compound A is optically active and rotates plane polarized light in a
(+) direction. After this compound reacts via nucleophilic
substitution 50% product is produced and of the 50% of the initial
material A remains. Would you expect the resulting mixture to still
rotate light and with what sign if the mechanism was (1)SN1 type and
(2)SN2 type.

Hello -

Compounds rotate plane polarized light due to their chirality, or
"handedness" - organic compounds which are chiral generally have four
separate groups around a central carbon.

In an SN1 substitution, an elimination is followed by the
substitution.  The intermediary compound formed by the elimination is
not chiral, in fact it is planar, therefore the nucleophile for the
substitution may attack either side of the compound producing a
racemic (equal stereoisomers) mixture of products.  Thus, 50% would be
still be (+) (unreacted initial material) and 50% would be products
with a 50% each (+) and (-) - overall, this gives 75% (+) and 25% (-).

In an SN2 substitution, the nucleophile attacks from the opposite side
of the molecule from the leaving group, yielding a pentagonal
intermediate which inverts the sterochemistry of the product.  In this
case, 50% would be (+) (the unreacted materials) and 50% would be (-)
(the inverted products) yielding a net of no rotation.


synarchy

References:
A nice shockwave showing chirality:
http://www.colby.edu/chemistry/OChem/DEMOS/Chirality.html

Another nice shockwave demonstrating substitution reactions (including
the effect on chirality)
http://www.colby.edu/chemistry/OChem/DEMOS/Substitution.html


Google search
chirality substitution reactions

While essentially correct, the above answer assumes the presence of a
single chiral center.  Compounds with multiple chiral centers will not
behave as described.  The question was framed pertaining to any
optically active compound, not simply those with a single chiral
center.

compounds with multiple chiral centers also would not be approachable
given the information available in the question - you would also need
to know the number and orientation of all the other chiral centers in
the molecule.

Actually, none of that is quite correct, although the problem is
likely due to a poorly phrased question.  The difficulty is that
optical rotation is a specific characteristic of a given molecule, so
once you change the molecule all bets are off with respect to both the
sign and the magnitude of the optical rotation.  So assuming that the
substitution does not regenerate the same molecule (which might be the
case), and assuming further that the stereocenter of interest is the
only source of chirality in the molecule, the answer would be as
follows:
SN1 -- the intermediate carbocation is indeed achiral, leading to
equal amounts of two enantiomeric products that by definition have
equal and opposite rotations.  So the net rotation is still (+), since
the starting material is the only thing present in any enantiomeric
excess (this is essentially the same as the original answer, but the
75(+)/25(-) bit is misleading since it suggests that all rotations are
created equal.

SN2-- here the answer is who knows, because while it is true that the
stereochemistry will be cleanly inverted, it *does not* follow that
the optical rotation will be the same and negative (or, for that
matter, even that it is negative).  Optical rotations have to be
measured empirically, and can't be predicted from structure (other
than the above guarantee that enantiomers have equal and opposite
reactions).

Of course, if the product of the reaction is the same as the starting
material, then the original answer is quite right (with the caveat
already posted about more than one source of chirality)...