Let M be a compact manifold. The diagonal $M \rightarrow M \times M$ induces 
co-product on singular cohomolgy $H^\*(M) \rightarrow H^\*(M) \times H^\*(M)$ via Poincare duality. 


I would like to know if there is a co-product on minimal Sullivan model $(\Lambda V, d)$ for M which induces the above co-product, at least in the special case when $C^\*(M) \cong H^\*(M)$. 



natural co-product on minimal Sullivan model
The short answer is: yes (in the very special case you mention).


Here is a longer answer. Note that the map you are talking about is not really a coproduct on $H^\ast(M)$, since it raises degree by $m$, the dimension of $M$. In fact you have a map
$$
\triangle\_!:H^\ast(M)\to H^{\ast+m}(M\times M).
$$ This map is given by $x\mapsto (1\times x)\cup d\_M$, where $d\_M$ is the *diagonal class*
$$
d\_M = \sum\_i^N (-1)^{|a\_i|}a\_i\times a\_i^\ast \in H^m(M\times M),
$$
which is Poincaré dual to the diagonal submanifold in $M\times M$. Here $\{a\_1,\ldots , a\_N\}$ is a homogeneous basis for $H^\ast(M)$ and $\{a\_1^\ast,\ldots , a\_N^\ast\}$ is thee Poincaré dual basis. Let's say $M$ is orientable and we're using rational coefficients. All of this is nicely explained in Chapter 11 of Milnor and Stasheff's book.


This map can be used to describe the cohomology of the two-point (unordered) configuration space $F(M,2)$. In fact, Cohen and Taylor showed that if $M$ is a closed oriented manifold, there is an isomorphism of algebras 
$$
H^\ast(F(M,2)) \cong \frac{H^\ast(M\times M)}{(d\_M)},
$$
where $(d\_M)$ denotes the ideal generated by the diagonal class.


Now, Lambrechts and Stanley (*The rational homotopy type of configuration spaces of two points,* 
Ann. Inst. Fourier, 54 (2004), no. 4, 1029–1052.) showed that the same thing happens at the level of rational models. More precisely, they showed that if $M$ admits a CDGA model $(A,d)$ for its rational homotopy type *which satisfies Poincaré duality*, then there exits a diagonal class $d\_A\in (A\otimes A)^M$ such that the quotient CDGA
$$
\frac{A\otimes A}{(d\_A)}
$$
is a CDGA model for $F(M,2)$, at least when $M$ is $2$-connected. They later showed in an important paper (*Poincaré duality and commutative differential graded algebras,* Ann. Sci. Éc. Norm. Supér. (4) 41 (2008), no. 4, 495–509) that $M$ always admits such a model.


All of this implies, I think, that $\triangle\_!$ is induced by the map $$
A\to A\otimes A,\qquad x\mapsto (1\otimes x)d\_A,
$$ where $(A,d)$ is a CDGA model for $M$ satisfying Poincaré duality and $d\_A$ is the diagonal class discussed above. In particular, if $M$ is formal we can take $(A,d)=(H^\ast(M),0)$ and $d\_A=d\_M$. 


