Nucleophilic addition of grignard reagent

Question

Here why isn't the phenyl group attacking the carbonyl carbon? Isn't the first step in addition of grignard reagent the attack by the carbanion on the carbonyl carbon ?

Answer 1

The attacking of the Grignard reagent in these kinds of $\alpha$-$\beta$ unsaturated carbonyl compounds depends on two factors mainly.
$1$) Hardness of the acid centre
$2$) Steric repulsion between the surrounding groups and incoming nucleophile.

When the attacking nucleophiles are Grignard reagents,as per your equation, they are generally hard bases.According to Pearson's HSAB Theory, Hard bases prefer to bind to the Hard acid centre and soft bases prefer to bind to the soft acid centre. In your compound i.e in $\alpha$-$\beta$ unsaturated carbonyl compounds, the carbonyl carbon centre is the Harder acid centre, whereas the $\ce{C=C}$ double bond centre is the softer acid centre.So, Grignard reagents try to first bind with the carbonyl carbons unless there is no other destabilising factor.
The second governing factor is the steric crowding, which can sometimes make the attacking nucleophile attach to the least sterically crowded region even though it is violating HSAB theory. So, the addition compound actually depends on these two competititive factors.

Now, in your given compound the attacking nucleophile is $\ce{PhMgBr}$.It is undoubtedly Hard Base. But if it were to bind to the carbonyl carbon i.e. $\ce{C=O}$ double bond, there will be huge repulsion between the lone pair of oxygen and $\pi$-electron cloud of Phenyl ring which actually makes the previously stated second factor to dominate. So, due to this huge steric repulsion, The attacking nucleophile has to attack the less sterically crowded(Note: The repulsion is also there at the double bonded carbon centre , but repulsion between methyl groups and phenyl rings are only due to inefficient spatial orientation which is definitely smaller than inter electronic strong columbic repulsion forces in close vicinity in the case of $\ce{C=O}$) softer basic $\ce{C=C}$ centre which infact gives the product.