neutralization

There is no reaction (no neutralization) between NaOH and 2-methylpropan-1-ol. Therefore, no equation can be written.

In HKAL, there is an experiment of determination of the partition coefficient (also known as distribution coefficient) for ethanoic acid between water and 2-methylpropan-1-ol. The experiment includes the titration of aqueous ethanoic acid solution against aqueous NaOH solution, and the titration of the ethanic acid in 2-methylpropan-1-ol solution against aqueous NaOH solution. If you are talking about this experiment, the titration will be between NaOH and ethanoic acid in 2-methylpropan-1-ol (the solvent), and the equation will be :
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
or H+(aq) + OH-(aq) → H2O(l)
The neutralization occurs in the aqueous layer. In the titration, the ethanoic acid diffuses to the aqueous layer, where the neutralization occurs. Therefore, water is added to the conical flask before the titration, and the reaction mixture has to be swirled well during the titration.

2007-04-22 12:52:13 補充：
The basic strength of NaOH is not strong enough to neutralize 2-methylpropan-1-ol.

2007-04-22 12:52:39 補充：
Pure 2-methylpropan-1-ol (or methylpropan-1-ol) can act as an acid only when it reacts with a very reactive metal (e.g. Na). However, the reaction rate is slow.2CH3CH(CH3)CH2OH 2Na → H2 2CH3CH(CH3)CH2O^-Na^

2-methylpropan-1-ol, also called isoputyl alcohol = C4H10O

Sodium Hydroxide = NaOH

As they both have the OH group on the molecule... I'm not sure if it's possible to create a neutralization reaction under normal circumstances.

A Neutralization reaction occurs when a base is reacted with an acid to form water and a salt.