Question

Given the quoted data, which alcohol is the LEAST stable of the three butyl alcohols?

`bb(color(white)(mmmmmmm)"Heat of Combustion"`
`bbul(color(white)(mmmm)"1-Butanol"color(white)(m)"2-Butanol"color(white)(m)"Isobutyl alc."`
`"Test 1"color(white)(mml)3236.20color(white)(mmm)3248.66color(white)(mmml)2663.82`
`"Test 2"color(white)(mml)3833.22color(white)(mmm)2846.88color(white)(mmml)2915.03`
`ul("Test 3"color(white)(mml)2823.51color(white)(mmm)2861.16color(white)(mmml)2992.80)`
`"Averages"color(white)(m)3300color(white)(mmmml)2990color(white)(mmmmm)2860`

Answer 1

Your data are very hard to interpret..........and if you want a complete answer you need to supply a complete question...you ain't even got units quoted.

Explanation:

I suspect that there is no difference in the enthalpies you measured; i.e. the variation in values stems from random error.

For the combustion of butanol, isobutanol, and tert-butanol, we gots....

`C_4H_9OH(l) + 6O_2(g) rarr 4CO_2(g) +5H_2O(l)+Delta`

I suspect that given complete combustion there would be very little difference in the absolute enthalpies. The branched alcohols MIGHT combust LESS efficiently, giving rise to reduced heat output, with respect to `"n-butanol"`, which should combust reasonably completely. Anyway, I cannot decipher your data.

Answer 2

Butan-1-ol is the least stable (highest energy) compound of the three alcohols.

Explanation:

Here is what I think your data are:

`bb(color(white)(mmmmmmm)"Heat of Combustion"`
`bbul(color(white)(mmmm)"1-Butanol"color(white)(m)"2-Butanol"color(white)(m)"Isobutyl alc."`
`"Test 1"color(white)(mml)3236.20color(white)(mmm)3248.66color(white)(mmml)2663.82`
`"Test 2"color(white)(mml)3833.22color(white)(mmm)2846.88color(white)(mmml)2915.03`
`ul("Test 3"color(white)(mml)2823.51color(white)(mmm)2861.16color(white)(mmml)2992.80)`
`"Averages"color(white)(m)3300color(white)(mmmml)2990color(white)(mmmmm)2860`

Now, let's display them on an energy level diagram:

Relative stabilities

All the compounds end up at the same energy level (as `"4CO"_2 + 5"H"_2"O"`), so the difference must be in the stabilities of the compounds.

The less stable compound is at a higher energy level and therefore releases the most energy on combustion.

Thus, the order of stabilities is

`"isobutyl > 2-butyl > 1-butyl"`

Why this order?

It appears to be related to the amount of branching in the alcohols.

If we count the `"C-C"` and `"C-O"` bonds off the middle of the chain as branches, the order is:

`"2 branches > 1 branch > 0 branches"`

Theoretical calculations suggest that the order is caused by overlap between a `"C-C"` σ bonding orbital on one carbon atom with a `"C-C"` σ* antibonding orbital on an adjacent carbon atom (as in `"C-H"` hyperconjugation).

The effect is small but, the more highly branched the molecule, the more of these interactions there will be, and the compound will be more stable.