Molecules having higher enthalpy of vaporization and boiling point than another but with lower entropy of vaporization?
The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes.  For low-molecular weight alcohols, this effect is pronounced.  The longer the alkane chain becomes, the more the compound behaves like a pure alkane.  For example, using data obtained from the NIST Webbook:  

Table 1. Enthalpy of vaporization in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  42}$
$\ce{n-propanol\ \  47}$
$\ce{n-butanol\ \ \ \  52}$
$\ce{n-pentanol\ \  57}$
$\ce{n-hexanol\ \ \ \  61}$
$\ce{n-heptanol\ \  67}$
$\ce{n-octanol\ \ \ \  71}$
$\ce{n-nonanol\ \ \  77}$
$\ce{n-decanol\ \ \ \  82}$  

Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like and more alkane-like.  This is much more obvious in the next table, which is simply Table 1 normalized to a per-carbon basis:  

Table 2. Enthalpy of vaporization on a per-carbon basis in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  21}$
$\ce{n-propanol\ \  16}$
$\ce{n-butanol\ \ \ \  13}$
$\ce{n-pentanol\ \  11}$
$\ce{n-hexanol\ \ \ \  10}$
$\ce{n-heptanol\ \  10}$
$\ce{n-octanol\ \ \ \  9}$
$\ce{n-nonanol\ \ \  9}$
$\ce{n-decanol\ \ \ \  8}$  

Table 2 even more drastically illustrates the deviation in $\mathrm{\Delta}H$ changes as we move from alcohol-like compounds to alkane-like compounds. So, given that the enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes, this should explain your findings.