Mistake in finding bond angles in VSEPR theory
Looking at compound D, one of the first things we notice is the carbonyl group. The carbon atom in a carbonyl is $\ce{sp^2}$ hybridized, so angle 6 involves an $\ce{sp^2}$ hybridized carbon. In such cases the $\ce{H-C-O}$ bond angle is ~ 120 degrees.  

Just to elaborate a bit more:


If you have four objects arranged in space about a central carbon atom (the $\ce{sp^3}$ hybridization case), the four bonds from carbon are going to arrange themselves spatially so that the electron - electron repulsions between the four bonds are minimized.  This is achieved by arranging the bonds such that they point towards the vertices of a tetrahedron.  In a case like $\ce{CH_4}$, where the 4 atoms bound to carbon are identical, the C-H bonds adopt the perfect tetrahedral angle of 109.5 degrees. In the case of $\ce{CH_2Cl_2}$ where the four bonds are not equivalent, the bond angles will differ somewhat from 109.5 degrees (can you guess which angle will be greater then 109.5 degrees and which angle smaller) - but still, the tetrahedral angle is a reasonable first approximation.  
In the case of an $\ce{sp^2}$ hybridized carbon, the 3 sigma bonds will arrange themselves in a plane with an approximate angle of 120 degrees between the bonds.  This arrangement again minimizes the electron - electron repulsion between the 3 sigma bonds in this geometry.  
Finally, what geometry will minimize electron - electron repulsion between the two sigma bonds in the $\ce{sp}$ hybridized case?


(mouse over for answer)


   Yes, the two bonds will be in a plane 180 degrees from one another, just like the $\ce{H-C-C}$ bond angle in acetylene.


Take home message:


$\ce{sp^3}$ hybridized carbon has angles ~109 degrees
$\ce{sp^2}$ ~ 120 degrees
and $\ce{sp}$ ~180 degrees.