happens 'At any one time, it may so happen that an iodine molecule(I-I) has more electrons at one end than the other, a dipole. This will induce dipoles in neighbouring molecule, the more electrons are present the larger the dipole. Dipoles which form randomly like this are called instantaneous dipoles'

These are the C-Hal bonds in the halogenoalkanes, i'll use I-C and Cl-C as an example.

C-F
C-Cl
C-Br
C-I

As you can see, the halogen atoms are getting bigger as you go down the group. Iodine has more electrons than chlorine. So if you have a load of iodoalkanes (Halogenoalkanes with iodine) and a load of chloroalkanes(Halogenoalkanes with chlorine). Then the instantaneous dipoles in the iodoalkanes will be stronger than the instantaneous dipoles in the chloroalkanes, because there are more electrons in I. More energy will have to be put in to break these interactions, so iodoalkanes will have higher boiling point.

It's the same as the part in italics, but just R-Hal molecules rather than Hal-Hal.

Reply 8

LearningMath

mchammer

OK so a reactivity increases down the group according to salters.
is this because the halogens have lowr charge density and so form weaker bonds???

As for reactivity... Flourine is the most electronegative atom in the group, it is the most reactive in the group. Iodine is the least electronegative, it is the least reactive. So when Flourine forms a bond, it forms a very strong covalent bond, it doesnt want to let go :yep:

. Iodine forms weak covalent bonds, the bond will break easily.

C-F is strong because F is very electronegative, the bond wont easily break, so Fluoroalkanes are unreactive

C-I is weak because I isnt very electronegative, the bond will break easily, so Iodoalkanes are reactive.