For the reaction sequence given below, the correct statement(s) is(are)

The first step is the addition of a Grignard reagent to an aldehyde.
Propanal, $$CH_3CH_2CHO$$, is treated with $$CH_3MgBr$$ followed by hydrolysis.

$$CH_3CH_2CHO \;\xrightarrow[\;H_2O\;]{CH_3MgBr}\; CH_3CH(OH)CH_2CH_3$$

The product P is $$2$$-butanol. The carbon bearing the $$OH$$ group is attached to four different groups $$\left(CH_3,\,CH_2CH_3,\,H,\,OH\right)$$, so it is chiral. Hence P is optically active. Therefore, statement A is correct.

Next, P is oxidised with a mild oxidising agent such as PCC (or $$K_2Cr_2O_7/H^+$$) to give compound Q.

$$CH_3CH(OH)CH_2CH_3 \;\xrightarrow{\text{PCC}}\; CH_3COCH_2CH_3$$

Q is 2-butanone, a neutral ketone. Ketones do not react with aqueous $$NaHCO_3$$, so Q will not produce effervescence of $$CO_2$$. Hence statement C is wrong.

The ketone Q is then completely reduced (Clemmensen or Wolff-Kishner reduction) to give compound R.

$$CH_3COCH_2CH_3 \;\xrightarrow{\text{Zn-Hg/HCl or N\!NH_2/KOH}}\; CH_3CH_2CH_2CH_3$$

R is n-butane, a saturated hydrocarbon containing no triple bond; therefore R is not an alkyne. Statement D is wrong.

Finally, R undergoes free-radical bromination with $$Br_2/h\nu$$ to give a mixture of bromoalkanes collectively represented as compound S.

Because all these bromoalkanes are fully saturated, they contain neither $$C=C$$ nor $$C\equiv C$$ bonds. Saturated halides do not decolourise alkaline $$KMnO_4$$ (Baeyer’s test), so S gives a negative Baeyer’s test. Statement B is wrong.

Thus, among the given statements only A is correct.

Option A which is: P is optically active.