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Order for solubility of isomeric alcohols in a book was given as:

Primary alcohol >Secondary alcohol >Tertiary alcohol After this I found a question as :

1) Which is more soluble in water?

(a) 3-ethyl-3-hexanol

(b) 2-octanol

Ans:- (a) & the reason was that (a) has more compact alkyl portion than (b). But as (a) is a tertiary alcohol while (b) is a secondary one ,so according to the given order answer should be reverse.

Also while comparing

2-methyl-2-propanol & 1-butanol

Here 1-butanol had higher solubility which matched the given order. I am not understanding what's wrong about the first case. Please explain me.

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The reasons that you cite are both important for the solubility of an alcohol. Generally speaking, if the molecule is branched its (hydrophobic) surface exposed to water is less than in a linear molecule. This effect is more pronounced when the number of atoms is bigger. On the other hand the effect of the degree of substitution on the solubility is less pronounced when the molecular size rise.

This would partially justify your second question, but the problem here is that your second question is wrong, 2-methyl-2-propanol is soluble with water in every ratio, while n-butanol is not.

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  • $\begingroup$ So what is the final conclusion? Where should I use that book's statement? $\endgroup$ – Avi Jan 31 '17 at 4:06
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    $\begingroup$ The statement primary > secondary > tertiary is valid when comparing ethanol > isopropanol > t-butanol or similar series. The hydrophobic surface in the series stays relatively unchanged this is why we can say that the effect on the solubility is caused by the degree of substitution of the alcohol. In the comparison between alcohols with the same number of carbons branched > linear is a much more important factor. $\endgroup$ – user288431 Jan 31 '17 at 14:01

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