Question

Could someone please explain (ii)* to me?

I don't get part ii), all of it is Greek to me TT I don't know why tertiary carbocations are more stable, how methyl groups stabilise charge, no idea what inductive effect or steric hindrance means. Could someone pls pls help me out here?

Answer 1

ISOMER 1: n-bromobutane

The primary (`1^@`) bromobutane (`"C"_4"H"_9"Br"`) has the electrophilic carbon marked below.

`"H"_3"C"-"CH"_2-"CH"_2-stackrel(delta^+)(stackrel("*")"C")"H"_2-stackrel(delta^-)"Br"`

You said that you are using `"OH"^(-)(aq)`, which would do a nucleophilic backside-attack on the primary carbon marked `"*"`.

Due to the low steric hindrance around that carbon (no non-H substituents are on `"C"^"*"`), it is susceptible to a second-order substitution (`"S"_N2`).

Therefore, the nucleophile and substrate will both participate in a reaction that has no intermediate, giving a rate law of:

`r(t) = k["OH"^(-)]["C"_4"H"_9"Br"]`

The reason why an intermediate doesn't form is that primary carbocations are very unstable.

The electron density from a `"C"-"H"` bond would spread out via hyperconjugation to stabilize the positive charge on the central carbon as shown below by sending electron density into an empty `2p_z` orbital:

The more alkyl groups you have surrounding the `"C"^"*"`, the more stabilized it is because alkyl groups are electron-releasing (electron-donating, giving a so-called "inductive effect").

Instead of an intermediate, a transition state forms, which is simply going to look halfway between the reactant and the product. It will be a trigonal bipyramidal geometry around the central carbon!

`" "" "" "" "" "" "" "" "color(white)(.)"Br"`
`" "" "" "" "" "" "" "" "color(white)(.)vdots`
`"H"_3"C"-"CH"_2-"CH"_2-stackrel("*")"C"^(delta^+)"H"_2`
`" "" "" "" "" "" "" "" "color(white)(.)vdots`
`" "" "" "" "" "" "" "color(white)(....):ddot"O":^(delta^(-))`
`" "" "" "" "" "" "" "" "" ""|"`
`" "" "" "" "" "" "" "color(white)(.....)"H"`

Of course, you should draw it with the proper bond angles. Be sure to include the partial charges.

ISOMER 2: tert-butyl bromide

You can tell then that tert-butyl bromide (`("H"_3"C")_3"CBr"`) is more substituted.

`" "" "" ""CH"_3`
`" "" "" ""|"`
`"H"_3"C"-"C"-"Br"`
`" "" "" ""|"`
`" "" "" ""CH"_3`

It has a tertiary (`3^@`) central carbon, so it has more steric hindrance, i.e. it has more things blocking a nucleophile from coming in for a backside-attack, so only one molecule can participate in the reaction (the substrate).

Hence, a first-order substitution (`"S"_N1`) will occur, and a planar carbocation intermediate will form, giving a rate law of:

`r(t) = k_1[("CH"_3)_3"CBr"]`

(where the first step is slow)

while the intermediate looks like this:

`" "" "" ""CH"_3`
`" "" "" ""|"`
`"H"_3"C"-"C"^(+)`
`" "" "" ""|"`
`" "" "" ""CH"_3`

(Of course, you should be drawing this with the proper bond angles!)

This is stabilized by the three `"CH"_3` electron-releasing alkyl groups around the central carbon, spreading electron density out to stabilize the positive charge via hyperconjugation as before.

This stabilizing effect is much stronger than in a primary carbocation.

[Being planar, the intermediate will lead to a racemic mixture of `R//S` stereoisomers if the alkyl group(s) around the central carbon are all different.]