Alcohols, Phenols and Ethers

(A) P is optically inactive

(B) $S$ is Alkene, and Alkene gives Bayer's test

(C) Q has carboxylic acid and it gives effervescence with $\mathrm{NaHCO}_3$

(D) R is Not Alkyne

For option (B) :

C₂H₅MgBr performs addition reaction with - C $ \equiv $ N.
C₂H₅MgBr performs substitution reaction with ester -COOCH₃.

Conjugated dienes (two double bonds separated by a single bond) gives Br water test. Either 1-2 or 1-4 addition can happen.

For option (C) :

C₂H₅MgBr reacts with -OH group and creates alkane C₂H₆.

As here C = C present, so this compound gives Br water tests.

Steric effect of halogens are as follows Cl₂ < Br₂ < I₂

Electronic effect of phenolic group directs the approaching electrophile towards ortho and para positions. Tertiary butyl group has large size so it causes steric effect around aromatic nucleus. On the basis of above factors the products of the given reactions are as follows


Hence, orientation in electrophilic substitution reaction is decided by

(a) The steric effect of the halogen

(b) The steric effect of the tert-butyl group

(c) The electronic effect of the phenolic group

So, (a), (b) and (c) are correct choices.

In presence of base, phenol forms phenoxide ion which is $ortho$-$para$ directing because in the resonating structures, the electron density increases at ortho and para positions. The phenoxide ion further undergoes electrophilic substitution with Br$_2$ at $ortho$ and $para$ positions.